Code 0 p2 Solution

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INSTRUCTIONS

Question Paper Format

The question paper consists of three parts (Physics, Chemistry and Mathematics). Each part consists

of three sections.

1. Section 1 contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and

(D) out of which ONE is correct.

2. Section 2 contains 3 paragraphs each describing theory, experiment and data etc. Six questions

relate to three paragraphs with two questions on each paragraph. Each question pertaining to a

particular passage should have only one correct answer among the four given choices (A), (B), (C)

and (D).3. Section 3 contains 4 multiple choice questions. Each question has two lists (List-1, P, Q, R and

S; List-2 : 1, 2, 3 and 4). The options for the correct match are provided as (A), (B), (C) and (D)

out of which ONLY ONE is correct.

Marking Scheme

4. For each question in Section 1, 2, and 3 you will be awarded 3 marks if you darken the bubble

corresponding to the correct answer and zero mark if no bubble is darkened. In all other cases, minus

one (–1) mark will be awarded.

Answers & Solutions for for for for for

JEE (Advanced)-2014

Time : 3 hrs. Max. Marks: 180

DATE : 25/05/2014 CODE

0

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075

Ph.: 011-47623456 Fax : 011-47623472

PAPER - 2 (Code - 0)



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PART–I : PHYSICS

SECTION - 1 : (Only One Option Correct Type)

This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of whichONLY ONE Option is correct.

1. If Cu

is the wavelength of K X-ray line of copper (atomic number 29) and

Mo is the wavelength of the K

X-ray line of molybdenum (atomic number 42), then the ratio Cu

/Mo

is close to

(A) 1.99 (B) 2.14

(C) 0.50 (D) 0.48

Answer (B)

Hint :

22

Cu Mo

2Mo Cu

( 1) 412.14

28( 1)

Z

Z

2. A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum

speeds of the photoelectrons corresponding to these wavelengths are u1 and u

2, respectively. If the ratio

u1 : u2 = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly

(A) 3.7 eV (B) 3.2 eV

(C) 2.8 eV (D) 2.5 eV

Answer (A)

Hint :

2

1

1

1

2

hcmu W

2

2

2

1

2

hcmu W

2

1 1

2

2

hcW

u

hcuW

∵ 1

2 1 2

44 ( 2)

uhc hcW W

u

2 1

43

hc hcW

⇒ 4 1240 1240 3310 248

W 16 – 5 = 3 W

11 = 3 W

W = 3.7 eV

3. Parallel rays of light of intensity I = 912 Wm –2 are incident on a spherical black body kept in surroundings

of temperature 300 K. Take Stefan-Boltzmann constant = 5.7 × 10 –8 Wm –2K –4 and assume that the energy

exchange with the surroundings is only through radiation. The final steady state temperature of the black

body is close to

(A) 330 K (B) 660 K

(C) 990 K (D) 1550 K

Answer (A)



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Hint : 2 4 4 2

04 .( ) 912R T T R

4 4 8

0 8

912 91240 10

4 4 5.7 10

T T

4 8 4 840 10 (300) (40 81) 10T

T = 330 K

4. During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed

at 40.0 cm using a standard resistance of 90 , as shown in the figure. The least count of the scale used in

the metre bridge is 1 mm. The unknown resistance is

R 90

40.0 cm

(A) 60 ± 0.15

(B) 135 ± 0.56

(C) 60 ± 0.25

(D) 135 ± 0.23

Answer (C)

Hint :

1 2

(100 )

R R

x x (Balanced Wheatstone)

For R,

90 90 40

40 60 60

RR

⇒

90 100

R x

x

For R,

ln ln ln(100 )R x x

(100 )

(100 )

R x x

R x x

(100 )

R x x

R x x

0.1 0.160

40 60R

R = 0.25

R = 60 ± 0.25



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5. A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire

is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and

it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is

B90°

A

(A) Always radially outwards

(B) Always radially inwards

(C) Radially outwards initially and radially inwards later

(D) Radially inwards initially and radially outwards later

Answer (D)

6. A planet of radius 1

(radius of Earth)10

R has the same mass density as Earth. Scientists dig a well of

depth5

R on it and lower a wire of the same length and of linear mass density 10 –3 kg m –1 into it. If the

wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take

the radius of Earth = 6 × 10

6

m and the acceleration due to gravity on Earth is 10 ms

–2

)(A) 96 N

(B) 108 N

(C) 120 N

(D) 150 N

Answer (B)

Hint :

6

56 106 10 m

10 10

e

P

RR

P = E

3

2 2

4

3

GM G g R

R R

⇒ 4

3 g G R

R P

x

dx

⇒ 21 m/s

10

e P P

P

e E

g g R g

g R

Acceleration due to gravity at a depth x

1

x P x g g R



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Force on a small segment of wire at a depth x from surface

( )dF dx gx

1

x dF gdx

R

∫

/5

01

R x F gdx

R

/52

02

R

x x g

R

3 3 59 910 10 6 10 1

5 50 50 50

R R g R g

540

108 N5

7. A glass capillary tube is of the shape of a truncated cone with an apex angle so that its two ends have cross

sections of different radii. When dipped in water vertically, water rises in it to a height h, where the radius

of its cross section is b. If the surface tension of water is S , its density is , and its contact angle with glassis . the value of h will be ( g is the acceleration due to gravity)

h

(A)

2cos( )

S

b g (B)

2cos( )

S

b g

(C)

2cos( /2)

S

b g (D)

2cos( /2)

S

b g

Answer (D)

Hint :

h

b

22 cos

2S b b h

2 cos ( )2S h

bg



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8. Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2, R

and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from

the centre of sphere 1, 2 and 3 are E 1, E 2 and E

3 respectively, then

R

P

2R

4Q

R

P

2QQ

P

R

R/2

Sphere 1 Sphere 2 Sphere 3

(A) E 1 > E 2 > E 3

(B) E 3 > E 1 > E 2

(C) E 2 > E 1 > E 3

(D) E 3 > E 2 > E 1

Answer (C)

Hint : 1 2 KQE R

2 2 2

(2 ) 2 K Q KQE

R R

Q

3 3 3

(4 )

8 2

K Q R KQE

R R

2 1 3

E E E

9. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72.

