Physics 1 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial speed

of 10 ms and 40 ms respectively Which of the following graph best represents the time

variation of relative position of the second stone with respect to the first

(Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =

10 119898119904minus2)

(the figure are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (C)

Solution 1198781 = 10119905 minus1

21198921199052

When 1198781 = minus240

rArr minus240 = 10119905 minus 51199052

rArr 119905 = 8119904

So at 119905 = 8 seconds first stone will reach ground

1198782 = 20119905 minus1

21198921199052

Till 119905 = 8 119904119890119888119900119899119889119904

1198782minus1198781 = 30119905

But after 8 second 1198781 is constant = minus240

Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240

rArr 1198782+240 = 20119905 minus1

21198921199052

And at t = 12s seconds stone will reach ground

The corresponding graph of relative position of second stone wrt first is

Topic Kinematics

Difficulty Moderate (Embibe predicted high weightage)

2 The period of oscillation of a simple pendulum is T = 2πradicL

g Measured value of L is 200 cm known to 1 mm

accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution

The accuracy in the determination of g is

(A) 2

(B) 3

(C) 1 (D) 5

Answer (B)

Solution there4 119879 = 2120587radic119871

119892rArr 119892 = 41205872

119871

1198792

there4 Error in g can be calculated as

Δ119892

119892=ΔL

119871+2ΔT

119879

there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation

rArrΔt

119905=Δ119879

119879

rArrΔ119892

119892=Δ119871

119871+2Δt

119905

Given that Δ119871 = 1 119898119898 = 10minus3 119888119898119871 = 20times 10minus2119898

Δ119905 = 1119904 119905 = 90119904

error in g

Δ119892

119892times 100 = (

Δ119871

119871+2Δ119905

119905) times100

=Δ119892

119892times 100 = (

Δ119871

119871+2Δ119905

119905) times100

= (10minus3

20times 10minus2+2 times1

90) times 100

=1

2+20

9

= 05 + 222

= 272

cong 3

Topic Unit amp Dimensions

Difficulty level Easy (embibe predicted easy to score)

3

Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed

against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between block

B and the wall is 015 the frictional force applied by the wall on block B is

(A) 100 N

(B) 80 N

(C) 120 N (D) 150 N

Answer (C)

Solution

For complete state equilibrium of the system The state friction on the block B by wall will balance the total

weight 120 N of the system

Topic Laws of Motion

Difficulty level Moderate (embibe predicted Low Weightage)

4 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in

the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy during the

collision is close to

(A) 44

(B) 50

(C) 56 (D) 65

Answer (C)

Solution

The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907

According to question as

On perfectly inelastic collision the particles stick to each other so

119875119891 = 3119898 119881119891

By conservation of linear momentum principle

119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881

rArr 119881119891 =2119881

3(119894 + ) rArr 119881119891 =

2radic2

3119881

there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897

=1

2119898(2119881)2+

1

2(2119898)1198812 minus

1

2(3119898)(

2radic2 119881

3)

2

= 21198981198812 +1198981198812 minus4

31198981198812 = 31198981198812 minus

41198981198812

3

=5

31198981198812

change in KE = 100 times∆119870

119870119894=

5

31198981199072

31198981198812=5

9times 100

=500

9= 56

Topic Conservation of Momentum

Difficulty level Moderate (embibe predicted easy to score (Must Do))

5 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its

height is h then z0 is equal to

(A) h2

4R

(B) 3h

4

(C) 5h

8

(D) 3h2

8R

Answer (B)

Solution

Topic Centre of Mass

Difficulty Easy (embibe predicted Low Weightage)

6 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut Moment

of inertia of cube about an axis passing through its centre and perpendicular to one of its faces

is

(A) 1198721198772

32radic2120587

(B) 1198721198772

16radic2120587

(C) 41198721198772

9radic3120587

(D) 41198721198772

3radic3120587

Answer (C)

Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R

rArr radic3119886 = 2119877 rArr 119886 =2119877

radic3

As densities of sphere and cube are equal Let Mrsquo be mass of cube

119872431205871198773 =

119872prime

431198863

119872prime =31198721198863

41205871198773

Moment of inertial of cube about an axis passing through centre

=119872prime(21198862)

12

=31198721198863

41205871198773times21198862

12

=1198721198865

81205871198773

119886 =2

radic3119877

=119872 times32 1198775

8120587 times9radic31198773

=41198721198772

9radic3120587

Topic Rotational Mechanics

Difficulty Moderate (embibe predicted Low Weightage)

7 From a solid sphere of mass M and radius R a spherical portion of radius 119877

2 is removed as shown

in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the center of the

cavity thus formed is

(G = gravitational constant)

(A) minus119866119872

2119877

(B) minus119866119872

119877

(C) minus2119866119872

3119877

(D) minus2119866119872

119877

Answer (B)

Solution

Topic Gravitation

Difficulty Moderate (embibe predicted high weightage)

8 A pendulum made of a uniform wire of cross sectional area A has time period T When and

additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos modulus of

the material of the wire is Y then 1

119884is equal to

(g = gravitational acceleration)

(A) [(119879119872

119879)2

minus 1]119860

119872119892

(B) [(119879119872

119879)2

minus 1]119872119892

119860

(C) [1minus (119879119872

119879)2]119860

119872119892

(D) [1minus (119879

119879119872)2]119860

119872119892

Answer (A)

Solution

Topic Simple Harmonic Motion

Difficulty Difficult (embibe predicted high weightage)

9 Consider a spherical shell of radius R at temperature T the black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume 119906 =119880

119881prop 1198794and

pressure 119901 =1

3(119880

119881) If the shell now undergoes an adiabatic expansion the relation between T

and R is

(A) 119879 prop 119890minus119877

(B) 119879 prop 119890minus3119877

(C) 119879 prop1

119877

(D) 119879 prop1

1198773

Answer (C)

Solution ∵ in an adiabatic process

119889119876 = 0

So by first law of thermodynamics

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

rArr minus240 = 10119905 minus 51199052

rArr 119905 = 8119904

So at 119905 = 8 seconds first stone will reach ground

1198782 = 20119905 minus1

21198921199052

Till 119905 = 8 119904119890119888119900119899119889119904

1198782minus1198781 = 30119905

But after 8 second 1198781 is constant = minus240

Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240

rArr 1198782+240 = 20119905 minus1

21198921199052

And at t = 12s seconds stone will reach ground

The corresponding graph of relative position of second stone wrt first is

Topic Kinematics

Difficulty Moderate (Embibe predicted high weightage)

2 The period of oscillation of a simple pendulum is T = 2πradicL

g Measured value of L is 200 cm known to 1 mm

accuracy and time for 100 oscillations of the pendulum is found to be 90s using a wrist watch of 1s resolution

The accuracy in the determination of g is

(A) 2

(B) 3

(C) 1 (D) 5

Answer (B)

Solution there4 119879 = 2120587radic119871

119892rArr 119892 = 41205872

119871

1198792

there4 Error in g can be calculated as

Δ119892

119892=ΔL

119871+2ΔT

119879

there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation

rArrΔt

119905=Δ119879

119879

rArrΔ119892

119892=Δ119871

119871+2Δt

119905

Given that Δ119871 = 1 119898119898 = 10minus3 119888119898119871 = 20times 10minus2119898

Δ119905 = 1119904 119905 = 90119904

error in g

Δ119892

119892times 100 = (

Δ119871

119871+2Δ119905

119905) times100

=Δ119892

119892times 100 = (

Δ119871

119871+2Δ119905

119905) times100

= (10minus3

20times 10minus2+2 times1

90) times 100

=1

2+20

9

= 05 + 222

= 272

cong 3

Topic Unit amp Dimensions

Difficulty level Easy (embibe predicted easy to score)

3

Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed

against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between block

B and the wall is 015 the frictional force applied by the wall on block B is

(A) 100 N

(B) 80 N

(C) 120 N (D) 150 N

Answer (C)

Solution

For complete state equilibrium of the system The state friction on the block B by wall will balance the total

weight 120 N of the system

Topic Laws of Motion

Difficulty level Moderate (embibe predicted Low Weightage)

4 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in

the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy during the

collision is close to

(A) 44

(B) 50

(C) 56 (D) 65

Answer (C)

Solution

The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907

According to question as

On perfectly inelastic collision the particles stick to each other so

119875119891 = 3119898 119881119891

By conservation of linear momentum principle

119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881

rArr 119881119891 =2119881

3(119894 + ) rArr 119881119891 =

2radic2

3119881

there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897

=1

2119898(2119881)2+

1

2(2119898)1198812 minus

1

2(3119898)(

2radic2 119881

3)

2

= 21198981198812 +1198981198812 minus4

31198981198812 = 31198981198812 minus

41198981198812

3

=5

31198981198812

change in KE = 100 times∆119870

119870119894=

5

31198981199072

31198981198812=5

9times 100

=500

9= 56

Topic Conservation of Momentum

Difficulty level Moderate (embibe predicted easy to score (Must Do))

5 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its

height is h then z0 is equal to

(A) h2

4R

(B) 3h

4

(C) 5h

8

(D) 3h2

8R

Answer (B)

Solution

Topic Centre of Mass

Difficulty Easy (embibe predicted Low Weightage)

6 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut Moment

of inertia of cube about an axis passing through its centre and perpendicular to one of its faces

is

(A) 1198721198772

32radic2120587

(B) 1198721198772

16radic2120587

(C) 41198721198772

9radic3120587

(D) 41198721198772

3radic3120587

Answer (C)

Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R

rArr radic3119886 = 2119877 rArr 119886 =2119877

radic3

As densities of sphere and cube are equal Let Mrsquo be mass of cube

119872431205871198773 =

119872prime

431198863

119872prime =31198721198863

41205871198773

Moment of inertial of cube about an axis passing through centre

=119872prime(21198862)

12

=31198721198863

41205871198773times21198862

12

=1198721198865

81205871198773

119886 =2

radic3119877

=119872 times32 1198775

8120587 times9radic31198773

=41198721198772

9radic3120587

Topic Rotational Mechanics

Difficulty Moderate (embibe predicted Low Weightage)

7 From a solid sphere of mass M and radius R a spherical portion of radius 119877

2 is removed as shown

in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the center of the

cavity thus formed is

(G = gravitational constant)

