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Paper-1 [ CODE – 05 ]

JEE Advance Exam 2015 (Solution)

PART I - PHYSICS

SECTION – 1 (Maximum Marks : 32)

• 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

• Marking scheme : + 4 If the bubble corresponding to the answer is darkened 0 In all other cases

Q.1 A bullet is fired vertically upwards with velocity ν from the surface of a spherical planet. When it reaches its maximum height, Its acceleration due to the planet's gravity is 1/4th of its value at the surface

of the planet. If the escape velocity from the planet is νesc = ν N , then the value of N is (ignore energy

loss due to atmosphere) Ans. [2] Sol. Applying conservation of mechanical energy

2m21

ν + R

GMm– = 0 – hR

GMm+

2m21

ν = GMm ⎥⎦⎤

⎢⎣⎡

+ hR1–

R1

2

21

ν = GM ⎥⎦

⎤⎢⎣

⎡+× )hR(R

h …….(i)

We have

g h = g 2–

Rh1 ⎟

⎠⎞

⎜⎝⎛ +

4g = g

2–

Rh1 ⎟

⎠⎞

⎜⎝⎛ +

g (2)–2 = g 2–

Rh1 ⎟

⎠⎞

⎜⎝⎛ +

h = R ..……(ii)

CAREER POINT

Date : 24 / 05 / 2015

CAREER POINT



From equation (i) & (ii)

2

21

ν = GM ⎥⎦⎤

⎢⎣⎡

2R2h = 2R2

GMR = R2

GM

ν2 = R

GM

ν = R

GM

Escape velocity at surface of planet

νesc = RGM2

νsec = ν2 [Given νsec = ν N ]

N = 2

Q.2 Two identical uniform discs roll without slipping on two different surfaces AB and CD (see figures) starting at A and C with linear speeds ν1 and ν2, respectively, and always remain in contact with the surfaces. If they reach B and D with the same linear speed and ν1 = 3m/s, then ν2 in m/s is (g = 10 m/s2)

30 m

B

ν1 = 3 m/s A

27 m

D

ν2 C

Ans. [7] Sol. For 1st body (Applying conservation of energy)

21m

21

ν ⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2

2

Rk1 = mg (30) + 2m

21

ν ⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2

2

Rk1 ……(i)

For 2nd body (Applying conservation of energy)

22m

21

ν ⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2

2

Rk1 = mg (27) + 2m

21

ν ⎟⎟⎠

⎞⎜⎜⎝

⎛+ 2

2

Rk1 ……(ii)

equation (i) & (ii)

CAREER POINT



21 m ( )2

221 – νν ⎟⎟

⎠

⎞⎜⎜⎝

⎛+ 2

2

Rk1 = mg (3)

21 ( )2

2–9 ν ⎟⎠⎞

⎜⎝⎛ +

211 = 30

( )22–9 ν ×

43 = 30

22–9 ν = 40

ν2 = 7 m/s

Q.3 Two spherical stars A and B emit blackbody radiation. The radius of A is 400 times that of B and A

emits 104 times the power emitted from B. The ratio ⎟⎟⎠

⎞⎜⎜⎝

⎛λλ

B

A of their wavelength λA and λB at which the

peaks occur in their respective radiation curves is Ans. [2] Sol. Emissive power

P = σAT4

2

1

PP = 4

22

411

TATA

σ

σ (A = 4πR2)

1

104 = 400 × 400 ×

4

A

B⎟⎟⎠

⎞⎜⎜⎝

⎛λλ [using Weins displacement law λmax ∝ T

1 ]

4

4

)20(10 =

4

A

B⎟⎟⎠

⎞⎜⎜⎝

⎛λλ ⇒

A

B

λλ =

2010

⇒ A

B

λλ =

21

⇒ B

A

λλ =

12

Q.4 A nuclear power plant supplying electrical power to a village uses a radioactive material of half life T

years as the fuel. The amount of fuel at the beginning is such that the total power requirement of the village is 12.5 % of the electrical power available from the plant at that time. If the plant is able to meet the total power needs of the village for a maximum period of n T years, then the value of n is

Ans. [3]

Sol. According to radioactivity N = n0

2N

(where n = HTt )

Requirement of power is 12.5% of total power. Then up to the time when radioactive nuclei are 12.5%

of initial number, plant will provide the energy requirement of village

CAREER POINT



N = 12.5 % of N0

N = ⎟⎠⎞

⎜⎝⎛

1005.12 N0

n0

2N

= 8

N0 ⇒ 2n = 8

⇒ n = 3 ⇒ HTt = 3

t = 3T

n = 3

Q.5 A Young's double slit interference arrangement with slits S1 and S2 is immersed in water (refractive

index = 4/3) as shown in figure. The positions of maxima on the surface of water are given by x2 = p2m2λ2 – d2, where λ is the wavelength of light in air (refractive index = 1), 2d is the separation between the slits and m is an integer. The value of p is

S1

Water

x Air

S2

d

d

Ans. [3] Sol.

S1

Water μ = 4/3

x Air

S2

d

d Airp

For maxima at point p

Path difference between waves Δx = mλ

S2 p – S1p = mλ

μ 22 dx + – 22 dx + = mλ

22 dx + (μ –1) = mλ

22 dx + = 3mλ (Q μ = 34 )

x2 = 32m2λ2 – d2

On comparing we get p = 3

CAREER POINT



Q.6 Consider a concave mirror and a convex lens (refractive index = 1. 5) of focal length 10 cm each , separated by a distance of 50 cm in air (refractive index = 1) as shown in the figure. An object is placed at a distance of 15 cm from the mirror. Its erect image formed by this combination has magnification M1. When the set-up is kept in a medium of refractive index 7/6, the magnification becomes M2. The

magnitude 1

2

MM is

15 cm

50 cm

Ans. [7] Sol. In vacuum

15 cm

50 cm

L

f2 = 10 cm f1 = 10 cm

M1

For mirror u = – 15, f = – 10

then v1 +

15–1 =

10–1 (According to mirror formula)

v = –30

m1 = – ⎟⎠⎞

⎜⎝⎛

15–30– = –2

ie 1st image is formed at 30 cm before mirror. Now it will behave as object for lens.

hence for lens u = – 20 , f = 10

v1 –

20–1 =

101

v = 20

m2 = ⎟⎠⎞

⎜⎝⎛

20–20 = – 1

overall magnification M1 = m1m2 = 2

CAREER POINT



In vacuum

f1 = (μ –1) ⎟⎟

⎠

⎞⎜⎜⎝

⎛

21 R1–

R1

101 = ⎟

⎠⎞

⎜⎝⎛ 1–

23 ⎟⎟

⎠

⎞⎜⎜⎝

⎛

21 R1–

R1

1R

1 – 2R

1 = 51

In medium for lens- ⎟⎠⎞

⎜⎝⎛ =μ

67

'f

1 = ⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

1–

6723

⎟⎟⎠

⎞⎜⎜⎝

⎛

21 R1–

R1

'f

1 = 72 ×

51

f' = 2

35

there is no change for mirror. hence for mirror u = –15, f = –10 then v = –30 m1 = –2 For Lens

u = –20, f = 2

35

v1 –

20–1 =

352

v = 140

m2 = ⎟⎠⎞

⎜⎝⎛

20–140 = –7

Overall magnification M2 = m1 m2 = 14

Now ⎟⎟⎠

⎞⎜⎜⎝

⎛

1

2

MM =

214 = 7

CAREER POINT



Q.7 An infinitely long uniform line charge distribution of charge per unit length λ lies parallel to the y-axis

in the y-z plane at z = 23 a (see figure). If the magnitude of the flux of the electric field through the

rectangular surface ABCD lying in the x-y plane with its centre at the origin is 0n

Lε

λ (ε0 = permittivity

of free space), then the value of n is

a

LD

O

BA

C a23

y

z

x Ans. [6] Sol.

