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JEE Main Exam 2018 Paper & Solution

| 8th April 2018 |

ALL THE GRAPHS/DIAGRAMS GIVEN ARE SCHEMATIC AND NOT DRAWN TO SCALE.

Part A – PHYSICS

Q.1 The density of a material in the shape of a cube is determined by measuring three sides of the cube and

its mass. If the relative errors in measuring the mass and length are respectively 1.5 % and 1%, the

maximum error in determining the density is -

(1) 2.5 %

(2) 3.5 %

(3) 4.5 %

(4) 6 %

Students may find similar question in CP exercise sheet :

[JEE Main, Chapter : Unit Dimension & Error, Based on Theory (6. 3)]

Ans. [3]

Sol. ρ = Vm

ρ = 3am

ρρd =

mdm +

ada3

% = 1.5 + 3 × 1

= 4.5 %

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Q.2 All the graphs below are intended to represent the same motion. One of them does it incorrectly. Pick it

up -

(1)

position

velocity

(2)

time

distance

(3)

time

position

(4)

time

velocity


[JEE Main, Chapter : Motion in One dimension, Based on theory Pg. (12)]

Ans. [2]

Sol. s = at – bt2 shows (3)

v = a – 2bt shows (4)

(1) is also correct because velocity at middle is zero.

(2) → incorrect.

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Q.3 Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m2 to stop the motion is -

m1

T

Tm

m2

m1g (1) 18.3 kg (2) 27.3 kg (3) 43.3 kg (4) 10.3 kg

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : Newtons laws of motion, Level # 1, Q. No. 57]

Ans. [2] Sol. m1g = μ(m2 + m) 5 = .15 (10 + m)

15500 = 10 + m

m = 15350 = 23.3

so answer will be (2).

Q.4 A particle is moving in a circular path of radius a under the action of an attractive potential U = – 22rk .

Its total energy is -

(1) – 24ak

(2) 22ak

(3) zero

(4) – 23 2a

k

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Work, Power & Energy, Similar in Exercise # 5 (B), Q. No. 3]

Ans. [3]

Sol. F = – drdU = –

2k (–2) r–2–1

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= 3rk

3rk =

rmv2

mv2 = 2rk

KE = K = 22rk

TE = 22rk− + 22r

k = 0

Q.5 In a collinear collision, a particle with an initial speed v0 strikes a stationary particle of the same mass. If

the final total kinetic energy is 50 % grater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is -

(1) 40v

(2) 2 v0

(3) 20v

(4) 20v

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Conservation laws]

Ans. [2] Sol. mv0 = mv1 + mv2 ….(1) (momentum conservation) KEf = KEi + 0.5 KEi

KEf = 1.5 KEi

21 m(v12 + v22) =

21 mv02 × 1.5 ….(2)

From (1) v1 + v2 = v0

From (2) v12 + v22 = 1.5 v22

v1 – v2 = 212

21 4)( vvvv −+

(v1 + v2)2 = v12 + v22 + 2v1v2 v02 = 1.5 v02 + 2v1v2

2v1v2 = –0.5 v02

v1 – v2 = )5.0(2 20

20 vv −−

= 2 v0

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Q.6 Seven identical circular planar disks, each of mass M and radius R are welded symmetrically as shown.

The moment of inertia of the arrangement about the axis normal to the plane and passing through the point P is -

P

O

(1) 2

19 MR2

(2) 2

55 MR2

(3) 2

73 MR2

(4) 2

181 MR2


[JEE Main, Chapter : Rotational Motion, Exercise # 1, Q. No. 90]

Ans. [4]

Sol. Iabout O = 2

2MR + ⎥⎥⎦

⎤

⎢⎢⎣

⎡+ 2

24

2RMMR × 6

= 2

2MR + 2

9 2MR × 6

= 2

55 MR2

Iabout P = 2

55 2MR + 7M (9R2)

= ⎟⎠

⎞⎜⎝

⎛ ×+2

26355 MR2

= 2

181 2MR

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Q.7 From a uniform circular disc of radius R and mass 9 M, a small disc of radius 3R is removed as shown

in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is -

R3

2R

(1) 4 MR2

(2) 9

40 MR2

(3) 10 MR2

(4) 9

37 MR2

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Rotational Motion, Exercise # 5(A), Q. No. 19]

Ans. [1] Sol. Iat centre = Ibig disc – Ismall disc

= 2

9 2MR – ⎥⎥⎦

⎤

⎜⎜

⎝

⎛⎟⎠

⎞⎜⎝

⎛+⎟⎠

⎞⎜⎝

⎛22

32

21

3RMRM

= 2

9 2MR – ⎥⎥⎦

⎤

⎢⎢⎣

⎡ ×+

94

18

22 RMMR

= 2

9 2MR – 2

2MR

= 4 MR2

Q.8 A particle is moving with a uniform speed in a circular orbit of radius R in a central force inversely

proportional to the nth power of R. If the period of rotation of the particle is T, then - (1) T ∝ R3/2 for any n

(2) T ∝ 1

2n

R+

(3) T ∝ R(n+1)/2

(4) T ∝ Rn/2


[JEE Main, Chapter : Gravitation, Exercise # 4, Q. No. 10]

Ans. [3]

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Sol. F ∝ nR1

F = nRC

R

mv2 = nR

C

v2 = mC R1–n

v = mC 2

1 n

R−

T = vRπ2

T = 2

12

n

RmC

R−

π

T ∝ 2)1(1 n

R−

−

∝ 212 n

R+−

∝ 21+n

R

Q.9 A solid sphere of radius r made of a soft material of bulk modulus K is surrounded by a liquid in a

cylindrical container. A massless piston of area a floats on the surface of the liquid, covering entire cross section of cylindrical container. When a mass m is placed on the surface of the piston to compress the

liquid, the fractional decrement in the radius of the sphere, ⎟⎠⎞

⎜⎝⎛

rdr , is -

(1) mgKa

(2) mgKa

3

(3) Ka

mg3

(4) Kamg


[JEE Main, Chapter : Properties of Matter (Elasticity)]

Ans. [3]

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Sol. Increase in pressure = Areamg =

amg

B = –

VdVdP

V

dV = – B

dP

V

dV = – Kamg

v = 334 Rπ

V

dV = RdR3

3 R

dR = – Kamg

R

dR = Ka

mg3

Q.10 Two moles of an ideal monoatomic gas occupies a volume V at 27ºC. The gas expands adiabatically to a

volume 2V. Calculate (a) the final temperature of the gas and (b) change in its internal energy -

(1) (a) 189 K (b) 2.7 kJ

(2) (a) 195 K (b) – 2.7 kJ

(3) (a) 189 K (b) –2.7 kJ

(4) (a) 195 K (b) 2.7 kJ


[JEE Main, Chapter : Thermodynamics, Exercise # 4, Q. No. 22]

Ans. [3] Sol. Ti = 273 + 27 = 300

PVγ = C

TVγ–1 = C

300 Vγ–1 = T(2V)γ–1

T = 12300

−γ

γ = 1 + 32

T = 3/22300 = 3/14

300

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T = 189 K

ΔU = 2f nRΔT

= )300189(314.8223

−××

= –2.7 kJ

Q.11 The mass of a hydrogen molecule is 3.32 × 10–27 kg. If 1023 hydrogen molecules strike, per second, a fixed wall of area 2 cm2 at an angle of 45° to the normal, and rebound elastically with a speed of 103 m/s, then the pressure on the wall is nearly :

(1) 2.35 × 103 N/m2

(2) 4.70 × 103 N/m2

(3) 2.35 × 102 N/m2

(4) 4.70 × 102 N/m2


[JEE Main, Chapter : K.T.G, Exercise # 1, Q.No.21]

Ans. [1] Sol.

v

45°

v Δp of one particle = mv cos 45 ° × 2

Force = pNΔ&

= 1023 × 2

mv × 2

= 1023 mv 2

Pressure = 4–

327–23

1022101032.310

××××

= 2

232.31023 ××

= 2

7.4103 ×

= 2.35 ×103 N/m2

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Q.12 A silver atom in a solid oscillates in simple harmonic motion in some direction with a frequency of 1012/sec. What is the force constant of the bonds connecting one atom with the other ? ( Mole wt. of silver = 108 and Avogadro number = 6.02 × 1023 gm mole–1)

(1) 6.4 N/m

(2) 7.1 N/m

(3) 2.2 N/m

(4) 5.5 N/m


[JEE Main, Chapter : S.H.M, Formula Based [Class Notes]

Ans. [2]

Sol.

m m k

T = 2π kμ

T = 2π k

m2

12101 = 2 × 3.14

10001002.6108

23 ×× k

24101 =

10001002.6108)14.3(4

23

2

××

××

k

k = 100002.6

10108)14.3(4 2

××××

k = 7.1 N/m

Q.13 A granite rod of 60 cm length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is 2.7 × 103 kg/m3 and its Young's modules is 9.27 × 1010 Pa. What will be the fundamental frequency of the longitudinal vibrations ?