(A) minus119866119872

2119877

(B) minus119866119872

119877

(C) minus2119866119872

3119877

(D) minus2119866119872

119877

Answer (B)

Solution

Topic Gravitation

Difficulty Moderate (embibe predicted high weightage)

8 A pendulum made of a uniform wire of cross sectional area A has time period T When and

additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos modulus of

the material of the wire is Y then 1

119884is equal to

(g = gravitational acceleration)

(A) [(119879119872

119879)2

minus 1]119860

119872119892

(B) [(119879119872

119879)2

minus 1]119872119892

119860

(C) [1minus (119879119872

119879)2]119860

119872119892

(D) [1minus (119879

119879119872)2]119860

119872119892

Answer (A)

Solution

Topic Simple Harmonic Motion

Difficulty Difficult (embibe predicted high weightage)

9 Consider a spherical shell of radius R at temperature T the black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume 119906 =119880

119881prop 1198794and

pressure 119901 =1

3(119880

119881) If the shell now undergoes an adiabatic expansion the relation between T

and R is

(A) 119879 prop 119890minus119877

(B) 119879 prop 119890minus3119877

(C) 119879 prop1

119877

(D) 119879 prop1

1198773

Answer (C)

Solution ∵ in an adiabatic process

119889119876 = 0

So by first law of thermodynamics

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Answer (B)

Solution there4 119879 = 2120587radic119871

119892rArr 119892 = 41205872

119871

1198792

there4 Error in g can be calculated as

Δ119892

119892=ΔL

119871+2ΔT

119879

there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation

rArrΔt

119905=Δ119879

119879

rArrΔ119892

119892=Δ119871

119871+2Δt

119905

Given that Δ119871 = 1 119898119898 = 10minus3 119888119898119871 = 20times 10minus2119898

Δ119905 = 1119904 119905 = 90119904

error in g

Δ119892

119892times 100 = (

Δ119871

119871+2Δ119905

119905) times100

=Δ119892

119892times 100 = (

Δ119871

119871+2Δ119905

119905) times100

= (10minus3

20times 10minus2+2 times1

90) times 100

=1

2+20

9

= 05 + 222

= 272

cong 3

Topic Unit amp Dimensions

Difficulty level Easy (embibe predicted easy to score)

3

Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed

against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between block

B and the wall is 015 the frictional force applied by the wall on block B is

(A) 100 N

(B) 80 N

(C) 120 N (D) 150 N

Answer (C)

Solution

For complete state equilibrium of the system The state friction on the block B by wall will balance the total

weight 120 N of the system

Topic Laws of Motion

Difficulty level Moderate (embibe predicted Low Weightage)

4 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in

the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy during the

collision is close to

(A) 44

(B) 50

(C) 56 (D) 65

Answer (C)

Solution

The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907

According to question as

On perfectly inelastic collision the particles stick to each other so

119875119891 = 3119898 119881119891

By conservation of linear momentum principle

119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881

rArr 119881119891 =2119881

3(119894 + ) rArr 119881119891 =

2radic2

3119881

there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897

=1

2119898(2119881)2+

1

2(2119898)1198812 minus

1

2(3119898)(

2radic2 119881

3)

2

= 21198981198812 +1198981198812 minus4

31198981198812 = 31198981198812 minus

41198981198812

3

=5

31198981198812

change in KE = 100 times∆119870

119870119894=

5

31198981199072

31198981198812=5

9times 100

=500

9= 56

Topic Conservation of Momentum

Difficulty level Moderate (embibe predicted easy to score (Must Do))

5 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its

height is h then z0 is equal to

(A) h2

4R

(B) 3h

4

(C) 5h

8

(D) 3h2

8R

Answer (B)

Solution

Topic Centre of Mass

Difficulty Easy (embibe predicted Low Weightage)

6 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut Moment

of inertia of cube about an axis passing through its centre and perpendicular to one of its faces

is

(A) 1198721198772

32radic2120587

(B) 1198721198772

16radic2120587

(C) 41198721198772

9radic3120587

(D) 41198721198772

3radic3120587

Answer (C)

Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R

rArr radic3119886 = 2119877 rArr 119886 =2119877

radic3

As densities of sphere and cube are equal Let Mrsquo be mass of cube

119872431205871198773 =

119872prime

431198863

119872prime =31198721198863

41205871198773

Moment of inertial of cube about an axis passing through centre

=119872prime(21198862)

12

=31198721198863

41205871198773times21198862

12

=1198721198865

81205871198773

119886 =2

radic3119877

=119872 times32 1198775

8120587 times9radic31198773

=41198721198772

9radic3120587

Topic Rotational Mechanics

Difficulty Moderate (embibe predicted Low Weightage)

7 From a solid sphere of mass M and radius R a spherical portion of radius 119877

2 is removed as shown

in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the center of the

cavity thus formed is

(G = gravitational constant)

(A) minus119866119872

2119877

(B) minus119866119872

119877

(C) minus2119866119872

3119877

(D) minus2119866119872

119877

Answer (B)

Solution

Topic Gravitation

Difficulty Moderate (embibe predicted high weightage)

8 A pendulum made of a uniform wire of cross sectional area A has time period T When and

additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos modulus of

the material of the wire is Y then 1

119884is equal to

(g = gravitational acceleration)

(A) [(119879119872

119879)2

minus 1]119860

119872119892

(B) [(119879119872

119879)2

minus 1]119872119892

119860

(C) [1minus (119879119872

119879)2]119860

119872119892

(D) [1minus (119879

119879119872)2]119860

119872119892

Answer (A)

Solution

Topic Simple Harmonic Motion

Difficulty Difficult (embibe predicted high weightage)

9 Consider a spherical shell of radius R at temperature T the black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume 119906 =119880

119881prop 1198794and

pressure 119901 =1

3(119880

119881) If the shell now undergoes an adiabatic expansion the relation between T

and R is

(A) 119879 prop 119890minus119877

(B) 119879 prop 119890minus3119877

(C) 119879 prop1

119877

(D) 119879 prop1

1198773

Answer (C)

Solution ∵ in an adiabatic process

119889119876 = 0

So by first law of thermodynamics

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

3

Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed

against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between block

B and the wall is 015 the frictional force applied by the wall on block B is

(A) 100 N

(B) 80 N

(C) 120 N (D) 150 N

Answer (C)

Solution

For complete state equilibrium of the system The state friction on the block B by wall will balance the total

weight 120 N of the system

Topic Laws of Motion

Difficulty level Moderate (embibe predicted Low Weightage)

4 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in

the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy during the

collision is close to

(A) 44

(B) 50

(C) 56 (D) 65

Answer (C)

Solution

The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907

According to question as

On perfectly inelastic collision the particles stick to each other so

119875119891 = 3119898 119881119891

By conservation of linear momentum principle

119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881

rArr 119881119891 =2119881

3(119894 + ) rArr 119881119891 =

2radic2

3119881

there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897

=1

2119898(2119881)2+

1

2(2119898)1198812 minus

1

2(3119898)(

2radic2 119881

3)

2

= 21198981198812 +1198981198812 minus4

31198981198812 = 31198981198812 minus

41198981198812

3

=5

31198981198812

change in KE = 100 times∆119870

119870119894=

5

31198981199072

31198981198812=5

9times 100

=500

9= 56

Topic Conservation of Momentum

Difficulty level Moderate (embibe predicted easy to score (Must Do))

5 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its

height is h then z0 is equal to

(A) h2

4R

(B) 3h

4

(C) 5h

8

(D) 3h2

8R

Answer (B)

Solution

Topic Centre of Mass

Difficulty Easy (embibe predicted Low Weightage)

6 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut Moment

of inertia of cube about an axis passing through its centre and perpendicular to one of its faces

is

(A) 1198721198772

32radic2120587

(B) 1198721198772

16radic2120587

(C) 41198721198772

9radic3120587

(D) 41198721198772

3radic3120587

Answer (C)

Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R

rArr radic3119886 = 2119877 rArr 119886 =2119877

radic3

As densities of sphere and cube are equal Let Mrsquo be mass of cube

119872431205871198773 =

119872prime

431198863

119872prime =31198721198863

41205871198773

Moment of inertial of cube about an axis passing through centre

=119872prime(21198862)

12

=31198721198863

41205871198773times21198862

12

=1198721198865

81205871198773

119886 =2

radic3119877

=119872 times32 1198775

8120587 times9radic31198773

=41198721198772

9radic3120587

Topic Rotational Mechanics

Difficulty Moderate (embibe predicted Low Weightage)

7 From a solid sphere of mass M and radius R a spherical portion of radius 119877

2 is removed as shown

in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the center of the

cavity thus formed is

(G = gravitational constant)

(A) minus119866119872

2119877

(B) minus119866119872

119877

(C) minus2119866119872

3119877

(D) minus2119866119872

119877

Answer (B)

Solution

Topic Gravitation

Difficulty Moderate (embibe predicted high weightage)

8 A pendulum made of a uniform wire of cross sectional area A has time period T When and

additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos modulus of

the material of the wire is Y then 1

119884is equal to

(g = gravitational acceleration)

(A) [(119879119872

119879)2

minus 1]119860

119872119892

(B) [(119879119872

119879)2

minus 1]119872119892

119860

(C) [1minus (119879119872

119879)2]119860

119872119892

(D) [1minus (119879

119879119872)2]119860

119872119892

Answer (A)

Solution

Topic Simple Harmonic Motion

Difficulty Difficult (embibe predicted high weightage)

9 Consider a spherical shell of radius R at temperature T the black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume 119906 =119880

119881prop 1198794and

pressure 119901 =1

3(119880

119881) If the shell now undergoes an adiabatic expansion the relation between T

and R is

(A) 119879 prop 119890minus119877

(B) 119879 prop 119890minus3119877

(C) 119879 prop1

119877

(D) 119879 prop1

1198773

Answer (C)

Solution ∵ in an adiabatic process

119889119876 = 0

So by first law of thermodynamics

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

4 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in

the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy during the

collision is close to

(A) 44

(B) 50

(C) 56 (D) 65

Answer (C)