X

x dx

Y

Electric flux though elemental part -

dφ = EdA cosθ

dφ =

4a3x2

22

0 +∈π

λ × Ldx ×

4a3x

2a3

22 +

04La3d

∈πλ

=φ∫ ∫−= ⎟⎟

⎠

⎞⎜⎜⎝

⎛+

2/a

2ax

22

4a3x

dx

CAREER POINT



φ = 24

La3

0×

∈πλ ∫

+

2/a

02

2

4a3x

dx

φ = 02La3

∈πλ ×

2/a

0

1

a3x2tan

a32

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛−

φ = ⎟⎟⎠

⎞⎜⎜⎝

⎛∈π

λ −

31tanL 1

0

φ = 06

L∈

λ

On comparing n = 6

Q.8 Consider a hydrogen atom with its electron in the nth orbital. An electromagnetic radiation of

wavelength 90 nm is used to ionize the atom. If the kinetic energy of the ejected electron is 10.4 eV, then value of n is (hc = 1242 eV nm)

Ans. [2]

Sol. Energy of photon EPh = )nmin(

1242λ

eV

λ = 90 nm

EPh = 90

1242 eV ⇒ EPh = 13.8 eV

K.E. of ejected electron K.E. = 10.4 eV

So let energy of the orbits is En

En + EPh = 10.4 eV

En + 13.8 = 10.4

eV4.3En −=

En = – 13.6 2

2

nZ eV

for H-atom (Z = 1)

– 3.4 = – 13.6 2

2

n)1(

n = 2

CAREER POINT




• This Section contains TEN 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.9 A container of fixed volume has a mixture of one mole of hydrogen and one mole of helium in equilibrium at temperature T. Assuming the gases are ideal, the correct statement (s) is (are)

(A) The average energy per mole of the gas mixture is 2RT.

(B) The ratio of speed of sound in the gas mixture to that in helium gas is 5/6

(C) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/2.

(D) The ratio of the rms speed of helium atoms to that of hydrogen molecules is 1/ 2 .

Ans. [A,B,D]

Sol. (A) (KEav)mix / mol = 2

RT)1(23RT)1(

15

+ = 2RT

(B) vsound = MRTγ

Here, γmix = ⎟⎠⎞

⎜⎝⎛ ×+⎟

⎠⎞

⎜⎝⎛ ×

⎟⎠⎞

⎜⎝⎛ ×+⎟

⎠⎞

⎜⎝⎛ ×

251

231

271

251

= 23

⇒ He

mix

vv =

He

mix

γγ

mix

He

MM

He

mix

vv =

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛

3523

× 34 =

56

CAREER POINT



Q.10 In an aluminum (Al) bar of square cross section, a square hole is drilled and is filled with iron (Fe) as

shown in the figure. The electrical resistivities of Al and Fe are 2.7 × 10–8 Ωm and 1.0 × 10–7 Ωm, respectively. The electrical resistance between the two faces P and Q of the composite bar is

Fe

2mm P

Q

Al50 mm

7mm

(A) Ωμ64

2475 (B) Ωμ64

1875

(C) Ωμ49

1875 (D) Ωμ1322475

Ans. [B]

Sol. For Aluminium

RAl = Al

lA

Alρ

= 6–

3–8–

10)4–49(1050107.2

××××

RAl = 30 × 10–6 Ω

For iron Fe

CAREER POINT



RFe = Fe

Fe

Alρ

RFe = 6–

3–7–

104105010

×××

RFe = 1250 ×10–6 Ω

Req. = Fe1A

Fe1A

RRRR

+

Req. = 6–

6–6–

10)125030(1012501030

×+×××

Req. = 6–101280

125030×

×

Req. = Ωμ64

1875

Q.11 For photo-electric effect with incident photon wavelength λ, the stopping potential is V0. Identify the

correct variation(s) of V0 with λ and 1 / λ.

(A)

λ

V0

(B)

λ

V0

(C)

1/λ

V0

(D)

1/ λ

V0

Ans. [A,C]

Sol. Einstein equation of photoelectric effect

λhc – φ = KEmax

λhc – φ = eV0

V0 = ehc ⋅

λ1 –

eφ

Straight line graph between V0 and λ1

CAREER POINT



λ1

V0

Ans. (C)

As λ increase V0 decreases

Check for the slope of V0 v/s λ graph

Slope = λd

dV0 = 2ehc–λ

as λ increase slope decreases and it is always negative

∴ Option (A) is correct

Q.12 Consider a Vernier calipers in which each 1 cm on the main scale is divided into 8 equal divisions and a

screw gauge with 100 divisions on its circular scale. In the Vernier calipers, 5 divisions of the Vernier

scale coincide with 4 divisions on the main scale and in the screw gauge, one complete rotation of the

circular scale moves it by two divisions on the linear scale. Then :

(A) If the pitch of the screw gauge is twice the least count of the Vernier calipers, the least count of the

screw gauge is 0.01 mm.

(B) If the pitch of the screw gauge is twice the least count of the Vernier calipers, the least count of the

screw gauge is 0.005 mm.

(C) If the least count of linear scale of the screw gauge is twice the least count of the Vernier calipers,

the least count of the screw gauge is 0.01 mm.

(D) If the least count of the linear scale of the screw gauge is twice the least count of the Vernier

calipers, the least count of the screw gauge is 0.005 mm.

Ans. [B,C]

Sol. Vernier scale least count calculations

M.S.D. = Main scale division = 81 cm

V.S.D. = Vernier scale division

L.C. = M.S.D. – V.S.D.

5 × V.S.D. = 4 × M.S.D.

V.S.D. = 54 M.S.D. =

54 ×

81 =

101 cm

CAREER POINT



L.C. = 81 –

101 ⇒

802 ⇒

401 cm ⇒ 0.25 cm

If pitch of screw gauge = 2 × 0.25 = 0.5 mm

then L.C. of screw gauge = 100

5.0 = 0.005 mm

option B is ans.

If L.C. of linear scale of screw gauge = 2 × 0.25 ⇒ 0.5 mm

then L.C. of screw gauge = 100

5.02 × = 0.01 mm

Q.13 Planck's constant h, speed of light c and gravitational constant G are used to form a unit of length L and

a unit of mass M. Then the correct option(s) is(are)

(A) M ∝ c

(B) M ∝ G

(C) L ∝ h

(D) L ∝ G

Ans. [A,C,D]

Sol. E = λhc

E = r

mGm– 21

λhc =

rGM 2

unit of λ and r is L

So M = 2/1

Ghc

⎟⎠⎞

⎜⎝⎛ …(1)

M ∝ c M ∝ G1 Ans. (A)

E = mc2

E = λhc

λhc = mc2

CAREER POINT



λ = mh

∴ L = Mh

L = 2/1)hc(h G1/2 from (1)

∴ L ∝ G L ∝ h

Q.14 Two independent harmonic oscillators of equal mass are oscillating about the origin with angular

frequencies ω1 and ω2 and have total energies E1 and E2, respectively. The variations of their momenta

p with positions x are shown in the figures. If ba = n2 and

Ra = n, then the correct equation(s) is(are)

b

P

Energy = E1

a x

P

Energy = E2

Rx

(A) E1ω1 = E2ω2 (B) 1

2

ωω = n2

(C) ω1ω2 = n2 (D) 1

1Eω

= 2

2Eω

Ans. [B,D] Sol. Ist harmonic oscillator.

b

P

ax

2

2

bP +

2

2

ax = 1

P2 = b2 ⎟⎟⎠

⎞⎜⎜⎝

⎛2

2

ax–1

P = ab 22 x–a

CAREER POINT



⇒ v = mP

v = am

b 22 x–a

Comparing v = ω 22 x–A

ω1 = am

b, A1 = a

& E1 = 21 m 2

1ω 21A

IInd harmonic oscillation P

Rx

P2 + x2 = R2

P = 22 x–R

v = mP

v = m1 22 x–R (v = ω 22 x–A )

Comparing ω2 = m1 , A2 = R1

E2 = 21 m 2

2ω 22A

(1) 1

2

ωω =

ambm1

⇒ 1

2

ωω =

ba (given

ba = n2)

1

2

ωω = n2

(2) 1

1Eω

= 21 mω1 2

1A

CAREER POINT



⇒ 1

1Eω

= 21 m ⎟

⎠⎞

⎜⎝⎛

amb (a)2

1

1Eω

= 21 ab

& 2

2Eω

= 21 m (ω2) 2

2A

2

2Eω

= 21 m ⎟

⎠⎞

⎜⎝⎛

m1 (R)2

2

2Eω

= 2

R 2

the value of ab = 2

2

na ⎟

⎠⎞

⎜⎝⎛ = 2n

ba

& R2 = 2n

a ⎟⎠⎞

⎜⎝⎛ = n

Ra

So 1

1Eω

= 2

2Eω

Q.15 A ring of mass M and radius R is rotating with angular speed ω about a fixed vertical axis passing

through its centre O with two point masses each of mass 8M at rest at O. These masses can move

radially outwards along two massless rods fixed on the ring as shown in the figure. At some instant the

angular speed of the system is 98

ω and one of the masses is at a distance of 53 R from O. At this instant

the distance of the other mass from O is

ω

O

(A) 32 R (B)

31 R (C)

53 R (D)

54 R

Ans. [C or D]

Sol. This question is based on angular momentum conservation

As given in the question initial angular velocity of ring is ω and final angular velocity of system is 98

ω.

so, Li = Lf Iω = Lf Iring ω = Lf

MR2 ω = MR2 × 98 ω +

8M ×

2

R53

⎟⎠⎞

⎜⎝⎛ ×

98 ω +

8M x2 ×

98 ω

CAREER POINT



R2 × 1 = 9R8 2

+ 81 ×

259 ×

9R8 2

+ 9

x 2

R2 ⎥⎦⎤

⎢⎣⎡

251–

98–1 =

9x 2

R2 ⎥⎦⎤

⎢⎣⎡

×××259

9–258–925 = 9

x 2

R2 2516 = x2

x = 54 R

So, option (D) is correct

OR Now we assume ring start to rotate at t = 0 by certain external agent.