(1) 5 kHz

(2) 2.5 kHz

(3) 10 kHz

(4) 7.5 kHz


[JEE Main, Chapter : Wave motion, Exercise # 2, Q. No.13]

Ans. [1]

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Sol.

4λ

4λ = 0.3

λ = 1.2 m v = nλ

n = λv =

2.11

dY

= 2.1

1 23

10

107.21027.9

×× = 4.8 kHz ≈ 5 kHz

Q.14 Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface

charge densities + σ, – σ and + σ respectively. The potential of shell B is :

(1) ⎥⎥⎦

⎤

⎢⎢⎣

⎡+

εσ c

aba 22

0

–

(2) ⎥⎥⎦

⎤

⎢⎢⎣

⎡+

εσ c

bba 22

0

–

(3) ⎥⎥⎦

⎤

⎢⎢⎣

⎡+

εσ a

bcb 22

0

–

(4) ⎥⎥⎦

⎤

⎢⎢⎣

⎡+

εσ a

ccb 22

0

–

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Electrostatics, Exercise # 5 (B), Q. No.18]

Ans. [2] Sol.

CB

A

a

c b

–4σπb24σπa24σπc2

VB =

bkqA +

bkqB +

ckqC

=04

1πε b

a24. πσ +04

1πε b

b24)( πσ− +04

1πε c

c24π×σ

= 0ε

σ ⎟⎟⎠

⎞⎜⎜⎝

⎛+ c

bba 22 –

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Q.15 A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric

material of dielectric constant K = 35 is inserted between the plates, the magnitude of the induced

charge will be : (1) 1.2 n C (2) 0.3 n C (3) 2.4 n C (4) 0.9 n C

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Capacitance, Formula Based [Class Notes]

Ans. [1]

Sol. qinduced = q ⎥⎦

⎤⎢⎣

⎡K1–1

Where q = KCV = KCV

q = 35 × 90 ×10–9 × 20

q = 3000 × 10–9

qinduced = q ⎥⎦

⎤⎢⎣

⎡ × 351–1

= 5

2×q

= 3 × 10–6 × 52

= 1.2 n C Q.16 In an a.c. circuit, the instantaneous e.m.f. and current are given by e = 100 sin 30 t

i = 20 sin ⎟⎠

⎞⎜⎝

⎛ π4

–30t

In one cycle of a.c., the average power consumed by the circuit and the wattless current are, respectively :

(1) 50, 10

(2) 2

1000 , 10

(3) 2

50 , 0

(4) 50, 0 Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Alternating Current, Exercise # 1, Q. No. 81]

Ans. [2] Sol. e = 100 sin 30 t

i = 20 sin ⎟⎠

⎞⎜⎝

⎛ π4

–30t

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P = 2cos00 φie

= 2

20100 × × cos 4π

P = 2

1000 watt

Wattless current

I = 2

)sin( 0 φI

= 2

20 = 10 A

Q.17 Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor of 10 Ω. The

internal resistances of the two batteries are 1 Ω and 2 Ω respectively. The voltage across the load lies between :

(1) 11.6 V and 11.7 V (2) 11.5 V and 11.6 V (3) 11.4 V and 11.5 V (4) 11.7 V and 11.8 V

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Current Electricity, Exercise # 4, Q. No.22 ]

Ans. [2] Sol.

12 V

13 V

0 V

2 Ω

1 Ω

10 Ω

VLoad =

101

21

11

213

112

++

+

=

101510

237

++ =

1610

237

×

≈ 11.56 volt

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Q.18 An electron , a proton and an alpha particle having the same kinetic energy are moving in circular orbits of radii re, rp, rα respectively in a uniform magnetic field B. The relation between re, rp, rα is :

(1) re > rp = rα (2) re < rp = rα

(3) re < rp < rα

(4) re < rα < rp Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Magnetic Effect of Current, Exercise # 5 (A),Q. No.4]

Ans. [2] Sol. e–, p+, α

r = qBmKE2

r ∝ qm

For electron qm =

eme

1

For proton qm =

emp

1

For α qm =

eu

24

= eu

212

re < rp = rα Q.19 The dipole moment of a circular loop carrying a current I, is m and the magnetic field at the centre

of the loop is B1. When the dipole moment is doubled by keeping the current constant, the magnetic

field at the centre of the loop is B2. The ratio 2

1BB is :

(1) 2 (2) 3 (3) 2 (4) 2

1

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Magnetic Effect of Current]

Ans. [3] Sol.

B1

I

m

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Q B = rNI

2)(0μ

∴ M = NIA = NI (πr2)

r = πNI

M

B =

π

μ

NIMIN

2

0

⇒ B ∝ M1

⇒ 2

1BB = 2

Q.20 For an RLC circuit driven with voltage of amplitude vm and frequency ω0 = LC1 the current

exhibits resonance. The quality factor, Q is given by :

(1) R

L0ω

(2) LR0ω

(3) )( 0C

Rω

(4) 0ω

CR

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Alternating Current, Theory Booklet point no. 11, Page No. 15]

Ans. [1]

Sol. ∴ ω0 = LC1

Q = R

XL

= R

L0ω

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Q.21 An EM wave from air enters a medium. The electric fields are ⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ −πν= tczxEE 2cosˆ011

r in air and

)]2(cos[ˆ022 ctzkxEE −=r

in medium, where the wave number k and frequency ν refer to their

values in air. The medium is non-magnetic. If 1r∈ and

2r∈ refer to relative permittivities of air and

medium respectively, which of the following options is correct ?

(1) 2

1

r

r

∈

∈ = 4

(2) 2

1

r

r

∈

∈ = 2

(3) 2

1

r

r

∈

∈ =

41

(4) 2

1

r

r

∈

∈ =

21


[JEE Main, Chapter : EMW, Theory, Page No.76, Point No. 5.1]

Ans. [3]

Sol. ⎥⎦

⎤⎢⎣

⎡⎟⎠

⎞⎜⎝

⎛ −πν= tczxEE 2cosˆ011

r

)]2(cos[ˆ022 ctzkxEE −=r

In Air

1r01

1cc/2

2v∈μ

==πν

πν= ...... (1)

In medium

2r02

12c

k2kcv

∈μ=== ...... (2)

by (1) & (2)

41

2

1

r

r =∈

∈

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Q.22 Unpolarized light of intensity I passes through an ideal polarizer A. Another identical polarizer B is

placed behind A. The intensity of light beyond B is found to be 2I . Now another identical polarizer C

is placed between A and B. The intensity beyond B is now found to be 8I . The angle between

polarizer A and C is (1) 0º (2) 30º (3) 45º (4) 60º

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Polarization, Ex.4, Q.No.8]

Ans. [3] Sol.

I 2I θ2cos

2I θθ 22 sincos

2I

8

sincos 2

22 II=θθ

4 cos2θ sin2θ = 1 sin22θ = 1 2θ = 90º θ = 45º Q.23 The angular width of the central maximum in a single slit diffraction pattern is 60º. The width of the slit

is 1 µm. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance ? (i.e. distance between the centres of each slit.)