Solution

The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907

According to question as

On perfectly inelastic collision the particles stick to each other so

119875119891 = 3119898 119881119891

By conservation of linear momentum principle

119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881

rArr 119881119891 =2119881

3(119894 + ) rArr 119881119891 =

2radic2

3119881

there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897

=1

2119898(2119881)2+

1

2(2119898)1198812 minus

1

2(3119898)(

2radic2 119881

3)

2

= 21198981198812 +1198981198812 minus4

31198981198812 = 31198981198812 minus

41198981198812

3

=5

31198981198812

change in KE = 100 times∆119870

119870119894=

5

31198981199072

31198981198812=5

9times 100

=500

9= 56

Topic Conservation of Momentum

Difficulty level Moderate (embibe predicted easy to score (Must Do))

5 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its

height is h then z0 is equal to

(A) h2

4R

(B) 3h

4

(C) 5h

8

(D) 3h2

8R

Answer (B)

Solution

Topic Centre of Mass

Difficulty Easy (embibe predicted Low Weightage)

6 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut Moment

of inertia of cube about an axis passing through its centre and perpendicular to one of its faces

is

(A) 1198721198772

32radic2120587

(B) 1198721198772

16radic2120587

(C) 41198721198772

9radic3120587

(D) 41198721198772

3radic3120587

Answer (C)

Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R

rArr radic3119886 = 2119877 rArr 119886 =2119877

radic3

As densities of sphere and cube are equal Let Mrsquo be mass of cube

119872431205871198773 =

119872prime

431198863

119872prime =31198721198863

41205871198773

Moment of inertial of cube about an axis passing through centre

=119872prime(21198862)

12

=31198721198863

41205871198773times21198862

12

=1198721198865

81205871198773

119886 =2

radic3119877

=119872 times32 1198775

8120587 times9radic31198773

=41198721198772

9radic3120587

Topic Rotational Mechanics

Difficulty Moderate (embibe predicted Low Weightage)

7 From a solid sphere of mass M and radius R a spherical portion of radius 119877

2 is removed as shown

in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the center of the

cavity thus formed is

(G = gravitational constant)

(A) minus119866119872

2119877

(B) minus119866119872

119877

(C) minus2119866119872

3119877

(D) minus2119866119872

119877

Answer (B)

Solution

Topic Gravitation

Difficulty Moderate (embibe predicted high weightage)

8 A pendulum made of a uniform wire of cross sectional area A has time period T When and

additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos modulus of

the material of the wire is Y then 1

119884is equal to

(g = gravitational acceleration)

(A) [(119879119872

119879)2

minus 1]119860

119872119892

(B) [(119879119872

119879)2

minus 1]119872119892

119860

(C) [1minus (119879119872

119879)2]119860

119872119892

(D) [1minus (119879

119879119872)2]119860

119872119892

Answer (A)

Solution

Topic Simple Harmonic Motion

Difficulty Difficult (embibe predicted high weightage)

9 Consider a spherical shell of radius R at temperature T the black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume 119906 =119880

119881prop 1198794and

pressure 119901 =1

3(119880

119881) If the shell now undergoes an adiabatic expansion the relation between T

and R is

(A) 119879 prop 119890minus119877

(B) 119879 prop 119890minus3119877

(C) 119879 prop1

119877

(D) 119879 prop1

1198773

Answer (C)

Solution ∵ in an adiabatic process

119889119876 = 0

So by first law of thermodynamics

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

=5

31198981198812

change in KE = 100 times∆119870

119870119894=

5

31198981199072

31198981198812=5

9times 100

=500

9= 56

Topic Conservation of Momentum

Difficulty level Moderate (embibe predicted easy to score (Must Do))

5 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its

height is h then z0 is equal to

(A) h2

4R

(B) 3h

4

(C) 5h

8

(D) 3h2

8R

Answer (B)

Solution

Topic Centre of Mass

Difficulty Easy (embibe predicted Low Weightage)

6 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut Moment

of inertia of cube about an axis passing through its centre and perpendicular to one of its faces

is

(A) 1198721198772

32radic2120587

(B) 1198721198772

16radic2120587

(C) 41198721198772

9radic3120587

(D) 41198721198772

3radic3120587

Answer (C)

Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R

rArr radic3119886 = 2119877 rArr 119886 =2119877

radic3

As densities of sphere and cube are equal Let Mrsquo be mass of cube

119872431205871198773 =

119872prime

431198863

119872prime =31198721198863

41205871198773

Moment of inertial of cube about an axis passing through centre

=119872prime(21198862)

12

=31198721198863

41205871198773times21198862

12

=1198721198865

81205871198773

119886 =2

radic3119877

=119872 times32 1198775

8120587 times9radic31198773

=41198721198772

9radic3120587

Topic Rotational Mechanics

Difficulty Moderate (embibe predicted Low Weightage)

7 From a solid sphere of mass M and radius R a spherical portion of radius 119877

2 is removed as shown

in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the center of the

cavity thus formed is

(G = gravitational constant)

(A) minus119866119872

2119877

(B) minus119866119872

119877

(C) minus2119866119872

3119877

(D) minus2119866119872

119877

Answer (B)

Solution

Topic Gravitation

Difficulty Moderate (embibe predicted high weightage)

8 A pendulum made of a uniform wire of cross sectional area A has time period T When and

additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos modulus of

the material of the wire is Y then 1

119884is equal to

(g = gravitational acceleration)

(A) [(119879119872

119879)2

minus 1]119860

119872119892

(B) [(119879119872

119879)2

minus 1]119872119892

119860

(C) [1minus (119879119872

119879)2]119860

119872119892

(D) [1minus (119879

119879119872)2]119860

119872119892

Answer (A)

Solution

Topic Simple Harmonic Motion

Difficulty Difficult (embibe predicted high weightage)

9 Consider a spherical shell of radius R at temperature T the black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume 119906 =119880

119881prop 1198794and

pressure 119901 =1

3(119880

119881) If the shell now undergoes an adiabatic expansion the relation between T

and R is

(A) 119879 prop 119890minus119877

(B) 119879 prop 119890minus3119877

(C) 119879 prop1

119877

(D) 119879 prop1

1198773

Answer (C)

Solution ∵ in an adiabatic process

119889119876 = 0

So by first law of thermodynamics

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Centre of Mass

Difficulty Easy (embibe predicted Low Weightage)

6 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut Moment

of inertia of cube about an axis passing through its centre and perpendicular to one of its faces

is

(A) 1198721198772

32radic2120587

(B) 1198721198772

16radic2120587

(C) 41198721198772

9radic3120587

(D) 41198721198772

3radic3120587

Answer (C)

Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R

rArr radic3119886 = 2119877 rArr 119886 =2119877

radic3

As densities of sphere and cube are equal Let Mrsquo be mass of cube

119872431205871198773 =

119872prime

431198863

119872prime =31198721198863

41205871198773

Moment of inertial of cube about an axis passing through centre

=119872prime(21198862)

12

=31198721198863

41205871198773times21198862

12

=1198721198865

81205871198773

119886 =2

radic3119877

=119872 times32 1198775

8120587 times9radic31198773

=41198721198772

9radic3120587

Topic Rotational Mechanics

Difficulty Moderate (embibe predicted Low Weightage)

7 From a solid sphere of mass M and radius R a spherical portion of radius 119877

2 is removed as shown

in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the center of the

cavity thus formed is

(G = gravitational constant)

(A) minus119866119872

2119877

(B) minus119866119872

119877

(C) minus2119866119872

3119877

(D) minus2119866119872

119877

Answer (B)

Solution

Topic Gravitation

Difficulty Moderate (embibe predicted high weightage)

8 A pendulum made of a uniform wire of cross sectional area A has time period T When and

additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos modulus of

the material of the wire is Y then 1

119884is equal to

(g = gravitational acceleration)

(A) [(119879119872

119879)2

minus 1]119860

119872119892

(B) [(119879119872

119879)2

minus 1]119872119892

119860

(C) [1minus (119879119872

119879)2]119860

119872119892

(D) [1minus (119879

119879119872)2]119860

119872119892

Answer (A)

Solution

Topic Simple Harmonic Motion

Difficulty Difficult (embibe predicted high weightage)

9 Consider a spherical shell of radius R at temperature T the black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume 119906 =119880

119881prop 1198794and

pressure 119901 =1

3(119880

119881) If the shell now undergoes an adiabatic expansion the relation between T

and R is

(A) 119879 prop 119890minus119877

(B) 119879 prop 119890minus3119877

(C) 119879 prop1

119877

(D) 119879 prop1

1198773

Answer (C)

Solution ∵ in an adiabatic process

119889119876 = 0

So by first law of thermodynamics

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Difficulty Easy (embibe predicted Low Weightage)

6 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut Moment

of inertia of cube about an axis passing through its centre and perpendicular to one of its faces

is

(A) 1198721198772

32radic2120587

(B) 1198721198772

16radic2120587

(C) 41198721198772

9radic3120587

(D) 41198721198772

3radic3120587

Answer (C)

Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R

rArr radic3119886 = 2119877 rArr 119886 =2119877

radic3

As densities of sphere and cube are equal Let Mrsquo be mass of cube

119872431205871198773 =

119872prime

431198863

119872prime =31198721198863

41205871198773

Moment of inertial of cube about an axis passing through centre

=119872prime(21198862)

12

=31198721198863

41205871198773times21198862

12

=1198721198865

81205871198773

119886 =2

radic3119877

=119872 times32 1198775

8120587 times9radic31198773

=41198721198772

9radic3120587

Topic Rotational Mechanics

Difficulty Moderate (embibe predicted Low Weightage)

7 From a solid sphere of mass M and radius R a spherical portion of radius 119877

2 is removed as shown

in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the center of the

cavity thus formed is

(G = gravitational constant)

(A) minus119866119872

2119877

(B) minus119866119872

119877

(C) minus2119866119872

3119877

(D) minus2119866119872

119877

Answer (B)

Solution

Topic Gravitation

Difficulty Moderate (embibe predicted high weightage)

8 A pendulum made of a uniform wire of cross sectional area A has time period T When and

additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos modulus of

the material of the wire is Y then 1

119884is equal to

(g = gravitational acceleration)