It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging

from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The

refractive index of the liquid is

Block

Liquid

S (A) 1.21 (B) 1.30

(C) 1.36 (D) 1.42

Answer (C)

Hint :

10 mm10 mm

= 30°

S

1sin

2.72

sin302.72

2.72

2

1.36 =



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10. A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting

the surface. The force on the ball during the collision is proportional to the length of compression of the ball.

Which one of the following sketches describes the variation of its kinetic energy K with time t most

appropriately? The figures are only illustrative and not to the scale.

(A)

K

t

(B)

K

t

(C)

K

t

(D)

K

t

Answer (B)

Hint :

t

v

t

v

K

t

v t

v2 t2

K.E. t2

SECTION - 2 : Comprehension Type (Only One Option Correct)

This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions related

to the three paragraphs with two questions on each paragraph. Each question has only one correct

answer among the four given options (A), (B), (C) and (D).

Paragraph For Questions 11 and 12

The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the plane

of the paper. The distance of each wire from the centre of the loop is d. The loop and the wires are carrying

the same current I . The current in the loop is in the counterclockwise direction if seen from above.

Q S dd

P R

Wire 2Wire 1 a



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11. When d a but wires are not touching the loop, it is found that the net magnetic field on the axis of the

loop is zero at a height h above the loop. In that case

(A) Current in wire 1 and wire 2 is the direction PQ and RS , respectively and h a

(B) Current in wire 1 and wire 2 is the direction PQ and SR, respectively and h a

(C) Current in wire 1 and wire 2 is the direction PQ and SR, respectively and h 1.2a

(D) Current in wire 1 and wire 2 is the direction PQ and RS , respectively and h 1.2a

Answer (C)

Hint : .

2

0 0

3 2 2

2 2 2

2

2 ( )2( )

ia ia

a ha h

2

2 2

2a a

a h

4 2

2 2 2

4a a

a h

2 2 210 4 4a a h

2 26 4a h

2

23

2

ah

1.2h a

12. Consider d a, and the loop is rotated about its diameter parallel to the wires by 30° from the position shown

in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new

position will be (assume that the net field due to the wires is constant over the loop)

(A)

2 2

0 I a

d

(B)

2 2

0

2

I a

d

(C)

2 2

03 I a

d

(D) 2 2

03

2

I a

d

Answer (B)

Hint : 0

22

I B

d

20 2 1

2 2

I I a

d

2 2

0

2

I a

d



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In the figure, a container is shown to have a movable (without friction) piston on top. The container and the

piston are all made of perfectly insulation material allowing no heat transfer between outside and inside the

container. The container is divided into two compartments by a rigid partition made of a thermally conducting

material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of

an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas

at 400 K. The heat capacities per mole of an ideal monatomic gas are , ,2 2

V P C R C R

3 5= = and those for an

ideal diatomic gas are5 7

,2 2

V P C R C R= = .

13. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final

temperature of the gases will be

(A) 550 K (B) 525 K

(C) 513 K (D) 490 K

Answer (D)

Hint : Q2 = Q1

3 72 ( 700) 2 ( 400)

2 2

R RT T 6 6 700 14 400 14T T

5600 + 4200 = 20 T T = 490 K

14. Now consider the partition to be free to move without friction so that the pressure of gases in both

compartments is the same. The total work done by the gases till the time they achieve equilibrium will be

(A) 250 R (B) 200 R

(C) 100 R (D) –100 R

Answer (D)

Hint :

1 21 2

( 700) (400 ) P P

n C T n C T

5 7

2 ( 700) 2 (400 )2 2

R RT T

5 3500 2800 7T T 12 3500 2 800T 6300

52512

T

– W = U = 1 2

1 2(525 700) (525 400)

V V n C n C

3 5

2 175 2 125

2 2

R R

= –525R + 625R

W = – U = –100 R



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Paragraph For Questions 15 & 16

A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross

section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston

pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the

spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm respectively. The upper end

of the container is open to the atmosphere.

15. If the piston is pushed at a speed of 5 mms –1, the air comes out of the nozzle with a speed of

(A) 0.1 ms –1 (B) 1 ms –1

(C) 2 ms –1 (D) 8 ms –1

Answer (C)Hint :

A1V 1 = A2V 2

2 2 2(0.2) 0.005L

V

2 20.04 0.005

LV

4 410

LV

V L = 2

16. If the density of air is a and that of the liquid

l, then for a given piston speed the rate (volume per unit

time) at which the liquid is sprayed will be proportional to

(A)

a

l

(B) a l

(C)

l

a

(D) l

Answer (A)

Hint :

2

1 2

1

2 a a

P P v

1 2

3

2

3 2

1

2 l l

P P v

P 3 = P 1

2 21 1

2 2l l a a

v v

a

l a

l

v v

Volume flow rate

a

l



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SECTION - 3 : Matching List Type (Only One Option Correct)

This section contains 4 questions, each having two matching lists. Choices for the correct combination of elements

from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct.

17. A person in lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the

water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the following, stateof the lift's motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match

the statements from List I with those in List II and select the correct answer using the code given below the lists.

List-I List-II

P. Lift is accelerating vertically up 1. d = 1.2 m

Q. Lift is accelerating vertically down 2. d > 1.2 m

with an acceleration less than the

gravitational acceleration

R. Lift is moving vertically up 3. d < 1.2 m

with constant speed

S. Lift is falling freely 4. No water leaks out of the jar

Code :

(A) P-2, Q-3, R-2, S-4 (B) P-2, Q-3, R-1, S-4

(C) P-1, Q-1, R-1, S-4 (D) P-2, Q-3, R-1, S-1

Answer (C)

Hint :

2

2 h

R gh g

x d v t

a

H

h

2H g+a( )

d

2 ( ) h

H g a g a

d Independent of acceleration of lift.

