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Paper-2 [ CODE – 5 ]

JEE Advance Exam 2015 (Solution)

Part I - PHYSICS

SECTION – 1 (Maximum Marks : 32)

• This Section contains EIGHT questions • The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive. • For each question, darken the bubble corresponding to the correct integer in the ORS • Marking scheme : + 4 If the bubble corresponding to the answer is darkened 0 In all other cases

Q.1 The densities of two solid spheres A and B of the same radii R vary with radial distance r as ρA(r) = k

⎟⎠⎞

⎜⎝⎛

Rr and ρB(r) = k

5

Rr

⎟⎠⎞

⎜⎝⎛ , respectively, where k is a constant. The moments of inertia of the individual

spheres about axes passing through their centres are IA and IB, respectively. If A

B

II =

10n , the value of n

is

Ans. [6]

Sol.

dr

r

Consider a shell of radius r and thickness dr

mass of element = ρ × 4πr2dr

Moment of inertial of shell = dI = 32 ρ × 4πr2dr r2

dI = 32 × ρ × 4πr4dr

CAREER POINT

Date : 24 / 05 / 2015

CAREER POINT



dI = 32 ×

Rkr × 4πr4 dr

dI = 32

Rk × 4πr5 dr

I = 32

Rk × ∫π

R

0

5 ]drr[4

IA = 32

Rk ×

6R4 6π =

32 k × 4π

6R5

For I2 = 32 × 5

5

Rkr × 4πr4dr

I2 ⇒ 32 × 5R

k × 4π ∫R

0

9drr

IB = 32 × 5R

k × 10R4 10π =

32 × k ×

10R4 5π

A

B

II =

106

Q.2 Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, π/3, 2π/3 and π.

When they are superposed, the intensity of the resulting wave is nI0. The value of n is

Ans. [3]

Sol.

60º

A A

AA

I0 = kA2

Resultant amplitude = 3 A

IR = k3A2

IR = 3I0

n = 3

Q.3 For a radioactive material, its activity A and rate of change of its activity R are defined as A = – dtdN

and R = – dtdA , where N(t) is the number of nuclei at time t. Two radioactive sources P (mean life τ) and

Q (mean life 2τ) have the same activity at t = 0. Their rates of change of activities at t = 2τ are RP and

RQ, respectively. If Q

P

RR =

en , then the value of n is

CAREER POINT



Ans. [2]

Sol. R = – dtdA

A = – dtdN

At time t

A = A0e–λt

R = – dtdA = – A0e–λt (–λ)

R = A0λe–λt

λP = τ1 λQ =

τ21

RP = A0 λP τ

τ− 21

e ⇒ A0λPe–2

RQ = A0λQ τ×

τ− 2

21

e = A0λQe–1

Q

P

RR =

Q

P

λλ 1

2

ee

−

−

⇒ e2

∴ n = 2

CAREER POINT



Q.4 A monochromatic beam of light is incident at 60º on one face of an equilateral prism of refractive index

n and emerges from the opposite face making an angle θ(n) with the normal (see the figure). For

n = 3 the value of θ is 60º and dndθ = m. The value of m is

60ºθ

Ans. [2]

Sol.

60ºθ r (60–r)

A

B C Applying snell law at face AB.

sin 60º = n × sin r

sin r = n23 ....(1)

Applying snell law at face AC

n × sin (60 – r) = 1 × sin θ ....(2)

if n = 3 , θ = 60º

3 sin (60 – r) = sin 60

sin (60 – r) = 21

60 – r = 30

r = 30º

from (1) sin 30º = n23

n = 3

differentiating (2) with respect to n

CAREER POINT



1 × sin (60 – r) + n × cos (60 – r) ⎥⎦⎤

⎢⎣⎡−

dndr = cos θ

dndθ ....(3)

differentiating (1) with respective to n

cos r dndr =

23 ⎥⎦

⎤⎢⎣⎡−

2n1 ....(4)

when n = 3 r = 30

cos 30º dndr =

23 ⎥⎦

⎤⎢⎣⎡−

31 Q –

dndr =

31

Put r = 30º, θ = 60º and – dndr =

31 in equation (3)

sin (60 – 30) + 3 × cos (60 – 30) ⎥⎦⎤

⎢⎣⎡31 =

21

dndθ ∴

dndθ = 2

Q.5 In the following circuit, the current through the resistor R (= 2Ω) is I Amperes. The value of I is

R (= 2Ω) 1Ω

2Ω 8Ω

6Ω4Ω

2Ω

10Ω

12Ω 4Ω

6.5V

Ans. [1]

Sol.

2Ω 1Ω

2Ω

6Ω4Ω

2Ω

10Ω

12Ω 4Ω

6.5V

balanced wheat stone bridge

CAREER POINT



2Ω 1Ω

2Ω

6Ω4Ω

2Ω

10Ω

12Ω 4Ω

6.5V

2Ω

2Ω 6Ω

10Ω

12Ω

6.5V

4Ω

balanced wheat stone bridge

2Ω

2Ω

6.5V

4Ω

6Ω

12Ω

i

Req. = 6.5Ω

.Amp15.65.6i ==

CAREER POINT



Q.6 An electron in an excited state of Li2+ ion has angular momentum 3h/2π. The de Broglie wavelength of

the electron in this state is pπa0 (where a0 is the Bohr radius). The value of p is

Ans. [2]

Sol. L = π2

nh

π2h3 =

π2nh

∴ n = 3

if n = 3

λ (λ = de Broglie wavelength)

3λ = length of orbit

length of orbit = 2πrn

= Z

na2 20π =

33a2 2

0π

= 6πa0

3λ = 6πa0

λ = 2πa0

So value of p = 2

Q.7 A large spherical mass M is fixed at one position and two identical point masses m are kept on a line

passing through the centre of M (see figure). The point masses are connected by a rigid massless rod of

length l and this assembly is free to move along the line connecting them. All three masses interact only

through their mutual gravitational interaction. When the point mass nearer to M is at a distance r = 3l

from M, the tension in the rod is zero for m = k ⎟⎠⎞

⎜⎝⎛

288M . The value of k is

r l

m mM

Ans. [7]

Sol. F1 = 29GMm

l – 2

2Gml

(Force between M and middle m)

F2 = 216GMm

l + 2

2Gml

(Force between M and extreme m)

CAREER POINT



F1 = F2

29GMm

l – 2

2Gml

= 216GMm

l + 2

2Gml

M ⎥⎦⎤

⎢⎣⎡ −

161

91 = 2m

M ⎥⎦⎤

⎢⎣⎡ −

144916 = 2m

m = 2887 M

m = 7

Q.8 The energy of a system as a function of time t is given as E(t) = A2 exp(–αt), where α = 0.2 s–1. The

measurement of A has an error of 1.25 %. If the error in the measurement of time is 1.50 %, the

percentage error in the value of E(t) at t = 5 s is

Ans. [4]

Sol. E = A2e–αt

E = A2e–0.2t

ln E = 2lnA – 0.2t

E

dE = 2 A

dA – 0.2 dt

E

dE = 2 A

dA – 0.2 dt

t

dt = 1.5 % ∴ dt = 1.5 × 5 % = 1.5 × 5 = 7.5

= 2 × 1.25 + 0.2 × 7.5

= 2.5 + 1.5

= 4 %


• This section contains EIGHT questions • Each questions has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

options(s) is(are) correct • For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS • Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened 0 If none of the bubbles is darkened –2 In all other cases

CAREER POINT



Q.9 A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air. When

two dielectrics of different relative permittivities (ε1 = 2 and ε2 = 4) are introduced between the two

plates as shown in the figure, the capacitance becomes C2. The ratio 1

2

CC is

S/2

S/2ε1

ε2

d/2

d

+ –

(A) 6/5 (B) 5/3 (C) 7/5 (D) 7/3

Ans. [D]