(A) [(119879119872

119879)2

minus 1]119860

119872119892

(B) [(119879119872

119879)2

minus 1]119872119892

119860

(C) [1minus (119879119872

119879)2]119860

119872119892

(D) [1minus (119879

119879119872)2]119860

119872119892

Answer (A)

Solution

Topic Simple Harmonic Motion

Difficulty Difficult (embibe predicted high weightage)

9 Consider a spherical shell of radius R at temperature T the black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume 119906 =119880

119881prop 1198794and

pressure 119901 =1

3(119880

119881) If the shell now undergoes an adiabatic expansion the relation between T

and R is

(A) 119879 prop 119890minus119877

(B) 119879 prop 119890minus3119877

(C) 119879 prop1

119877

(D) 119879 prop1

1198773

Answer (C)

Solution ∵ in an adiabatic process

119889119876 = 0

So by first law of thermodynamics

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

=119872 times32 1198775

8120587 times9radic31198773

=41198721198772

9radic3120587

Topic Rotational Mechanics

Difficulty Moderate (embibe predicted Low Weightage)

7 From a solid sphere of mass M and radius R a spherical portion of radius 119877

2 is removed as shown

in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the center of the

cavity thus formed is

(G = gravitational constant)

(A) minus119866119872

2119877

(B) minus119866119872

119877

(C) minus2119866119872

3119877

(D) minus2119866119872

119877

Answer (B)

Solution

Topic Gravitation

Difficulty Moderate (embibe predicted high weightage)

8 A pendulum made of a uniform wire of cross sectional area A has time period T When and

additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos modulus of

the material of the wire is Y then 1

119884is equal to

(g = gravitational acceleration)

(A) [(119879119872

119879)2

minus 1]119860

119872119892

(B) [(119879119872

119879)2

minus 1]119872119892

119860

(C) [1minus (119879119872

119879)2]119860

119872119892

(D) [1minus (119879

119879119872)2]119860

119872119892

Answer (A)

Solution

Topic Simple Harmonic Motion

Difficulty Difficult (embibe predicted high weightage)

9 Consider a spherical shell of radius R at temperature T the black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume 119906 =119880

119881prop 1198794and

pressure 119901 =1

3(119880

119881) If the shell now undergoes an adiabatic expansion the relation between T

and R is

(A) 119879 prop 119890minus119877

(B) 119879 prop 119890minus3119877

(C) 119879 prop1

119877

(D) 119879 prop1

1198773

Answer (C)

Solution ∵ in an adiabatic process

119889119876 = 0

So by first law of thermodynamics

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Gravitation

Difficulty Moderate (embibe predicted high weightage)

8 A pendulum made of a uniform wire of cross sectional area A has time period T When and

additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos modulus of

the material of the wire is Y then 1

119884is equal to

(g = gravitational acceleration)

(A) [(119879119872

119879)2

minus 1]119860

119872119892

(B) [(119879119872

119879)2

minus 1]119872119892

119860

(C) [1minus (119879119872

119879)2]119860

119872119892

(D) [1minus (119879

119879119872)2]119860

119872119892

Answer (A)

Solution

Topic Simple Harmonic Motion

Difficulty Difficult (embibe predicted high weightage)

9 Consider a spherical shell of radius R at temperature T the black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume 119906 =119880

119881prop 1198794and

pressure 119901 =1

3(119880

119881) If the shell now undergoes an adiabatic expansion the relation between T

and R is

(A) 119879 prop 119890minus119877

(B) 119879 prop 119890minus3119877

(C) 119879 prop1

119877

(D) 119879 prop1

1198773

Answer (C)

Solution ∵ in an adiabatic process

119889119876 = 0

So by first law of thermodynamics

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Simple Harmonic Motion

Difficulty Difficult (embibe predicted high weightage)

9 Consider a spherical shell of radius R at temperature T the black body radiation inside it can be

considered as an ideal gas of photons with internal energy per unit volume 119906 =119880

119881prop 1198794and

pressure 119901 =1

3(119880

119881) If the shell now undergoes an adiabatic expansion the relation between T

and R is

(A) 119879 prop 119890minus119877

(B) 119879 prop 119890minus3119877

(C) 119879 prop1

119877

(D) 119879 prop1

1198773

Answer (C)

Solution ∵ in an adiabatic process

119889119876 = 0

So by first law of thermodynamics

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

119889119876 = 119889119880+119889119882

rArr 0 = 119889119880+119889119882

rArr 119889119882 = minus119889119880

there4 119889119882 = 119875119889119881

rArr 119875119889119881 = minus119889119880 hellip(i)

Given that 119880

119881prop 1198794rArr 119880 = 1198961198811198794

rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)

Also 119875 =1

3

119880

119881=1

3

1198961198811198794

119881=1198701198794

3

Putting these values in equation

rArr1198701198794

3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)

rArr119879119889119881

3= minus119879119889119881minus4119881119889119879

rArr4119879

3119889119881 = minus4119881119889119879

rArr

13119889119881

119881= minus

119889119879

119879

rArr1

3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3

rArr 1198811198793 = 119888119900119899119904119905119886119899119905

4

31205871198773 1198793 = 119888119900119899119904119905119886119899119905

119877119879 = 119888119900119899119904119905119886119899119905

rArr 119879 prop1

119877

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

10 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with

reservoirs in two ways

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same

amount of heat

(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat

In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862

Entropy change of the body in the two cases respectively is

(A) 119897119899241198971198992 (B) 1198971198992 1198971198992

(C) 119897119899221198971198992

(D) 2119897119899281198971198992

Answer (B)

Solution

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Heat ampThermodynamics

Difficulty Moderate (embibe predicted high weightage)

11 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic expansion

the average time of collision between molecules increase as Vq where V is the volume of the gas The value

of q is

(γ =CPCv)

(A) 3γ+5

6

(B) 3γminus5

6

(C) γ+1

2

(D)γminus1

2

Answer (C)

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Heat ampThermodynamics

Difficulty Difficult (embibe predicted high weightage)

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

12 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential energy

(PE) against its displacement d which one of the following represents these correctly

(Graphs are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Simple Harmonic Motion

Difficulty Easy (embibe predicted high weightage)

13 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the

frequency of 1000 Hz The percentage change in the frequency heard by a person standing near

the track as the train passes him is (speed of sound = 320 119898119904minus1) close to

(A) 6

(B) 12

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(C) 18 (D) 24

Answer (B)

Solution

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Wave amp Sound

Difficulty Moderate (embibe predicted high weightage)

14 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative surface

charge ndash120590 in the lower half The electric field lines around the cylinder will look like figure given

in

(Figures are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(C)

(D)

Answer (A)

Solution

Consider cross section of cylinders which is circle the half part of circle which has positive charge can be

assume that total positive charge is at centre of mass of semicircle In the same way we can assume that

negative charge is at centre of mass of that semicircle

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should

not intersect the graph would be

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

15 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin) on

its surface For this sphere the equipotential surfaces with potential 31198810

251198810

431198810

4 and

1198810

4 have

radius 1198771 1198772 1198773 and 1198774 respectively Then

(A) 1198771 = 0 and 1198772 gt (1198774minus1198773)

(B) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)

(C) 1198771 = 0 and 1198772 lt (1198774minus1198773)

(D) 2119877 lt 1198774

Answer (C)

Solution

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

16 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a function

of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(A)

(B)

(C)

(D)

Answer (B)

Solution

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

∵ 1 amp 2 120583119891 are in parallel

there4 Equivalent capacitance of the series combination is

119862119890119902 is=3119888

119862+3

So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864

119862+3

there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is

∆119881 =119876

3=

119862119864

119862+3

So charge on 2120583119891 capacitor is

1198762 = 1198622∆119881=2119862119864

119862+3

rArr1198762

2119864=

119862

119862+3rArr1198762

2119864=119862+3minus3

119862+3

rArr1198762

2119864= 1 minus

3

119862+3rArr (

1198762

2119864minus 1) = minus

3

119862+3

rArr (1198762minus 2119864)(119862 + 3) = minus6119864

Which is of the form (119910 minus 120572)(119909 + 120573) lt 0

So the graph in hyperbola With down ward curve line ie

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Electrostatics

Difficulty Moderate (embibe predicted high weightage)

17 When 5V potential difference is applied across a wire of length 01 m the drift speed of

electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the resistivity

of the material is close to

(A) 16 times10minus8Ω119898 (B) 16 times10minus7Ω119898

(C) 16 times10minus6 Ω119898

(D) 16 times10minus5 Ω119898

Answer (D)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

18

In the circuit shown the current in the 1Ω resistor is

(A) 13 A from P to Q

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(B) 0 A

(C) 013 A from Q to P

(D) 013 A from P to Q

Answer (C)

Solution

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Electrostatics

Difficulty Easy (embibe predicted high weightage)

19 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the

magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on the

outer solenoid due to the inner one Then

(A) 1198651 = 1198652 = 0

(B) 1198651 is radially inwards and 1198652 is radially outwards

(C) 1198651 is radially inwards and 1198652 = 0

(D) 1198651 is radially outwards and 1198652 = 0

Answer (A)

Solution

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o

there4 force on 1198782 by 1198781 = 0

In the uniform field of 1198782 1198781 behaves as a magnetic dipole

there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

20

Two long current carrying thin wires both with current I are held by insulating threads of length L and are in

equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have mass λ

per unit length then the value of I is

(g = gravitational acceleration)

(A) sinθradicπλgL

μ0 cosθ

(B) 2sinθradicπλgL

μ0 cosθ

(C) 2radicπgL

μ0tanθ

(D) radicπλgL

μ0tanθ

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Answer (B)

Solution

By equilibrium of mix

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

21 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different

orientations as shown in the figure below

If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the

loop would be in (i) stable equilibrium and (ii) unstable equilibrium

(A) (a) and (b) respectively

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(B) (a) and (c) respectively

(C) (b) and (d) respectively

(D) (b) and (c) respectively

Answer (C)

Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =

times and potential energy U is given as

119880 = minus = minus119872 119861 119888119900119904120579

When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900

rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB

As 120579 = 180119900

Unstable equilibrium in (d)

Topic Magnetism

Difficulty Easy (embibe predicted high weightage)

22 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery of

15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then at 119905 =

0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the circuit will

be (ℓ5 cong 150)

