Physics - champsquare.co.in - 2 (ADVANCED) 21.11.2018.pdf · CHAMP SQUARE, Sunrise Forum, 2nd Floor (Debuka Nursing Home Lane), Circular Road, Ranchi – 834 001, Ph.: 0651-2562276

CHAMP SQUARE, Sunrise Forum, 2nd Floor (Debuka Nursing Home Lane), Circular Road, Ranchi – 834 001, Ph.: 0651-2562276

Physics Section – A

(One or More Than One Options Correct Type)

This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which only ONE or MORE THAN ONE is/are correct.

1. A particle (mass m and charge q) moves inside a fixed smooth groove (horizontal plane) in the shape of a sine

curve. In space, there exists a uniform electric field E as shown. The particle is projected with

m

qE3v

00

along the groove. Then

(A) The number of times it crosses the x-axis before coming to rest for the first time is 3

(B) The number of times it crosses the x-axis before coming to rest for the first time is 4

(C) The number of times it crosses the x-axis before coming to rest for the first time will be same for any acute

angle of the projected velocity with the groove

(D) The number of times it crosses the x-axis before coming to rest for the first time will be different for different

acute angles of the projected velocity with the groove

Sol. BC

Vtotal = K

Normal and mg does not work and friction = 0

200 mv

21)qE(

2. The switch S, after being closed for a very long time is now opened. Then

(A) The total amount of heat generated in the resistor R1 after that operation is

2

2

R2L

(B) The total amount of heat generated in the resistor R1 after that operation is

2

2

R4L

(C) The direction of current in L1 always remains the same

(D) The direction of current in L1 changes direction

R

S

RR1 LL2

R

LL1

x

yE

q,m

Origin

0v

0 02


Sol. BC

Initially the circuit is

Immediately after opening S, the circuit is as shown.

2

22

R4L2

R221Heat

3. A particle of mass m and charge q is located at origin. An uncharged

particle of mass m is moving along x-axis. They collide elastically such that

the uncharged particle makes an angle 60o with x-axis after collision. A

cylindrically symmetrical (radius R) magnetic field B = 0.01 T/s exists as

shown. (R = 2 m, 3

4V0 m/s, q = 1 C, m = 10 grams)

(A) The co-ordinates of charged particle at the moment it comes out of the

magnetic field are 34,134

(B) The co-ordinates of charged particle at the moment it comes out of the


(C) The co-ordinates of charged particle at the moment it comes out of the


(D) Net deflection of the charge particle is 45o

Sol. C

After collision, charged particle has speed v0cos 30o

along the centre of the field.

o90deflectionRm2BqmvRadius

34,134ordinatesCo

m1

m3

0v

o60o30

m

0v

y

x

R

)4,34( )q,m(

R

RR1

R/

L

R2/

R

RR1

L


4. The circuit shown in figure (i) is driven by an a.c. supply of voltage 3

t5sin16 volts. The capacitor (area of each

plate of capacitor = 10-4

m2) is suddenly removed at

310

7t second and then two geometrically identical

dielectrics are inserted as shown in figure (ii):

(A) The net charge density at the interface of the two dielectrics is – 3000 C/m

2

(B) The net charge density at the interface of the two dielectrics is – 1500 C/m2

(C) Negative charges reside on the immediate left of the interface

(D) Negative charges reside on the immediate right of the interface

Sol. BD

6

5

3

t5sin34VC

2net m/C1500

43

2

Where /2 is on the immediate left and - 3/4 is on the immediate right 5. The circuit shown in the figure has reached its steady state.

(A) The charge on the capacitor is 0.1 C

(B) The charge on the capacitor is 0.3 C

(C) Power developed by the 2 Volts battery is 1 W.

(D) Power developed by the 2 Volts battery is 2 W.