Sol. When no dielectric is filled then C1 = dS0ε

when dielectric is filled

ε1 = 2 ε2 = 42d

2d

2S

2S

ε1 = 2

d

C2 = d2

S10εε +

⎥⎦

⎤⎢⎣

⎡ε

+ε

ε

21

0

2d

2d2

S

= d2

S20 ×ε +

⎥⎦⎤

⎢⎣⎡

×+

×

ε

42d

22d2

S0

⇒ dS0ε +

d328S0

××ε

⇒ dS0ε ⎥⎦

⎤⎢⎣⎡

614 ⇒

dS0ε ⎥⎦

⎤⎢⎣⎡

37

1

2

CC =

37

CAREER POINT



Q.10 An ideal monoatomic gas is confined in a horizontal cylinder by a spring loaded piston (as shown in the

figure). Initially the gas is at temperature T1, pressure P1 and volume V1 and the spring is in its relaxed

state. The gas is then heated very slowly to temperature T2, pressure P2 and volume V2. During this

process the piston moves out by a distance x. Ignoring the friction between the piston and the cylinder.

the correct statements (s) is (are)

(A) If V2 = 2V1 and T2 = 3T1, then the energy stored in the spring is 41 P1V1

(B) If V2 = 2V1 and T2 = 3T1, then the change in internal energy is 3P1V1

(C) If V2 = 3V1 and T2 = 4T1, then the work done by the gas is37 P1V1

(D) If V2 = 3V1 and T2 = 4T1, then the heat supplied to the gas is 6

17 P1V1

Ans. [A,B,C]

Sol.

P1 = P0

P0

x0

PP0

x0 + x

kx

(A) P = P1 + A

Kx

P2 = 23 P1 ⇒ x =

AV1

2P3 1 = P1 +

AKx

Kx = 2AP1

CAREER POINT



Energy of spring

21 Kx2 =

4AP1 x =

4VP 11 Ans. A

(B) ΔU = 2f (P2V2 – P1V1)

= 3P1V1 Ans. B

(C) Pf = 3P4 1 KX =

3P1 A , X =

AV2 1

Wgas = – (atmPW + Wspring)

= (P1Ax + 21 Kx . x)

= + ⎟⎠⎞

⎜⎝⎛ +

AV2.

3AP.

21

AV2.AP 111

1

= 2P1V1 + 3VP 11 =

3VP7 11

(D) ΔQ = W + ΔU

= 3VP7 11 +

23 (P2V2 – P1V1)

= 3VP7 11 +

23 ⎟

⎠⎞

⎜⎝⎛ − 1111 VPV3.P

34

= 3VP7 11 +

29 P1V1 =

6VP41 11

Q.11 A fission reaction is given by U23692 → Xe140

54 + Sr9438 + x + y, where x and y are two particles. Considering

U23692 to be at rest, the kinetic energies of the products are denoted by KXe KSr, Kx(2MeV) and

Ky(2MeV), respectively. Let the binding energies per nucleon of U23692 , Xe140

54 and Sr9438 be 7.5 MeV,

8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option (s) is (are)

(A) x = n , y = n, KSr = 129 MeV, KXe = 86 MeV

(B) x = p , y = e–, KSr = 129 MeV, KXe = 86MeV

(C) x = p , y = n, KSr = 129 MeV, KXe = 86 MeV

(D) x = n , y = n, KSr = 86 MeV, KXe = 129 MeV

Ans. [A]

Sol. Q value of reaction, Kx + Ky + KXe + KSr – UK23692

Q = 2 + 2 + KXe + KSr – 0

Q = 4 + KXe + KSr

Q = 8.5 × 140 + 8.5 × 94 – 7.5 × 236

CAREER POINT



Q = 219 MeV

219 – 4 = KXe + KSr

KXe + KSr = 215 MeV

From momentum conservation

heavy particle must have small kinetic energy.

Q.12 Two spheres P and Q of equal radii have densities ρ1 and ρ2, respectively. The spheres are connected by

a massless string and placed in liquids L1 and L2 of densities σ1 and σ2 and viscosities η1 and η2,

respectively. They float in equilibrium with the sphere P in L1 and sphere Q in L2 and the string being

taut (see figure). If sphere P alone in L2 has terminal velocity PVr

and Q alone in L1 has terminal

velocity QVr

, then

Q

PL1

L2

(A) |V||V|

Q

Pr

r

= 2

1

ηη (B)

|V||V|

Q

Pr

r

= 1

2

ηη

(C) PVr

. QVr

> 0 (D) PVr

. QVr

< 0

Ans. [A,D]

Sol.

T

σ1Vg

ρ1Vg

T

ρ2Vg

σ2 Vg

(σ1 + σ2) Vg = (ρ1 + ρ2) Vg σ1 + σ2 = ρ1 + ρ2

ρ1 – σ2 = σ1 – ρ2

If ρ1 > σ2 then σ1 > ρ2

CAREER POINT



In separate liquid one body will get terminal velocity upward then other one will get downward terminal

velocity.

PV→

. QV→

< 0 (angle between pV→

and QV→

is π)

& for terminal velocity V ∝ η1

|V|

|V|

Q

P→

→

= ⎟⎟⎠

⎞⎜⎜⎝

⎛ηη

2

1 [because P in liquid L2 and Q in liquid L1]

Q.13 In terms of potential difference V, electric current I, permittivity ε0, permeability µ0, and speed of light

c, the dimensionally correct equation (s) is (are)

(A) µ0I2 = ε0V2 (B) ε0I = µ0V (C) I = ε0cV (D) µ0cI = ε0V

Ans. [A,C]

Sol. W = qV

ML2T–2 = ATV

V = M1L2T–3A–1

I ⇒ A

F = 04

1πε

2

2

rq

ε0 = 2

2

Frq

ε0 = 22

22

LMLTTA

×−

ε0 = M–1L–3T4A2

c = LT–1 Similarly µ0I2 = ε0V2

I = ε0cV

A = M0L0T0A1

CAREER POINT



Q.14 Consider a uniform spherical charge distribution of radius R1 centred at the origin O. In this distribution,

a spherical cavity of radius R2, centred at P with distance OP = a = R1 – R2 (see figure) is made. If the

electric field inside the cavity at position rr is E

r( rr ), then the correct statement (s) is (are)

R1

O

aPR2

(A) Er

is uniform, its magnitude is independent of R2 but its direction depends on rr

(B) Er

is uniform, its magnitude depends on R2 and its direction depends on rr

(C) Er

is uniform, its magnitude is independent of a but its direction depends on ar

(D) Er

is uniform and both its magnitude and direction depend on ar

Ans. [D]

Sol.

ar

r1

a

r2

P

Strength of electric field at point P is given by Er

= )rr(3 21

0

rr−

ερ .... (1)

because in solid sphere Er

= r3 0

r

ερ

Q ar = 21 rr

rr−

∴by eq(1)

Er

= a3 0

r

ερ

Q | ar | = R1 – R2 = constant

∴ magnitude of Er

is constant and direction of Er

is depends on direction of ar

CAREER POINT



Q.15 In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the

y-axis and stress on the x-axis as shown in the figure. Then the correct statements (s) is (are)

PQ

Stra

in

Stress (A) P has more tensile strength than Q

(B) P is more ductile than Q

(C) P is more brittle than Q

(D) The Young's modulus of P is more than that of Q

Ans. [A,B]

Sol. Actual graph is

PQSt

ress

Strain

Q Length of linear part of P is large then Q so P has more tensile strength than Q.

Q Curve part of P is large the curve part of Q so P is more ductile than Q.