(A) 100 mA

(B) 67 mA

(C) 67 mA (D) 067 mA

Answer (D)

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Solution

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Magnetism

Difficulty Moderate (embibe predicted high weightage)

23 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a

distance of 1 m from the diode is

(A) 173 Vm

(B) 245 Vm

(C) 548 Vm

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(D) 775 Vm

Answer (B)

Solution

Topic Optics

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Difficulty Easy (embibe predicted high weightage)

24 Monochromatic light is incident on a glass prism of angle A If the refractive index of the

material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get transmitted

through the face AC of the prism provided

(A) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1

120583))]

(B) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1

120583))]

(C) 120579 gt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

(D) 120579 lt cosminus1 [120583 sin (119860 + sinminus1 (1

120583))]

Answer (A)

Solution

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Difficulty Moderate (embibe predicted high weightage)

25 On a hot summer night the refractive index of air is smallest near the ground and increases with

height from the ground When a light beam is directed horizontally the Huygensrsquo principle leads

us to conclude that as it travels the light beam

(A) Becomes narrower

(B) Goes horizontally without any deflection

(C) Bends downwards (D) Bends upwards

Answer (D)

Solution Consider air layers with increasing refractive index

At critical angle it will bend upwards at interface This process continues at each layer and light ray bends

upwards continuously

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

26 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25 cm

the minimum separation between two objects that human eye can resolve at 500 nm

wavelength is

(A) 1 120583119898 (B) 30 120583119898

(C) 100 120583119898

(D) 300 120583119898

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Answer (B)

Topic Optics

Difficulty Moderate (embibe predicted high weightage)

27 As an electron makes transition from an excited state to the ground state of a hydrogen like atomion

(A) Its kinetic energy increases but potential energy and total energy decrease

(B) Kinetic energy potential energy and total energy decrease

(C) Kinetic energy decreases potential energy increases but total energy remains same

(D) Kinetic energy and total energy decreases but potential energy increases

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Answer (A)

Solution 119880 = minus1198902

41205871205760119903

U = potential energy

119896 =1198902

81205871205760119903

K = kinetic energy

119864 = 119880+ 119896 = minus1198902

81205871205760119903

119864 = Total energy

there4 as electron excites from excited state to ground state k increases U amp E decreases

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

28 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the

choices given below the list

List-I List-II A Franck-Hertz Experiment (i) Particle nature of light

B Photo-electric experiment (ii) Discrete energy levels of atom C Davison-Germer Experiment (iii) Wave nature of electron

D (iv) Structure of atom

(A) Aminus (i) B minus (iv)Cminus (iii)

(B) Aminus (ii) B minus (iv)Cminus (iii)

(C) Aminus (ii) B minus (i) C minus (iii)

(D) Aminus (iv) B minus (iii) Cminus (ii)

Answer (C)

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to

confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states

Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer

momentum and energy due to collision

Davisson-Germer experment = this experiment shows the wave nature of electron

Topic Modern Physics

Difficulty Easy (embibe predicted high weightage)

29 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz The

frequency of the resultant signal isare

(A) 2 MHz only

(B) 2005 kHz and 1995 kHz

(C) 2005 kHz 2000 kHz and 1995 kHz

(D) 2000 kHz and 1995 kHz

Answer (C)

Solution

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Communication Systems

Difficulty Easy (embibe predicted high weightage)

30 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged to

1198760 and then connected to the L and R as shown below

If a student plots graphs of the square of maximum charge (119876119872119886119909

2 ) on the capacitor with time (t)

for two different values 1198711 and 1198712(1198711 gt 1198712) of then which of the following represents this graph

correctly (plots are schematic and not drawn to scale)

(A)

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(B)

(C)

(D)

Answer (A)

Solution

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Magnetism

Difficulty Difficult (embibe predicted high weightage)

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Chemistry

31 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin

(A) 1

103

(B) 1

206

(C) 2

309

(D) 1

412

Solution (D) 2C8H7SO3minusNa+ +Ca(aq)

2+ rarr (C8H7SO3)2Ca + 2Na+

2 moles (resin) = 412 g equiv 40 g Ca2+ ions

= 1 moles of Ca2+ ions

Moles of Ca2+ gfrasl of resin =1

412

Topic s-Block elements and Hydrogen Difficulty Level Moderate (Embibe predicted Low Weightage) 32 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å

The radius of sodium atom is approximately

(A) 186 Å

(B) 322 Å

(C) 572 Å

(D) 093 Å

Solution (A) For BCC

4r = radic3a

r =radic3

4 a =

1732

4times 429

= 186 Å

Topic Solid State

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Difficulty Level Easy [Embibe predicted Low Weightage]

33 Which of the following is the energy of a possible excited state of hydrogen

a +136 eV

b -68 eV

c -34 eV

d +68 eV

Solution (C) En = minus136

n2eV

Where n = 2 rArr E2 = minus340 eV

Topic Atomic Structure

Difficulty Level Easy [Embibe predicted high weightage]

34 The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is

(A) Ion-ion interaction (B) Ion-dipole interaction

(C) London force (D) Hydrogen bond

Solution (B) For Ion-dipole interaction

F prop μ (dE

dr)

μ is dipole moment of dipole

r is distance between dipole and ion

prop μd

dr(1

r2)prop

μ

r3

Ionminus ion prop1

r2

London force prop1

r6

Hydrogen bond minusDipole minus dipole prop1

r4

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Chemical Bonding

Difficulty Level Easy [Embibe predicted high weightage]

35 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)

The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is

the standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)

(A) R(298)ln(16 times 1012) = 86600 (B) 86600 + R(298)ln(16 times 1012)

(C) 86600 minusln(16times1012 )

R(298)

(D)05[2 times 86600 minus R(298)ln(16 times 1012)]

Solution (D) ΔGreaco = minus2303 RTlogKP

= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac

o = 2ΔGf(NO2)o minus2ΔGf(NO)g

o

= 2ΔGf(NO2)o minus2 times 866 times 103

2ΔGf(NO2)

o = minusR(298)ln(16 times 1012) + 2 times 86600

ΔGf(NO2)o = 86600 minus

R(298)

2ln(16 times 1012)

= 05[2times 86600 minus R(298)ln(16times 1012)] Topic Thermochemistry and thermodynamics Difficulty Level Moderate [Embibe predicted high weightage]

36 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was

dissolved in 100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1)

of the substance is

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(A) 32

(B) 64

(C) 128

(D) 488

Solution (B) ΔP = 185 minus 183 = 2 torr

ΔP

Po=2

185= XB =

12M

(12M)+

10058

M(CH3)2CO = 15 times 2+ 16 + 12 = 58 g molefrasl

12

M≪100

58

rArr2

185=12

Mtimes58

100

M =58 times 12

100times185

2

= 6438 asymp 64 g molefrasl

Topic Liquid solution

Difficulty Level Easy [Embibe predicted high weightage]

37 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given time

the composition of the reaction mixture is [A] =1

2 [B] = 2 and [C] =

1

2 The reaction proceeds in

the [R = 8314 J K minusmol e = 2718]frasl

(A) Forward direction because Q gt KC (B) Reverse direction because Q gt KC

(C) Forward direction because Q lt KC (D) Reverse direction because Q lt KC

Solution (B) ∆Go = minusRT In KC

24942 = minus8314times 300 in KC

24942 = minus8314times 300 times 2303log KC

minus24942

2303times300times8314= minus044 = log KC

logKC = minus044 = 1 56

KC = 036

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Chemical Equilibrium

Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]

38 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper

deposited at the cathode is (Atomic mass of Cu = 635 amu) (A) 0 g

(B) 635 g (C) 2 g

(D) 127 g

Solution (B) 2F equiv 2 Eqs of Cu

= 2 times635

2= 635 g

Topic Electrochemistry

Difficulty Level Easy [Embibe predicted high weightage]

39 Higher order (gt 3) reactions are rare due to

(A) Low probability of simultaneous collision of all the reacting species

(B) Increase in entropy and activation energy as more molecules are involved (C) Shifting of equilibrium towards reactants due to elastic collisions

(D) Loss of active species on collision

Solution (A) Probability of an event involving more than three molecules in a collision are

remote

Topic Chemical kinetics

Difficulty Level Easy [Embibe predicted high weightage]

40 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was filtered and the strength of the filtrate was found to be 0042 N The

amount of acetic acid adsorbed (per gram of charcoal) is

(A) 18 mg (B) 36 mg

(C) 42 mg (D) 54 mg

Solution (D) Meqs of CH3COOH (initial) = 50times 006 = 3 Meqs

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Meqs CH3COOH (final) = 50times 0042 = 21 Meqs

CH3 COOH adsorbed = 3 minus 21 = 09 Meqs

= 9 times10minus1times 60 g Eq frasl times 10minus3 g

= 540 times 10minus4 = 0054 g

= 54 mg

Per gram =54

3= 18 mg gfrasl of Charcoal

Topic Surface chemistry

Difficulty Level Moderate [Embibe predicted high weightage]

41 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively

(A) 136 140 and 171

(B) 136 171 and 140 (C) 171 140 and 136

(D) 171 136 and 140

Solution (C) As Z

euarr ionic radius decreases for isoelectronic species

N3minus(Z

e)=

7

10

O2minus(Z

e)=

8

10

Fminus(Z

e) =

9

10

N3minus gt O2minusgt Fminus

Topic Periodicity

Difficulty Level Easy [Embibe predicted Low Weightage]

42 In the context of the Hall-Heroult process for the extraction of Al which of the following statements

is false (A) CO and CO2 are produced in this process (B) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity

(C) Al3+ is reduced at the cathode to form Al (D) Na3AlF6 serves as the electrolyte

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Solution (D) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion

temperature and provides good conductivity

Topic Metallurgy

Difficulty Level Moderate [Embibe predicted high weightage]

43 From the following statements regarding H2O2 choose the incorrect statement

(A) It can act only as an oxidizing agent (B) It decomposes on exposure to light

(C) It has to be stored in plastic or wax lined glass bottles in dark (D) It has to be kept away from dust

Solution (A) It can act both as oxidizing agent and reducing agent

Topic s-Block elements and hydrogen

Difficulty Level Easy [Embibe predicted Low Weightage]