(1) 25 µm (2) 50 µm (3) 75 µm (4) 100 µm

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Wave Optics (Diffraction), Formula based in Class Notes]

Ans. [1]

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Sol. For Ist minima

θ = 30

a

nsin λ=θ

m1

121

μλ×

=

λ = 0.5 μm

D = 0.5 m

β = 1 × 10–2 m

d = ?

d

Dλ=β

d2

105.01106

2−

− ××=

2

6

101025.0d

−

−×=

d = 0.25 × 10–4 m

d = 25 ×10–6 m

d = 25 μm

Q.24 An electron from various excited states of hydrogen atom emit radiation to come to the ground state. Let

λn, λg be the de Broglie wavelength of the electron in the nth state and the ground state respectively. Let nΛ be the wavelength of the emitted photon in the transition from the nth state to the ground state.

For large n, (A, B are constants)

(1) nΛ ≈ A + 2n

Bλ

(2) nΛ ≈ A + B λn

(3) 2nΛ ≈ A + B 2

nλ

(4) 2nΛ ≈ λ

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Atomic Structure]

Ans. [1]

Sol. )eV6.13(n1E 2−⇒

2n

22

22

n cm)Ve6.13(h)137(E

λ−⇒ ,

mc137hn

mvh ×

==λ , h137

mcn λ=

CAREER POINT


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n

1nhcEEλ

⇒−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

λ−

λ=

λ 2n

2g

22

22

n

11cm

)eV6.13(h)137(hc

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

λ−

λ=

λ 2n

2g

32

2

n

11cm

)eV6.13(h)137(1

So, 1

2n

2g

2

32

n11

)eV6.13(h)137(cm

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

λ−

λ=λ

using binomial approximation

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

λ+

λ=λ 2

n2g

2

32

n11

)eV6.13(h)137(cm

Let )eV6.13(h)137(

cm2

32 be a constant Let it be B

then ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

λ+

λ=λ 2

n2g

nBB

2gλ is also constant hence

2g

Bλ

is another constant let called it A then

⎟⎟⎠

⎞⎜⎜⎝

⎛

λ+=λ 2

nn

BA

Q.25 If the series limit frequency of the Lyman series is νL, then the series limit frequency of the Pfund series

is : (1) 25 νL

(2) 16 νL (3) νL/16 (4) νL/25

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Atomic Structure, Theory Point No.7, Page No.12 &13]

Ans. [4] Sol. Z = 1 n → ∞ to n = 1

⎟⎟⎠

⎞⎜⎜⎝

⎛−=ν 2

221

2

n1

n1RcZ

⎟⎠

⎞⎜⎝

⎛∞

−=ν1

11Rc 2L

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νL = Rc n → ∞ to n = 5

⎟⎠

⎞⎜⎝

⎛∞

−=ν1

51Rc 2P

25Rc

P =ν

25

LP

ν=ν

Q.26 It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of

its energy is pd ; while for its similar collision with carbon nucleus at rest, fractional loss of energy is pc. The value of pd and pc are respectively :

(1) (0.89, 0.28) (2) (0.28, 0.89) (3) (0, 0) (4) (0, 1)

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Work-Power-Energy, Ex.4, Q.No.25]

Ans. [1] Sol.

m1

u1

m2

u2 = 0 ⇒

m1

v1

m2

v2+

v1 = ⎟⎟⎠

⎞⎜⎜⎝

⎛+−

21

21mmmm u1

fractional loss of energy

= 211

21

2

21

211

211

21

21

21

um

ummmmmum ⎟⎟

⎠

⎞⎜⎜⎝

⎛+−

−

= 1 – 2

21

21⎟⎟⎠

⎞⎜⎜⎝

⎛+−

mmmm

For neutron deutron collision : m1 = m and m2 = 2m

pd = 1 – 2

22

⎟⎠

⎞⎜⎝

⎛+−

mmmm = 1 –

91 =

98 = 0.89

For neutron – carbon collision : m1 = m and m2 = 12m

pc = 1 – 2

1212

⎟⎠

⎞⎜⎝

⎛+−

mmmm = 1 –

169121 = 0.28

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Q.27 The reading of the ammeter for a silicon diode in the given circuit is

3 V

200 Ω

(1) 0 (2) 15 mA (3) 11.5 mA (4) 13.5 mA

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Semi Conductor & Electronics, Ex.5, Q.No.1]

Ans. [3] Sol.

2.3V

0.7V

3V

200Ω

i = A200

3.2 = mA200

2300

i = 11.5 mA Q.28 A telephonic communication service is working at carrier frequency of 10 GHz. Only 10 % of it is utilized

for transmission. How many telephonic channels can be transmitted simultaneously if each channel requires a bandwidth of 5 kHz ?

(1) 2 × 103 (2) 2 × 104 (3) 2 × 105 (4) 2 × 106

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Communication System, Ex.4, Q.No.8]

Ans. [3] Sol. Working frequency range

= 10010 × 10 × 109 = 109 Hz

Band width of each channel = 5 × 103 Hz

No. of channels = 3

9

10510×

= 2 × 105

CAREER POINT


[CODE – A]

Q.29 In a potentiometer experiment, it is found that no current passes through the galvanometer when the terminals of the cell are connected across 52 cm of the potentiometer wire. If the cell is shunted by a resistance of 5 Ω, a balance is found when the cell is connected across 40 cm of the wire. Find the internal resistance of the cell.

(1) 1 Ω (2) 1.5 Ω (3) 2 Ω (4) 2.5 Ω

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Current Electricity, Ex.2, Q.No.52]

Ans. [2] Sol.

G

r ε

i = 0

l1 = 52cm

ε0

G

rε

R = 5Ω

l2 = 40cm

ε0

r = R ⎟⎟⎠

⎞⎜⎜⎝

⎛−1

2

1l

l

r = 5 ⎟⎠

⎞⎜⎝

⎛ −14052

r = 8

52 – 5 = 1.5 Ω

Q.30 On interchanging the resistances, the balance point of a meter bridge shifts to the left by 10 cm. The

resistance of their series combination is 1 kΩ. How much was the resistance on the left slot before interchanging the resistances ?

(1) 990 Ω (2) 505 Ω (3) 550 Ω (4) 910 Ω

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Current Electricity, Ex.1, Q.No.75]

Ans. [3]

CAREER POINT


[CODE – A]

Sol. As on interchanging resistors balance point move towards left by 10 cm.

55cm 45cm 45cm 55cm

P Q Q P

QP =

4555

QP =

911

⇒ 9P = 11Q

P + Q = 1000

9

11Q + Q = 1000

9

20Q = 1000

Q = 50 × 9 = 450 Ω

∴ P = 550 Ω

CAREER POINT


[CODE – A]

Part B – CHEMISTRY

Q.31 The ratio of mass percent of C and H of an organic compound (CxHyOz) is 6 : 1. If one molecule of

the above compound (CxHyOz) contains half as much oxygen as required to burn one molecule of

compound CxHy completely to CO2 and H2O. The empirical formula of compound CxHyOz is -

(1) C3H6O3

(2) C2H4O

(3) C3H4O2

(4) C2H4O3


[JEE Main, Chapter : Mole Concept, Ex. # 4, , Q. No. 19]

Ans. [4]

Sol. CxHy + ⎟⎠

⎞⎜⎝

⎛ +4yx O2 ⎯→ OH

2yxCO 22 +

1 mol ⎟⎠

⎞⎜⎝

⎛ +4yx

amount of oxygen in the compound = zz

4yx

2 =⎟⎠⎞

⎜⎝⎛ +

× or z4

yx=

+

z4yx =+ On putting the value of y in above equation

As C : H mass ratio 6 : 1 ∴ z4x2x =+

∴ mole ratio21

11

126

yx

== z2x3

=

21

=yx

y = 2x

and x : y : z

putting y and z in terms of x

x : 2x : 2x3

1 : 2 : 23

2 : 4 : 3

CAREER POINT


[CODE – A]

Q.32 Which type of 'defect' has the presence of cations in the interstitials sites ? (1) Schottky defect (2) Vacancy defect (3) Frenkel defect (4) Metal deficiency defect

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Solid State, Theory notes, Page no. 14]

Ans. [3] Sol. Frenkel defect result in arrangement of cation at interstitial site Q.33 According to molecular orbital theory, which of the following will not be a viable molecule ?