Q (Slope)Q > (Slope)P

∴ Young's modulus of Q > Young's modulus of P

Q.16 A spherical body of radius R consists of a fluid of constant density and is in equilibrium under its own

gravity. If P(r), is the pressure at r(r < R), then the correct option (s) is (are)

(A) P(r = 0) =0 (B) )3/R2r(P)4/R3r(P

== =

8063

(C) )5/R2r(P)5/R3r(P

== =

2116 (D)

)3/Rr(P)2/Rr(P

== =

2720

Ans. [B,C]

CAREER POINT



Sol.

dm EG

r

Pds

(P + dp)ds

R

Gravitational field

GE = 34 πρGr

EG ∝ r s

EG = Cr ⎥⎥⎦

⎤

⎢⎢⎣

⎡

=

πρ=

.constCG

34C

(P + dp) ds + dm EG = Pds

dpds = – dm EG

dpds = – ρ dsdr . EG

dp = – ρEGdr

∫P

0

dp = ∫ ρ−r

R

drCr [pressure at surface is zero]

P = ρC r

R

2

2r

⎥⎦

⎤⎢⎣

⎡−

P = 2Cρ [R2 – r2]

(1) P(r = 0) ≠ 0

(ii) ⎟⎠⎞

⎜⎝⎛ =

⎟⎠⎞

⎜⎝⎛ =

3R2rP

4R3rP

=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛−

⎟⎠⎞

⎜⎝⎛−

22

22

3R2R

4R3R

= 8063

CAREER POINT



(iii) ⎟⎠⎞

⎜⎝⎛ =

⎟⎠⎞

⎜⎝⎛ =

5R2rP

5R3rP

= 22

22

5R2R

5R3R

⎟⎠⎞

⎜⎝⎛−

⎟⎠⎞

⎜⎝⎛−

= 2116

(iv) ⎟⎠⎞

⎜⎝⎛ =

⎟⎠⎞

⎜⎝⎛ =

3RrP

2RrP

= 22

22

3RR

2RR

⎟⎠⎞

⎜⎝⎛−

⎟⎠⎞

⎜⎝⎛−

= 3227


• This section contains TWO paragraphs • Based on each paragraph, there will be TWO questions • Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

option(s) is(are) correct • For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS • Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is(are) darkened 0 If none of the bubbles is darkened –2 In all other cases

PARAGRAPH 1

In a thin rectangular metallic strip a constant current I flows along the positive x-direction, as shown in

the figure. The length, width and thickness of the strip are l, w and d, respectively. A uniform magnetic

field Br

is applied on the strip along the positive y-direction. Due to this, the charge carriers experience

a net deflection along the z-direction. This results in accumulation of charge carriers on the surface

PQRS and appearance of equal and opposite charges on the face opposite to PQRS. A potential

difference along the z-direction is thus developed. Charge accumulation continues until the magnetic

force is balanced by the electric force. The current is assumed to be uniformly distributed on the cross

section of the strip and carried by electrons.

l

I

Q

I K

M z

x

y

RS

P d

w

CAREER POINT



Q.17 Consider two different metallic strips (1 and 2) of the same material. Their lengths are the same, widths

are w1 and w2 and thicknesses are d1 and d2, respectively. Two points K and M are symmetrically

located on the opposite faces parallel to the x-y plane (see figure). V1 and V2 are the potential

differences between K and M in strips 1 and 2, respectively. Then, for a given current I flowing through

them in a given magnetic field strength B, the correct statement(s) is(are)

(A) If w1 = w2 and d1 = 2d2, then V2 = 2V1

(B) If w1 = w2 and d1 = 2d2, then V2 = V1

(C) If w1 = 2w2 and d1 = d2, then V2 = 2V1

(D) If w1 = 2w2 and d1 = d2, then V2 = V1

Ans. [A,D]

Sol. i = neAVd

Vd = neA

i

e–

B

Magnetic force on e = eVdB

due to magnetic force e drift towards face M. due to which an electric field is produce between two

faces.

Vd = BE

E = VdB

Potential difference between k & M = V

V = EW

V = VdBW

V = neAiBW ⇒

neWdiBW =

nediB

V∝ d1

∴ if d1 = 2d2 then V1 = 2

V2 or V2 = 2V1 Ans. (A)

if d1 = d2 then V1 = V2 Ans. (D)

CAREER POINT



Q.18 Consider two different metallic strips (1 and 2) of same dimensions (length l, width w and thickness d)

with carrier densities n1 and n2, respectively. Strip 1 is placed in magnetic field B1 and strip 2 is placed

in magnetic field B2, both along positive y-directions. Then V1 and V2 are the potential differences

developed between K and M in strips 1 and 2, respectively. Assuming that the current I is the same for

both the strips, the correct option(s) is(are)

(A) If B1 = B2 and n1 = 2n2, then V2 = 2V1 (B) If B1 = B2 and n1 = 2n2, then V2 = V1

(C) If B1 = 2B2 and n1 = n2, then V2 = 0.5V1 (D) If B1 = 2B2 and n1 = n2, then V2 = V1

Ans. [A,C]

Sol. V = nediB

d is same for both strip

V ∝ nB

if B1 = B2

V ∝ n1

n1 = 2n2

V1 = 2

V2

V2 = 2V1 Ans. (A)

if n1 = n2

V ∝ B

if B1 = 2B2 then V1 = 2V2

V2 = 0.5 V1 Ans. (C)

CAREER POINT



PARAGRAPH 2

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid

glass cylinder of refractive index n1 surrounded by a medium of lower refractive index n2. The light

guidance in the structure takes place due to successive total internal reflections at the interface of the

media n1 and n2 as shown in the figure. All rays with the angle of incidence i less than a particular value

im are confined in the medium of refractive index n1. The numerical aperture (NA) of the structure is

defined as sin im.

Air Cladding

Coreθ n1

n2

i

n1 > n2

Q.19 For two structures namely S1 with n1 = 4/45 and n2 = 3/2, and S2 with n1 = 8/5 and n2 = 7/5 and

taking the refractive index of water to be 4/3 and that of air to be 1, the correct option(s) is(are)

(A) NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index 153

16

(B) NA of S1 immersed in liquid of refractive index 15

16 is the same as that of S2 immersed in water

(C) NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index 154

(D) NA of S1 placed in air is the same as that of S2 placed in water

Ans. [A,C]

Sol.

n2 B Cladding

r'

rA

i

Medium (μ)

n1

at B

sin C = 1

2

nn

CAREER POINT



cos C = 1

22

21

nnn −

For TIR at B –

r' ≥ C & r + r' = 90º

or cosr' ≤ cos C r = 90 – r'

cosr' ≤ 1

22

21

nnn −

sinr = cosr'

Now at A – (using snell's Laws)

μ × sin i = n1sinr

= n1 cos r'

sini ≤ μ− 2

221 nn

i.e. sin im = μ− 2

221 nn

& i ≤ im

i.e. NA = μ− 2

221 nn

using this relation we get

Ans. A & C

Q.20 If two structures of same cross-sectional area, but different numerical apertures NA1 and NA2

(NA2 < NA1) are joined longitudinally, the numerical aperture of the combined structure is

(A) 21

21

NANANANA

+ (B) NA1 + NA2 (C) NA1 (D) NA2

Ans. [D]

Sol. On the basis of discussion in Q.19, we can conclude that

if NA2 < NA1 then 2mi <

1mi

for transmission (i ≤ im)

hence i ≤ 2mi

i.e. NA should be NA2 for combined structure.