Because both rods are frictionless and mass of the particles are equal. At t = 0, both particle starts to move from centre so they will experience same force at all the time. Hence position of particle will be same. If first

particle is at a distance R53 from centre then other will be also at a distance R

53 .

So, option (C) is also correct.

Q.16 The figures below depict two situations in which two infinitely long static line charges of constant positive line charge density λ are kept parallel to each other. In their resulting electric field, point charges q and –q are kept in equilibrium between them. The point charges are confined to move in the x direction only. If they are given a small displacement about their equilibrium positions, then the correct statements(s) is(are)

+q

λ λ

x

–q

λ λ

x v

(A) Both charges execute simple harmonic motion. (B) Both charges will continue moving in the direction of their displacement. (C) Charge +q executes simple harmonic motion while charge –q continues moving in the direction of

its displacement. (D) Charge –q executes simple harmonic motion while charge +q continues moving in the direction of

its displacement.

CAREER POINT



Ans. [C]

Sol.

q

λ λ

x q

r x r

r–x

If q is displaced slightly along x axis it will come back to original position

Fnet on q on displacing by small displacement x = restoring force = )x–r(2

q

0πελ –

)xr(2q

0 +πελ

⇒ F = 02

qπελ ⎥⎦

⎤⎢⎣⎡

+ xr1–

x–r1

⇒ F = )x–r(2

x2q22

0πε

×λ neglecting x

Frestoring = 20rxq

πελ

a = 2

0rmxq

πε

λ

a ∝ x so q will do SHM

– q will not able to oscillation as Fnet on – q will not send it back to its original equilibrium position. – q

continue moving in direction of its displacement.

Q.17 Two identical glass rods S1 and S2 (refractive index = 1.5) have one convex end of radius of curvature

10 cm. They are placed with the curved surfaces at a distance d as shown in the figure, with their axes (shown by the dashed line) aligned. When a point source of light P is placed inside rod S1 on its axis at a distance of 50 cm from the curved face, the light rays emanating from it are found to be parallel to the axis inside S2. The distance d is

50 cm

P S1

d

S2S1

(A) 60 cm (B) 70 cm (C) 80 cm (D) 90 cm

CAREER POINT



Ans. [B]

Sol.

50 cm

P I1

50 d-50 S1 S2

This question is of multiple event.

First event on curved surface of S1 & Second on curved surface of S2

Event – 1 n2 = 1

n1 = 1.5

v1 –

u5.1 =

R5.1–1

v1 –

50–5.1 =

10–5.0–

v1 +

505.1 =

201

v1 =

201 –

505.1

v1 =

1003–5 =

1002 ∴ v = 50 cm

for second event u = – (d – 50)

v = ∞

R = + 10 cm

v5.1 –

u1 =

R1–5.1

∞5.1 –

)50–d(–1 =

105.0

50–d

1 = 201

20 = d – 50

d = 70 cm

CAREER POINT



Q.18 A conductor (shown in the figure) carrying constant current I is kept in the x-y plane in a uniform

magnetic field Br

. If F is the magnitude of the total magnetic force acting on the conductor, then the

correct statement(s) is(are)

x I π/6 R R

π/4

L R R L

y

(A) If B is along z , F ∝ (L + R)

(B) If Br

is along x , F = 0

(C) If Br

is along y , F ∝ (L + R)

(D) If Br

is along z , F = 0

Ans. [A,B,C]

Sol. Option A if B is along z axis i.e B k

li = 2i (L + R) i

Fr

= li × Br

⇒ F = 2i (L + R) i × Br

k

∴ F ∝ (L + R) option A is correct answer.

Option B if B is along x axis i.e B i

then F = 2i (L + R) i × (B i ) = 0 option B is correct answer

Option C if B is along y axis i.e. B j

then F = 2i (L + R) i × B j

| F | ⇒ 2i (L + R) B

∴ F ∝ (L + R) option (C) is correct answer.

CAREER POINT



SECTION – 3 (Maximum Marks 16)

• This section contains TWO questions

• Each question contains two columns, Column I and Column II

• Column I has four entries (A), (B), (C) and (D)

• Column II has five entries (P), (Q), (R), (S) and (T)

• Match the entries in Column I with the entries in Column II

• One or more entries in Column I may match with one or more entries in Column II

• The ORS contains a 4 × 5 matrix whose layout will be similar to the one shown below:

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)

(D) (P) (Q) (R) (S) (T) • For each entry in Column I, darken the bubbles of all the matching entries. For example, if entry (A) in

Column I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS. Similarly, for entries (B), (C) and (D).

• Marking scheme: For each entry in Column I. +2 If only the bubble(s) corresponding to all the correct match(es) is(are) darkened 0 If none of the bubbles is darkened


Q.19 Match the nuclear processes given in column I with the appropriate option(s) in column II.

Column I Column II

(A) Nuclear fusion (P) Absorption of thermal neutrons by U23592

(B) Fission in a nuclear reactor (Q) Co6027 nucleus

(C) β-decay (R) Energy production in stars via hydrogen conversion to helium

(D) γ-ray emission (S) Heavy water (T) Neutrino emission Ans. [A→R ; B → P,S ; C→ T ; D → P,Q,R,T ] Sol. Above matching is based on theoretical concept of nuclear reaction and radioactivity.

CAREER POINT



Q.20 A particle of unit mass is moving along the x-axis under the influence of a force and its total energy is conserved. Four possible forms of the potential energy of the particle are given in column I (a and U0 are constants). Match the potential energies in column I to the corresponding statements(s) in column II.

Column I Column II

(A) U1(x) = 2

U0

22

ax–1

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ (P) The force acting on the particle is zero at x = a.

(B) U2(x) = 2

U02

ax

⎟⎠⎞

⎜⎝⎛ (Q) The force acting on the particle is zero at x = 0.

(C) U3(x) = 2

U02

ax

⎟⎠⎞

⎜⎝⎛ exp

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

2

ax– (R) The force acting on the particle is zero at x = –a.

(D) U4(x) = 2

U0

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

3

ax

31–

ax (S) The particle experiences an attractive force towards

x = 0 in the region |x| < a.

(T) The particle with total energy 4

U0 can oscillate

about the point x = –a. Ans. [((A → P,Q,R,T) (B →Q,S) (C → P,Q,R,S) (D → P,R,T))]

Sol. (A) U1 = 2

U0

22

ax–1

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞

⎜⎝⎛

F = – dx

dU1 = 2U– 0

⎥⎥⎦

⎤

⎢⎢⎣

⎡

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛

2

ax–12 ⎟

⎠⎞

⎜⎝⎛

2ax2–

F = 2

0

axU2 ⎟⎟

⎠

⎞⎜⎜⎝

⎛2

2

ax–1

at x = a, x = 0, x = – a F = 0 if x = + a/2 F = + ve

4

U0 A

– a

U = 0 Fr

Fr

Fr

B

4U0

O a

Fr

2

U0

Particle oscillating about x = – a. Shown energy in graph are potential energy and A and B are extreme point of oscillation where total energy is potential energy.

Option → P, Q, R, T correct

(B) U2 = 2

U0 2

ax

⎥⎦⎤

⎢⎣⎡

F = – dx

dU 2 = – 20

a2U (2x)

at x = 0 F = 0

if x = + a/2 F = – ve

CAREER POINT



towards origin

Option (Q, S) correct.