(1) +22He

(2) +2He

(3) −2H

(4) −22H

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical Bonding, Ex. # 4 Q. No. 34]

Ans. [4]

Sol. −22H 4e– = σ 1s2, σ∗ 1s2

B.O. = 02

22=

−

∴ It does not exist as B.O. is zero

Q.34 Which of following lines correctly show the temperature dependence of equilibrium constant K, for an exothermic reaction ?

)K(T1

A

B

C

D

(0,0)

in K

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

CAREER POINT


[CODE – A]

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical Equilibrium, Ex # 4, Page 45, Q. No. 10]

Ans. [1]

Sol. .constRT

HKIn +Δ

−=

Since ΔH = –ve for exothermic reaction Slope is +ve i.e. (A) & (B)

Q.35 The combustion of benzene (l) gives CO2(g) and H2O(l). Given that heat of combustion of benzene at constant volume is –3263.9 kJ mol–1 at 25°C; heat of combustion (in kJ mol–1) of benzene at constant pressure will be - (R = 8.314 JK–1 mol–1)

(1) 4152.6 (2) –452.46 (3) 3260 (4) –3267.6

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical Energetic, Ex # 4, Page 35, Q. No. 30]

Ans. [4]

Sol. )(OH3)g(CO6)g(O2

15)(HC 22266 ll +⎯→⎯+

2156ng −

=Δ = 23

−

ΔH = ΔE + Δng RT

= 1000

298314.8239.3263 ×

×−−

= –3263.9 – 3.716358 = – 3267.616 kJ / mol Q.36 For 1 molal aqueous solution of the following compounds, which one will show the highest freezing

point ? (1) [Co(H2O)6]Cl3 (2) [Co(H2O)5Cl]Cl2.H2O (3) [Co(H2O)4Cl2]Cl.2H2O (4) [Co(H2O)3Cl3].3H2O

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Solution and Colligative Properties, Ex # 2, Page 35 Q. No. 85]

CAREER POINT


[CODE – A]

Ans. [1] Sol. fTΔ (Freezing point depression) ∝ n(no. of particles)

∴ If no. of particles ↓ , freezing point is more Q.37 An aqueous solution contains 0.10 M H2S and 0.20 M HCl. If the equilibrium constants for the

formation of HS– from H2S is 1.0 × 10–7 and that of S2– from HS– ions is 1.2 × 10–13 then the concentration of S2– ions in aqueous solution is -

(1) 5 × 10–8 (2) 3 × 10–20 (3) 6 × 10–21 (4) 5 × 10–19

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical Equilibrium, Solved Example, Page 33 Q. No. 61]

Ans. [2]

Sol. H2S 2H+ + 2

S−

Keq. = 1 × 10–7 × 1.2 × 10–13

= ]SH[

]S[]H[2

22 −+

1.0

S)2.0(102.122

20−

− ×=×

04.0

102.11.0]S[20

2−

− ××=

M103 20−×=

Q.38 An aqueous solution contains an unknown concentration of Ba2+. When 50 mL of a 1 M solution of

Na2SO4 is added, BaSO4 just begins to precipitate. The final volume is 500 mL. The solubility product of BaSO4 is 1 × 10–10. What is the original concentration of Ba2+ ?

(1) 5 × 10–9 M (2) 2 × 10–9 M (3) 1.1 × 10–9 M (4) 1.0 × 10–10 M

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Ionic Equilibrium, Ex #4, Page 50, Q. No. 9]

CAREER POINT


[CODE – A]

Ans. [3]

Sol. Ksp = [Ba+2] [ 24SO− ], In 500 ml volume 2

4SO− conc. is

M1V1 = M2V2

1 × 50 = M2 × 500 M2 = 0.1 M 1 × 10–10 [Ba+2] [0.1] [Ba+2] = 1 × 10–9 M Initial conc. will be, M1V1 = M2V2 M1 × 450 = 1× 10–9 × 500 M1 = 1.11 × 10–9 M

Q.39 At 518°C, the rate of decomposition of sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr s–1 when 5% had reacted and 0.5 Torr s–1 when 33% had reacted. The order of the reaction is -

(1) 2 (2) 3 (3) 1 (4) 0

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical Kinetics, Page 15, Example No. 8]

Ans. [1] Sol.

CH3CHO ⎯→ CH4 + CO 363

– + + x x x

363 –x x x

if 63.35363100

5x ×=×=

= 18.15

rate of decomposition X3 ]A[Kdt

)CHOCH(dr =−=

1.00 = K ⋅ (363 – 18.15)X

similarly 0.5 = K (363 – 119.29)X

2 = X

21.24385.344

⎟⎠

⎞⎜⎝

⎛

2 = (1.41)X

or x = 2

CAREER POINT


[CODE – A]

Q.40 How long (approximate) should water be electrolysed by passing through 100 amperes current so that the oxygen released can completely burn 27.66 g of diborane ? (Atomic weight of B = 10.8 u)

(1) 6.4 hours (2) 0.8 hours (3) 3.2 hours (4) 1.6 hours

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Electro Chemistry, Ex # 4, Q. No.19]

Ans. [3] Sol. B2H6 + 3O2 ⎯→ B2O3 + 3H2O

6.2766.27 ↓

= 1 mole

3 mole

= 3 × 22.4 (at N.T.P. volume in litre)

96500

tiVV

e

×=

96500

t1006.5

4.223 ×=

×

t = 11580 sec = 3.21 hours Q.41 The recommended concentration of fluoride ion in drinking water is up to 1 ppm as fluoride ion is

required to make teeth enamel harder by converting [3Ca3(PO4)2.Ca(OH)2] to : (1) [CaF2] (2) [3(CaF2).Ca(OH)2] (3) [3Ca3(PO4)2.CaF2] (4) [3{Ca(OH)2}.CaF2]

Ans. [3]

Sol. [3Ca3(PO4)2.Ca(OH)2] Hydroxyapatite

FΘ 1 ppm

or 1 mg dm–3

[3Ca3(PO4)2.CaF2] Fluorapatite

[HARD ENAMEL]

Q.42 Which of the following compounds contains(s) no covalent bond(s) ? KCl, PH3, O2, B2H6, H2SO4

(1) KCl, B2H6, PH3 (2) KCl, H2SO4 (3) KCl

(4) KCl, B2H6 Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical Bonding, Ex # 1, Q. No. 37]

Ans. [3] Sol. KCl is an ionic compound it does not content covalent bond.

CAREER POINT


[CODE – A]

Q.43 Which of the following are Lewis acids ? (1) PH3 and BCl3 (2) AlCl3 and SiCl4

(3) PH3 and SiCl4

(4) BCl3 and AlCl3

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Boron Carbon Family Q. No. 7 Ex#2, page 14, Page 10 Theory notes]

Ans. [2, 4] Sol. BCl3 , AlCl3 and SiCl4 acts as Lewis acid.

Q.44 Total number of lone pair of electrons in −3I ion is

(1) 3 (2) 6 (3) 9 (4) 12

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Chemical Bonding Q. No. 11, Ex#5 (IIT 2005)]

Ans. [3]

Sol.

I ..

::

I .

: ...

I :: ..

–

(9 pair of e–)

Q.45 Which of the following salts is the most basic in aqueous solution ?