Sol. BD

In steady state, the circuit reduces to

By KVL,

VBA = 3 V

All power by 2 Volt battery = 2 1 = 2 W

6. An infinite line charge carry a uniform line charge of density is placed along the body

diagonal AG of a cube of side ‘a’ as shown. Then

(A) The flux of the electric field of this charge through one of the face of the cube is

032

a

A B

CD

E F

GH

V2

1

A1

A

B

V4

1

H1

V2

V4

1

F1.0

H5.0 2

2

4 H6.3F1.0

~

)3/t5(sin16

)i(Figure )ii(Figure

1K 2K

2K1

4K2


(B) The flux of the electric field of this charge through one of the face of the cube is

02

a3

(C) The flux of the electric field of this charge through one of the face of the cube is

034

a

(D) The flux of the electric field through the face ABFE is same as that through BCGF

Sol. AD

Each surface carries the same flux, the net flux is (total charge/0). 7. A direct current flows in a solenoid of length L and radius R(<< L), producing

a magnetic field of magnitude B0 inside the solenoid. P is a point at the axis of

solenoid and Q be a point well inside the solenoid. C1 be a circular loop of

radius r (<< R) in the vicinity of P such that its plane is perpendicular to the

axis of solenoid and C2 is another loop of same radius with its plane parallel to

C1 in the vicinity of Q, then

(A) Magnetic flux through C1 and C2 are equal.

(B) Magnetic flux through C1 and half that of through C2

(C) Direction of magnetic field at P and Q are same

(D) Direction of magnetic field at P and Q are opposite.

Sol. BC

If magnetic field at Q is B0, then magnetic field at P should be 2

B0 and direction of magnetic fields at point P and

Q are towards left. 8. A conducting frame in the shape of an equilatent triangle (mass m, side a)

carrying a current I is placed vertically on a horizontal rough surface

(coefficient of friction is ). A magnetic field exists such that ^

0 iyBB

. Then

(A) The maximum value B0 so that the frame does note rotate 2Ia

mg2

(B) The maximum value of B0 so that the frame does not rotate 2Ia

mg

(C) If seen from the top, the frame will have a tendency to rotate counter clockwise

(D) If seen from the top, the frame will have a tendency to rotate clockwise

O x

y

X X X X X X X X X X X X X X X X

R

P Q


Sol. AD

dF on each of the two symmetric elements 2

3yB

3

dy2I 0

a2

3

0

0

ya2

3

3

lydyB2

8aIBydyya

2

3IB

3

2 30

a2

3

0

0

4

mgaxdx

a

mg2

2/a

0

friction

For the frame to not rotate

4

Mga

8aIB 3

0

20

Ia

mg2B

9. Three identical resistors R1, R2 and R3 are connected as shown in figure. When

the switch S is closed

(A) Power dissipated in R1 increases

(B) Power dissipated in R2 increases

(C) Total power delivered by the cell increases

(D) Power dissipated in R2 is less than that in R1

ANS. ACD

10. During an experiment, an ideal gas is found to obey a condition constant2

[ = density of the gas]. The gas

is initially at temperature T, pressure P and density. The gas expands such that density changes to /2.

(A) The pressure of the gas changes to P2 .

(B) The temperature of the gas changes to T2 .

(C) The graph of the above process on the P – T diagram is parabola.

(D) The graph of the above process on the P – T diagram is hyperbola.

Sol. BD

2

22

1

2i

2 Pp constant

p

so A is wrong

By gas equation M

RTP

KTKT

222

so B is correct

KPTM

RTKPPRTM

P2

so D is correct

S

E

1R 2R

3R

y

dy


Section – C

(Only Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive).

1. A wire (of diameter D) is wound in the form of square with side varying continuously from ‘a’ to

‘b’ (virtually having no gap between two adjacent wire parts). A single turn of very small area A

is placed at the geometric centre of the square. The mutual inductance of the arrangement is

,abln

Dn

A2 0

then n is

Sol. 1

No. of turns in thickness dx

DdxdN

D2

abN

Ddx21

2

1x4

i4dB

0

abn

D

i2B

0

abn

d

iA2 0

abn

D

A2M

0

2. We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a

unit scale of this type whose length remains, says 10 cm. We can use a bimetallic strip made of brass and iron

each of different length whose length (both components) would change in such in way that difference between

their lengths remain constant. If αiron = 1.2 10-5

/ K and αbrass = 1.8 10-5

/K, if we take as length of each strip

10 n cm. Find the value of n.