18. Four charges Q1, Q

2, Q

3 and Q

4 of same magnitude are fixed along the x axis at x = –2a, – a, +a and +2a,

respectively. A positive charge q is placed on the positive y axis at a distance b > 0. Four options of the signs

of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match

List I with List II and select the correct answer using the code given below the lists.

Q

a1

(–2 ,0)

Q

a2

(– ,0)

Q

a3

(+ ,0)

Q

a4

(+2 ,0)

q b(0, )

List-I List-II

P. Q1, Q

2, Q

3, Q

4 all positive 1. +x

Q. Q1, Q

2 positive; Q

3, Q

4 negative 2. – x

R. Q1, Q4 positive; Q2, Q3 negative 3. + y

S. Q1, Q

3 positive; Q

2, Q

4 negative 4. – y



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Code :

(A) P-3, Q-1, R-4, S-2 (B) P-4, Q-2, R-3, S-1

(C) P-3, Q-1, R-2, S-4 (D) P-4, Q-2, R-1, S-3

Answer (A)

Hint:

P. Not along + y Q. Not along +x

Q1

Q2

Q3

Q4

Q1

Q2

Q3

Q4

R. Not along – y S. Not along – x

Q1

Q2

Q3

Q4 Q

1 Q

2 Q

3 Q

4

19. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is

r and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length

in List II and select the correct answer using the code given below the lists.

List-I List-II

P. 1. 2r

Q. 2.2

r

R. 3. – r

S. 4. r

Code :

(A) P-1, Q-2, R-3, S-4 (B) P-2, Q-4, R-3, S-1

(C) P-4, Q-1, R-2, S-3 (D) P-2, Q-1, R-3, S-4



(13)

Answer (B)

Hint : (P)

1 1 1 2 1( 1) (1.5 1)u

f r r r r ⇒ f r

⇒ 1 2

1 1 1'

' 2

r f

f f f

P 2

(Q)

1

1 1 1 0.5(1.5 1)

f r r ⇒

1 22 f r f

⇒ ⇒ 1 2

1 1 1 1 1 1

2 2 f r

f f f f r r

Q 4

(R)

⇒

1

1

1 1 1 0.5(1.5 1) 2 f r

f r r

⇒

2

2

1 1 1 0.51.5 1 2 f r

f r r

1 2

1 1 1 1 1

2 2 f f f r r

f r

R 3

(S)

⇒

1

1

1 1 1 2 1(1.5 1) 0.5 f r

f r r r r

⇒ ⇒

2

2 2

1 1 1 1 0.5(1.5 1) 2 f r

f r f r

1 2

1 1 1 1 1 1

2 2 f f f r r r ⇒ 2 f r

S 1



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20. A block of mass m1= 1 kg another mass m

2 = 2 kg, are placed together (see figure) on an inclined plane

with angle of inclination . Various values of are given in List I. The coefficient of friction between the

block m1 and the plane is always zero. The coefficient of static and dynamic friction between the block m

2

and the plane are equal to = 0.3. In List II expressions for the friction on block m2 are given. Match the

correct expression of the friction in List II with the angles given in List I, and choose the correct option.

The acceleration due to gravity is denoted by g.

[Useful information : tan(5.5°) 0.1; tan(11.5°) 0.2; tan(16.5°) 0.3]

m

m

List-I List-II

P. = 5° 1. m2 g sin

Q. = 10° 2. (m1

+ m2) g sin

R. = 15° 3. m2 g cos

S. = 20° 4. (m1 + m2) g cos

Code :

(A) P-1, Q-1, R-1, S-3 (B) P-2, Q-2, R-2, S-3

(C) P-2, Q-2, R-2, S-4 (D) P-2, Q-2, R-3, S-3

Answer (D)

Hint :

m

m

2 1 2cos ( ) sinm g m m g

2

1 2

tan 0.2m

m m

So block m2 starts sliding at > 11.5°.

When = 5° and = 10°, then block will be rest, so friction will be1 2

( ) sinm m g

For = 15° and = 20°, block starts sliding.

So friction will be2

cosm g .

So option (D) is correct.



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PART–II : CHEMISTRY



21. For the identification of -naphthol using dye test, it is necessary to use

(A) Dichloromethane solution of -naphthol (B) Acidic solution of -naphthol

(C) Neutral solution of -naphthol (D) Alkaline solution of -naphthol

Answer (D)

Hint :

O HNaOH

O

N Cl2

+ –

O H

N N

22. Assuming 2s-2 p mixing is NOT operative, the paramagnetic species among the following is

(A) Be2 (B) B2

(C) C2

(D) N2

Answer (C)

Hint : Be2 = 1s2*1s22s2*2s2

B2 = 1s2*1s22s2*2s22 pz 2

C2 = 1s2*1s22s2*2s22 p

z 22 p

x 12 p

y1

N2 = 1s2*1s22s2*2s22 pz 22 px 22 p y2

23. For the elementary reaction M N, the rate of disappearance of M increases by a factor of 8 upon doublingthe concentration of M. The order of the reaction with respect to M is

(A) 4 (B) 3

(C) 2 (D) 1

Answer (B)

Hint :

2 2

1 1

r [M]

r [M]

8 = (2)

= 3

So, order of reaction is 3.24. Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure.

and and

(I) (II) (III)

The correct order of their boiling point is

(A) I > II > III (B) III > II > I

(C) II > III > I (D) III > I > II

Answer (B)

Hint : As branching increases boiling point decreases, so order of boiling point is III > II > I.



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25. The acidic hydrolysis of ether (X) shown below is fastest when

[Figure]

Acid

OH + ROHOR

(A) One phenyl group is replaced by a methyl group

(B) One phenyl group is replaced by a para-methoxyphenyl group

(C) Two phenyl groups are replaced by two para-methoxyphenyl groups

(D) No structural change is made to X

Answer (C)

Hint : –OCH3 group has +R effect. It increases the stability of the carbocation.