(1) Al(CN)3 (2) CH3COOK (3) FeCl3 (4) Pb(CH3COO)2

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Ionic Equilibrium Q. No. 6, Ex # 5]

Ans. [2] Sol. CH3COOK WASB salt Upon hydrolysis KOH gives strongest basic solution

CAREER POINT


[CODE – A]

Q.46 Hydrogen peroxide oxidises [Fe(CN)6]4– to [Fe(CN)6]3– in acidic medium but reduces [Fe(CN)6]3– to [Fe(CN)6]4– in alkaline medium. The other products formed are, respectively.

(1) (H2O + O2) and H2O (2) (H2O + O2) and (H2O + OH–) (3) H2O and (H2O + O2) (4) H2O and (H2O + OH–)

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Hydrogen & Its compounds, Theory notes page 7]

Ans. [3]

Sol. (a) H2O2 ⎯→⎯ [Fe(CN)6]–4 H+ H2O + [Fe(CN)6]–3

–1 +2 –2 +3

Reduction

Oxidation

(b) H2O2 ⎯→⎯ [Fe(CN)6]–3 Alkaline O2 + [Fe(CN)6]–4–1 +3 0 +2

Reduction

Oxidation

Q.47 The oxidation states of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2], and K2[Cr(CN)2(O)2(O2)(NH3)] respectively are :

(1) +3, +4, and +6 (2) +3, +2, and +4

(3) +3, 0, and +6 (4) +3, 0, and +4

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Coordination Compounds Q. No. 45, Ex # 1]

Ans. [3] Sol. [Cr(H2O)6]Cl3 x + 6(0) + 3(–1) = 0 x = +3 [Cr(C6H6)2] x = 2(0) = 0 x = 0 K2[Cr(CN)2(O)2(O2)(NH3)]

2 + x + 2(–1) + 2(–2) + (–2) + (0) = 0

x = +6

CAREER POINT


[CODE – A]

Q.48 The compound that does not produce nitrogen gas by the thermal decomposition is :

(1) Ba(N3)2

(2) (NH4)2Cr2O7

(3) NH4NO2

(4) (NH4)2SO4


[JEE Main, Chapter : p-Block Q. No. 2, Ex # 5, IIT 2011, 1991]

Ans. [4]

Sol. (1) BaN3 ⎯→⎯ Ba + 3N2

(2) (NH4)2Cr2O7 ⎯→⎯ N2 + Cr2O3 + N2O All these gives N2

(3) NH4NO2 ⎯→⎯ N2 + 2H2O

Q.49 When metal ‘M’ is treated with NaOH, a white gelatinous precipitate ‘X’ is obtained, which is soluble in

excess of NaOH. Compound ‘X’ when heated strongly gives an oxide which is used in chromatography as

an adsorbent. The metal ‘M’ is :

(1) Zn

(2) Ca

(3) Al

(4) Fe


[JEE Main, Chapter : Boron family & Metallurgy Q. No. 5, Board Pattern

Exercise]

Ans. [3]

Sol. M + NaOH NaOHAl(OH)3 ↓ NaAlO2 (x)

Al2O3

∴ Element x = Al

CAREER POINT


[CODE – A]

Q.50 Consider the following reaction and statements :

[CO(NH3)4Br2]+ + Br– → [CO(NH3)3Br3] + NH3

(I) Two isomers are produced if the reactant complex ion is a cis-isomer.

(II) Two isomers are produced if the reactant complex ion is a trans-isomer.

(III) Only one isomer is produced if the reactant complex ion is a trans-isomer.

(IV) Only one isomers is produced if the reactant complex ion is a cis-isomer.

(1) (I) and (II)

(2) (I) and (III)

(3) (III) and (IV)

(4) (II) and (IV)


[JEE Main, Chapter : Coordination Compound, Theory notes page 15]

Ans. [4]

Sol. When cis-isomer reacts -

NH3

Br

NH3

Co Br

NH3

NH3

+

+ Br–

BrBr

NH3

CoBr

NH3

NH3

NH3

Br

NH3

CoBr

Br

NH3

Facial isomer

Meridional isomer

cis-isomer

When trans-isomer reacts -

Br NH3

Br

Co NH3

NH3

NH3 + Br–

BrBr

Br

CoNH3

NH3

NH3

Only one isomertrans-isomer ∴ cis isomer gives two isomer where as trans isomer gives one isomer

CAREER POINT


[CODE – A]

Q.51 Glucose on prolonged heating with HI gives :

(1) n-Hexane

(2) 1-Hexene

(3) Hexanoic acid

(4) 6-iodohexanal


[JEE Main, Chapter : Biomolecules notes]

Ans. [1]

Sol.

CHO

(CHOH)4

CH2OH

olongedHI

Pr/⎯⎯⎯ →⎯ Δ

CH3

CH2

CH2

CH2

CH2

CH3

n-Hexane This reaction gives the proof about straight chain structure of Glucose.

Q.52 The trans-alkenes are formed by the reduction of alkynes with :

(1) H2 – Pd/C, BaSO4

(2) NaBH4

(3) Na/liq. NH3

(4) Sn – HCl


[JEE Main, Chapter : Hydrocarbon, Q. No. 21 & 28 (solved) Ex # 4]

Ans. [3]

Sol.

R – C ≡ C – R ⎯⎯⎯⎯⎯⎯ →⎯ 3NHLiquidNa

Na → Na+ + e–

C = CH R

R Htrans-alkene

C = C R

R ⎯⎯⎯⎯ →⎯ 2—NHH C = C

H R

R Na ⎯→ Na⊕ + eΘ

H ⎯ NH2 Radical anion

CAREER POINT


[CODE – A]

Q.53 Which of the following compounds will be suitable for Kjeldahl's method for nitrogen estimation?

(1)

N

(2) NH2

(3) NO2

(4) –

2ClN +

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Purification & Characterisation of organic compounds Q. No. 10, Ex # 3]

Ans. [2] Sol. In Kjeldahl's method nitrogen is estimated in the form of NH3. We can not use pyridine, Nitro and

diazo compound for this

NH2

Δ⎯⎯⎯⎯⎯ →⎯ NaOHExcess NH2 ↑

Q.54 Phenol on treatment with CO2 in the presence of NaOH followed by acidification produces compound X

as the major product. X on treatment with (CH3CO)2O in the presence of catalytic amount of H2SO4 produces :

(1) O

CO2H

O

CH3

(2)

CO2H

O

CH3O

(3)

O CH3C

O

O OH

(4)

CH3

CO2H

O

CO2HO

CAREER POINT


[CODE – A]

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Oxygen containing compounds-I, Q. No.12, Ex # 4]

Ans. [1]

Sol.

OH

2CONaOH⎯⎯⎯ →⎯

Kolbe schmidt reaction

OΘNa⊕ C–OΘNa⊕

O ⎯⎯ →⎯⊕H

OHCOOH

Salicylic acidAcetylation (CH3CO)2O

O–C–CH3

COOH

O

ASPIRIN

Mechanism :

OΘ OΘ C = O

O OHC–OO

OH

C–OHO ⎯⎯ ⎯←

⊕H

OΘ

C–OΘ

O

(CH3CO)2O

O–C–CH3

COOH

O

Θ

Tautomerism

Acetylation

Q.55 An alkali is titrated against an acid with methyl orange as indicator, which of the following is a correct

combination ? Base Acid End point (1) Weak Strong Colourless to pink (2) Strong Strong Pinkish red to yellow (3) Weak Strong Yellow to pinkish red (4) Strong Strong Pink to colourless

Ans. [3]

CAREER POINT


[CODE – A]

Sol.

Yellow to red

alkali

Acid

Methyl orange

Q.56 The predominant form of histamine present in human blood is (pKa′ Histidine = 6.0)

(1)

NH2

N

N H

(2)

NH3

N

N H

⊕

⊕ H

(3)

NH2N

N H

⊕ H

(4)

NH3

N

N H

⊕

Ans. [4]

Sol.