(C) U3 = 2

U02

ax

⎟⎠⎞

⎜⎝⎛

2

ax–

e⎟⎠⎞

⎜⎝⎛

F = – dx

dU3 = ⎟⎟⎠

⎞⎜⎜⎝

⎛2

0

a2U

– dxd

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡2

2

ax–

2ex

F = – 2

0

a2U

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡+ 2

2

2

2

ax–

2ax–

2 ex2a

x2–ex

F = – 2

0

a2U 2

2

ax–

e ⎥⎦

⎤⎢⎣

⎡+ x2

ax2–2

3

F = – 20

a2U 2

2

ax–

e (+ 2x) ⎥⎦

⎤⎢⎣

⎡2

2

ax–1

at x = 0, a, – a F = 0

if x = + a/2 F = – ve attractive

⇒ P, Q, R, S Option correct

(D) U4 = 2

U0⎥⎦

⎤⎢⎣

⎡3

3

ax

31–

ax

F = – dx

dU 4 = – 2

U0⎥⎦

⎤⎢⎣

⎡)x3(

a31–

a1 2

3

F = – 2

U0⎥⎦

⎤⎢⎣

⎡3

2

ax–

a1

F = – 30

a2U (a2 – x2)

at x = ± a ⇒ F = 0

at ⎪⎭

⎪⎬⎫

==

=+=

ve–F2a–x

ve–F2ax

P, R, T correct

4

U0 A

– a

U = 0 Fr

O a

3U0

B

3U– 0

4U0

Fr

Fr

Oscillation will be above x = – a. For Total energy

4U0 . The extreme points will be A & B where

potential energy will be maximum. (In diagram P.E. are shown).

CAREER POINT



PART II - CHEMISTRY


• This section contains EIGHT questions.

• The answer to each questions 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.21 The total number of lone pairs of electrons in N2O3 is -

Ans. [8]

Sol. N2O3

N–N

O O

O

Number of lone pairs = 8

Q.22 For the octahedral complexes of Fe3+ in SCN– (thiocyanato-S) and in CN– ligand environments, the

difference between the spin-only magnetic moments in Bohr Magneton (when approximated to the

nearest integer) is -

[Atomic number of Fe = 26]

Ans. [4]

Sol. Fe+3 + SCN– → [Fe(SCN)6]3–

Fe+3 ⇒ [Ar] 3d5

eg

t2g

Here Δ0 < Pd-orbitals

Number of unpaired electrons = 5

μ = 5.9

Fe+3 + CN– → [Fe(CN)6]3–

Fe+3 = [Ar]3d5

CAREER POINT



eg

t2g

Δ0 > P

d-orbitals

Number of unpaired electrons = 1

μ = 1.732

Difference = 5.91 – 1.73 = 4.18

~– 4

Q.23 Among the triatomic molecules/ions, BeCl2, −3N , N2O, NO2

+, O3, SCl2, ICI2–, I3

–- and XeF2, the total

number of linear molecule(s)/ion(s) where the hybridization of the central atom does not have

contribution from the d-orbital(s) is.

[Atomic number : S = 16, Cl = 17, I = 53 and Xe = 54]

Ans. [4]

Sol. Cl–Be–Cl N≡N:→O

sp, linear sp, linear

–N =

+N =

–N O =

+N = O

sp, linear sp, linear

OO O

sp2, V-shape

SCl Cl

V-shape

IΘ Cl

Cl

sp3d, Linear

like I3

–, XeF2 → sp3d, Linear

Ans. = 4

Q.24 Not considering the electronic spin, the degeneracy of the second excited state (n = 3) of H atom is 9,

while the degeneracy of the second excited state of H– is -

Ans. [3]

Sol. 2nd excited state H– = 2p state

∴ degeneracy of second excited state of H– = 3

(This is not is jee advance syllabus)

CAREER POINT



Q.25 All the energy released from the reaction X → Y, ΔrGº = –193 kJ mol– is used for oxidizing M+ as

M+ → M3+ + 2e–, Eº = –0.25 V. Under standard conditions, the number of moles of M+ oxidized when

one mole of X is converted to Y is [F = 96500 C mol–1] Ans. [4]

Sol. M+ → M3+ + 2e–

nM+ → nM3+ + 2ne– ΔGº = –2n FEº

And :

|ΔrGº| = |ΔGº|

∴ 193 × 1000 = 2 × n × 96500 × 0.25

4n =

Q.26 If the freezing point of a 0.01 molal aqueous solution of a cobalt(III) chloride-ammonia complex

(which behaves as a strong electrolyte) is – 0.0558ºC, the number of chloride(s) in the coordination sphere of the complex is

[Kf of water = 1.86 K kg mol–1] Ans. [1]

Sol. ΔTf = i.Kf.m

0.0558 = n × 1.86 × 0.01 n = 3

∴ [Co(NH3)5Cl]Cl2

i.e. number of chloride ion in complex is 1. Q.27 The total number of stereoisomers that can exist for M is

H3C CH3

H3COM

Ans. [2] Sol. Due to it's rigid structure camphor form only two optical isomers (stereoisomer)

H3C CH3

H3C O

*

*

CAREER POINT



Q.28 The number of resonance structures for N is

OHNaOH N

Ans. [9]

Sol.

OH+ NaOH –H2O

O NaΘ ⊕

N

OΘ OΘ

(2)(1)

O

Θ

O

Θ

OΘ

(5) (4) (3)

OΘ

(6)

OΘ

(7)

O

Θ

(8)

R.S. of N is :

OΘ

(9)


• This section contains TEN 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


–2 In all other cases.

CAREER POINT



Q.29 Fe3+ is reduced to Fe2+ by using - (A) H2O2 in presence of NaOH (B) Na2O2 in water (C) H2O2 in presence of H2SO4

(D) Na2O2 in presence of H2SO4

Ans. [A,B] Sol. Fe+3 + H2O2 + NaOH → Fe+2 + O2

Fe+3 + ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+↓

+

22

222

OHNaOH

OHONa → Fe+2 + O2

Q.30 The %yield of ammonia as a function of time in the reaction N2(g) + 3H2(g) 2NH3(g), ΔH < 0 at (P, T1) is given below.

T

time

% y

ield

If this reaction is conducted at (P, T2), with T2 > T1, the %yield of ammonia as a function of time is

represented by -

(A)

T1

time

% y

ield

T2

(B)

T2

time

% y

ield

T1

(C)

T2

time

% y

ield

T1

(D)

T1

time

% y

ield

T2

Ans. [B] Sol. Initially larger is the temperature greater is the yield but due to exothermic reaction as the time lapse

yield become less.

CAREER POINT



Q.31 If the unit cell of a mineral has cubic close packed (ccp) array of oxygen atoms with m fraction of

octahedral holes occupied by aluminium ions and n fraction tetrahedral holes occupied by magnesium

ions, m and n, respectively, are -

(A) 21 ,

81 (B) 1,

41

(C) 21 ,

21 (D)

41 ¸

81

Ans. [A] Sol. T.V. O.V. c.c.p.

Mg2+ Al3+ O2–

8n 4m 4

∴ 8n × (+2) + 4m × (+3) + 4 × (–2) = 0

On verifying option, we get

n = 81 , m =

21

Q.32 Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is / are -

(A)

H3C

BrH

CH3 (B)

H2C

H BrCH3

(C)

H2C

H Br

CH3

CH3

(D) H2C

Br HCH3

Ans. [B,D]

Sol. (A)

CH3

H Br

CH3 H2

H Br

CH3*

Optically active

(B) H2C

H BrCH3

H2 H Br

H3C CH3

Optically inactive

(C) H2C

Br

CH3 CH3

H2 H3CBr

CH3CH3

H

*

H

Optically active

(D) H2C

Br HCH3

H2 H3C

Br HCH3

Optically inactive

CAREER POINT



Q.33 The major product of the following reaction is - O

OCH3

Heat,H)ii(

OH,KOH)i( 2+

⎯⎯⎯⎯ →⎯

(A)

CH3 O

(B)

CH3

O

(C)

O CH3

(D)

CH3

O

Ans. [A] Sol.

O

O CH3

KOH O

OCH3

–Intramolecular

Aldol attack

O–

H2O

CH3HO

Aldol–H2O

ΔO

CH3

OCH3

O

–H2O

H

Q.34 In the following reaction, the major product is -

CH3

H2C CH2 ⎯⎯⎯⎯⎯⎯⎯ →⎯ HBrEquivalent1

(A)

CH3

H2C CH3

Br (B) H3C Br

CH3

(C)

H2C

CH3

Br (D)

H3C

CH3

Br

Ans. [D] Sol.