26. The major product in the following reaction is

[ Figure ]

Cl

O

CH3

1. CH MgBr, dry ether, 0 °C

2. aq. acid

3

(A) H C3

O

CH3

(B)HC

2

OH

CH3

CH3

(C)O

CH2

(D)O

CH3

CH3

Answer (D)

Hint : Cl

O

CH – MgI3

Cl

O –

CH3 O

CH3

CH3

27. Hydrogen peroxide in its reaction with KIO4 and NH

2OH respectively, is acting as a

(A) Reducing agent, oxidising agent

(B) Reducing agent, reducing agent

(C) Oxidising agent, oxidising agent

(D) Oxidising agent, reducing agent

Answer (A)

Hint : H2O2 acting as a reducing agent with KIO4 and H2O2 acting as an oxidising agent with NH2OH.



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28. The product formed in the reaction of SOCl2 with white phosphorous is

(A) PCl3

(B) SO2Cl2

(C) SCl2

(D) POCl3

Answer (A)

Hint : 4 2 3 2 2 2(White)

P 8 SOCl 4PCl 4SO 2S Cl

29. Under ambient conditions, the total number of gases released as products in the final step of the reaction

scheme shown below is

XeF6

Complete

HydrolysisP + Other product

Q

OH /H O –

2

Products

Slow disproportionation in OH /H O –

2

(A) 0 (B) 1

(C) 2 (D) 3

Answer (C)

Hint : XeF6

Complete

HydrolysisXeO + H F

3 2 2

HXeO4

–

OH /H O –

2

XeO + Xe(g) + H O + O (g)6 2 2

–4

Slow disproportionation in OH /H O –

2

30. For the process

H2O(I) H

2O(g)

at T = 100 ºC and 1 atmosphere pressure, the correct choice is

(A) Ssystem

> 0 and Ssurroundings

> 0

(B) Ssystem

> 0 and Ssurroundings

< 0

(C) Ssystem

< 0 and Ssurroundings

> 0

(D) Ssystem

< 0 and Ssurroundings

< 0

Answer (B)

Hint : Given conditions are boiling conditions for water due to which

Stotal = 0

Ssystem + Ssurroundings = 0

Ssystem = – Ssurroundings

For process, Ssystem > 0

Ssurroundings

< 0



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This section contains 3 paragraphs, each describing theory, experiments, data etc. 6 questions related

to the three paragraphs with two questions on each paragraph. Each question has only one correct

answer among the four given options (A), (B), (C) and (D).


X and Y are two volatile liquids with molar weights of 10 g mol –1 and 40 g mol –1 respectively. Two cotton plugs,

one soaked in X and the other soaked in Y, are simultaneously placed at the ends of a tube of length L = 24 cm,

as shown in the figure. The tube is filled with an inert gas at 1 atmosphere pressure and a temperature of

300 K. Vapours of X and Y react to form a product which is first observed at a distance

d cm from the plug soaked in X. Take X and Y to have equal molecular diameters and assume ideal behaviour

for the inert gas and the two vapours.

L = 24 cm

dCotton wool

soaked in X Initial formation

of the product

Cotton wool

soaked in Y

31. The value of d in cm (shown in the figure), as estimated from Graham's law, is

(A) 8

(B) 12

(C) 16

(D) 20

Answer (C)

Hint : x 40

24 – x 10

x

224 – x

x = 16

32. The experimental value of d is found to be smaller than the estimate obtained using Graham's law. This is

due to

(A) Larger mean free path for X as compared to that of Y

(B) Larger mean free path for Y as compared to that of X

(C) Increased collision frequency of Y with the inert gas as compared to that of X with the inert gas

(D) Increased collision frequency of X with the inert gas as compared to that of Y with the inert gas

Answer (D)

Hint : Increased collision frequency of X with the inert gas as compared to that of y with the inert gas. Therefore,

the experimental value of d is found to be smaller than the estimate obtained using Graham's law.



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Schemes 1 and 2 describe sequential transformation of alkynes M and N. Consider only the major products formed

in each step for both the schemes.

HO

1. NaNH (excess)

2. CH CH I (1 equivalent)

2

3 2

3. CH I (1 equivalent)4. H , Lindlar’s catalyst

3

2

H X

M

Scheme-1

1. NaNH ( )2

2 equivalent

H Y

N

Scheme-2

Br

OH2.

3. H O , (mild)3

4. H , Pd/C2

5. CrO3

33. The product X is

(A)

H H

H CO3

(B)

H

H

H CO3

(C)

H H

CH C O3

H2

(D)

H

H

CH C O3

H2

Answer (A)

Hint :

H NaNH

2

Na+ –

O

C Na – +

CH2

ICH3

(1 eq)C

ICH3

OCH3

H2

Lindlar's

catalystH H

CH3

O

HO Na+

O –

34. The correct statement with respect to product Y is

(A) It gives a positive Tollens test and is a functional isomer of X

(B) It gives a positive Tollens test and is a geometrical isomer of X

(C) It gives a positive iodoform test and is a functional isomer of X

(D) It gives a positive iodoform test and is a geometrical isomer of X

Answer (C)

Hint :

1. NaNH ( )2

2 eq

H

Br

OH2.

3. H O (mild)3

+

4. H , Pd/C2

5. CrO3

O



(20)


An aqueous solution of metal ion M1 reacts separately with reagents Q and R in excess to give tetrahedral and

square planar complexes, respectively. An aqueous solution of another metal ion M2 always forms tetrahedral

complexes with these reagents. Aqueous solution of M2 on reaction with reagent S gives white precipitate which

dissolves in excess of S. The reactions are summarized in the scheme given below :

Scheme :

Tetrahedral

Tetrahedral

Excess

Excess

Q

Q

M1

M2

Excess

Excess

S, stoichiometric amount

R

R

Square planar

Tetrahedral

White precipitateExcess

RPrecipitate dissolves

35. M1, Q and R, respectively are

(A) Zn

2+

, KCN and HCl (B) Ni

2+

, HCl and KCN(C) Cd2+, KCN and HCl (D) Co2+, HCl and KCN

Answer (B)

Hint : Ni+2 + HCl [NiCl4] –2

Ni+2 + KCN [Ni(CN)4] –2

36. Reagent S is

(A) K 4[Fe(CN)

6] (B) Na

2HPO4

(C) K 2CrO4

(D) KOH

Answer (D)

Hint :

OH2 2

2 4white ppt SolubleZn OH Zn(OH) [Zn(OH) ]


This section contains 4 questions, each having two matching lists. Choice for the correct combination of elements

from List-I and List-II are given as options (A), (B), (C) and (D), out of which one is correct.

37. Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select

the correct answer using the code given below the lists.

{en = H2NCH2CH2NH2; atomic numbers : Ti = 22, Cr = 24; Co = 27; Pt = 78}

List-I List-II

P. [Cr(NH3)4Cl2]Cl 1. Paramagnetic and exhibits ionisation isomerism

Q. [Ti(H2O)5Cl](NO

3)2

2. Diamagnetic and exhibits cis-trans isomerism

R. [Pt(en)(NH3)Cl]NO

33. Paramagnetic and exhibits cis-trans isomerism

S. [Co(NH3)4(NO3)2]NO3

4. Diamagnetic and exhibits ionisation isomerism

Code :

P Q R S

(A) 4 2 3 1

(B) 3 1 4 2

(C) 2 1 3 4

(D) 1 3 4 2



(21)

Answer (B)

Hint : Magnetic character Isomerism

P. [Cr(NH3)4Cl2]Cl Paramagnetic cis/trans

Q. [Ti(H2O)5Cl](NO

3)2

Paramagnetic ionization

R. [Pt(en)(NH3)Cl]NO

3Diamagnetic ionization

S. [Co(NH3)4(NO3)2]NO3 Diamagnetic cis/trans

38. Match the orbital overlap figures shown in List-I with the description given in List-II and select the correct

answer using the code given below the lists.

List-I List-II

P. 1. p – d antibonding

Q. 2. d – d bonding

R. 3. p – d bonding

S. 4. d – d antibonding

Code :

P Q R S

(A) 2 1 3 4

(B) 4 3 1 2

(C) 2 3 1 4

(D) 4 1 3 2

Answer (C)

Hint :

P. d – d ( bonding)

Q. p – d ( bonding)

R. p – d ( antibonding)

S. d – d ( antibonding)



(22)

39. Different possible thermal decomposition pathways for peroxyesters are shown below. Match each pathway

from List I with an appropriate structure from List II and select the correct answer using the code given

below the lists.

R

O

O

O

R

R + R O

R + R O

R + X

+ carbonyl compound

RCO + R O2

R + X

+ carbonyl compound –CO

2

RCO + R O2

R + R O

–CO

2

–CO2

P

–CO2

Q

R

S

(Peroxyester)

List-I List-II

P. Pathway P 1.CHCH6 5 2

O

O

O

CH3

Q. Pathway Q 2.CH

6 5

O

OO

CH3

R. Pathway R 3. C H CH6 5 2

O

OO CH

3

CH3

CH C H2 6 5

S. Pathway S 4. C H6 5

O

OO CH

3

CH3

C H6 5

Code :

P Q R S

(A) 1 3 4 2

(B) 2 4 3 1

(C) 4 1 2 3

(D) 3 2 1 4

Answer (A)

Hint : (1)C H CH – C – O

6 5 2CH

3

OO

C H CH + CO + CH – O6 5 2 2 3

(3) C H CH – C – O6 5 2

CH3

OO

CO +2

C H CH + CH – C – O6 5 2 3

CH3

CH3

CH – C H2 6 5

CH – C H2 6 5

C

(4) C H6 5

C – CH3

C

O

C H – C – O + O – C – CH6 5 3

CH3

CH3

O

CH3

CH3

O

O

(2)C H

6 5CH

3

C C

O O

O + CH – O3

C H6 5

O

O



(23)

40. Match the four starting materials (P, Q, R, S) given in List I with the corresponding reaction schemes (I,

II, III, IV ) provided in List II and select the correct answer using the code given below the lists.

List-I List-II

P. H H 1. Scheme I

(i) KMnO , HO, heat (ii) H, H O(iii) SOCl (iv) NH

4 2

2 3

? C H N O7 6 2 3

Q.

OH

OH

2.

Scheme II

(i) Sn/HCl (ii) CH COCl (iii) conc. H SO

(iv) HNO (v) dil. H SO , heat (vi) HO3 2 4

3 2 4

? C H N O6 6 2 2

R.

NO2

3.

Scheme III

(i) red hot iron, 873 K (ii) fuming HNO , H SO , heat

(iii) H S.NH (iv) NaNO , H SO (v) hydrolysis3 2 4

2 3 2 2 4

? C HNO6 5 3

S.

NO2

CH3

4.

Scheme IV

(i) conc. H SO , 60ºC

(ii) conc. HNO , conc. H SO (iii) dil. H SO , heat2 4

3 2 4 2 4

? C HNO6 5 4

Code :

P Q R S

(A) 1 4 2 3

(B) 3 1 4 2

(C) 3 4 2 1

(D) 4 1 3 2

Answer (C)

Hint : Scheme I

NO2

CH3

(S)

Scheme II

NO2

(R)

Scheme III H – C C – H (P)

Scheme IV

OH

OH



(24)

PART–III : MATHEMATICS



41. Six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6 and cards are to be placed in envelopes so that each

envelope contains exactly one card and no card is placed in the envelope bearing the same number and

moreover the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it can

be done is

(A) 264 (B) 265

(C) 53 (D) 67

Answer (C)

Hint :1 1 1 1 1 16! 1

1! 2! 3! 4 ! 5! 6!