NH2

N

N H

⎯⎯ →⎯⊕H

NH3

N

N H

⊕

Histamine

Blood pH = 7.3

1st pKa of N = 9.9

2nd pKa of N = 5.88

Histamine : It has three nitrogen

In normal physiological condition it exist as monoprotonated and aliphatic nitrogen is most basic so

it will be protonated.

CAREER POINT


[CODE – A]

Q.57 Phenol reacts with methyl chloroformate in the presence of NaOH to form product A, A reacts with Br2 to form product B. A and B are respectively :

(1)

andOH

O

OCH3

OH

O

OCH3

Br and

(2)

and O

O

O

Oand

Br

OO

(3)

andO

O

O

Oand

OO

Br

(4)

and OH

O

OCH3

OH

O

OCH3and

Br

Ans. [3]

Sol.

O–H O

⎯⎯⎯ →⎯NaOH

OΘ

Cl – C – OMe

O–C–OMeO

Br

⎯⎯ →⎯ 2Br

O–C–OMeO

Q.58 The increasing order of basicity of the following compounds is :

(a) NH2

(b) NH

(c)

NH2

NH (d) NHCH3 (1) (a) < (b) < (c) < (d) (2) (b) < (a) < (c) < (d) (3) (b) < (a) < (d) < (c) (4) (d) < (b) < (a) < (c)

Students may find similar question in CP exercise sheet : [JEE Advance, Chapter : GOC, Q. No.17, Ex # 5]

Ans. [3]

CAREER POINT


[CODE – A]

Sol.

NH2sp3

1º Amine

⎯⎯ →⎯⊕H

NH3⊕

NHsp2

More % s character so less basic

⎯⎯ →⎯⊕H

NH2⊕

⎯⎯ →⎯⊕H

NH2

NH

NH—CH3

NH2

NH2⊕

NH2

NH2

⊕

Most stabilized cation

sp3

2º Amine

⎯⎯ →⎯⊕H

N⊕—CH3

H

H

Basic strength C > d > a > b Q.59 The major product formed in the following reaction is :

HeatHI⎯⎯→⎯

O

O

(1)

OH

OH

(2)

I

I

(3)

OH

I

(4)

I

OH

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Oxygen containing compound-I, Q. No. 6, Ex-4]

Ans. [4]

Sol.

O

O

SN1 reaction at this centre

⎯⎯→⎯HI

SN2 at this centre

I

O–H+ CH3OH + CH3 – I

CAREER POINT


[CODE – A]

Q.60 The major product of the following reaction is :

MeOHNaOMe⎯⎯⎯ →⎯

Br

(1)

OMe

(2)

(3)

(4)

OMe

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Hydrocarbon, Q. No. 11, Ex # 3]

Ans. [2]

Sol.

Brα

β MeOHNaOMe⎯⎯⎯ →⎯

β

Strong base sodium methoxide is used so E2 elimination will occur and removed of Br and β-H

occur simultaneously C anti elimination

Br α

β

NOβ-H

H

CAREER POINT


[CODE – A]

Part C – MATHEMATICS Q.61 Two sets A and B are as under: A = {(a, b) ∈ R × R : |a – 5| < 1 and |b – 5| < 1} ; B = {(a, b) ∈ R × R : 4(a – 6)2 + 9(b – 5)2 ≤ 36}. Then

(1) B ⊂ Α

(2) A ⊂ B (3) A ∩ B = φ (an empty set) (4) neither A ⊂ B nor B ⊂ A

Ans. [2] Sol. From set A |a – 5| < 1 |b – 5| < 1 –1 < a – 5 < 1 –1 < b – 5 < 1 4 < a < 6 4 < b < 6 A = { 4 < a < 6, 4 < b < 6} From set B

4

)5(9

)6( 22 −+

− ba ≤ 1

(6, 7)

(6, 5)

(6, 3)

(3, 5) (9, 5)

B = {3 < a < 9, 3 < b < 7} A ⊂ B Q.62 Let S = {x ∈ R : x ≥ 0 and 06)6(|3|2 =+−+− xxx } . Then S :

(1) is an empty set (2) contains exactly one element (3) contains exactly two elements (4) contains exactly four elements

Ans. [3]

Sol. 06)6(|3|2 =+−+− xxx

Let x = t

2|t – 3| + t2 – 6t + 6 = 0 case (i) t < 3 ...(1) –2t + 6 + t2 – 6t + 6 = 0

CAREER POINT


[CODE – A]

t2 – 8t + 12 = 0 t = 2, 6 Since t < 3, hence t = 2

x = 2

x = 4 case (ii) t ≥ 3 2t – 6 + t2 – 6t + 6 = 0 t2 – 4t = 0 t = 4, 0 Since t ≥ 3 hence t = 4

x = 4

x = 16

Q.63 If α, β ∈ C are the distinct roots, of the equation x2 – x + 1 = 0, then α101 + β107 is equal to

(1) –1 (2) 0 (3) 1 (4) 2

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Quadratic equation, Ex. 2, Q. No. 20]

Ans. [3] Sol. We have, x2 – x + 1 = 0

α, β = 2

411 −± = 2

31 i±

α = 2

31 i+ = –ω2

and β = 2

31 i− = –ω

α101 + β107 = (–ω2)101 + (–ω)107 = –ω202 – ω107 = –ω – ω2 = –(ω + ω2) = 1

CAREER POINT


[CODE – A]

Q.64 If 422

242224

−−

−

xxxxxxxxx

= (A + Bx) (x – A)2 , then the ordered pair (A, B) is equal to :

(1) (–4, –5) (2) (–4, 3) (3) (–4, 5) (4) (4, 5)

Ans. [3]

Sol. 422

242224

−−

−

xxxxxxxxx

= (A + Bx) (x – A)2

Put x = 0 both side, we get

400

040004

−−

− = (A + 0) (0 – A)2

– 43 = A3 A = –4 Put x = 1 both side, we get

322

232223

−−

− = (A + B) (1 – A)2

–3(9 – 4) – 2(–6 – 4) + 2(4 + 6) = (–4 + B) (1 + 4)2 –15 + 20 + 20 = (–4 + B) . 52 25 = (–4 + B) . 25 –4 + B = 1 B = 5 Q.65 If the system of linear equations x + ky + 3z = 0 3x + ky – 2z = 0 2x + 4y – 3z = 0

has a non-zero solution (x, y, z) then 2yxz is equal to

(1) –10 (2) 10 (3) –30 (4) 30

Ans. [2] Sol. Linear equation has non zero solution

CAREER POINT


[CODE – A]

Hence 34223

31

−−k

k = 0

1(–3k + 8) – k(–9 + 4) + 3(12 – 2k) = 0 –3k + 8 + 9k – 4k + 36 – 6k = 0 –4k = –44 k = 11 Hence equations are x + 11y + 3z = 0 ...(1) 3x + 11y – 2z = 0 ...(2) 2x + 4y – 3z = 0 ...(3) add (1) and (3) 3x + 15y = 0 x = –5y from (3) 2x + 4y – 3z = 0

Q x = –5y –10y + 4y – 3z = 0 –6y – 3z = 0 z = –2y

consider 2yxz = 2

)2)(5(y

yy −− = 10

Q.66 From 6 different novels and 3 different dictionaries, 4 novels and 1 dictionary are to be selected and

arranged in a row on a shelf so that the dictionary is always in the middle. The number of such arrangements is :

(1) at least 1000 (2) less than 500 (3) at least 500 but less than 750 (4) at least 750 but less than 1000

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : P & C, Ex. 4, Q. No. 14]

Ans. [1] Sol. 6C4 × 3C1 × 4!