CH3 CH2

Conjugated Diene

CH3

⊕H3C

Resonance ⊕ Br –

Br1, 4 Addition

Product Major product

H⊕

CH3

CH3

CAREER POINT



Q.35 The structure of D-(+)-glucose is CHO

H OHHO H

H OHH OH

CH2OH The structure of L-(–)-glucose is :

(A)

CHO HO H

H OH HO H HO H

CH2OH

(B)

CHOH OH

HO H H OH

HO H

CH2OH

(C)

CHOHO HHO H

H OHHO H

CH2OH

(D)

CHOHO HHO HHO H

H OH

CH2OH

Ans. [A]

Sol.

CHO H OH

HO H H OH H OH

CH2OH

D(+) glucose

CHOHO H

H OHHO HHO H

L(–) glucose

CH2OH

CAREER POINT



Q.36 The major product of the reaction is

H3C

CH3 NH2

CO2H NaNO2, aqueous HCl

0°C

(A) H3C

CH3 OH

NH2 (B)

H3C

CH3OH

CO2H

(C) H3C

CH3 OH

CO2H (D) H3C

CH3OH

NH2

Ans. [C]

Sol.

NH2

C–OH NaNO2/ aq. HCl

0°C

O

N≡N

C–OH

O

⊕

H2O (NuΘ)

–N2 (LG) OH

C–OH

O

Q.37 The correct statement(s) about Cr2+ and Mn3+ is (are) -

[Atomic numbers of Cr = 24 and Mn = 25]

(A) Cr2+ is a reducing agent

(B) Mn3+ is an oxidizing agent

(C) Both Cr2+ and Mn3+ exhibit d4 electronic configuration

(D) When Cr2+ is used as a reducing agent, the chromium ion attains d5 electronic configuration

Ans. [A,B,C]

Sol. Cr+2 ⎯→ Cr+3

d4 d3 (stable)

So, act as a reducing agent

Mn+3 ⎯→ Mn+2

d4 d5 (stable)

So, act as a oxidising agent

CAREER POINT



Q.38 Copper is purified by electrolytic refining of blister copper. The correct statement(s) about this process

is (are) :

(A) Impure Cu strip is used as cathode

(B) Acidified aqueous CuSO4 is used as electrolyte

(C) Pure Cu deposits at cathode

(D) Impurities settle as anode mud

Ans. [B,C,D]

Sol. (i) Due to acidified aqueous CuSO4, conduction is increased

(ii) Cathode is made by pure Cu

(iii) Anode is made by Impure Cu

(iv) Ag and Au are settle down as a anode mud


• This section contains TWO questions

• Each question contains two columns, Column I and Column II

• Column I has four entries (A), (B), (C) and (D)

• Column II has five entries (P), (Q), (R), (S) and (T)

• Match the entries in Column I with the entries in Column II

• One or more entries in Column I may match with one or more entries in Column II

• The ORS contains a 4 × 5 matrix whose layout will be similar to the one shown below:

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)


Column I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS. Similarly,

for entries (B), (C) and (D).

• Marking scheme:

For each entry in Column I.

+2 If only the bubble(s) corresponding to all the correct match(es) is(are) darkened



CAREER POINT



Q.39 Match the anionic species given in Column - I that are present in the ore(s) given in Column - II.

Column - I Column - II

(A) Carbonate (P) Siderite

(B) Sulphide (Q) Malachite

(C) Hydroxide (R) Bauxite

(D) Oxide (S) Calamine

(T) Argentite

Ans. [A → P,Q,S ; B → T ; C → Q,R ; D → R]

Sol. Siderite → FeCO3

Malachite → CuCO3 . Cu(OH)2

Bauxite → AlOx (OH)3 – 2x (0 < x < 1)

Calamine → ZnCO3

Argentite → Ag2S

Q.40 Match the thermodynamic processes given under Column - I with the expressions given under Column - II

Column - I Column - II

(A) Freezing of water at 273 K and 1 atm (P) q = 0

(B) Expansion of 1 mol of an ideal gas into a (Q) w = 0

vacuum under isolated conditions

(C) Mixing of equal volumes of two ideal gases (R) ΔSsys < 0

at constant temperature and pressure in an

isolated container

(D) Reversible heating of H2(g) at 1 atm from (S) ΔU = 0

300 K to 600 K, followed by reversible

cooling to 300 K at 1 atm

(T) ΔG = 0

Ans. [A → R,T ; B → P,Q,S ; C → P,Q,S ; D → P,Q,S,T]

Sol. (A) Water Ice 1atm

273K

* From liquid to solid entropy decreases

∴ ΔS < 0

* Liquid is present under equilibrium with solid

∴ ΔG = 0

Therefore A → R,T

CAREER POINT



(B) Expansion of 1 mole gas into vacuum under isolated condition * Isolated system q = 0 * During expansion against vacuum no work is done W = U * No change in internal energy ΔE = 0

Therefore B → P,Q,S

(C) Volume = V

gas Volume = V

gas

Constant, T Constant P

∴ q = 0, W = 0, ΔU = 0

∴ C → P,Q,S

(D) Reversible cyclic process

A B

(1 atm, 300K) (1 atm, 600K) WAB = – nR (600 – 300) = – 300 nR WBA = – nR (300 – 600) = 300 nR ∴ ∑W = 0

* For cyclic process change in the value of state functions are zero i.e. ΔU = 0, ΔG = 0

* From 1st law : ∑ W = 0, ΔU = 0, ∴ ∑ q = 0

∴ D → P, Q, S, T

CAREER POINT



PART III : MATHEMATICS

SECTION 1 (Maximum Marks : 32)

• 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

• Marking scheme : + 4 If the bubble corresponding to the answer is darkened 0 In all other cases

Q.41 The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two

heads is at least 0.96, is

Ans. [8]

Sol. 1 – P(no head) – P(one head) ≥ 0.96

1 – nC0

n

21

⎟⎠⎞

⎜⎝⎛ – nC1

n

21

⎟⎠⎞

⎜⎝⎛ ≥ 0.96

n21n +

≤ 0.04

n21n +

≤251

Minimum value of n is 8.

Q.42 Let n be the number of ways in which 5 boys and 5 girls can stand in a queue is such a way that all the girls

stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue

in such a way that exactly four girls stand consecutively in the queue. Then the value of nm is

Ans. [5]

Sol. n = 5! 6!

selecting 2 gaps in between 5 boysarranging 5 boysselecting 4 girls

4! 2! C 5! C m 26

45 ××××=

m = 5! × 5! × 6C2 × 2!

∴ nm =

6!2C2

6 × = 5

CAREER POINT



Q.43 If the normals of the parabola y2 = 4x drawn at the end points of its latus rectum are tangents to the circle

(x – 3)2 + (y + 2)2 = r2, then the value of r2 is

Ans. [2]

Sol. One end point of the latus rectum is (1, 2)

Slope of the normal = –)2,1(2

y⎟⎠⎞

⎜⎝⎛ = –1

So its equation will be

y – 2 = – (x – 1)

x + y = 3

it is tangent to the circle

so2

3–2–3 = r

∴ r2 = 2

Q.44 Let f : R → R be a function defined by ⎩⎨⎧

>≤

=2x,02x],x[

)x(f ,

where [x] is the greatest integer less than or equal to x. If ∫−

++=

2

1

2,dx

)1x(f2)x(xfI then the

value of (4I – 1) is

Ans. [0]

Sol. I = dx)1x(f2

]x[x2

1–

2

∫ +++ dx0

2

2∫

= dx02)0(x0

1–∫ +

+ dx12)0(x1

0∫ +

+ ∫ +

2

102)1(x dx

= dxx21 2

1∫ =

2

1

2

4x

⎥⎦

⎤⎢⎣

⎡=

41

∴ 4I – 1 = 0

CAREER POINT



Q.45 A cylindrical container is to be made from certain solide material with the following constraints : It has a

fixed inner volume of V mm3, has a 2 mm thick solid wall and is open at the top. The bottom of the container

is a solid circular disc of thickness 2 mm and is of radius equal to the outer radius of the container.

If the volume of the material used to make the container is minimum when the inner radius of the container is 10

mm, then the value of π250

V is

Ans. [4]

Sol.