CAREER POINT



Part II - CHEMISTRY

SECTION – 1 (Maximum Marks: 32)

• This Section contains EIGHT questions

• This answer to each questions is a SINGLE DIGIT INTEGER ranging from 0 to 9. both inclusive

• For each questions, darken the bubble corresponding to the correct integer in the ORS

• Marking scheme :

+ 4 If the bubbler corresponding to the answer is darkened 0 In all other cases

Q.21 In the complex acetylbromidodicarbonylbis(triethylphosphine)iron(II), the number of Fe – C bond (s) is Ans. [3]

Sol. 1232

|

3

2])PEt()CO)(Br)(OCCH(Fe[ +

+

Fe

BrPEt3

OH – CH

H

O+ ≡ C–

C– ≡ O+

PEt3

C

Q.22 Among the complex ions, [Co(NH2 – CH2 – CH2 – NH2)2Cl2]+, [CrCl2(C2O4)2]3–, [Fe(H2O)4(OH)2]+,

[Fe(NH3)2(CN)4]–, [Co(NH2 – CH2 – CH2 – NH2)2(NH3)Cl]2+ and [Co(NH3)4(H2O)Cl]2+, the number of complex ion(s) that show(s) cis-trans isomerism is

Ans. [6] Sol. [Co(NH2 – CH2 – CH2 – NH2)2Cl2]+ shows one cis and one trans isomer [CrCl2(C2O4)2]3– shows one cis and one trans isomer [Fe(H2O)4(OH)2]+ shows one cis and one trans isomer [Fe(NH3)2(CN)4]– shows one cis and one trans isomer [Co(NH2 – CH2 – CH2 – NH2)2(NH3)Cl]2+ shows one cis and one trans isomer [Co(NH3)4(H2O)Cl]2+ shows one cis and one trans isomer Q.23 Three moles of B2H6 are completely reacted with methanol. The number of moles of boron containing

product formed is Ans. [6]

Sol. 3B2H6 + 18MeOH → 6B(OMe)3 + 18H2

CAREER POINT



Q.24 The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar

conductivity of a solution of a weak acid HY (0.10 M). If 0Y

0X −− λ≈λ , the difference in their pKa values,

pKa(HX) – pKa(HY), is (consider-degree of ionization of both acids to be << 1) Ans. [3]

Sol. Degree of ionization of weak acid HX(αHX) = ºHX

HX

ΛΛ

Degree of ionization of weak acid HY(αHY) = ºHY

HY

ΛΛ

∴ αHX = 101 αHY

⎭⎬⎫

⎩⎨⎧ Λ

=Λ10

HYHXQ

Ka = α2C {for α << 1)

HXaK = 2

HYα × 100

1 × C1 = 2HYα ×

1001 × 0.01 {Q C1 = 0.01 M}

HYaK = 2

HYα × C2 = 2HYα × 0.1 {Q C2 = 0.1 M}

HXaK =

10001

HYaK

HYapK = – log

HXaK = – log 1000K

HYa = 3 – log HYaK

HYapK = – log

HYaK = – logHYaK

HYkapK –

HYapK = 3

Q.25 A closed vessel with rigid walls contains 1 mole of U23892 and 1 mole of air at 298 K. Considering

complete decay of U23892 to Pb206

82 , the ratio of the final pressure to the initial pressure of the system at

298 K is Ans. [9] Sol. Initial mole of gas ni = 1 92U238 → 82Pb206 + 8 2He4 + 6 –1e0 Final mole of gas nf = 8 + 1 = 9 At const T, & V ; p ∝ n

19

nn

PP

i

f

i

f == ∴ Answer is 9

Q.26 In dilute aqueous H2SO4, the complex diaquodioxalatoferrate(II) is oxidized by −4MnO , For this

reaction, the ratio of the rate of change of [H+] to the rate of change of [ −4MnO ] is

Ans. [8]

CAREER POINT



Sol. OH6CO4Fe1Mn1H8])OX()OH(Fe[1OMn1 2

4

23

3222

2

3

22

)2(

4

)7(+++→++

++

++++−

++−

+

5 × 1 = 5 (1 × 1) + (1 × 4) = 51 1

∴ 18

]MnO[ofchangeofrate]H[ofchangeofrate

4

=−

+

Q.27 The number of hydroxyl group(s) in Q is

H

HO H3C CH3

heat

H⎯→⎯+

P Cº0

)excess(KMnOdiluteaqueous 4 ⎯⎯⎯⎯⎯⎯⎯⎯ →⎯ Q

Ans. [4]

Sol. First reaction involve dehydration of alcohol and second reaction is formation of syndiol rearrangement

H

HO H3C CH3

heat

H⎯→⎯+

HO

H

H3C CH3

+

H

Δ

–H2O +

H3C CH3CH3 CH3

–H⊕

H3C

HO HO CH3

OH

OH

(Q)

Rearrangement 1-2 Methyl Shift ⊕

CH3 CH3

(P)

0ºC, aq. & dil. KMnO4 (excess) Baeyer reagent (syn add.)

Total number of hydroxyl groups in Q ⇒ 4

CAREER POINT



Q.28 Among the following, the number of reaction(s) that produce(s) benzaldehyde is

I.

CuCl/AlClAnhydrous

HCl,CO

3

⎯⎯⎯⎯⎯ →⎯

II. CHCl2

Cº100

OH2⎯⎯ →⎯

III. COCl

4

2

BaSOPd

H

−⎯⎯ →⎯

IV CO2Me

OH

Cº78,–Toluene

HDIBAL

2

⎯⎯⎯⎯ →⎯ −

Ans. [4]

Sol. All four reaction produce benzaldehyde

I.

(Benzene) + CO + HCl

CuCl

AlClAnhydrous 3⎯⎯⎯⎯⎯ →⎯

C – H

O

Benzaldehyde

(Gatter mann's koch aldehyde syn.)

II.

CH – Cl

Cl

Cº100

OH2⎯⎯ →⎯

CH – OH

OH

⎯⎯ →⎯− OH2

C – H

O

Benzaldehyde

III.

C – Cl

O

Benzoyal chloride

4

2

BaSOPd

HcatalystLindlar

−⎯⎯⎯⎯⎯⎯ →⎯

C – H

O

Benzaldehyde [Rosenmund reduction]

IV.

—C — O — Me

O

Methyl Benzoate

reductionerestrictivOH

Cº78,–Toluene

HDIBAL

2

⎯⎯⎯⎯ →⎯ −

C – H

O

Benzaldehyde

CAREER POINT




• This Section contains EIGHT questions

• Each question has FOUR options (A), (B), (C) and (D) ONE OR MORE THAN ONE of these four option

(s) is (are) correct

• For each questions, darken the bubble (s) corresponding to all the correct option (s) in the ORS


+4 If only the bubble (s) corresponding to all the correct option (s) is (are) darkened

0 If none of the bubbles is darkened

–2 In all other cases

Q.29 Under hydrolytic conditions, the compounds used for preparation of linear polymer and for chain

termination, respectively, are

(A) CH3SiCl3 and Si(CH3)4 (B) (CH3)2SiCl2 and (CH3)3SiCl

(C) (CH3)2SiCl2and CH3SiCl3 (D) SiCl4 and (CH3)3SiCl

Ans. [B]

Sol. (i) (CH3)2SiCl2 undergoes hydrolysis by two sides so it form linear polymer

(ii) (CH3)3 SiCl undergoes hydrolysis by one side so it is chain termination.

Q.30 When O2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O2. The TRUE

statement(s) regarding this adsorption is(are)

(A) O2 is physisorbed

(B) heat is released

(C) occupancy of *p2π of O2 is increased

(D) bond length of O2 is increased

Ans. [B,C,D]

Sol. M O = O

Vacant π* ABMO

(i) Due to bonding, energy is released

(ii) Due to back donation bond length of O2 is increased

(iii) Due to presence of vacant π*2p orbital it accept some electron cloud from metal.

CAREER POINT



Q.31 One mole of a monoatomic real gas satisfies the equation p(V – b) = RT where b is a constant. The

relationship of interatomic potential V(r) and interatomic distance r for the gas is given by-

(A)

V(r)

0 r (B)

V(r)

0 r

(C)

V(r)

0 r (D)

V(r)

0 r

Ans. [C]

Sol. when only repulsive force act.

V(r) = ∞ r < σ

& V(r) = 0 r > σ

V(r)

0 rσ

Q.32 In the following reaction, the product S is –

H3C i. O3

ii. Zn, H2O

NH3R S

(A)

H3C N

(B)

H3C N

(C) H3C

N

(D)

H3C

N

CAREER POINT



Ans. [A]

Sol.