44 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater

than its lattice enthalpy

(A) CaSO4

(B) BeSO4 (C) BaSO4

(D) SrSO4

Solution (B) ΔHHydration gt ΔHLattice

Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp

for BeSO4 is very high

Topic s-Block elements and hydrogen Difficulty Level Easy [Embibe predicted Low Weightage]

45 Which among the following is the most reactive

(A) Cl2 (B) Br2 (C) I2

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(D)ICI

Solution (D) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive

Topic Group 15 ndash 18

Difficulty Level Moderate [Embibe predicted high weightage]

46 Match the catalysts to the correct process

Catalyst Process

A TiCl3 i Wacker process B PdCl2 ii Ziegler ndash Natta

polymerization C CuCl2 iii Contact process

D V2O5 iv Deaconrsquos process

(A) A rarr iii B rarr ii C rarr iv D rarr i

(B) A rarr ii B rarr i C rarr iv D rarr iii

(C) A rarr ii B rarr iii C rarr iv D rarr i

(D) A rarr iii B rarr i C rarr ii D rarr iv

Solution (B) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in

water in the presence of tetrachloropalladate (II) as the catalyst In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred

In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts including copper chloride(CuCl2)

In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are

usually used in combination with different organo aluminium co-catalyst

Topic Transition metals

Difficulty Level Easy [Embibe predicted high weightage]

47 Which one has the highest boiling point

(A) He

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(B) Ne (C) Kr

(D) Xe

Solution (D) Due to higher Vander Waalrsquos forces Xe has the highest boiling point

Topic Group 15 ndash 18

Difficulty Level Easy [Embibe predicted high weightage]

48 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]

+ is (py = pyridine)

(A) 2 (B) 3

(C) 4 (D) 6

Solution (B) Complexes with general formula [Mabcd]+ square planar complex can have three isomers

Topic Coordination compounds

Difficulty Level Moderate [Embibe predicted high weightage]

49 The color of KMnO4 is due to (A) M rarr L charge transfer transition

(B) dminus d transition (C) L rarr M charge transfer transition

(D) σ minus σlowast transition

Solution (C) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)

Topic Coodination compounds

Difficulty Level Easy [Embibe predicted high weightage]

50 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these do not react to form oxides of nitrogen

Reason The reaction between nitrogen and oxygen requires high temperature

(A) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(B) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion

(C) The Assertion is incorrect but the Reason is correct (D)Both the Assertion and Reason are incorrect

Solution (B) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion

and reason are correct and reason is correct explanation Topic Group 15 ndash 18 Difficulty Level Moderate [Embibe predicted high weightage]

51 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg

of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80)

(A) 24

(B) 36 (C) 48

(D) 60

Solution (A) Rminus BrCarius methodrarr AgBr

250 mg organic compound is RBr

141 mg AgBr rArr 141 times80

188mg Br

Br in organic compound

= 141 times80

188times

1

250times 100 = 24

Topic Stoichiometry Difficulty Level Moderate [Embibe predicted high weightage] 52 Which of the following compounds will exhibit geometrical isomerism (A) 1 - Phenyl - 2 - butane (B) 3 - Phenyl - 1 ndash butane (C) 2 - Phenyl - 1 ndash butane (D)1 1 - Diphenyl - 1 ndash propane

Solution (A) 1 - Phenyl - 2 - butene

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic General Organic Chemistry and Isomerism Difficulty Level Easy [Embibe predicted high weightage]

53 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis

(A)

(B)

(C)

(D)

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Solution (B)

Topic Hydrocarbons Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

54 The synthesis of alkyl fluorides is best accomplished by (A) Free radical fluorination (B) Sandmeyerrsquos reaction (C) Finkelstein reaction (D)Swarts reaction

Solution (D) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2CoF2 SbF3

CH3 Br+AgF rarr CH3 minusF+ AgBr

The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro

hydrocarbons such as CCl2F2 is known as swarts reaction

Topic Alkyl and Aryl halides

Difficulty Level Easy [Embibe predicted high weightage]

55 In the following sequence of reactions

Toluene KMnO4rarr A

SOCl2rarr B

H2 Pdfrasl BaSO4rarr C

The Product C is

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(A) C6H5COOH

(B) C6H5CH3 (C) C6H5CH2OH

(D) C6H5CHO

Solution (D)

Topic Aldehyde and Ketone

Difficulty Level Easy [Embibe predicted high weightage]

56 In the reaction

The product E is

(A)

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(B)

(C)

(D)

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Solution (C)

Topic Amines

Difficulty Level Easy [Embibe predicted Low Weightage]

57 Which polymer is used in the manufacture of paints and lacquers

(A) Bakelite

(B) Glyptal (C) Polypropene

(D) Poly vinyl chloride

Solution (B) Glyptal is polymer of glycerol and phthalic anhydride

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Polymers

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

58 Which of the vitamins given below is water soluble

(A) Vitamin C

(B) Vitamin D

(C) Vitamin E

(D) Vitamin K

Solution (A) B complex vitamins and vitamin C are water soluble vitamins that are not stored in the body and

must be replaced each day

Topic Biomolecules

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

59 Which of the following compounds is not an antacid

(A) Aluminium hydroxide

(B) Cimetidine

(C) Phenelzine (D) Rantidine

Solution (C) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a

type of medication that can control the acid levels in stomach Working of antacids Antacids counteract (neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the

symptoms of heartburn and relieves pain

Topic Chemistry in everyday life

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

60 Which of the following compounds is not colored yellow (A) Zn2[Fe(CN)6] (B) K3[Co(NO2)6]

(C) (NH4)3[As (Mo3O10)4]

(D) BaCrO4

Solution (A) Cyanides not yellow

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

BaCrO4 minusYellow

K3[Co(NO2)6]minus Yellow(NH4)3[As(Mo3O10)4] minus Yellow

Zn2[Fe(CN)6] is bluish white

Topic Salt Analysis

Difficulty Level Easy [Embibe predicted easy to score (Must Do)]

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Mathematics 61 Let A and B be two sets containing four and two elements respectively Then the number of

subsets of the set 119860 times 119861 each having at least three elements is

(A) 219

(B) 256

(C) 275

(D) 510

Answer (A)

Solution

119899(119860) = 4

119899(119861) = 2

Number of elements in 119860 times 119861 = 2 sdot 4 = 8

Number of subsets having at least 3 elements

= 28minus81198620 minus 1198628 1 minus 1198628 2

= 256 minus 1minus 8minus 28 = 219

Topic Set and Relations

Difficulty level Moderate(embibe predicted high weightage)

62 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex

numbers such that 1199111minus21199112

2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius 2

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(D) Circle of radius radic2

Answer (C)

Solution

Given 1199111minus 211988522minus11991111199112

is unimodular

rArr | 1199111minus 211988522minus11991111199112

| = 1

|1199111minus 21198852 | =|2 minus 11991111199112|

Squaring on both sides

|1199111minus 21198852 |2= |2 minus 11991111199112|

2

(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)

Topic Complex Number

Difficulty level Easy (embibe predicted high weightage)

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

63 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the

value of 11988610minus21198868

21198869 is equal to

(A) 6

(B) -6

(C) 3

(D) -3

Answer (C)

Solution

Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0

rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0

rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)

11988610 minus2119886821198869

= (12057210 minus12057310) minus 2(1205728 minus 1205738)

2(1205729 minus1205739)

= (12057210minus21205728 )minus(12057310minus21205738)

2(1205729minus1205739)

= 1205728 (1205722minus2)minus1205738(1205732minus2)

2(1205729minus1205739)

= 1205728 (6120572)minus1205738(6120573)

2(1205729minus1205739) [∵ 119865119903119900119898 (1)]

= 61205729minus61205739

2(1205729minus1205739)= 3

Topic Quadratic Equation amp Expression

Difficulty level Moderate (embibe predicted high weightage)

64 If 119860 = [1 2 22 1 minus2119886 2 119887

] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity

matrix then the ordered pairs (a b) is equal to

(A) (2minus1)

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(B) (minus21)

(C) (21)

(D) (minus2 minus1)

Answer (D)

Solution

119860119860119879 = 9119868

[1 2 22 1 minus2119886 2 119887

] [1 2 1198862 1 22 minus2 119887

] = [9 0 00 9 00 0 9

]

[

1(1) + 2(2)

2(1) + 1(2)119886(1) + 2(2)

+ 2(2)

minus 2(2)+ 119887(2)

1(2) + 2(1)

2(2) + 1(1)119886(2) + 2(1)

+ 2(minus2)

minus 2(minus2)+ 119887(minus2)

1(119886) +

2(119886) +119886(119886) +

2(2) + 2(119887)

1(2) minus 2(119887)2(2) + 119887(119887)

]

= [9 0 00 9 00 0 9

]

[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2

119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4] = [

9 0 00 9 00 0 9

]

119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904

119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0

and 1198862 + 1198872 +4 = 9

____________________

3119886 + 6 = 0

119886 = minus2 rArr minus2 + 2119887 + 4 = 0

119887 = minus1

The third equation is useful to verify whether this multiplication is possible

Topic Matrices

Difficulty level Easy (embibe predicted high weightage)

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

65 The set of all values of 120582 for which the system of linear equations

21199091 minus21199092 + 1199093 = 1205821199091

21199091 minus31199092 + 21199093 = 1205821199092

minus1199091 +21199092 = 1205821199093 has a non-trivial solution

(A) Is an empty set

(B) Is a singleton

(C) Contains two elements

(D) Contains more than two elements

Answer (C)

Solution

(2 minus 120582)1199091 = 21199092 +1199093 = 0

21199091 minus (120582 + 3)1199092 + 21199093 = 0

minus 1199091 + 21199092 minus 1205821199093 = 0

The systems of linear equations will have a non-trivial solution

rArr |2 minus 1205822minus1

minus2

minus120582 minus 32

12minus120582| = 0

rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0 21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0

minus1205823 minus 1205822 + 5120582 minus 3 = 0 1205823 + 1205822 minus 5120582 + 3 = 0 (120582 minus 1) (1205822 +2120582 minus 3) = 0

(120582 minus 1) (120582+ 3) (120582minus 1) = 0 rArr 120582 = 3minus1minus1

Topic Determinants

Difficulty level Easy (embibe predicted high weightage)