5

265

535

42. In a triangle the sum of two sides is x and the product of the same two sides is y. If x 2 – c2 = y, where c is

the third side of the triangle, then the ratio of the in radius to the circum-radius of the triangle is

(A)3

2 ( )

y

x x c(B)

3

2 ( )

y

c x c

(C)3

4 ( )

y

x x c(D)

3

4 ( )

y

c x c

Answer (B)

Hint : a + b = x , ab = y, x 2 – c2 = y

Now, (a + b)2 – c2 = ab

a2 + b2 – c2 = – ab

2 2 21

2 2

a b c

ab

1

cos2

c c = 120°

22

sin 3

c cR

c

3

c

R

( ) tan2

c

r s c

32

x c

r c

( ) 3

2

x c

r



(25)

( ) 3 3( )3

2 2

r x c x c

R c c

3( )( )

2 ( )

x c x c

c x c

2 23( )

2 ( )

x c

c x c

3

2 ( )

y

c x c

43. The common tangents to the circle x 2 + y2 = 2 and the parabola y2 = 8x touch the circle at the points P , Q

and the parabola at the poitns R, S . Then the area of the quadrilateral PQRS is

(A) 3 (B) 6

(C) 9 (D) 15

Answer (D)

Hint : x 2 + y2 = 2

Equation of tangent to circle x 2 + y2 = 2 is2

2 1 y mx m

Equation of tangent to y2 = 8x is2

y mx m

2 2

2( 1 ) mm

(2, –4)

(–1, –1)

(–1, 1)

(2, 4)

2

2

42(1 ) m

m

m2(1 + m2) = 2

m4 + m2 – 2 = 0

(m2 + 2)(m2 – 1) = 0

m = 1

y = x + 2 y = – x – 2

Area1

3 1 3 3 22

= 6 + 9 = 15 sq. unit

44. Three boys and two girls stand in a queue. The probability, that the number of boys ahead of every girl is

at least one more than the number of girls ahead of her, is

(A) 1

2(B)

1

3

(C)2

3(D)

3

4



(26)

Answer (A)

Hint : n(S ) = 5!

Case I :G

1

2 × 4! = 48

Case II : G1

2 × 3! = 12

n(E ) = 60

60 1( )

5! 2 P E

45. The quadratic equation p(x ) = 0 with real coefficients has purely imaginary roots. Then the equation

p( p(x )) = 0 has

(A) Only purely imaginary roots (B) All real roots

(C) Two real and two purely imaginary roots (D) Neither real nor purely imaginary roots Answer (D)

Hint : Let p(x ) = x 2 + a (a > 0) (∵ roots are purely imaginary)

p( p(x )) = (x 2 + a)2 + a (a R)

x 4 + 2a(x 2) + a2 + a = 0

x 2 = 2 2

2 4 4 4

2

a a a a

= a ai

x = 1 1

a ai x iy

46. For x (0, ), the equation sinx + 2sin2x – sin3x = 3 has

(A) Infinitely many solutions (B) Three solutions

(C) One solution (D) No solution

Answer (D)

Hint : sinx + 2sin2x – sin3x = 3

2cos2x (–sinx ) + 4sinx cosx = 3

2sinx[2cosx – cos2x ] = 3

2sinx (2cosx – (2cos2x – 1)) = 3

2sinx (1 + 2cosx – 2cos2

x ) = 32

3 12 sin 2 cos 3

2 2

x x

Possible only when

sinx = 1 …(i)

and

21

cos 02

x

1

cos2

x …(ii)

From (i) and (ii)No solution.



(27)

47. The following integral2

17

4

(2cosec )

∫ x dx is equal to

(A) log(1 2 ) 160

2( ) ∫ u ue e du (B) log(1 2 ) 17

0( ) ∫

u ue e du

(C) log(1 2 ) 17

0( )

∫ u u

e e du (D) l og(1 2 ) 16

02( )

∫ u u

e e du

Answer (A)

Hint :2

17

4

(2cosec )

∫ I x dx

162

4

1cosec cot 2 cosec

cosec cot

∫ x x xdx

x x

Let cosecx + cotx = eu

–cosecxdx = du

016

ln( 2 1)

2 ( )

∫ u u

I e e du

∫

ln( 2 1 )16

0

2( )u ue e du

48. Coefficient of x 11 in the expansion of (1 + x 2)4(1 + x 3)7(1 + x 4)12 is

(A) 1051 (B) 1106

(C) 1113 (D) 1120

Answer (C)

Hint : Power of Coefficient of x 11

2 3 4

0 1 2

2 1 1

4 1 0

1 3 0

x x x 4 7 12

4 7 12

4 7 12

4 7 12

C × C × C

C × C × C

C × C × C

C × C × C

0 1 2

2 1 1

4 1 0

1 3 0

1113

49. Let f : [0, 2] be a function which is continuous on [0, 2] and is differentiable on (0, 2) with f (0) = 1.

Let

2

0

( ) ( ) ∫ x

F x f t dt for x [0, 2]. If F (x ) = f (x ) for all x (0, 2), then F (2) equals

(A) e2 – 1 (B) e4 – 1

(C) e – 1 (D) e4



(28)

Answer (B)

Hint : F (x ) = f (x ).2x = f (x )

'( )

2( )

f x x

f x

2

( ) x f x ke

Given f (0) = 1 2

( ) x f x e

So

2

0

( ) ∫ x

x F x e dx

2

1 x e

So F (2) = e4 – 1

50. The function y = f (x ) is the solution of the differential equation4

2 2

2

1 1

dy xy x x

dx x x

in (–1, 1) satisfying

f (0) = 0. Then

3

2

3

2

( )

∫ f x d x is

(A) 3

3 2

(B)

3

3 4

(C) 3

6 4

(D)

3

6 2

Answer (B)

Hint :

4

2 2

2

1 1

dy x x x y

dx x x

I.F. 2 21 1

∫

x dx

x e x

So5

2 4 21 ( 2 )5

∫ x

y x x x dx x c

Given f (0) = 0 c = 0

So

52

2

5 ( )

1

x x

f x

x

53 3 32

22 2 2

2 203 3

2 2

5( ) 2

1 1

∫ ∫ ∫

x x

x dx I f x dx dx

x x

Let x = sin

32

0

2 sin

∫ I d 3

0

(1 cos2 )

∫ d

3

0

sin 2 3

2 3 4



(29)


This section contains 3 paragraphs, each describing theory, experiments, data etc. six questions

related to the three paragraphs with two questions on each paragraph. Each question has only one

correct answer among the four given options (A), (B), (C) and (D).