H3C i. O3

ii. Zn, H2O

NH3R S

H3C

i. O3 O

OO

H3C

Zn, H2O , Reductive cleavage

HHO

O

(R)

H3CNH3 O H NH

C H3C H

H3C OH

–H2OH3C

NN H

Q.33 The major product U in the following reactions is –

CH2=CH–CH3, H+

high pressure heat

radical initiator, O2T U

(A)

O O

CH3

H

(B)

O

H3C CH3O

H

(C)

O O

CH2

H

(D)

O

CH2

OH

||

Ans. [B]

CAREER POINT



Sol. This reaction is a part of formation of phenol

CH2=CH–CH3, H+

high pressure heat

radical initiator, O2

Free radical intermediate

CH3 CH3

H

C

CH3–CH–CH3 Carbocation Intermadiate

⊕

H3C CH3

O

C

O H

Cumene

[Cumene hydroperoxide]

Q.34 In the following reactions, the major product W is

0° C V

NH2

NaNO2, HCl

OH

, NaOHW

(A)

N=N OH (B)

N=N

OH

(C)

N=N

OH

(D)

N=N

HO

Ans. [A]

CAREER POINT



Sol.

N=N OH

NaNO2 + HCl HO–N=O + NaCl

N= O ⊕

(Attacking species)[Nitrosonium ion]

| NH2

+ N= O ⊕ O°C

–H2O

N≡N ⊕

|

ClΘ

Benzene Diazonium chloride (BDC)

(coupling reaction)

V OH

, NaOH

W In coupling reaction if para position is blocked then attack occur at ortho position

Q.35 The correct statement(s) regarding, (i) HClO, (ii) HClO2, (iii) HClO3, and (iv) HClO4, is (are)

(A) The number of Cl = O bonds in (ii) and (iii) together is two

(B) The number of lone pairs of electrons on Cl in (ii) and (iii) together is three

(C) The hybridization of Cl in (iv) is sp3

(D) Amongst (i) to (iv), the strongest acid is (i)

Ans. [B, C]

Sol. (i)

H–O Cl

(ii)

H–O Cl||O

(iii) H–O

Cl||O

||O

(iv)

H–O Cl = O|| O

|| O

sp3 , hybridizationstrongest acid

CAREER POINT



Q.36 The pair(s) of ions where BOTH the ions are precipitated upon passing H2S gas in presence of dilute HCl, is(are) -

(A) Ba2+, Zn2+ (B) Bi3+, Fe3+ (C) Cu2+, Pb2+ (D) Hg2+, Bi3+ Ans. [C, D] Sol. Group number (I) and (II) radicals give precipitate


• This section contains TWO Paragraphs • Based on each paragraph, there will be TWO questions • Each question has FOUR options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four

option(s) is (are) correct • Four each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS. • Marking scheme : + 4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened 0 If none of the bubbles is darkened –2 In all other cases

PARAGRAPH 1

In the following reactions

C8H6Pd-BaSO4

C8H8i. B2H6

ii. H2O2, NaOH, H2OX

H2O

HgSO4,H2SO4

C8H8O i. EtMgBr, H2Oii. H+, heat

H2

CAREER POINT



Q.37 Compound X is -

(A)

O

CH3 (B)

OH

CH3

(C) OH

(D) CHO

Ans. [C]

Q.38 The major compound Y is -

(A) CH3 (B) CH3

(C)

CH2 CH3

(D)

CH3

CH3

Ans. [D]

Sol. The following reactions

88

2

468 HC

HBaSOPdHC −

DBE = (x + 1) –1/2 = (8 + 1) –6/2 = 9 – 3 = 6

Compound - C8H6 is

C≡C–H

Now,

CAREER POINT



C ≡ C–H

+ H2 Pd

BaSO4

C = C H H

H

HgSO4.H2SO4 Hydration

HO–C=CH2 |

CH2–CH2 |

Reduction ( Lindlar-catalyst)

Syn Addition B2H6

B

3

H–OH

Tautomerism

O=C–CH3 |

EtMgBr, H2O

HO–C–CH3 |

Et |

HO–OH NaOH, H2O

CH2– CH2 –OH|

H+ Δ

C |

Et |Me OH2

⊕

H3C

H3C–HC–C–CH3 |

⊕ |H

CH= C |

CH3

(Saytzeff product)

[X] Antimarkownikov alcohol

–H2O

–H⊕

PARAGRAPH 2

When 100 mL of 1.0 M HCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant

pressure, a temperature increase of 5.7ºC was measured for the beaker and its contents (Expt.1).

Because the enthalpy of neutralization of a strong acid with a strong base is a constant (–57.0 kJ mol–1),

this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2),

100 mL of 2.0 M acetic acid (Ka = 2.0 × 10–5) was mixed with 100 mL of 1.0 M NaOH (under identical

conditions to Expt. 1) where a temperature rise of 5.6 ºC was measured.

(Consider heat capacity of all solutions as 4.2 J g–1 K–1 and density of all solutions as 1.0 g mL–1)

CAREER POINT



Q.39 Enthalpy of dissociation (in kJ mol–1) of acetic acid obtained from the Expt. 2 is -

(A) 1.0 (B) 10.0 (C) 24.5 (D) 51.4

Ans. [A]

Sol. After experiment 1

mass of solution = 200× 1.0 = 200 g {because density of all solution = 1g/ml}

let C is calorimetric constant

Q = m.s.ΔT + C.ΔT.

7.5C1000

7.52.42007.5 ×+××

= {for 0.1 mol}

⇒ C = 0.16

After experiment 2

6.516.01000

6.52.420010x

×+××

=

x = 56

ΔHN(WASB) = ΔHN(SASB) + ΔHdiss. of WA

57 = 56 + ΔHdiss. of WA {for 1 mol}

ΔHdiss. of WA = 1 KJ /mol.

Q.40 The pH of the solution after Expt. 2 is -

(A) 2.8 (B) 4.7 (C) 5.0 (D) 7.0

Ans. [B]

Sol. CH3COOH + NaOH CH3COONa + H2O

Initial 0.2 0.1 0

Final 0.1 0 0.1

pH = pKa + log ]acid[

]baseconjugate[

pH = pKa = 4.7

CAREER POINT



Part III - MATHEMATICS


• This section contains EIGHT questions

• The answer to each questions is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive

• For each questions, darken the bubble corresponding to the correct integer in the ORS


+ 4 If the bubble corresponding to the answer is darkened

0 In all other cases

Q.41 The coefficient of x9 in the expansion of (1 + x) (1 + x2) (1 + x3) ……. (1 + x100) is

Ans. [8]

Sol. Required will be same as that of coefficient of x9 in (1 + x) (1 + x2) …….(1 + x9)

Possible cases are

(i) x0 x9

(ii) x1 x8

(iii) x2 x7

(iv) x3 x6

(v) x4 x5

(vi) x1 x2 x6

(vii) x2 x3 x4

(viii) x1 x3 x5

Each give (+1) to coefficient of x9 = 8

Q.42 Suppose that the foci of the ellipse 5

y9

x 22+ = 1 are (f1, 0 ) and (f2, 0) where f1 > 0 and f2 < 0. Let P1 and

P2 be two parabolas with a common vertex at (0, 0) and with foci at (f1, 0) and (2f2 , 0), respectively.

Let T1 be a tangent to P1 which passes through (2f2, 0) and T2 be a tangent to P2 which passes through

(f1, 0). If m1 is the slope of T1 and m2 is the slope of T2, then the value of ⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2

221

mm1 is

CAREER POINT



Ans. [4]

Sol. Foci of the ellipse are (2, 0) and (–2, 0) So f1 = 2, f2 = – 2 Equation of T1 :

y2 = –16x y2 = 8x

(–4, 0) (2, 0)

T1 : y = m1x + 1m

2

Q T1 passes through (– 4, 0)

so 21m =

21

T2 : y = 2

2 m4xm −

T2 passes through (2, 0) ∴ 2

2m = 2

∴ 21m

1 + 22m = 4

Q.43 Let m and n be two positive integers greater than 1.