66 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and

8 without repetition is

(A) 216

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(B) 192

(C) 120

(D) 72

Answer (B)

Solution

4 digit and 5 digit numbers are possible

4 digit numbers

678

uarr uarr uarr3 4 3

uarr2

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72

5 digit numbers

uarr uarr uarr5 4 3

uarr uarr2 1

119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120

there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192

Topic Permutation amp Combination

Difficulty level Easy (embibe predicted Low Weightage)

67 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)50

is

(A) 1

2(350 +1)

(B) 1

2(350)

(C) 1

2(350 minus1)

(D) 1

2(250 + 1)

Answer (A)

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Solution

Integral powers rArr odd terms

119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )

50+ (1 + 2radic119909 )

50

2

119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50

2

= 1 + 350

2

Topic Binomial Theorem

Difficulty level Moderate (embibe predicted high weightage)

68 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three

geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662

4 + 11986634 equals

(A) 4 1198972119898119899

(B) 4 1198971198982119899

(C) 4 1198971198981198992

(D) 4 119897211989821198992

Answer (B)

Solution

m is the AM of 119897 119899

rArr 119898 =119897+119899

2 hellip(1)

1198661 1198662 1198663 are GM of between l and n

rArr 1198971198661 11986621198663 119899 are in GP

119899 = 119897(119903)4 rArr 1199034 =119899

119897 hellip(2)

1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903

3

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

1198661

4 +211986624 + 1198663

4 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312

= 11989741199034[1 + 21199034 + 1199038]

= 1198974 times119899

119897[1 + 1199034]2 = 1198973119899 times [1 +

119899

119897]2

= 1198973 times 119899 times(119897+119899)2

1198972

= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]

= 41198971198982119899

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

69 The sum of first 9 terms of the series 13

1+13+23

1+3+13+23+33

1+3+5+⋯ 119894119904

(A) 71

(B) 96

(C) 142

(D) 192

Answer (B)

Solution

119879119899 =13+23+⋯ +1198993

1+3+⋯+(2119899minus1)=

1198992 (119899+1)2

4

1198992=(119899+1)2

4

Sum of 9 terms = sum(119899+1)2

4

9119899=1

=1

4times [22 +32 +⋯+102]

=1

4[(12 +22 +⋯+102) minus 12]

=1

4[101121

6minus 1]

=1

4[384] = 96

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Topic Progression amp Series

Difficulty level Moderate (embibe predicted high weightage)

70 119897119894119898119909 rarr 0

(1minuscos 2119909) (3+cos 119909)

119909 tan 4119909 is equal to

(A) 2

(B) 3

(C) 2

(D) 1

2

Answer (C)

Solutions

lim119909rarr119900

(1 minus cos2119909) (3+ cos119909)

119909 tan 4119909

= lim119909rarr119900

2 sin2 119909 sdot (3 + cos119909)

119909 sdot tan 4119909

= lim119909rarr119900

2 sdot (sin 119909119909 )

2

sdot (3 + cos119909)

tan 41199094119909 sdot 4

= 2 sdot (1)2 sdot (3 + 1)

1 sdot 4

= 2

Topic LimitContinuity amp Differentiability

Difficulty level Easy (embibe predicted high weightage)

71 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5

is differentiable then the value of k + m is

(A) 2

(B) 16

5

(C) 10

3

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(D) 4

Answer (A)

Solution

119892(119909) is differentiable at x = 3

rArr 119892 (119909) is continuous at x = 3

there4 LHL = RHL = 119892(3)

lim119909rarr3minus

119892(119909) = lim119909rarr3+

119892(119909) = 119892(3)

119896radic3+ 1 = 119898(3) + 2

2119896 = 3119898 + 2

119896 = 3

2 119898 + 1 helliphellip(1)

119892(119909) is differentiable

rArr 119871119867119863 = 119877119867119863

119892prime(119909) = 119896

2radic119909 + 1119898

0 lt 119909 lt 33 lt 119909 lt 5

lim119909rarr3minus

119892prime(119909) = lim119909rarr3+

119892prime(119909)

119896

2radic3+ 1= 119898

119896 = 4119898

From (1) 4119898 = 3

2119898+ 1

5119898

2= 1

rArr 119898 = 2

5 119896 =

8

5

119896 +119898 = 2

5+

8

5= 2

Topic LimitContinuity amp Differentiability

Difficulty level Moderate (embibe predicted high weightage)

72 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)

(A) Does not meet the curve again

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(B) Meets the curve again in the second quadrant

(C) Meets the curve again in the third quadrant

(D) Meets the curve again in the fourth quadrant

Answer (D)

Solution

1199092 +2119909119910minus 31199102 = 0 (119909 + 3119910) (119909minus 119910) = 0

Pair of straight lines passing through origin

119909 + 3119910 = 0 or 119909 minus 119910 = 0

Normal exists at (1 1) which is on 119909 minus 119910 = 0

rArr Slope of normal at (1 1) = minus1

there4 Equation of normal will be

119910 minus 1 = minus1 (119909 minus 1) 119910 minus 1 = minus119909 + 1

119909 + 119910 = 2

Now find the point of intersection with 119909 + 3119910 = 0

119909 + 119910 = 2

119909 + 3119910 = 0 ____________

minus 2119910 = 2 rArr 119910 = minus1 119909 = 3 (3 1) lies in fourth quadrant

Topic Straight Lines

Difficulty level Easy (embibe predicted high weightage)

73 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0

[1 +119891(119909)

1199092] = 3 119905ℎ119890119899 119891(2) is equal to

(A) minus8

(B) minus 4

(C) 0

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(D) 4

Answer (C)

Solution

Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890

119897119894119898119909 ⟶ 0

[1 +119891(119909)

1199092] = 3

119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888 +119889

119909+

119890

1199092] = 3

This limit exists when d = e = 0

So 119897119894119898119909 ⟶ 0

[1 + 1198861199092 +119887119909 + 119888] = 3

rArr 119888 + 1 = 3

119888 = 2

It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0

119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909

119909(41198861199092 +3119887119909 + 2119888) = 0

12 are roots of quadratic equation

rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887

4 119886= 1+ 2 = 3

rArr 119887 = minus4119886

Product of roots = 2119888

4119886= 12 = 2

rArr 119886 =119888

4

119886 =1

2 119887 = minus2

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

there4 119891(119909) =

1

2 1199094 minus 21199093 + 21199092

119891 (2) = 8minus 16+ 8

= 0

Topic Application of Derivatives

Difficulty level Moderate ( embibe predicted high weightage)

74 The integral int119889119909

1199092 (1199094+1)34

equals

(A) (1199094+1

1199094)

1

4+ 119888

(B) (1199094+ 1)1

4 + 119888

(C) minus(1199094 +1)1

4 + 119888

(D) minus (1199094+1

1199094)

1

4+ 119888

Answer (D)

Solution

119868 = int119889119909

1199092(1199094+1)34

= int119889119909

1199092times 1199093 (1+1

1199094)

34

119875119906119905 1 +1

1199094= 119905

rArr minus4

1199095119889119909 = 119889119905

there4 119868 = intminus119889119905

4

11990534

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

= minus1

4times int 119905

minus3

4 119889119905

=minus1

4times (

11990514

1

4

)+ 119888

= minus (1+1

1199094)

1

4 + c

Topic Indefinite Integral

Difficulty level Moderate (embibe predicted Low Weightage)

75 The integral intlog 1199092

log 1199092+log (36 minus 12119909 + 1199092 )119889119909

4

2 is equal to

(A) 2

(B) 4

(C) 1

(D) 6

Answer (C)

Solution

119868 = intlog 1199092

log1199092 + log(6 minus 119909)2

4

2

119889119909

119868 = intlog(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909

119887

119886

119887

119886

]

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092

4

2

119889119909

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

2119868 = intlog 1199092 + log(6 minus 119909)2

log(6 minus 119909)2 + log1199092 119889119909

4

2

2119868 = int1 119889119909

4

2

2119868 = [119909]24

2119868 = 4 minus 2

119868 = 1

Topic Definite Integral

Difficulty level Easy (embibe predicted high weightage)

76 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is

(A) 7

32

(B) 5

64

(C) 15

64

(D) 9

32

Answer (D)

Solution

Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1

(4119909 minus 1)2 = 2119909

161199092 minus10119909 + 1 = 0

(8119909 minus 1) (2119909 minus 1) = 0

119909 =1

21

8

119909 =1

2 rArr 119910 = 4 (

1

2)minus 1 = 1

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

119909 = 1

8 rArr 119910 = 4 (

1

8)minus 1 =

minus1

2

Required area = int (1199091minus1

2 2

minus 1199091) 119889119910

= int [1199102

2minus (119910 + 1

4)]

1

minus12

119889119910

=1

6times lfloor1199103rfloorminus1

2

1 minus 1

8 lfloor1199102rfloorminus1

2

1 minus1

4 lfloor119910rfloorminus1

2

1

=1

6 [1 +

1

8] minus

1

8 [1 minus

1

4] minus

1

4 [1 +

1

2]

=1

6times9

8minus1

8times3

4minus1

4times3

2

=3

16minus3

32minus3

8

=6 minus 3 minus 12

32=minus9

32 squnits

Topic Area under Curves

Difficulty level Easy (embibe predicted high weightage)

77 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910

119889119909+119910 = 2119909 log119909 (119909 ge 1)

Then 119910 (119890) is equal to

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(A) e

(B) 0

(C) 2

(D) 2e

Answer (B)(incorrect Question)

Solution 119889119910

119889119909+ (

1

119909 log 119909) 119910 = 2

Integrating factor = 119890int119875119889119909

= 119890int

1

119909 log119909 119889119909

= 119890 log (log 119909)

= log 119909

The solution will be

119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888

119910 sdot log119909 = int2 sdot log 119909 + 119888

119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888

Let 119875(11199101) be any point on the curve

1199101(0) = 2(0 minus 1) + 119888

119888 = 2

When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2

119910 = 2

Topic Differential Equation

Difficulty level Easy (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

78 The number of points having both co-ordinates as integers that lie in the interior of the

triangle with vertices (0 0) (0 41) and (41 0) is

(A) 901

(B) 861

(C) 820

(D) 780

Answer (D)