2mm

V = πr2h

Let S denotes volume of the material used to make the container

S = πr2 × 2 + 2πrh × 2

S = 2πr2 + 4πr × 2rV

π

S = 2πr2 +rV4

Now drdS = 4πr – 2r

V4

r3 = πV

r = 3/1V

⎟⎠⎞

⎜⎝⎛

π

Given r = 10 mm

V = 1000π

∴ π250

V = 4

CAREER POINT



Q.46 Let ∫π

+

=6

x

x

2

2

dttcos2)x(F for all x ∈ R and f : ),0[21,0 ∞→⎥⎦

⎤⎢⎣⎡ be a continuous function. For a ∈ ⎥⎦

⎤⎢⎣⎡

21,0 ,

if F'(a) + 2 is the area of the region bounded by x = 0, y = 0, y = f(x) and x = a, then f(0) is

Ans. [3]

Sol. F′(x) = 2 cos2 ⎟⎠⎞

⎜⎝⎛ π

+6

x2 .2x – 2 cos2x

Now ∫a

0

dx)x(f = 4a cos2 ⎟⎠⎞

⎜⎝⎛ π

+6

a2 – 2 cos2a + 2

Differentiate with respect to 'a'

⇒ f(a) = 4 cos2 ⎟⎠⎞

⎜⎝⎛ π

+6

a2 + 4a ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ π

+6

acos2 2⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ π

+6

asin– 2 .2a + 4 cos a sin a

∴ f(0) = 4 cos2

6π + 0 + 0

= 4 × 43 = 3

Q.47 The number of distinct solutions of the equation

45 cos2 2x + cos4 x + sin4 x + cos6 x + sin6 x = 2

in the interval [0, 2π] is

Ans. [8]

Sol. 45 cos22x + 1 – 2 sin2x cos2x + 1 – 3 sin2x cos2x = 2

⇒ 5 cos22x – 5 sin22x = 0

⇒ 5 cos 4x = 0

⇒ 4x = 2nπ ± 2π , n ∈ I

⇒ x = 2

nπ ±8π , n ∈ I

In the interval [0, 2π]

Possible solutions are

8π ,

83π ,

85π ,

87π ,

89π ,

811π ,

813π ,

815π

So total number of solutions in [0, 2π] is 8.

CAREER POINT



Q.48 Let the curve C be the mirror image of the parabola y2 = 4x with respect of the line x + y + 4 = 0.

If A and B are the points of intersection of C with the line y = –5, then the distance between A and B is

Ans. [4]

Sol. Let any point on the parabola y2 = 4x is (t2, 2t)

Let its mirror image with respect to the line x + y + 4 = 0 is (h, k) then

1

t–h 2=

1t2–k =

2)4t2t(2– 2 ++

∴ h = –2t – 4, k = – t2 – 4

So k + 4 = –2

24h

⎟⎠⎞

⎜⎝⎛ +

⇒ (h + 4)2 = – 4(k + 4)

So locus of C is

(x + 4)2 = – 4(y + 4)

It intersects y = –5

So (x + 4)2 = 4

⇒ x + 4 = ± 2

⇒ x = –2, –6

∴ |x1 – x2| = 4

CAREER POINT



SECTION 2 (Maximum Marks : 40)

• This Section contains TEN 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



Q.49 Let g : R → R be a differentiable function with g(0) = 0, g'(0) = 0 and g'(1) ≠ 0. Let

⎜⎜⎜

⎝

⎛

=

≠=

0x,0

0x),x(g|x|

x)x(f

and h(x) = e|x| for all x ∈ R. Let (f o h) (x) denote f(h(x)) and (h o f)(x) denote h(f(x)). Then which of the following is (are) true ?

(A) f is differentiable at x = 0 (B) h is differentiable at x = 0 (C) f o h is differentiable at x = 0 (D) h o f is differentiable at x = 0 Ans. [A,D]

Sol. g(0) = 0, g′(0) = 0, g′(1) ≠ 0

f(x) = ⎪⎩

⎪⎨

⎧

<=>

0x)x(g–0x00x)x(g

f ′(x) = ⎪⎩

⎪⎨

⎧

<′=>′

0x)x(g–0x00x)x(g

f(x) is diff. at x = 0, as g′(0) = 0 So option (A) is true. h(x) = e|x|

not diff. at x = 0 x

y

O

So option (B) is not true f o h(x) = f(h(x)) as h(x) > 0 So f(h(x)) = g(h(x))

CAREER POINT



dx

))x(h(df = g′(h(x)) h′(x) since g′(1) ≠ 0

at x = 0, f(h(x)) is not differentiable

So option (C) is not true

h o f (x) = h(f(x))

= e|f(x)|

Differentiable at x = 0

So option (D) is true.

Q.50 Let ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ππ

= xsin2

sin6

sin)x(f for all x ∈ R and xsin2

)x(g π= for all x ∈ R. Let (f o g)(x) denote f(g(x)) and

(g o f)(x) denote g(f(x)). Then which of the following is (are) true ?

(A) Range of f is ⎥⎦⎤

⎢⎣⎡−

21,

21 (B) Range of f o g is ⎥⎦

⎤⎢⎣⎡−

21,

21

(C) 6)x(g

)x(flim0x

π=

→ (D) There is an x ∈ R such that (g o f)(x) = 1

Ans. [A,B,C]

Sol. ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ππ

= xsin2

sin6

sin)x(f

– 1 ≤ sin x ≤ 1 as x ∈ R

–2π

≤ 2π sinx ≤

2π

–1 ≤ sin ⎟⎠⎞

⎜⎝⎛ π xsin

2≤ 1

–6π

≤ 6π sin ⎟

⎠⎞

⎜⎝⎛ π xsin

2≤

6π

–21

≤ sin ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ππ xsin

2sin

6 ≤

21

Range ∈ ⎥⎦⎤

⎢⎣⎡

21,

21–

So option (A) is correct.

f o g(x)

f g

g(x) fog(x)x

CAREER POINT



Range of g(x) = ⎥⎦⎤

⎢⎣⎡ ππ

2,

2– ∩ Domain of f(x) = R

Common = ⎥⎦⎤

⎢⎣⎡ ππ

2,

2–

Range of fog(x) = Range of f(x) when input ⎥⎦⎤

⎢⎣⎡ ππ

2,

2–

= ⎥⎦⎤

⎢⎣⎡

21,

21–

So option (B) is correct

0x

lim→ xsin

2

xsin2

sin6

sin

π

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ππ

=0x

lim→

⎟⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛ ππ

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ππ

xsin2

sin6

xsin2

sin6

sin

⎟⎟⎟⎟

⎠

⎞

⎜⎜⎜⎜

⎝

⎛

⎟⎠⎞

⎜⎝⎛ π

⎟⎠⎞

⎜⎝⎛ ππ

xsin2

xsin2

sin6

= 0x

lim→ 6

π .(1) = 6π

So option (B) is correct

g o f(x)

gf

f(x) gof(x)x

Range of f(x) = ⎥⎦⎤

⎢⎣⎡

21,

21–

g o f(x) = 2π sin f(x)

= 2π sin

444 3444 21

21to

21–

xsin2

sin6

sin ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎠⎞

⎜⎝⎛ ππ = 1

sin ⎟⎠⎞

⎜⎝⎛

21to

21– =

π2

≈ 0.6379 (not possible)

21

≈ 28.5º

So option (D) is not correct.

CAREER POINT



Q.51 Let ΔPQR be a triangle Let ,QRa =r

RPb =r

and .PQc =r

If ,12|a| =r

34|b| =r

and ,24cb =rr

. then which of

the following is (are) true ?

(A) 12|a|2|c| 2

=−r

r

(B) 30|a|2|c| 2

=+r

r

(C) 348|acba| =×+×rrrr

(D) 72ba −=⋅rr

Ans. [A, C, D]

Sol.

br

cr

R Q ar

P

24cb =⋅rr

0cba =++rrr

cbarrr

−−=

cb2|c|48144 2 rrr⋅++=

)24(2|c|96 2 +=r

48|c| 2 =r

(A) 1212248|a|

2|c| 2

=−=−r

r

So option (A) is correct

(B) 2412248|a|

2|c| 2

=+=+r

r

So option (B) is not correct

(C) |caba||acba|rrrrrrrr

×−×=×+×

|)cb(a|rrr

−×=

|)cb()cb(|rrrr

−×−−=

|0bccb0| +×−×+=rrrr

|cb|2rr

×=

⎟⎠⎞⎜

⎝⎛ ⋅−= 222 )cb(|c||b|2

rr

= 2)24(–)48(482

= 2 × 24 × 3 = 348

So option (C) is correct

CAREER POINT



(D) 0cba =++rrr

cbarrr

−=+

48ba248144 =⋅++rr

72ba −=⋅rr

So option (D) is correct

Q.52 Let X and Y be two arbitrary, 3 × 3, non-zero, skew-symmetric matrices and Z be an arbitrary 3 × 3, non-

zero, symmetric matrix. Then which of the following matrices is (are) skew symmetric ?