If ⎟⎠⎞

⎜⎝⎛−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

α

−α

→α 2eeelim m

)cos(

0

n

then the value of nm is

Ans. [2]

Sol. ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

α

−α

→α m

cos

0

eelimn

= ⎥⎥⎦

⎤

⎢⎢⎣

⎡

−α

−−α

1cos1ee n

1cos n

. 2n

n

)(1cos

α

−α m

2n )(α

α

= e(lne) ⎟⎠⎞

⎜⎝⎛ −

21 a2n – m =

2e−

So 2n – m = 0 m = 2n

∴ nm = 2

CAREER POINT



Q.44 If

α = ∫ ⎟⎟⎠

⎞⎜⎜⎝

⎛

+

+−+1

02

2xtan3x9 dx

x1x912)e(

1

where tan–1x takes only principal values, then the value of ⎟⎠⎞

⎜⎝⎛ π

−α+4

3|1|loge is

Ans. [9]

Sol. Let x = tan θ

dx = sec2 θ dθ

α = ( )∫π

θ+θ θθ+4/

0

2tan93 d)tan912.(e

α = ( )∫π

θ+θ θ+θ4/

0

2tan93 d)3sec9.(e

Let 3θ + 9tanθ = t

(3 + 9 sec2 θ) dθ = dt

α = ∫

π+ π

+

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡−=

439

0

439t 1edte

α + 1 = 439

eπ

+

ln |1 + α | – 4

3π = 9

Q.45 Let f : R → R be a continuous odd function, which vanishes exactly at one point and 21)1(f = . Suppose

that F(x) = ∫−

x

1

dt)t(f for all x ∈ [–1, 2] and G(x) = dt|))t(f(f|tx

1∫−

for all x ∈ [–1, 2]. If )x(G)x(Flim

1x→=

141 ,

then the value of f ⎟⎠⎞

⎜⎝⎛

21 is

Ans. [7]

Sol. As f(x) is continuous odd function

So f(0) = 0

Given f(1) = 21

CAREER POINT



F(x) = ∫−

x

1

dt)t(f ∀ x ∈ [– 1, 2]

G(x) = ∫−

x

1

dt|))t(f(f|t ∀ x ∈ [– 1, 2]

)x(G)x(Flim

1x→=

141

⇒

∫

∫

−

−→ x

1

x

11x

dt|))t(f(f|t

dt)t(f

lim = |))x(f(f|x

)x(flim1x→

{Using L' Hospital's Rule}

⇒ ⎟⎠⎞

⎜⎝⎛

21f

21

= 141

⇒ ⎟⎠⎞

⎜⎝⎛

21f = 7

Q.46 Suppose that randq,prrr are three non-coplanar vectors in R3. Let the components of a vector s

r along

randq,prrr be 4, 3 and 5, respectively. If the components of this vector s

r along )rqp(rrr

++− , )rqp(rrr

+−

and )rqp(rrr

+−− are x, y and z, respectively, then the value of 2x + y + z is

Ans. [9]

Sol. r5q3p4srrrr

++=

)rq–p(z)rq–p(y)rqp(xsrrrrrrrrrr

+−+++++−=

r)zyx(q)z–y–x(p)z–yx(srrrr

+++++−=

∴ – x + y – z = 4 ….(1)

x – y – z = 3 ….(2)

x + y + z = 5 ….(3)

Solving (1), (2) & (3)

x = 4 ; y – z = 8 and y + z = 1

∴ 2x + y + z = 8 + 1 = 9

CAREER POINT



Q.47 For any integer k, let αk = cos ⎟⎠⎞

⎜⎝⎛ π

7k + i sin ⎟

⎠⎞

⎜⎝⎛ π

7k , where i = 1− . The value of the expression

∑

∑

=−

=+

α−α

α−α

3

1k2k41–k4

12

1kk1k

||

|| is

Ans. [4]

Sol. αk = cos ⎟⎠⎞

⎜⎝⎛ π

7k + i sin ⎟

⎠⎞

⎜⎝⎛ π

7k = 7

kie

π

Note :

|| n1n α−α + = 7ni

7)1n(i

eeππ

+−

= ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−

ππ

1ee 7i

7ni

= ⎟⎟⎠

⎞⎜⎜⎝

⎛−

π−

πππ14

i14

i14

i7

nieee.e

= ⎟⎠⎞

⎜⎝⎛ π

+π

147ni

e i14

sin2 π

= 2 sin 14π

∴

∑

∑

=−−

=+

α−α

α−α

3

1K2k41k4

12

1Kk1k

||

||=

14sin23

14sin212

π⋅

π⋅

= 4

CAREER POINT



Q.48 Suppose that all the terms of an arithmetic progression (A.P.) are natural numbers. If the ratio of the

sum of the first seven terms to the sum of the first eleven terms is 6 : 11 and the seventh term lies in

between 130 and 140, then the common difference of this A.P. is

Ans. [9]

Sol. An A.P. of natural numbers

116

SS

11

7 = ⇒ 116

]d10a2[2

11

]d6a2[27

=+

+

⇒ 116

)d10a2(11)d6a2(7

=++

⇒ 14a + 42d = 12a + 60 d

⇒ 2a = 18 d

⇒ a = 9d

and 130 < a7 < 140

⇒ 130 < a + 6d < 140

⇒ 130 < 15 d < 140

⇒ 3

26 < d < 328

⇒ d = 9 (as 'd' is a natural number)

CAREER POINT




• This Section contains EIGHT questions • Each question has FOUR options (A), (B), (C) and (D) ONE OR MORE THAN ONE of these four option(s) is (are) correct • For each questions, darken the bubble (s) corresponding to all the correct option (s) in the ORS • Marking scheme : +4 If only the bubble (s) corresponding to all the correct option (s) is (are) darkened 0 If none of the bubbles is darkened –2 In all other cases

Q.49 Let f, g : [–1, 2] → R be continuous functions which are twice differentiable on the interval (–1, 2). Let

the values of f and g at the points – 1, 0 and 2 be as given in the following table :

x = – 1 x = 0 x = 2

f(x) 3 6 0

g(x) 0 1 –1

In each of the intervals (–1, 0) and (0, 2) the function (f – 3g)" never vanishes. Then the correct

statement(s) is (are)

(A) f'(x) – 3g'(x) = 0 has exactly three solutions in (– 1, 0) ∪ (0, 2)

(B) f'(x) – 3g'(x) = 0 has exactly one solution in (– 1, 0)

(C) f'(x) – 3g'(x) = 0 has exactly one solution in (0, 2)

(D) f'(x) – 3g'(x) = 0 has exactly two solutions in (– 1, 0) and exactly two solutions in (0, 2)

Ans. [B, C]

Sol. f(–1) = 3 f(0) = 6 f(2) = 0

g(–1) = 0 g(0) = 1 g(2) = – 1

h(x) = f(x) – 3g(x)

h (–1) = f(–1) – 3g (–1) = 3 – 3 (0) = 3

h (0) = f(0) – 3g (0) = 6 – 3 (1) = 3

h (2) = f(2) – 3g (2) = 0 – 3 (–1) = 3

By rolles theorem atleast one solution for h'(x) = 0 in (–1, 0)

& atleast one solution for h'(x) = 0 in (0, 2)

Now given at (–1, 0) & (0, 3)

(f – 3g)" = h"(x) ≠ 0

h'(x) is monotonic

so exactly one solution in (–1, 0)

& exactly one solution in (0, 2)

CAREER POINT



Q.50 Let f(x) = 7 tan8 x + 7tan6 x – 3 tan4 x – 3tan2 x for all x ∈ ⎟⎠⎞

⎜⎝⎛ ππ−

2,

2. Then the correct expression(s) is

(are)

(A) ∫π

=4/

0121dx)x(xf (B) ∫

π

=4/

0

0dx)x(f

(C) ∫π

=4/

061dx)x(xf (D) ∫

π

=4/

0

1dx)x(f

Ans. [A, B]