Sol. 3

cm10ii brassiron at all temperature

cm10)t1(i)t1(i brassobrassiron

oiron

brassobrassiron

oiron ii

23

2.18.1

i

iobrass

oiron

cm20icm10i21 o

brassobrass

cm30iand oiron

o

a

b


3. In a perfectly insulating but viscous medium, a system consists of an

insulating charged hollow sphere (having charge Q) and a small dipole as

shown. The sphere is fixed and the dipole is movable, which is released

when – q is at a distance L(>> ) from O (centre of the sphere). The

speed of the dipole when the distance reduces to L/2 (assume drag force

to be a constant of magnitude F0 acting on each charge) is

,mLF

mL

nKQq 0

2

then n is

Sol. 3

dUdKdVdrag

2/L

L

v

0

2/L

L

30 dx

x

KQq2mvdv2dxF2

4. There are two very small dipoles (will masses and charges shown) of length d each. The lower

dipole is fixed while the upper is released from towards the lower with a very small speed.

The minimum distance between the dipole (04

1K,d

) is ,1GM

Kq

nd3

2

2

then n is

Sol. 6

From energy conservation Ei = Er

0 = UG + UE

- UG = UE

22

22

22

22

d

Kqkq2

d

GmGm2

2

2

2

2

22

22

2

2

d21

d2111

d

11d

11

Gm

kq

(by using binomial)

1d4

Gm

kq2

2

2

2

1Gm

kq

2d

2

2

5. An infinite wire carries a current I which is changing with time. If a square loop of side ‘a’

is placed at a distance ‘a’ from the wire (i.e. nearest side at a distance ‘a’ from the wire),

the induced emf is 1. If the loop is placed at a distance ‘a/3’. The emf induced in the

loop is 2. The ratio 1/2 is

Sol. 2

The ratio of the emf is the ratio of flux of the fields in the two cases. 6. Two point charges q each are placed at a distance R from each other in vacuum. The force

between them is F1. One of the charges is spread uniformly over the surface of a

hemisphere bowl of radius R, while the other is spread uniformly over the volume of a

sphere of radius ;2Rrr

and both have a common centre O. The force on the sphere is

,nF1 then n is

Rr

a

a

q,m

q,m

q,m

q,m

O

q

04

1K

q,m q,m


Sol. 2

For sphere of radius r, every point on the shell is external it behaves as a point charge. Now field due to

shell at centre 2Fq

4

R4/qF

41

0

2

2

0

7. In a gravity free space, two charged insulating wires (one finite and other finite) with

linear charge densities (- , + ) and + respectively are arranged as shown with a

common plane containing them. Infinite wire is fixed while the finite is movable

(initially at rest) with moment interia I. Its initial angular acceleration is

,2

cotlnsec1I9

tann0

2

then n is

ANS. 9 8. In the circuit shown, all the switches are open initially. Now they are closed one at a time, starting from S1, such

that when a switch is closed, its preceding switch is simultaneously opened. After closing the fifth switch, the

fourth capacitor (8C) is removed and connected (t = 0) to an uncharged capacitor of original capacity C (plate

separation d) which now has a dielectric (dielectric constant 2) filling half its volume, through a resister R:

The time t0 at which a proton should be placed at point P so that its net acceleration at that moment is zero

(given : In SI units, RCD = 10 and I0 < < RC, mp = 1.6 10-27

kg) is sec10 6 . Find the value of ( - ).

Sol. 5

Fourth capacitor has a potential difference of

V8V2732

3

immediately after removing.

Capacity of the capacitor with dielectric 2C3

2C2

2C

Circuit is

RC24/t190 e119Q3q Where Q0 = 64 C

00

P

)2/A(

3/qE

eEP = mg

R

qC64

)qC64(

q

q

PC8

R

C4

C2

C

V27

1S

2S

3S

2


9. A hollow conducting cone (mass m, radius R, height h) has a current I flowing

uniformly along its curved surface area. This cone is kept on a very rough surface and

a uniform magnetic field ^

0 iBB

exists in space. The maximum value of I so that the

cone does not topple about point P is given by ,RBn

mg14

0then n is

Sol. 7

Current in the element elementofarea)eccosR(R

i

02

B Br)dec(cosR2)eccosR(R

IBdd

2

IRBd

2I0h

m

BB

For not topple

mg R s

RB

mgh2I

0

10. Three material points of masses 10 kg , 20 kg and 30 kg are at the vertices of an equilateral triangle of side d.

The system is rotating in free space in such a way that under the mutual gravitational interaction of the three

particles, the system is neither expanding nor contracting. Velocity of this rotation is given by n 10-5

rad/sec. Universal gravitational constant is 2

11

2

M.mG 6.0 10 .

kg

Find n.