= 45 × 24 = 1080

CAREER POINT


[CODE – A]

Q.67 The sum of the co-efficients of all odd degree terms in the expansion of 5

35

3 11 ⎟⎠⎞⎜

⎝⎛ −−+⎟

⎠⎞⎜

⎝⎛ −+ xxxx , (x > 1) is

(1) –1 (2) 0 (3) 1 (4) 2

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Binomial Theorem, Ex.5 Q. No.1]

Ans. [4]

Sol. 5

35

3 11 ⎟⎠⎞⎜

⎝⎛ −−+⎟

⎠⎞⎜

⎝⎛ −+ xxxx

= ⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎠⎞⎜

⎝⎛ −+⎟

⎠⎞⎜

⎝⎛ −+

431

45

233

255

05 1)(1)()(2 xxCxxCxC

= 2[x5 + 10x3(x3 – 1) + 5x(x3 – 1)2] = 2[x5 + 10x6 – 10x3 + 5x(x6 – 2x3 + 1)] = 2[x5 + 10x6 – 10x3 + 5x7 – 10x4 + 5x] = 2[1 – 10 + 5 + 5] = 2

Q.68 Let a1, a2, a3, ………, a49 be in A.P. such that ∑=

+ =12

014 416

kka and a9 + a43 = 66.

If 140...... 217

22

21 =+++ aaa m, then m is equal to :

(1) 66 (2) 68 (3) 34 (4) 33

Ans. [3] Sol. If a1 + a5 + a9 + a13 +……… + a49 = 416 6(a1 + a49) + a25 = 416 6(a1 + a1 + 48d) + (a1 + 24d) = 416 12(a1 + 24d) + (a1 + 24d) = 416 13(a1 + 24d) = 416 a1 + 24d = 32 ...(1) and a9 + a43 = 66 a1 + 8d + a1 + 42d = 66 2a1 + 50d = 66 (a1 + 25d) = 33 ...(2) By (1) and (2) d = 1

CAREER POINT


[CODE – A]

a1 = 8

(a12 + a22 + .... + a172) = 140 m

a12 + (a1 + 1)2 + (a1 + 2)2 +.......+ (a1 + 16)2 = 140 m

17a12 + 2a1 (1 + 2 + 3 + ... + 16) + (12 + 22 +... + 162) = 140 m

17 × 64 + 2 × 8 ⎟⎠

⎞⎜⎝

⎛ ×2

1716 + 6

173316 ×× = 140 m

1088 + 2176 + 1496 = 140 m 4760 = 140 m 34 = m Q.69 Let A be the sum of the first 20 terms and B be the sum of the first 40 terms of the series 12 + 2.22 + 32 + 2.42 + 52 + 2.62 + …. If B – 2A = 100λ, then λ is equal to :

(1) 232 (2) 248 (3) 464 (4) 496

Ans. [2] Sol. Sum of first 20 terms A = (12 + 32 + .... + 192) + 2(22 + 42 + ... + 202) A = (12 + 22 + 32 + .... + 192 + 202) + (22 + 42 + .... + 202)

A = 6

214120 ×× + 22(12 + 22 + .... + 102)

A = 2870 + 4 × 6

211110 ××

A = 2870 + 1540 A = 4410 Now, sum of first 40 terms B = (12 + 22 + .... + 402) + (22 + 42 + .... + 402)

= 6

814140 ×× + 22(12 + 22 + .... + 202)

= 6

814140 ×× + 4 ⎟⎠

⎞⎜⎝

⎛ ××6

412120

= 22140 + 11480 = 33620 So, B – 2A = 100 λ 33620 – 8820 = 100 λ 24800 = 100 λ λ = 248

CAREER POINT


[CODE – A]

Q.70 For each t ∈ R, let [t] be the greatest integer less than or equal to t.

Then ⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦⎤

⎢⎣⎡++⎥⎦

⎤⎢⎣⎡+⎥⎦

⎤⎢⎣⎡

+→ xxxx

x

15......21lim0

(1) is equal to 0. (2) is equal to 15. (3) is equal to 120. (4) does not exist (in R)

Ans. [3]

Sol. ⎟⎟⎠

⎞⎜⎜⎝

⎛⎥⎦⎤

⎢⎣⎡++⎥⎦

⎤⎢⎣⎡+⎥⎦

⎤⎢⎣⎡

+→ xxxx

x

15......21lim0

= ⎪⎭

⎪⎬⎫

⎪⎩

⎪⎨⎧

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎭⎬⎫

⎩⎨⎧+

⎭⎬⎫

⎩⎨⎧+

⎭⎬⎫

⎩⎨⎧−⎟

⎠

⎞⎜⎝

⎛ ++++→ xxxxxx

xx

15....2115....21lim0

where { } denotes fractional part of x

⎟⎠

⎞⎜⎝

⎛ −++++→

015...21lim0 xxx

xx

= 1 + 2 + ..... + 15

= 2

)115(15 +

= 15 × 8 = 120 Q.71 Let S = {t ∈ R : f(x) = |x – π|⋅ (e|x| – 1) sin|x| is not differentiable at t}. Then the set S is equal to :

(1) φ (an empty set) (2) {0} (3) {π} (4) {0, π}

Ans. [1] Sol. S = {t ∈ R, f(x) = |x – π|⋅ (e|x| – 1) ⋅ sin|x|} Doubtful points, x = 0, π

At x = π

f(x) = |x – π|(ex – 1) sin x

⎪⎩

⎪⎨⎧

π<−π−−π>−π−=

xxexxxex

x

x

sin)1()(sin)1()(

f '(π + h) = (x – π) (ex – 1) cos x + (x – π) sin x . ex + (ex – 1) sin x . 1

at x = π = 0 + 0 + 0 similarly f '(π – h) = 0 hence at x = π, differentiable at x = 0

CAREER POINT


[CODE – A]

⎪⎩

⎪⎨⎧

<−π−+>−π−−=

− 0sin)1()(0sin)1()()(

xxexxxexxf x

x

f '(0 + h) = – {(x – π) (ex – 1) cos x + (ex – 1) sin x ⋅ 1 + (x – π) sin x ⋅ ex} x = 0 = – {(–π) ⋅ 0 + 0 + 0} similarly f '(0 – h) = 0 hence differentiable at x = 0 also differentiable every where

Q.72 If the curves y2 = 6x, 9x2 + by2 = 16 intersect each other at right angles, then the value of b is :

(1) 6

(2) 27

(3) 4

(4) 29

Students may find same question in CP exercise sheet : [JEE Main, Chapter : Tangent & Normal, Ex. 2, Q. No. 10]

Ans. [4] Sol. y2 = 6x …(1) and 9x2 + by2 = 16 …(2) slope of tangent of first curve

62 =⋅dxdyy

ydx

dym26

1 == …(3)

slope of tangent of second curve

0218 =+dxdybyx

by

xby

xdxdym 9

218

2−

=−

== …(4)

Since curve intersest at right angle ∴ m1 m2 = –1

⇒ ⎟⎟⎠

⎞⎜⎜⎝

⎛y2

6 19−=⎟⎟

⎠

⎞⎜⎜⎝

⎛−

byx

⇒ – 27 x = – by2 ⇒ – 27 x = – b(6x) (from (1))

⇒ 29

627

==b

CAREER POINT


[CODE – A]

Q.73 Let f(x) = x2 + 21x

and g(x) = x – x1 , x ∈ R – {–1, 0, 1}. If h(x) =

)()(

xgxf , then the local minimum value

of h(x) is (1) 3 (2) – 3

(3) 22−

(4) 22 Ans. [4]

Sol. f(x) = x2 + 21x

, g(x) = x – x1

h(x) =

xx

xx

1

12

2

−

+ =

xx

xx

1

21 2

−

+⎟⎠⎞

⎜⎝⎛ −

h(x) = x

x 1− +

xx 1

2

−

Differentiate both side

h′(x) = 1 + 21x

– 212

⎟⎠⎞

⎜⎝⎛ −

xx

⎟⎠

⎞⎜⎝

⎛ + 211x

=

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

⎟⎠⎞

⎜⎝⎛ −

−⎟⎠⎞

⎜⎝⎛ −

⎟⎠

⎞⎜⎝

⎛ + 2

2

2 1

2111

xx

xx

x

= ⎥⎥⎦

⎤

⎢⎢⎣

⎡ −−

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ +

2

222

2

2 2)1(1

11

xxx

xx

x

= ( ) 0]41[1

1 24

42

2=−+

⎟⎠⎞

⎜⎝⎛ −

+ xxx

xx

x

= (x2 – 2)2 – 3 = 0

= x2 = 2 ± 3

= x2 = 2

)13( 2±

= x = ± ⎟⎟⎠

⎞⎜⎜⎝

⎛ ±

213

CAREER POINT


[CODE – A]