Sol. ∫π 4/

0

dx)x(f

= ∫π 4/

0

26 xdxsecxtan7 – ∫π 4/

0

22 xdxsec.xtan3

Let tan x = t , sec2 xdx = dt

= ∫1

0

6 –dtt7 ∫1

0

2dtt3

= 10

7 ]t[ – 10

3]t[ = 1 – 1 = 0

Now

∫π 4/

0

dx)x(xf

= ∫π 4/

0 II

26I

dx)xsecxtan7(x – ∫π 4/

0 II

22I

dx)xsecxtan3(.x

= 4/0

7 ]xtan.x[ π – 4/0

34/

0

7 ]xtanx[–xdxtan ππ

∫ + ∫π 4/

0

3 xdxtan

=4

–dx)xtan–x(tan4

4/

0

37 π−

π∫

π

= ∫π

−−4/

0

43 dx)1x(tanxtan

∫π

−−4/

0

223 xdxsec)1x(tanxtan

Let tan x = t

∴ sec2xdx = dt

CAREER POINT



= ∫ −−1

0

23 dt)1t(t

= – 1

0

46

4t

6t

⎥⎥⎦

⎤

⎢⎢⎣

⎡−

= – ⎥⎦⎤

⎢⎣⎡ −

41

61 =

121

So options (A, B) are correct

Q.51 Let f ′(x) = xsin2

x1924

3

π+for all x ∈ R with f ⎟

⎠⎞

⎜⎝⎛

21 = 0. If m ≤ ∫

1

2/1

dx)x(f ≤ M, then the possible values of m

and M are -

(A) m = 13, M = 24 (B) m = 41 , M =

21

(C) m = –11, M = 0 (D) m = 1, M = 12

Ans. [D]

Sol. f ′(x) = xsin2

x1924

3

π+ > 0 x ∈ ⎟

⎠⎞

⎜⎝⎛ 1,

21

f(x) monotonic increasing

min f ′(x) = 12

x192 3

+ =

3x192 3

(when sin4 πx = 1)

min f (x) = dxx64 3∫ = 4x64 4

= 16x4 at x =21

max f ′(x) = 02

x192 3

+ = 96x3 (when sin4 πx = 0)

max f(x) = 4

96 x4 = 24x4 at x = 1

16 ∫1

2/14 dx

21 ≤ ∫

1

2/1

dx)x(f ≤ ∫1

2/1

dx24

⎟⎠⎞

⎜⎝⎛ −

211 ≤ ∫

1

0

dx)x(f ≤ 24 ⎟⎠⎞

⎜⎝⎛ −

211

21 ≤ ∫

1

2/1

dx)x(f ≤ 12

CAREER POINT



Q.52 Let S be the set of all non-zero real numbers α such that the quadratic equation αx2 – x + α = 0 has two

distinct real roots x1 and x2 satisfying the inequality |x1 – x2| < 1. Which of the following intervals is(are)

a subset(s) of S ?

(A) ⎟⎟⎠

⎞⎜⎜⎝

⎛5

1,–21– (B) ⎟⎟

⎠

⎞⎜⎜⎝

⎛0,

51–

(C) ⎟⎟⎠

⎞⎜⎜⎝

⎛5

1,0 (D) ⎟⎟⎠

⎞⎜⎜⎝

⎛21,

51

Ans. Ans. [A, D]

Sol. Roots are real so

D > 0

1 – 4α2 > 0

⇒ –21 < α <

21 ...(i)

also, x1 + x2 = α1 , x1x2 = 1

Given |x1 – x2| < 1

⇒ 212

21 xx4)xx( −+ < 1

⇒ 412 −

α < 1

⇒ 21

α – 4 < 1

⇒ 21

α – 5 < 0

⇒ (5α2 – 1) > 0

( 5 α – 1) ( 5 α + 1) > 0

⇒ α < 51− or α >

51 ...(ii)

from (i) & (ii)

α ∈ ⎟⎟⎠

⎞⎜⎜⎝

⎛ −−51,

21 ∪ ⎟⎟

⎠

⎞⎜⎜⎝

⎛21,

51

CAREER POINT



Q.53 If α = 3sin–1 ⎟⎠⎞

⎜⎝⎛116 and β = 3cos–1 ⎟

⎠⎞

⎜⎝⎛

94 , where the inverse trigonometric functions take only the principal

values, then the correct option(s) is(are) -

(A) cos β > 0 (B) sin β < 0

(C) cos(α + β) > 0 (D) cos α < 0

Ans. [B,C,D]

Sol.

6 11

85

α/3

3α = tan–1

856

as 3

1 < 856 < 1

30º < 3α < 45º

⇒ 90º < α < 135º

So, cos α < 0 (Option (D) is correct)

Also

65 9

β/3

4

⇒ 3β = tan–1

465

as 465 > 3

So, 60º < 3β < 90º

∴ 180º < β < 270º

∴ cos β < 0, sin β < 0

CAREER POINT



so option (A) is incorrect and option (B) is correct.

(α + β) will lie in 4th quadrant.

So, cos (α + β) > 0

Hence option (C) is correct.

Q.54 Let E1 and E2 be two ellipses whose centers are at the origin. The major axes of E1 and E2 lie along the

x-axis and the y-axis, respectively. Let S be the circle x2 + (y – 1)2 = 2. The straight line x + y = 3

touches the curves S, E1 and E2 at P, Q and R, respectively. Suppose that PQ = PR =3

22 . If e1 and e2

are the eccentricites of E1 and E2, respectively, then the correct expression(s) is (are) -

(A) 4043ee 2

221 =+ (B) e1e2 =

1027 (C)

85e–e 2

221 = (D) e1e2 = 4

3

Ans. [A, B]

Sol.

P

R

Q

E1

E2

x + y = 3

(0, 1)x + y = 3

P (1, 2)

Given PQ = PR = 3

22

∴ using parametric equation of line

∴ Q = ⎟⎟⎠

⎞⎜⎜⎝

⎛×−×+

21

3222,

21

3221

⇒ Q = ⎟⎠⎞

⎜⎝⎛

34,

35

Similarly; R = ⎟⎠⎞

⎜⎝⎛

38,

31

Now :

Let E1 : 21

2

21

2

by

ax

+ = 1

CAREER POINT



Let point of contact is Q(x1, y1) then equation of tangent is 21

1

axx + 2

1

1

byy = 1

Comparing it with x + y = 3

Q(x1, y1) = ⎟⎟⎠

⎞⎜⎜⎝

⎛

3b,

3a 2

121 ≡ ⎟

⎠⎞

⎜⎝⎛

34,

35

21a = 5, 2

1b = 4

⇒ e1 = 5

1

Similarly let E2 : 22

2

ax + 2

2

2

by = 1

Then R ⎟⎟⎠

⎞⎜⎜⎝

⎛

3b,

3a 2

222 = ⎟

⎠⎞

⎜⎝⎛

38,

31

22a = 1, 2

2b = 8

⇒ e2 22

7

Now checking option

(A) 21e + 2

2e = 51 +

87 =

4043 (Correct)

(B) e1e2 = 1027 (Correct) & (D) Incorrect

(C) | 21e – 2

2e |= 87

51

− = 4027 (Incorrect)

Q.55 Consider the hyperbola H : x2 – y2 = 1 and a circle S with center N(x2, 0). Suppose that H and S touch

each other at a point P(x1, y1) with x1 > 1 and y1 > 0. The common tangent to H and S at P intersects the

x-axis at point M. If (l, m) is the centroid of the triangle ΔPMN, then the correct expression(s) is(are) -

(A) 1dx

dl = 1 – 21x3

1 for x1 > 1 (B) 1dx

dm = ⎟⎠⎞⎜

⎝⎛ 1–x3

x21

1 for x1 > 1

(C) 1dx

dl = 1 + 21x3

1 for x1 > 1 (D) 1dy

dm = 31 for y1 > 0

Ans. [A,B,D]

Sol. PN is normal to hyperbola & circle

CAREER POINT



p(x1, y1)

x

y

OM

⎟⎟⎠

⎞⎜⎜⎝

⎛0,

x1

1

N(x2, 0)

Now normal at P(x1, y1) on hyperbola is

y – y1 = 1

1xy− (x – x1)

put y = 0

x = 2x1

∴ x2 = 2x1

N(2x1, 0)

Equation of tangent at P (x1, y1) on hyperbola is xx1 – yy1 = 1

Put y = 0

x = 1x

1

Point M ⎟⎟⎠

⎞⎜⎜⎝

⎛0,

x1

1

∴ l = 3

x1xx21

11 ++ & m =

30y0 1 ++

1dy

dm = 31 )1x( 1 >

l = x1 + 1x3

1

1dx

dl = 1 – 21x3

1 (x1 > 1)

Also m = 31 1x2

1 −

∴ 1dx

dm = ⎟⎠⎞⎜

⎝⎛ −1x3

x21

1 if x1 > 1

CAREER POINT



Q.56 The option(s) with the values of a and L that satisfy the following equation is(are)

∫

∫π

π

+

+

0

46t

4

0

46t

dt)atcosat(sine

dt)atcosat(sine

= L ?