Ans. 6

x

y

P

r

cos

x

y

P

0B


Chemistry Section – A

(One or More Than One Options Correct Type)

This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of which only ONE or MORE THAN ONE is/are correct.

1. Which of the following compounds gives blood red colour with FeCl3 in the Lassaigne’s test?

(A)

NH2

SO3H

(B)

N2

+Cl

-

(C)

C NH2

S

NH2

(D)

SO3H

Ans. A, C Sol. Compounds containing both N and S give blood red colouration in the Lassaigne’s test. 2. Which of the following compounds are more basic than aniline?

(A) C

NH2

NH

NH2

(B) C

NH2

NH

CH3

(C)

NH2

OH

(D)

N..

Ans. A, B, C, D

Sol.

C NH

NH2

NH2

(pKa = 13.6) (12.4) (4.72) (10.95) (4.62)

C NH

CH3

NH2

NH2

OH

N

NH2

3. Which of the following can be resolved?

(A)

COOH

COOH

(B)

COOH

NO2 COOH

O2N


(C) N

CH3

C2H5. .

(D)

O

O

O

O

H

CH3

CH3

H

Ans. B, C

Sol.

COOH

COOH

O

O

O

O

H

CH3

CH3

H

and

are achiral and optically inactive.

4. Which of the following statements is/are correct for the compound (A)?

O

O

O

HOH2C

HOH2C

OH

OH

OH

OH

OH

OH

H

H(A)

(A) The compound (A) is a reducing sugar

(B) The compound (A) shows mutarotation

(C) The compound (A) can be obtained by the hydrolysis of cellulose

(D) The compound (A) has - 1, 4’ – glycosidic linkage

Ans. A, B, C, D

Sol. Compound (A) is cellobiose. It is a disaccharide 5. The keto form of which of the following compound is more stable than their enol form?

(A)

O

(B) N

H

O

(C)

O

(D)

H5C2O OC2H5

OO (protic solvent)

Ans. A, C, D

Sol.


N

H

O N OH

(Keto form) (Enol form)

Enol form is more stable than keto form because the enol form is aromatic. 6. Consider the following reaction

N

CH3

H5C6

OH

H2SO4(A) (B)

(major product)

LiAlH4

The incorrect statement is/are:

(A) the product (A) on nitration gives the major product

O2N

C

O

NH CH3

(B) the product (B) liberates N2 gas when reacted with NaNO2/HCl

(C) the substitution on the benzene ring of the product (A) is meta directing and deactivating

(D) the substituent on the benzene ring of the product (B) is ortho, para directing and activating

Ans. A, B, C

Sol.

C N

CH3

H5C6

OH

H2SO4

(A)

(B)

LiAlH4

C NH

O

CH3 C6H5

CH2 NHCH3 C6H5

7. The compounds among the following which exhibit geometrical isomerism is/are:

(A) C C C

CH3

H H

CH3

(B) C C C

CH3

H

C

H

CH3

(C)

CH3

CH3

CH3

CH3

(D) CH3CH3

Ans. B, C, D Sol.

C C C

CH3

H H

CH3

cannot exhibit geometrical isomerism


8. In the following reaction:

C

C

CH3

C

C

OH

CH3 OH

H OH

OHIO4 (excess)

Product

The product in this reaction is/are

(A) C OH

O

H

(B) C H

O

H

(C) C CH3

O

CH3

(D) C OH

O

CH3 Ans. A, C Sol.

C

C

CH3

C

C

OH

CH3 OH

H OH

OHIO4 (excess)

C OH

O

H2 + CO2C CH3

O

CH3+

9. Which of the following reactions along with major product is/are correct?

(A)

CH3 CH3 OH OH

CH3 CH3

(i) mCPBA

(ii) H2O+

(Major product)

(B)

CH3 CH3 OH OH

CH3 CH3KMnO4/OH

-

(Major product)

(C)

CH3 D H OH

CH3 D(i) BH3 - THF.H2O

(ii) H2O2/OH -

(Major product)

(D)

CH3 D CH3 OH

D(i) Hg(OAc)2 - THF

(ii) NaBH4/OH -

(Major product) Ans. B, C

Sol. (A) Anti hydroxylation

(D) Oxymercuration demercuration.