(A) Y3Z4 – Z4Y3 (B) X44 + Y44

(C) X4Z3 – Z3X4 (D) X23 + Y23

Ans. [C,D]

Sol. XT = –X YT = –Y ZT = Z

(A) (Y3 Z4 – Z4 Y3)T = (Y3 Z4)T – (Z4 Y3)T

= (Z4)T (Y3)T – (Y3)T (Z4)T

= (ZT)4 (YT)3 – (YT)3 (ZT)4

= Z4 (–Y)3 – (–Y)3 (Z)4

= –Z4 Y3 + Y3 Z4

= Y3Z4 – Z4Y3

Hence it is symmetric matrix.

(B) (X44 + Y44)T = (XT)44 + (YT)44

= X44 + Y44

Hence it is symmetric matrix.

(C) (X4 Z3 – Z3 X4)T = (X4 Z3)T – (Z3 X4)T

= (ZT)3 (XT)4 – (XT)4 (ZT)3

= Z3 X4 – X4 Z3

= – (X4 Z3 – Z3 X4)

Hence it is skew symmetric matrix.

(D) (X23 + Y23)T = (XT)23 + (YT)23

= – (X23 + Y23)

Hence it is skew symmetric matrix.

CAREER POINT



Q.53 Which of the following values of α satisfy the equation.

?648)33()23()3()32()22()2()31()21()1(

222

222

222

α−=α+α+α+α+α+α+α+α+α+

(A) – 4 (B) 9

(C) – 9 (D) 4

Ans. [B,C]

Sol. 169144121

222 9432111

αααααα = – 648 α

⇒ (–4) (2α3) = – 648 α

⇒ α2 = 81

⇒ α = ±9

Q.54 In R3, consider the planes P1 : y = 0 and P2 : x + z = 1. Let P3 be a plane, different from P1 and P2, which passes

through the intersection of P1 and P2. If the distance of the point (0, 1, 0) from P3 is 1 and the distance of a point

(α, β, γ) from P3 is 2, then which of the following relations is (are) true ?

(A) 2α + β + 2γ + 2 = 0 (B) 2α – β + 2γ + 4 = 0

(C) 2α + β – 2γ – 10 = 0 (D) 2α – β + 2γ – 8 = 0

Ans. [B,D]

Sol. Equation of P3 will be

(x + z – 1) + λy = 0

x + λy + z – 1 = 0

Its disantace from (0, 1, 0) will be

11

1–002 +λ+

+λ+ = 1

⇒ (λ – 1)2 = 1 + λ2 + 1

⇒ λ2 – 2λ + 1 = λ2 + 2

⇒ λ = –1/2

∴ Equation of P3 is 2x – y + 2z – 2 = 0

Its distance from (α, β, γ) is

3

|2–2–2| γ+βα = 2

CAREER POINT



2α – β + 2γ – 2 = ± 6 ⇒ 2α – β + 2γ – 8 = 0 and 2α – β + 2γ + 4 = 0

Q.55 In R3, Let be a straight line passing through the origin. Suppose that all the points on L are at a constant distance

from the two planes P1 : x + 2y – z + 1 = 0 and P2 : 2x – y + z – 1 = 0. Let M be the locus of the feet of the perpendiculars drawn from the points on L to the plane P1. Which of the following points lie(s) on M ?

(A) ⎟⎠⎞

⎜⎝⎛ −−

32,

65,0 (B) ⎟

⎠⎞

⎜⎝⎛ −−

61,

31,

61

(C) ⎟⎠⎞

⎜⎝⎛−

61,0,

65 (D) ⎟

⎠⎞

⎜⎝⎛−

32,0,

31

Ans. [A,B] Sol. P1 : x + 2y – z + 1 = 0 & P2 : 2x – y + z – 1 = 0

Direction Ratios of common line (1, –3, –5) ⇒ k5–j3–i

L : 1x =

3–y =

5–z = t

Let M(α, β, γ) is feet of perpendicular from (t, –3t, – 5t) on P1

1

t–α = 2

t3+β = 1–

t5+γ = – ⎟⎠⎞

⎜⎝⎛ ++

61t5t6–t

61–t=α

31–t3–=β γ = – 5t +

61

Only option (A) & (B) satisfies.

Q.56 Let P and Q be distinct points on the parabola y2 = 2x such that a circle with PQ as diameter passes through the

vertex O of the parabola. If P lies in the first quadrant and the area of the triangle ΔOPQ is ,23 then which of the

following is (are) the coordinates of P ?

(A) ( )22,4 (B) ( )23,9

(C) ⎟⎟⎠

⎞⎜⎜⎝

⎛2

1,41 (D) ( )2,1

Ans. [A,D]

Sol. Let coordinates of P and Q are P(2t12, 2t1), Q(2t2

2, 2t2)

As the circle with PQ as diameter passes through the vertex O.

So ∠POQ = 90º

mOP × mOQ = –1

21

1

t2t2 × 2

2

2

t2t2 = –1 ⇒ t1t2 = –1

Now Area of ΔOPQ will be

CAREER POINT



21

222

121

t2t21t2t21001

= 23

⇒ 4|t1 t2 (t1 – t2)| = 6 2

⇒ |t1 – t2| =2

3

⇒ t1 + 1t1 =

23

⇒ 2 21t – 3t1 + 2 = 0

⇒ ( 2 t1 – 1) (t1 – 2 ) = 0

⇒ t1 = 2

1 , 2

∴ Coordinates of P can be (4, 2 2 ), (1, 2 )

Q.57 Let y(x) be a solution of the differential equation (1 + ex)y' + yex = 1. If y(0) = 2, then which of the following

statements is (are) true ?

(A) y(–4) = 0

(B) y(–2) = 0

(C) y(x) has a critical point in the interval (–1, 0)

(D) y(x) has no critical point in the interval (–1, 0)

Ans. [A,C]

Sol. (1 + ex)y′ + yex =1

⇒ d((1 + ex)y) = 1

⇒ y(1 + ex) = x + C

Q y(0) = 2

∴ 2 × 2 = C ⇒ C = 4

∴ y(1 + ex) = x + 4

∴ y(–4) = 0, y(–2) = 2–e12

+≠ 0

For critical point y′ = 0

⇒ yex = 1

CAREER POINT



Now let g(x) = yex – 1 = x

x

e1)4x(e

++ – 1

g(–1) = 1–

1–

e1e3

+–1 =

1e3+

–1 < 0

g(0) = 2 – 1 > 0

So there exists one value of x in (–1, 0) for which g(x) = 0 ⇒ y′ = 0

⇒ there exist a critical point of y(x) in (–1, 0) Q.58 Consider the family of all circles whose centers lie on the straight line y = x. If this family of circles is

represented by the differential equation Py" + Qy' + 1 = 0, where P, Q are functions of x, y and y'

(here ,dxdy'y = 2

2

dxyd"y = ), then which of the following statements is (are) true ?

(A) P = y + x

(B) P = y – x

(C) P + Q = 1 – x + y + y' + (y')2

(D) P – Q = x + y – y' – (y')2

Ans. [B, C] Sol. Let the equation of circle is (x – α)2 + (y – α)2 = r2 ⇒ x2 + y2 – 2 α x – 2 α y + 2 α 2 – r2 = 0 differentiate w.r.t. x ⇒ 2x + 2yy' – 2 α – 2 α y' = 0

⇒ 'y1'yyx

++

=α ...(i)

differentiate again w.r.t. x 2 + 2(y')2 + 2yy" – 2 α y" = 0

⇒ "y

"yy)'y(1 2 ++=α ...(ii)

from (i) & (ii) xy" + yy'y" = 1 + (y')2 + yy" + y' + (y')3 + yy'y" ⇒ (y – x) y" + y' [y' + 1 + (y')2] + 1 = 0 P = y – x Q = y' + 1 + (y')2

CAREER POINT



SECTION 3 (Maximum Marks: 16)

• This section contains TWO questions • Each question contains two columns, Column I and Column II • Column I has four entries (A), (B), (C) and (D) • Column II has five entries (P), (Q), (R), (S) and (T) • Match the entries in Column I with the entries in Column II • One or more entries in Column I may match with one or more entries in Column II • The ORS contains a 4 × 5 matrix whose layout will be similar to the one shown below:

(A) (P) (Q) (R) (S) (T)

(B) (P) (Q) (R) (S) (T)

(C) (P) (Q) (R) (S) (T)


Column I matches with entries (Q), (R) and (T), then darken these three bubbles in the ORS. Similarly, for entries (B), (C) and (D).