(A) a = 2, L =1–e1–e4

π

π

(B) a = 2, L = 1e1e4

++

π

π

(C) a = 4, L = 1–e1–e4

π

π

(D) a = 4, L =1e1e4

++

π

π

Ans. [A,C]

Sol.

1

0

46t

I

dt)atcosat(sine∫π

+ +

2

246t

I


π

+ +

3

3

2

46t

I


π

+ +

4

4

3

46t

I


π

+

I2 put t = π + α I3 put t = 2π + α I4 put t = 3π + α Numerator =

∫π

+0

46t dt)atcosat(sine + ∫π

π +0

46t dt)atcosat(sinee + ∫π

π +0

46t2 dt)atcosat(sinee + ∫π

π +0

46t3 dt)atcosat(sinee

= (1 + eπ+ e2π + e3π) ∫π

+0

46t dt)atcosat(sine

Now,

∫

∫π

ππππ

+

++++

0

46t

0

46t32

dt)atcosat(sine

dt)atcosat(sine)eee1(

L = 1. 1e1e4

−−

π

π

, a = 2, 4


• This Section contains Two paragraphs • Based on each paragraph, there will be Two questions • Each question has Four options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s)

is (are) correct • For each question, darken the bubble(s) corresponding to all the correct option(s) in the ORS • Marking scheme : +4 If only the bubble(s) corresponding to all the correct option(s) is (are) darkened 0 If none of the bubbles is darkened –2 In all other cases

CAREER POINT



Paragraph # 1 (Questions 57 & 58)

Let n1 and n2 be the number of red and black balls, respectively, in box I. Let n3 and n4 be the number of

red and black balls, respectively, in box II.

Q.57 One of the two boxes, box I and box II, was selected at random and a ball was drawn randomly out of

this box. The ball was found to be red. If the probability that this red ball was drawn from box II is 31 ,

then the correct option(s) with the possible values of n1, n2, n3 and n4 is(are) -

(A) n1 = 3, n2 = 3, n3 = 5, n4 = 15 (B) n1 = 3, n2 = 6, n3 = 10, n4 = 50

(C) n1 = 8, n2 = 6, n3 = 5, n4 = 20 (D) n1 = 6, n2 = 12, n3 = 5, n4 = 20

Ans. [A, B]

Sol. Box I Box II

Red → n1 Red → n3

Black → n2 Black → n4

Event A = Red ball is drawn.

E1 : Box I selected

E2 : Box II selected

P ⎟⎠⎞

⎜⎝⎛

AE2 =

⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟⎟

⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛

22

11

22

EAP)E(P

EAP)E(P

EAP)E(P

31 =

43

3

21

1

43

3

nnn

21

nnn

21

nnn.

21

++

+

+

21

1

nnn+

+ 43

3

nnn+

= 43

3

nnn3+

(A) 43

205.3

41

21

205

63

==+=+ correct

(B) 60

)10(36010

93

=+ ⇒ 21

61

31

=+ ⇒ 21

21

= correct

(C) 25

5.3255

148

=+ ⇒ 53

51

74

≠+ incorrect

(D) 25

5.3255

186

=+ ⇒ 53

51

31

≠+ incorrect

CAREER POINT



Q.58 A ball is drawn at random from box I and transferred to box II. If the probability of drawing a red ball

from box I, after this transfer, is 31 , then the correct option(s) with the possible values of n1 and n2

is (are) -

(A) n1 = 4 and n2 = 6 (B) n1 = 2 and n2 = 3 (C) n1 = 10 and n2 = 20 (D) n1 = 3 and n2 = 6

Ans. [C, D]

Sol. Case (I) : Red ball is transferred

Red → n1 – 1

Black → n2

Case (II) : Black ball is transferred

Red → n1

Black → n2 – 1

P = ⎟⎟⎠

⎞⎜⎜⎝

⎛+ 21

1

nnn

⎟⎟⎠

⎞⎜⎜⎝

⎛−+

−1nn

1n

21

1 + ⎟⎟⎠

⎞⎜⎜⎝

⎛+ 21

2

nnn ⎟⎟

⎠

⎞⎜⎜⎝

⎛−+ 1nn

n

21

1

↓ ↓ ↓ ↓

Probability of Red ball transferred

Probability of selecting Red ball

Probability of Black ball transferred

Probability of selecting Red ball

P = )1nn)(nn(

nnnn

2121

21121

−+++− =

)1nn)(nn()1nn(n

2121

211

−++−+ =

)nn(n

21

1

+=

31

(A) 104 ≠

31 Incorrect

(B) 52 ≠

31 Incorrect

(C) 3010 =

31 Correct

(D) 93 =

31 Correct

Paragraph For Questions 59 and 60

Let F : R → R be a thrice differentiable function. Suppose that F(1) = 0, F(3) = –4 and F′(x) < 0 for all x ∈ (1/2, 3). Let f(x) = xF(x) for all x ∈ R.

Q.59 The correct statement(s) is(are) -

(A) f ′(1) < 0 (B) f(2) < 0

(C) f ′(x) ≠ 0 for any x ∈ (1, 3) (D) f ′(x) = 0 for some x ∈ (1, 3)

CAREER POINT



Ans. [A,B,C]

Sol. F(1) = 0 F(3) = –4 F′(x) < 0, x ∈ ⎟⎠⎞

⎜⎝⎛ 3,

21

F(x) is decreasing in ⎟⎠⎞

⎜⎝⎛ 3,

21

f(x) = xF(x)

f ′(x) = F(x) + xF′(x) < 0, x ∈ (1, 3)

so f(x) is decreasing

(A) f ′(1) = F(1) + F′(1) = F′(1) < 0

So option (A) is correct

(B) f(1) = 1F(1) = 0

f(3) = 3F(3) = –12

f(2) < 0 correct

(C) f ′(x) < 0 so f ′(x) ≠ 0 any x ∈ (1, 3) correct

Q.60 If ∫ ′3

1

2 dx)x(Fx = –12 and ∫ ″3

1

3 dx)x(Fx = 40, then the correct expression(s) is(are) -

(A) 9f ′(3) + f ′(1) – 32 = 0 (B) ∫3

1

dx)x(f = 12

(C) 9f ′(3) – f ′(1) + 32 = 0 (D) ∫3

1

dx)x(f = –12

Ans. [C,D]

Sol. ∫ ″3

1

3 dx)x(Fx = 3

13 xFx )′( – ∫ ′

3

1

2 dx)x(Fx3

40 = 27F′(3) – F′(1) – 3(–12)

27F′(3) – F′(1) = 4 … (1)

f ′(x) = F(x) + xF′(x)

f ′(1) = F(1) + F′(1) ⇒ F′(1) = f ′(1)

f ′(3) = F(3) + 3F′(3)

f ′(3) = – 4 + 3F′(3) ⇒ 3F′(3) = f ′(3) + 4

Eqn (1) 9(f ′(3) + 4) – f ′(1) = 4

⇒ 9f ′(3) – f ′(1) + 32 = 0 so option (C) is correct

CAREER POINT



∫3

1

dx)x(f = ∫3

1

dx)x(xF = 3

1

2)x(F

2x

– ∫ ′3

1

2dx)x(F

2x

= 29 F(3) –

2)1(F –

21

∫ ′3

1

2 dx)x(Fx

=29 (–4) – 0 –

21 (–12)

= –18 + 6 = – 12 so option (D) is correct

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