10. The correct reactions among the following is/are

(A)

OH

CH3

O

CH3

CHO

O

CHCl2CH3

+(i) CHCl3 / KOH

(ii) H+

(B)

O O

CH3

(i) (CH3)2CuLi

(ii) EtOH

(Major product)

(C) C

O

CHCH2

(i) Ph MgBr(ii) EtOH

CH2 C

O

CH2

(Major product)

(D)

O

CH2 C OCH3

O

(i) NaBH4

(ii) H2O

OH

OH

(Major product) Ans. A, C

Sol.

(B)

O O

CH3

(i) (CH3)2CuLi

(ii) EtOH

(D)

O

CH2 C OCH3

O

(i) NaBH4

(ii) H2O

OH

CH2 C OCH3

O

Section – C

(Only Integer Value Correct Type)

This section contains 10 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive).

1. Total number of isomers containing six membered ring (including stereoisomers) that are possible of dimethyl

cyclohexane is

Ans. 9

Sol.


CH3 CH3

H

CH3CH3

H CH3

HCH3

H, ,

CH3

H

H

CH3

CH3

H

CH3

H

CH3

H

H

CH3

CH3

H

CH3

H, ,

(& enantiomers) (& enantiomers)

2. Total number of aromatic ion/molecule among the following is:

N

H

B

CH3

N N

H

NN

S NN

H

, , , ,, , , ,

Ans. 6

Sol. The following compounds are aromatic.

N

H

N N

H

NN

SN

, , , ,,

3. Total number of compounds among the following which liberatesCO2 gas from NaHCO3 solution is

SO3H

CH3

COOH

NO2

OH

OCH3

OHO

OOH

OH

, , , ,

Ans. 3

Sol.

SO3H

CH3

COOH

NO2

O

OOH

OH

, ,


4. Total number of compounds among the following which undergo decarboxylation on heating is:

OH

O O

OH

O O

COOH

COOH

COOH

COOH

H

H

CH3 CH

COOH

COOH

CH2

CH2

COOH

COOH

, ,

,

,

Ans. 4

Sol.

COOH

COOH

H

H

CH2

CH2

COOH

COOH

&

Undergo dehydration on heating.

5. Total number of compounds among the following which gives carbylamines reaction is

CH NH2

CH3

CH3 C NH

O

CH3 CH3 N

CH3

H

CH3 NH2

NH CH3N

H

H5C6 CH3 CH NH2

CH3

, , ,

, ,

Ans. 3

Sol. 1o amines give carbylamines reaction.

6. Consider the following reaction

CH3 CH CH CH3

N

CH3 CH3

CH3

(i) CH3I

(ii) AgOH, (A)

(Major product)

The number of hyperconjugative structure of the product (A) is

Ans. 1 Sol.

CH3 CH CH CH3

N

CH3 CH3

CH3

(i) CH3I

(ii) AgOH, CH3 CH CH CH2

CH3

(A)

H = 1

Number of hyperconjugative structure of the compound (A) = 1


7. The number of compounds among the following which give positive iodoform test:

CH3 C CH2 CH3

O

C CH2 CH3

O

CH2CHCH3

OH

CH3 CH2 OH CH2 CH CH2 CH3CH3

OH

OH

I2HC C CH3

O

O

O

C

O

CH2

CH3 CH3

CH2

CH3 CH2 CH

Cl

CH3

, ,

, ,,

,,

Ans. 6

Sol. The following compounds give positive iodoform test.

CH3 C CH2 CH3

O

CH2CHCH3

OH

CH3 CH2 OH I2HC C CH3

O

O

O

CH3 CH2 CH

Cl

CH3

,

, , ,

8. Total number of compounds among them the following for which their gauche form is more stable than anti

form in aqueous medium is

CH3 – CH2 – CH2 – CH3, HO – CH2 – CH2 – OH, HO – CH2 – CH2 – F, HO – CH2 – CH2 – Cl

F – CH2 – CH2 – F

Ans. 4

Sol. Gauche form of the following compounds is more stable than anti form.

HO – CH2 – CH2 – OH, HO – CH2 – CH2 – F, HO – CH2 – CH2 – Cl, F – CH2 – CH2 – F 9. Total number of compounds among the following which are more acidic than benzoic acid is:

COOH

CH3

COOH

Cl

COOH

OCH3

H C OH

O

CH3 C OH

O

, ,, , ,

COOH

CH3

Ans. 3

Sol.