• Marking scheme: For each entry in Column I. +2 If only the bubble(s) corresponding to all the correct match(es) is(are) darkened 0 If none of the bubbles is darkened

–1 In all other cases Q.59 Column I Column II

(A) In R2, if the magnitude of the projection vector of (P) 1

the vector ji β+α on ji3 + is 3 and if α = 2 + β3 ,

then possible value(s) of |α| is (are)

(B) Let a and b be real numbers such that the function (Q) 2

f(x) = ⎩⎨⎧

≥+<

1x,abx1x,2–ax3–

2

2 is differentiable for

all x ∈ R. Then possible value(s) of a is (are)

(C) Let ω ≠ 1 be a complex cube root of unity. If (R) 3

(3 – 3ω + 2ω2)4n + 3 + (2 + 3ω – 3ω2)4n+3

+ (–3 + 2ω + 3ω2)4n+3 = 0 then possible value(s)

of n is (are)

(D) Let the harmonic mean of two positive real numbers a and (S) 4

b be 4. If q is a positive real number such that a, 5, q, b is

an arithmetic progression, then the value(s) of |q – a| is (are) (T) 5

Ans. [(A) → (P,Q); (B) → (P,Q); (C) → (P,Q,S,T) D → (Q,T)]

CAREER POINT



Sol. (A) Projection of ji β+α on ji3 + will be

2

)ji3).(ji( +β+α = 3

⇒ 2

3 β+α = 3

as α = 2 + 3 β

So 3

2–3 α+α = 2 3

⇒ |3α + α – 2| = 6

⇒ 4α – 2 = ±6

α = 2, – 1

∴ |α| = 1, 2

So (A) → (P, Q)

(B) f(x) = ⎩⎨⎧

≥+<

1x,abx1x,2–ax3–

2

2

as f(x) is differentiable for x ∈ R. So it will also be continuous for x ∈ R.

So –3a – 2 = b + a2

a2 + 3a + 2 + b = 0 … (i)

f ′(x) = ⎩⎨⎧

≥<

1x,b1x,ax6–

So –6a = b … (ii)

From (i) & (ii)

a2 + 3a – 6a + 2 = 0

⇒ a = 1, 2

(B) → (P,Q)

(C) (3 – 3ω + 2ω2)4n+3 + (2 + 3ω – 3ω2)4n+3 + (–3 + 2ω + 3ω2)4n+3 = 0

⇒ (1 – 5ω)4n+3 + (ω – 5ω2)4n+3 + (– 5 + ω2)4n+3 = 0

⇒ (1 – 5ω)4n+3 + ω4n + 3 (1 – 5ω)4n+3 + (ω2)4n+3 (1 – 5ω)4n+3 = 0

⇒ (1 – 5ω)4n+3 + (1 + ω4n+3 + (ω2)4n+3) = 0

⇒ 1 + ω4n+3 + (ω2)4n+3 = 0

⇒ 4n + 3 should not be integral multiple of 3

So n can be 1, 2, 4, 5

So (C) → (P,Q,S,T)

CAREER POINT



(D) ba

ab2+

= 4

also a, 5, q, b are in A.P.

then a = 5 – d, b = 5 + 2d,

So d25d–5

)d25)(d–5(2+++ = 4

⇒ 40 + 4d = 2(25 + 5d – 2d2)

⇒ 2d2 – 3d – 5 = 0

⇒ d = – 1, 5/2

|q – a| = |2d| = 2, 5

(D) → (Q,T)

Q.60 Column I Column II

(A) In a triangle ΔXYZ, let a, b and c be the lengths (P) 1

of the sides opposite to the angles X, Y and Z,

respectively. If 2(a2 – b2) = c2 and λ =Zsin

)Y–Xsin( ,

then possible values of n for which cos(nπλ) = 0

is (are)

(B) In a triangle ΔXYZ, let a, b and c be the lengths (Q) 2

of the sides opposite to the angles X, Y and Z,

repsectively. If 1 + cos 2X – 2 cos 2Y = 2 sin X sin Y,

then possible value(s) of ba is (are)

(C) In R2, let j3i,ji3 ++ and j)–1(i β+β be the position (R) 3

vectors of X, Y and Z with respect to the origin O,

respectively. If the distance of Z from the bisector of

the acute angle of ⎯→OX with

⎯→OY is

23 , then possible

value(s) of |β| is (are)

CAREER POINT



(D) Suppose that F(α) denotes the area of the region (S) 5

bounded by x = 0, x = 2, y2 = 4x and

y = |αx – 1| + |αx – 2| + αx, where α ∈ {0, 1}. Then the

value(s) of F(α) + 238 , when α = 0 and α = 1, is (are)

(T) 6

Ans. [A → (P, R, S); B →(P); C →(P, Q); D →(S, T)]

Sol. (A)

b c

ZY a

X

2(a2 – b2) = c2

Using Sine Rule

⇒ 2(sin2 X – sin2 Y) = sin2 Z

⇒ 2sin (X – Y) sin (X + Y) = sin2 Z

⇒ 2sin (X – Y) sin Z – sin2 Z = 0

⇒ sin Z (2 sin (X – Y) – sin Z) = 0

⇒ sin Z = 0 or sin(X – Y)21

= sin z

∴ 21

Zsin)YXsin(

=−

=λ

Now 02

ncos =⎟⎠⎞

⎜⎝⎛ π

For n = 1, 3, 5

∴ (A) → (P, R, S)

(B) 1 + cos 2X – 2 cos 2Y = 2 sin X sinY

⇒ 2 cos2 X – 2(2 cos2 Y – 1) = 2 sin x sinY

⇒ 2 cos2 X + 2 – 4 cos2Y = 2 sinX sinY

⇒ 2 – 2 sin2 X + 2 – 4 + 4 sin2 Y = 2 sinX sinY

⇒ 2 sin2 X + 2 sin X sin Y – 4 sin2 Y = 0

Using Sine Rule

CAREER POINT



a2 + ab – 2b2 = 0

⇒ 02ba

ba 2

=−+⎟⎠⎞

⎜⎝⎛ ⇒ 2,1

ba

−=

∴ (B) ⎯→ (P)

(C) Bisector of OX and OY will be along the vector j)31(i)13( +++

∴ x = y

⇒ x – y = 0

2

32

)1(=

β−−β

∴ | 2β – 1| = 3

2β – 1 = ± 3

∴ β = 2, –1

∴ | β | = 1, 2

(C) → (P, Q)

(D) When α = 0

y = | 0 – 1 | + | 0 – 2 | + 0 = 3

then Area bounded by x = 0, x = 2, y2 = 4x, and y = 3 will be

x = 2

y = 3 y2 = 4x

x = 0

O X

Y

∫−=2

0

x46)0(F dx

3

2862/3

x262

0

2/3−=⎥

⎦

⎤⎢⎣

⎡−=

∴ 63

28)0(F =+

when α = 1

y = | x – 1 | + | x – 2 | + x

CAREER POINT



⎪⎩

⎪⎨

⎧

−+−

=;3x3;1x;x3

2x2x11x0

≥<≤<≤

then Area bounded by x = 0, x = 2, y2 = 4x and

y = | x – 1 | + | x – 2 | + x will be

x = 2 x = 0

O X

Y

1 2

F(1) = ∫−2

0

x45 dx

3

285 −=

∴ 53

28)1(F =+

(D) → (S, T)

CAREER POINT Paper-1 [ CODE – 05 ] CAREER POINT POINT Kota H.O. : Career Point Ltd., CP Tower,...

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Transcript of CAREER POINT Paper-1 [ CODE – 05 ] CAREER POINT POINT Kota H.O. : Career Point Ltd., CP Tower,...