Solution

119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873

1199091 +1199101 lt 41

there4 2 le 1199091 +1199101 le 40

Number of points inside = Number of solutions of the equation

1199091 +1199101 = 119899

2 le 119899 le 40 1199091 ge 1 1199101 ge 1

(1199091 minus1) + (1199101 minus 1) = (119899 minus 2) (119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904

119899+119903minus1119862 119903 )

Number of solutions = 2+(119899minus2)minus1119862119899minus2

= 119899minus1 119862119899minus2

= 119899 minus 1

We have 2 le 119899 le 40

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

there4 Number of solutions = 1 + 2 +hellip+39

= 39 times40

2

= 780

79 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin

119877 is a

(A) Straight line parallel to x-axis

(B) Straight line parallel to y-axis

(C) Circle of radius radic2

(D) Circle of radius radic3

Answer (C)

Solution

Given family of lines 1198711 + 1198712 = 0

Let us take the lines to be

1198712+ 120582(1198711 minus 1198712) = 0 (119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0 (1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0

Let Image of (2 3) be (h k) ℎ minus 2

1 + 120582=

119896 minus 3

minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)

(1 + 120582)2 + (2 + 120582)2

ℎminus2

120582+1=

119896minus3

minus(120582+2)=

minus2(minus1)

(1+120582)2+(2+120582)2 (1)

ℎ minus 2

120582 + 1=

119896 minus 3

minus(120582 + 2)rArr 120582 + 2

120582 + 1= 3 minus 119896

ℎ minus 2

1 + 1

120582+1=

3minus119896

ℎminus2

1

120582+1=

5minusℎminus119896

ℎminus2 (2)

From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4

(120582+1)2+(120582+2)2 (3)

From (1) and (3)

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

ℎ minus 2

120582 + 1=

2

(120582 + 1)2 + (120582 + 2)2

5 minus ℎ minus 119896 = 1

2 [(ℎ minus 2)2 + (119896 minus 3)2]

10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13

1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0

Radius = radic1 + 4minus 3

= radic2

Topic Straight lines

Difficulty level Difficult (embibe predicted high weightage)

80 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +

1199102 +6119909 + 18119910+ 26 = 0 is

(A) 1

(B) 2

(C) 3

(D) 4

Answer (C)

Solution

1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0

1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5

1199092 + 1199102 + 6119909 + 18119910+ 26 = 0

1198622(minus3minus9) 1199032 = radic32 +92 minus 26

= radic90 minus 26 = 8

11986211198622 = radic52 + 122 = 13

11986211198622 = 1199031 + 1199032

rArr Externally touching circles

rArr 3 common tangents

Topic Circle

Difficulty level Easy (embibe predicted high weightage)

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

81 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the

latera recta to the ellipse 1199092

9+1199102

5= 1 is

(A) 27

4

(B) 18

(C) 27

2

(D) 27

Answer (D)

Solution

1199092

9+1199102

5= 1

119890 = radic1 minus5

9=2

3

1198862 = 9 1198872 = 5

Focii = (plusmn 1198861198900) = (plusmn 20)

Ends of latus recta = (plusmn 119886119890 plusmn 1198872

119886)

= (plusmn 2plusmn5

3)

Tangent at lsquoLrsquo is T = 0

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

2 sdot 119909

9+ 5

3 sdot 119910

5= 1

It cut coordinates axes at 119875 (9

2 0) 119886119899119889 119876(03)

Area of quadrilateral PQRS = 4(Area of triangle OPQ)

= 4 (1

2 sdot 9

2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904

Topic Ellipse

Difficulty level Easy (embibe predicted high weightage)

82 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the

line segment OQ internally in the ratio 1 3 then the locus of P is

(A) 1199092 = 119910

(B) 1199102 = 119909

(C) 1199102 = 2119909

(D) 1199092 = 2119910

Answer (D)

Solution

General point on 1199092 = 8119910 is 119876 (4119905 21199052)

Let P(h k) divide OQ in ratio 13

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(h k) = (1 (4119905)+3(0)

1+3 1(21199052 )+3(0)

1+3)

(h k) = (11990521199052

4)

h = t and 119896 =1199052

2

119896 =ℎ2

2

rArr 1199092 = 2119910 is required locus

Topic Parabola

Difficulty level Moderate (embibe predicted high weightage)

83 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2

3=119910+1

4=119911minus2

12

and the plane 119909 minus 119910 + 119911 = 16 is

(A) 2radic14

(B) 3radic21

(C) 8

(D) 13

Answer (D)

Solution

119909minus2

3=119910+1

4=119911minus1

12= 119905 (119904119886119910)

rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)

It lies on the plane 119909 minus 119910 + 119911 = 16

rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16

rArr 11119905 = 11

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

rArr 119905 = 1

there4 The point of intersection will be

(2 + 3(1)minus1 + 4(1)2 + 12(1))

= (5 3 14)

Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2

= radic42 +32 +122

= 13

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

84 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and

parallel to the plane 119909 + 3119910 + 6119911 = 1 is

(A) 2119909 + 6119910+ 12119911 = 13

(B) 119909 + 3119910 + 6119911 = minus7

(C) 119909 + 3119910 + 6119911 = 7

(D) 2119909 + 6119910 + 12119911 = minus13

Answer (C)

Solution

Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of the form

(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

It is parallel to plane 119909 + 3119910+ 6119911 = 1

rArr 2 + 120582

1= 120582 minus 5

3= 1 + 4120582

6 ne

minus(3 + 5120582)

11

2 + 120582

1= 120582 minus 5

3rArr 6 + 3120582 = 120582 minus 5

2120582 = minus11

120582 = minus11

2

there4 Required plane is

(2119909 minus 5119910+ 119911 minus 3) minus11

2 (119909 + 119910 + 4119911 minus 5) = 0

2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0

4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0

minus7119909 minus 21119910minus 42119911 + 49 = 0

rArr 119909 + 3119910 + 6119911 minus 7 = 0

Topic 3 - D Geometry

Difficulty level Easy (embibe predicted high weightage)

85 Let 119886 119888 be three non-zero vectors such that no two of them are collinear and (119886 times ) times 119888

=1

3 | | |119888 | |119886 | If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is

(A) 2radic2

3

(B) minusradic2

3

(C) 2

3

(D) minus2radic3

3

Answer (A)

Solution

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

(119886 times ) times 119888 = 1

3 | | |119888 |119886

(119886 sdot 119888 ) minus ( sdot 119888 ) 119886 = 1

3 | | |119888 | 119886

rArr 119886 sdot 119888 = 0 119886119899119889 sdot 119888 = minus1

3 | | |119888 |

| | |119888 | 119888119900119904120579 = minus1

3 | | |119888 |

119888119900119904120579 = minus1

3

119904119894119899120579 = radic1 minus1

9= radic

8

9 =

2radic2

3

Topic Vector Algebra

Difficulty level Moderate (embibe predicted high weightage)

86 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the

boxes contains exactly 3 balls is

(A) 22(1

3)11

(B) 55

3(2

3)11

(C) 55(2

3)10

(D) 220(1

3)12

Answer (A)(incorrect question)

Solution

B1 B2 B3

12 0 0

11 1 0

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

10 1 1

10 2 0

9 3 0

9 2 1

8 4 0

8 3 1

8 2 2

7 5 0

7 4 1

7 3 2

6 6 0

6 5 1

6 4 2

6 3 3

5 5 2

5 4 3

4 4 4

Total number of possibilities = 19

Number of favorable case = 5

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

Required probability =

5

19

INCORRECT SOLUTION

Required Probability = 121198629 sdot (1

3)3

sdot (2

3)9

= 55 sdot (2

3)11

The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo

Topic Probability

Difficulty level Difficult (embibe predicted high weightage)

87 The mean of the data set comprising of 16 observations is 16 If one of the observation

valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then

the mean of the resultant data is

(A) 168

(B) 160

(C) 158

(D) 140

Answer (D)

Solution

Given sum 119909119894+1615119894=1

16= 16

rArr sum119909119894+ 16 = 256

15

119894=1

sum119909119894 = 240

15

119894=1

Required mean = sum 119909119894+3+4+515119894=1

18=240+3+4+5

18

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

=

252

18= 14

Topic Statistics

Difficulty level Moderate (embibe predicted easy to score (Must Do))

88 If the angles of elevation of the top of a tower from three collinear points A B and C on a

line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶

119861119862 is

(A) radic3 ∶ 1

(B) radic3 ∶ radic2

(C) 1 ∶ radic3

(D) 2 ∶ 3

Answer (A)

Solution

tan 30119900 = ℎ

119909 + 119910+ 119911tan 450 =

ℎ

119910 + 119911 tan600 =

ℎ

119911

rArr 119909 + 119910 + 119911 = radic3 ℎ

119910 + 119911 = ℎ

119911 =ℎ

radic3

119910 = ℎ (1 minus 1

radic3)

119909 = (radic3minus 1)ℎ

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

119909

119910= (radic3 minus 1)ℎ

ℎ(radic3minus 1

radic3)

= radic3

1

Topic Height amp Distance

Difficulty level Moderate (embibe predicted easy to score (Must Do))

89 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) where |119909| lt

1

radic3 Then a value of y is

(A) 3119909minus1199093

1minus31199092

(B) 3119909+1199093

1minus31199092

(C) 3119909minus1199092

1+31199092

(D) 3119909+1199093

1+31199092

Answer (A)

Solution

tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909

1minus1199092) |119909| lt

1

radic3

= tanminus1 119909 + 2 tanminus1 119909

= 3tanminus1 119909

tanminus1 119910 = tanminus1 (3119909minus1199093

1minus31199092)

rArr 119910 =3119909minus1199093

1minus31199092

Topic Inverse Trigonometric Function

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))

90 The negation of ~119904 or (~119903 and 119904) is equivalent to

(A) 119904 and sim 119903

(B) 119904 and (119903 and sim 119904)

(C) 119904 or (119903 or sim 119904)

(D) 119904 and 119903

Answer (D)

Solution

~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]

= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]

= (119904 and 119903) or (119904 and (sim 119904))

= (119904 and 119903) or 119865

= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)

Topic Mathematical Reasoning

Difficulty level Moderate (embibe predicted easy to score (Must Do))