Box 1 contains three cards bearing numbers 1, 2, 3; box 2 contains five cards bearing numbers 1, 2, 3, 4, 5;

and box 3 contains seven cards bearing numbers 1, 2, 3, 4, 5, 6, 7. A card is drawn from each of the boxes.

Let x i be number on the card drawn from the ith box, i = 1, 2, 3.

51. The probability that x 1 + x 2 + x 3 is odd, is

(A)29

105(B)

53

105

(C)57

105(D)

1

2

Answer (B)

Hint :

1 1

2 2B1

B2

3 3

4

5

1

2B3

3

4

5

6

7

x 1 + x 2 + x 3 = odd

Case–I : One odd (2 × 3 × 2) + (4 × 1 × 2) + (3 × 3 × 1)

Two even = 12 + 8 + 9 = 29

Case–II : 3 odd 2 × 4 × 3 = 24

= 29 + 24 = 53

Total cases = 3 × 5 × 7 = 105

Probability =53

105

52. The probability that x 1, x 2, x 3 are in an arithmetic progression, is

(A)9

105(B)

10

105

(C)11

105(D)

7

105

Answer (C)

Hint : 2x 2 = x 1 + x 3

x 1 and x

3 must be even number.

both x 1 {1, 3}

x 1 and x 3 odd x 3 {1, 3, 5, 7}

also x 1 + x 3 {2, 4, 6, 8, 10}



(30)

x 1

x 3

1

3

(1, 3, 5, 7)

(1, 3, 5, 7)

8 cases

x 1 and x

3 both even

x 2 = 2

x 3 = (2, 4, 6) 3 cases

Total = 11 cases

Required probability =11

105


Let a, r, s, t be nonzero real numbers. Let P (at2, 2at), Q, R(ar2, 2ar) and S (as2, 2as) be distinct points on

the parabola y2 = 4ax . Suppose that PQ is the focal chord and lines QR and PK are parallel, where K is the

point (2a, 0).

53. The value of r is

(A) – 1

t(B)

+2

1t

t

(C)1

t (D) −2

1t

t

Answer (D)

Hint :

P a t

a t

(

, 2 )

2

S

( 0)a,R

( 2 )ar , ar2

2a

t

a

t2

,

–2a

t2

a

t2

,

Q

K a,(2 0)



(31)

Let co-ordinates of Q be (at22, 2at2)

Since PQ is focal chord

t t2 = –1

2

1t

t= −

Co-ordinates of Q2

2,

a a

tt

−

Slope of PK = 2 2

2 0 2

2 2

at t

at a t

−

=

− −

Slope of RQ 2 2

2 2

1222

1

arar

tt

aar r

t t

++ = =

− −

2

1r

t

= −

Slope of PK = Slope of RQ

2

2 2

12

t

tr

t

=

−−

2 21 2 1t t

r rt t t

− −− = ⇒ =

54. If st = 1, then the tangent at P and the normal at S to the parabola meet at a point whose ordinate is

(A)+

2 2

3

( 1)

2

t

t(B)

+2 2

3

( 1)

2

a t

t

(C)+

2 2

3

( 1)a t

t(D)

+2 2

3

( 2)a t

t

Answer (B)

Hint :

Since st = 1

s =1

t

Co-ordinates of S =2

2,

a a

tt

Equation of tangent at P

y =1

t(x + at2)

2x yt at= − ....(i)

Slope of normal at S – s

1

t= −



(32)

Equation of normal

2

2 1a a y x

t t t

− − = −

22

aty a x

t− + = −

22

ax ty a

t= − + + ....(ii)

From (i) and (ii)

y.t – at2 = – ty + 2a + 2

a

t

2ty = at2 + 2

a

t + 2a

2ty = a

21

t

t

+

y =

21

2

a tt

t

+

2 2

3

( 1)

2

a t y

t

+=


Given that for each a (0, 1),

+

−

− −

→

− ∫ 1

1

0

lim (1 )

h

a a

hh

t t dt

exists. Let this limit be g (a). In addition, it is given that the function g (a) is differentiable on (0, 1).

55. The value of g 1

2 is

(A) (B) 2

(C)π

2(D)

π

4

Answer (A)

Hint : g (a) =

(1 )

1

0lim (1 )

h

a a

hh

t t dt+

−

− −

→

− ∫

g

1

2

= ( )

(1 )1/21/2

0lim 1

h

hh

t t dt+

−

−−

→

−

∫ (1 )

0

1lim

(1 )

h

hh

dtt t

+

−

→

=

− ∫

1

1

0

1

2lim sin

1

2

h

h

h

t

+

−

−

→

− =

1 1

0

lim [sin (2(1 ) 1)) sin (2 1)]h

h h+

− −

→

= − − − −

2 2

π π= + = π



(33)

56. The value of g' 1

2 is

(A)π

2(B)

(C) – π

2(D) 0

Answer (D)

Hint : g (a) =

1

1

0

lim (1 )

h

a a

hh

t t dt+

−

− −

→

− ∫

∫

1

1

0

1'( ) lim (1 ) ln

h

a a

hh

t g a t t dt

t

∫

1 11

2 2

0

1 1

' lim (1 ) ln2

h

hh

t

g t t dtt

1' 0

2 g


This section contains four questions, each having two matching lists. Choices for the correct

combination of elements from List-I and List-II are given as options (A), (B), (C) and (D), out of which

one is correct.

57. List-I List-II

P. The number of polynomials f (x ) with non-negative integer 1. 8

coefficients of degree 2, satisfying f (0) = 0 and

∫ 1

0( ) 1, is f x dx

Q. The number of points in the interval [ 13, 13] at 2. 2

which f (x ) = sin(x 2) + cos(x 2) attains its maximum value, is

R.

∫ 2

2

2

3

(1 )x

x dx

e equals 3. 4

S.