COOH

CH3

COOH

OCH3

CH3 C OH

O

,,

are less acidic than benzoic acid. 10. Total number of compounds among the following which gives positive Fehling test is

C CH3

O

C H

O

CH3

OC H

O

CH3 C H

O

CH3 C CH3

O

CH3 C OH

OC OH

OCHO

,

, ,

, ,

Ans. 2

Sol.

C OH

O

,CH3COOH

Ketones, Aromatic aldehyde do not give positive Fehling test.


Mathematics

(Straight Objective Type) This section contains 10 multiple choice questions numbered 1 to 10. Each question has 4 choices (A), (B), (C) and (D) out of which only ONE is correct

1. Let α,, , be natural numbers. Now, consider the equation x4 – (α + + + ) x

3 – (α + + )x

2 – (α + )x – α = 0,

then the number of integer roots in the equation is

(A) 0 (B) 1 (C) 2 (D) None of these

Ans. A

Sol. Let ‘m’ be the integer root suppose m > 0, m < 0 the proof can be concluded

2. Consider a polynomial ,1|x|1|)x(f|,1xxx)x(f,)0,(23

then

||1

||111Max is


Ans. B

Sol. 01,1

1)1(f

01,01

3)0(f2)1(f1111

01,01

21f

38

21f

38)1(f

31)1(f

341111

738

38

31

34

01,1

21f

34

21f4)1(f

351111

7344

35

3. Consider an equation with α, real roots 0b1

axx 2 . Also I

b1,

a1 and ,)Integer(

aa.2

a

12

if

and only if

(A) α is an integer, may not be an integer (B) is an integer, α may not be an integer

(C) α, are both integers (D) both α, may not be integers

Ans. C

Sol. (α + )2 – 2 < [α(α + )] + [(α + )] (α + )

2

α and are both integers

4. Let A(3i) and B(- i) be two points and CD is a given line as shown z(1 – 2i) – z (1 + 2i) – 2ic = 0. Then, smallest

positive integral value of ‘c’ for which there exist exactly 2 points on CD at which AB subtends a right angle is


Ans. A

Sol. The circle through AB as diameter should cut in two points


5. Consider lines z(1 – mi) – z (1 + mi) = 0 and z(1 – ni) – z (1 + ni) = 0. Let, arg(z’) be the angle between them

such that .2

,0)'zarg(

There is another line z(1 + i) – z (- 1 + i) + 2i = 0 which intersects the previous lines in

A and B such that AB – OA = OB – AB, Then, Max(arg(z’)) is k where k is


Ans. D

Sol. )'z(arg

OB.OA.2ABOBOAcos

222

41

OAOB

OBOA

83

OB.OA.2

)OBOA(41OBOA

cos

222

21cos)Maximum(For

6. Consider numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, then probability of constructing a 5 – digit number a1a2a3a4a5 of the

form (1) a1 a2 a3 a4 a5 or (2) a1 a2 a3 a4 a5 is

(A) 592565

(B) 592574

(C) 595148

(D) None of these

Ans. A

Sol. 5

513

992C

7. If in a triangle ABC, A B C

tan ,tan ,tan2 2 2

are in H.P. then minimum value of B

cot2

is

(A) 1

3 (B) 1 (C) 3 (D) 3 1

Ans. C

Hint: A B C

cot ,cot ,cot2 2 2

are in A.P.

8. In triangle ABC, 2

C3

, then the value of 2 2cos A cos B cosA cosB is equal to:

(A) 3

4 (B)

3

2 (C)

1

2 (D)

1

4

Ans. A

9. If sinx cosx tanx cot x sec x cosecx 7 and sin2x a b c , then a – b + 2c is:

(A) 0 (B) 14 (C) 28 (D) 42

Ans. C

10. If A1, A2, A3,………., An be a regular polygon of n sides and 1 2 1 3 1 4

1 1 1

A A A A A A then

(A) n = 5 (B )n = 6 (C) n = 7 (D) None of these

Ans. C

O

A

B


Section – B

Multiple Correct Choice Types

The sections 5 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) for its answer, out of which only ONE OR MORE THAN ONE is/are correct