∫

∫

1

2

1

2

1

2

0

1cos 2 log

1

1cos 2 log

1

x x dx

x

x x dx

x

equals 4. 0

P Q R S

(A) 3 2 4 1

(B) 2 3 4 1

(C) 3 2 1 4

(D) 2 3 1 4



(34)

Answer (D)

Hint : (P) Let f (x ) = a0x 2 + a

1x + a2

; a0, a1, a2 > 0

Now, a2 = 0 and

1

0

( ) 1∫ f x dx

13 2

0 1

0

13 2

a x a x

0 1

13 2

a a

2a0 + 3a

1 = 6

i.e., for integral solution

0 1

0 1

either 3, 0

or 0, 2

a a

a a

i.e., f (x ) = 3x 2 or f (x ) = 2x

P (2)

(Q)max

( ) 2 f x at2 9

,4 4

x

9,

4 4

x

So, number of points = 4

Q (3)

(R)

2 2

2

3

1

∫ x

x I dx

e

again,2 2

2

3

1

∫ x

x I dx

e

Adding,2

2 3 2

2

2

2 3 { } 16

∫ I x dx x

I = 8

R (1)

(S) As1

cos 2 log1

x

x

x

is an odd function

So,

1

2

1

2

1cos 2 log 0

1

∫ x

x dx x

S (4)



(35)

58. List-I List-II

P. Let y(x ) = cos(3 cos –1 x ), 3

[ 1,1],2

x x . Then 1. 1

2

2

2

1 ( ) ( )

( 1)( )

d y x dy x

x x y x dx dx equals

Q. Let A1, A2, ... , A

n (n > 2) be the vertices of a regular 2. 2

polygon of n sides with its centre at the origin.

Let

ka be the position vector of the point A

k,

k = 1, 2, ... , n. If ∑

1 1

1 1 1 1( ) ( . )n n

k k k k k ka a a a ,

then the minimum value of n is

R. If the normal from the point P (h, 1) on the ellipse 3. 8

2 2

16 3

x y is perpendicular to the line x + y = 8,

then the value of h is

S. Number of positive solutions satisfying the equation 4. 9

1 1 1

2

1 1 2tan tan tan

2 1 4 1x x x

is

P Q R S

(A) 4 3 2 1

(B) 2 4 3 1

(C) 4 3 1 2

(D) 2 4 1 3

Answer (A)

Hint : (P) 1

2

3sin(3 cos )

1

dyx

dx x

2

2 2 1 2(1 ) 9(1 cos (3 cos )) 9(1 )

dyx x y

dx

222 2

2(1 ) ( 2 ) 18

dy d y dy dyx x y

dx dx dx dx

22

2( 1) 9

d y dyx x y

dx dx

(Q) 2 22 2( 1) sin ( 1) cos

n R n R

n n

2

tan 1

n

n = 8



(36)

(R) As, equation of normal is

6 33

cos sin

x y…(i)

Given its slope = 1 1

tan

2

Also (i) passes through (h, 1)

So,6 3

312

33

h h = 2

(S) 1 1

2

1 122 1 4 1tan tan

11

(2 1)(4 1)

x x

x

x x

2 2

6 2 2

8 6

x

x x x

3 1 2

4 3

x

x x

(where x 0)

3x 2 – 7x – 6 = 0

2

3,3

x

But 2x + 1 > 0 and 4x + 1 > 0

So, solution are x = 3

59. Let 1

: f , f 2 : [0, ) , f

3 : and f

4 : [0, ) be defined by

1

| | if 0,( )

if 0;x

x x f x

e x

f 2(x ) = x 2;

f 3(x ) =

sin if 0,

if 0

x x

x x

and

2 1

4

2 1

( ( )) if 0,( )

( ( )) 1 if 0.

f f x x f x

f f x x

List-I List-II

P. f 4 is 1. Onto but not one-one

Q. f 3 is 2. Neither continuous nor one-one

R. f 2of 1 is 3. Differentiable but not one-one

S. f 2 is 4. Continuous and one-one

P Q R S

(A) 3 1 4 2

(B) 1 3 4 2

(C) 3 1 2 4

(D) 1 3 2 4



(37)

Answer (D)

Hint :

(P) f 4(x ) =

2 1

2 1

[ ( )], 0

[ ( )] 1, 0

f f x x

f f x x

<

− ≥

Onto but not one-one =

2

2

, 0

1, 0x

x x

e x

<

− ≥

Now f (x ) is not differentiable at x = 0, not one-one but continuous.

(P) 1

(Q) Now, f 3 =

sin , 0

, 0

x x

x x

<

≥

Differentiable but not one-one.

(Q) 3

(R) f 2

of 1

=

2

2

, 0

, 0x

x x

e x

<

≥

(R) 2

(S) f 2 : [0, ] R, f

2(x ) = x 2

(S) 4

60. Let

2 2cos sin

10 10k

k kz i ; k = 1, 2, ..., 9.

List-I List-II

P. For each z k there exists a z

j such that z

k . z

j = 1 1. True

Q. There exists a k {1, 2, ..., 9} such that z 1.z = z

k has 2. False

no solution z in the set of complex numbers

R.

1 2 9|1 ||1 |...|1 |

10

z z z

equals 3. 1

S.

91

21 cos10

kk equals 4. 2



P Q R S

(A) 1 2 4 3

(B) 2 1 3 4

(C) 1 2 3 4

(D) 2 1 4 3

Answer (C)

Hint : z k =

2

102 2

cos sin10 10

i kk

i e

ππ π + =

(P) z k.z j

= 1

2

( )10 1

i k je

π+

=

2 ( ) 2 ( )

cos 1 and sin 010 10

k j k jπ + π + = =

k + j = 10n and k + j = 5m; so true

(Q)

2( 1)

10

1

i kk

z

z e

z

π

−

= =

So, if k {1, 2, .....9}, z has solution; i.e., false

(Q) 2

(R) z 10 – 1 = 0

(z – z 1)(z – z

2).....(z – z

9) = 1 + z + z 2 + ..... + z 9

So, |1 – z 1||1 – z

2|.......|1 – z 9| = 10

i.e., (R) 3

(S) 2 4 18

1 cos cos ..... cos10 10 10

π π π − + + +

4 28

10 101 cos 2

2

π π + × = − =

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