11. Consider a line (1 – 3i)z – z (1 + 3i) + 6i = 0 and curve iz2 + i z

2 + z(- 2 + 14i) + z (2 + 14i) + 2iz z + 8i = 0, then

(a + ib) is point on the curve at a shortest distance from line, then

(A) |a| + |b| = 10 (B) 4ab (C) b – a = 6 (D) |a| - |b| = 6

Ans. A, B

Sol. General point on the curve can be taken as (α, α2 + 7α + 2)

Distance from line

10

3273 2

For minimum value α = - 2 and point is (- 2, - 8) (a, b)

12. Consider 0pqpxpqxqx,01qxx

px 232

3

where, p, q R and roots of the equations belong to set of

positive real numbers, then of pq can be

(A) 1 (B) 2 (C) 3 (D) 4

Ans. C, D

Sol. Let, α, , are roots

α + + = p, α + α + = q

p

α = p

(α + + )2 3 (α + + α)

q

p3p2

3pq

13. Consider a finite sequence of real numbers such that sum of any 5 consecutive term is negative and the sum of

any 9 – consecutive term is positive, then number of terms can be


Ans. A, B

Sol. a1 + a2 + a3 + a4 + a5 < 0

a2 + a3 + a4 + a5 + a6 < 0

……………………………

……………………………

a9 + a10 + a11 + a12 + a13 < 0

If we sum horizontally and vertically we get different sums 14. Consider ‘n’ persons participating in race where reaching at finishing the together or one after the other is

considered completion of race. If k(n) denotes the number of ways in which ‘n’ persons can complete the race,

then

(A) k(2) = 3 (B) k(3) = 13 (C) k(4) = 74 (D) k(1) = 1

Ans. A, B, D


Sol. k(2) = 1 + 2 + 3

k(3) = 1 + 2! + 3C2 + 3! = 13

!4!3!2!2!1!1

!4!2

!2!2!2

!4!2

!1!3

!41)4(k

= 1 + 8 + 6 + 36 + 24 = 75 15. Consider the following seating arrangements. Six persons A, B, C, D, E, F are to be seated at the marked

vertices such that in any two arrangements all does not have the same neighbours

(A) 360 ways for figure (I) (B) 360 ways for figure (II)

(C) 720 ways for figure (III) (D) 540 ways for figure (II)

Ans. A, B

Sol. (1) 300!2

!6Ways

(2) 360!3C6Ways 35

(3) 360!2!2!2!2!2

!456Ways

Section – C

One Integer Value Correct Types

The sections 5 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive)

1. Given a, b, c, x, y, z > 0; a – x, b – y, c – z > 0, then the greatest value of (a – x) (b – y) (c – z) (ax + by + cz) is

42 2 21 a b c

abc N

where N is

Ans. 4

Sol. Using the inequality 4

)czbyax()zc(c)yb(b)xa(a

4Ngetwe)]czbyax()zc(c)yb(b)xa(a[ 4/1

2. The greatest value of (a + x)3 (a – x)

4 for x R, a R

+, |x| < a is ,23

7a2

then α + + = 9k (k N) where

‘k’ is

Ans. 2

Sol. ,43

)xa()xa(

7

4xa4

3xa3

7

1

43

43

we get α + + = 7 + 3 + 8 = 18

)I( )II( )III(


3. Consider a circle C1: |z| = 4 intersects another circle C2 of radius 5 such that common chord has maximum

length and equation as ,0i2i431zi

431z

then centre of C2is (a + ib), where 5(|b| - |a|) is

Ans. 3

Sol. For maximum length of chord use, 0i431zi

431zk)16zz(

[Here = 0]

Equation radius = 5, we get

512,

59)b,a(

4. If P is any point on the curve z2(- 2 – 3i) + z

2(- 2 + 3i) + 8z (z) - 4 = 0 and origin is centre of curve, then

N,,OPMin2

1

where last digit of α + + is

Ans. 9

Sol. Consider a general point as (r cos, r sin) on the curve, then with given equation

cossin3sin3cos

1r22

2

21

min

134

2r

5. Consider the curves z2 - z

2 = 4ic, (c R) and z . z = 1. If they touch at exactly two points A, B, then |AB| is

Ans. 2

Sol. Substituting xcy in x

2 + y

2 = 1, we get x

4 – x

2 + c

2 = 0 (which has equal roots)

41c 2 and points can be

i

2

1

2

1andi2

1

2

1

2|AB|

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