– + – – +

⎟⎟⎠

⎞⎜⎜⎝

⎛ +−

213

–1 0 1⎟⎟⎠

⎞⎜⎜⎝

⎛ −−

213

2

13 −2

13 +

x = 2

13 +

2

131313

1321 −

=−

−

+=

x

x

x 1− =

22 = 2

2 + 2

2 = 22

Q.74 The integral ∫ +++dx

xxxxxxxx

2523235

22

)coscossinsincos(sincossin is equal to

(1) )tan1(3

13 x+

+ C

(2) )tan1(3

13 x+

− + C

(3) x3cot1

1+

+ C

(4) x3cot1

1+

− + C

(where C is a constant of integration) Ans. [2]

Sol. ∫ +++ 2523235

22

)coscossinsincos(sincossin

xxxxxxdxxx

∫ +++⋅

= 2332332

22

)}cos(sincos)cos(sin{(sincossin

xxxxxxdxxx

∫ ++⋅

= 23322

22

)}cos(sin)cos{(sincossin

xxxxdxxx

∫ += 233

22

)cos(sincossin

xxdxxx

divide by cos3x in nominator and denominator we get

∫ +⋅

= dxx

xx23

22

)1(tantansec

Let 1 + tan3x = t 3 tan2 x sec2 x dx = dt

∫ +−== Ctt

dt 131

31

2

Cx

++

−=)tan1(3

13

CAREER POINT


[CODE – A]

Q.75 The value of ∫π

π−

+

2

2

2

21sin dxx

x is :

(1) 8π

(2) 2π

(3) 4 π

(4) 4π

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Definite integration Q. No. 41]

Ans. [4]

Sol. I = ∫π

π−

+

2

2

2

21sin

xdxx

I = dxxx∫

π

π−

−+−2

2

0

2

21)0(sin

⎥⎥⎦

⎤

⎢⎢⎣

⎡−+=∫ ∫

b

a

b

a

dxxbafdxxfUse )()(

2I = dxxxx

x

x 12sin2

21sin 22

2

2

+⋅

++∫

π

π−

2I = ∫π

π−

++

2

2

2

)21(21

sin dxx xx

= ∫π

π−

2

2

2sin dxx [by using property ∫ ∫−

=a

a

a

dxxfdxxf0

)(2)( , when f(x) is even]

2I = 2 ∫π2

0

2sin dxx

= 4π

CAREER POINT


[CODE – A]

Q.76 Let g(x) = cos x2, f(x) = x , and α, β (α < β) be the roots of the quadratic equation 18x2 – 9πx + π2 = 0. Then the area (in sq. units) bounded by the curve y = (gof)(x) and the lines x = α , x = β and y = 0, is:

(1) )13(21

−

(2) )13(21

+

(3) )23(21

−

(4) )12(21

−

Ans. [1] Sol. 18x2 – 9πx + π2 = 0 18x2 – 3πx – 6πx + π2 = 0 3x(6x – π) –π(6x – π) = 0 (6x – π) (3x – π) = 0

x = 3

,6

ππ

α < β

α = 6π , β =

3π

Now y = gof(x)

g(f(x)) = g( x ) = cosx

6π

3π

y

x O

Area = ∫π

π

ππ=

3

6

3/6/)(sincos xdxx

= 21

23

−

CAREER POINT


[CODE – A]

Q.77 Let y = y(x) be the solution of the differential equation sinx dxdy + y cos x = 4x, x ∈ (0, π). If ⎟

⎠

⎞⎜⎝

⎛ π2

y = 0,

then ⎟⎠

⎞⎜⎝

⎛ π6

y is equal to:

(1) 2

394

π

(2) 2

398

π−

(3) 2

98

π−

(4) 2

94

π−

Ans. [3]

Sol. sinx dxdy + y • cosx = 4x

Divide by sinx

⇒ 0,sin4cot ≠=⋅+ x

xxxy

dxdy

Now integrating factor

I.F. = xdxxee sinlogcot

=∫ = sinx ∴ solution

y • sinx = ∫ ⋅ dxxx

x sinsin4

y • sinx = ∫ dxx4

y • sinx = 2x2 + c ….(1)

at x = 2π , y = 0

∴ c = 2

2π−

From (1)

⇒ y • sinx = 2x2 – 2

2π

at x = 6π

y • 236

221 22 π

−π

×=

y = 22

9π−

π

= 9

8 2π−

CAREER POINT


[CODE – A]

Q.78 A straight line through a fixed point (2, 3) intersects the coordinate axes at distinct points P and Q. If O is the origin and the rectangle OPRQ is completed, then the locus of R is:

(1) 3x + 2y = 6 (2) 2x + 3y = xy (3) 3x + 2y = xy (4) 3x + 2y = 6xy

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Straight line Q. No. 12]

Ans. [3] Sol.

R(h, k)

(2, 3)

P O

Q

Let R be (h, k) ∴ equation of line PQ

1=+ky

hx

It passes through (2, 3)

∴ 132=+

kh

∴ locus 132=+

yx

2y + 3x = xy Q.79 Let the orthocenter an centroid of a triangle be A(–3, 5) and B(3, 3) respectively. If C is the circumcentre

of this triangle, then the radius of the circle having line segment AC as diameter, is:

(1) 10

(2) 102

(3) 253

(4) 253

Ans. [3] Sol. orthocentre A(–3 , 5) Centroid B(3, 3)

CAREER POINT


[CODE – A]

A B

(h, k)

2 1orthocentre

(–3, 5) centroid

(3, 3)

circumcentre

By section formula

6

9323

3)3(2

==−

=−+

hh

h

2

9523

352

==+

=+

kk

k

C(6, 2) A(–3, 5)

diameter AC = 10319339 22 =+=+

Radius = 253

2103

=

Q.80 If the tangent at (1, 7) to the curve x2 = y – 6 touches the circle x2 + y2 + 16x + 12y + c = 0 then the

value of c is: (1) 195 (2) 185 (3) 85 (4) 95

Ans. [4] Sol. Curve x2 = y – 6

⇒ 2x = 0−dxdy

⇒ 2)7,1(

=⎟⎠

⎞⎜⎝

⎛dxdy

Equation of tangent at (1, 7) y – 7 = 2 (x – 1) ⇒ 2x – y + 5 = 0 ….(1) Line (1) touches the circle x2 + y2 + 16x + 12y + c = 0

∴ rp =

c−+=+

+−−− 366414

5)6()8(2

⇒ 525 = 100 – c

⇒ 5 = 100 – c ⇒ c = 95

CAREER POINT


[CODE – A]

Q.81 Tangent and normal are drawn at P(16, 16) on the parabola y2 = 16x, which intersect the axis of the parabola at A and B, respectively. If C is the centre of the circle through the points P, A and B and ∠CPB = θ, then a value of tan θ is:

(1) 21

(2) 2 (3) 3

(4) 34

Ans. [2] Sol. y2 = 16x

Slope of tangent yydx

dy 8216

==

21

)16,16(=⎟

⎠

⎞⎜⎝

⎛dxdy

Equation of tangent

y – 16 = 21 (x – 16)

⇒ 2y – 32 = x – 16 ⇒ x – 2y + 16 = 0 ….(1) Equation of normal y – 16 = –2(x – 16) 2x + y – 48 = 0 ….(2) Tangent & normal intersect the axis of parabola ∴ A(–16, 0) B(24, 0)

(4, 0) C A B(–16, 0)

(24, 0)

P (16, 16)

θ

(slope of AP) (slope of PB)

⇒ 12416016

1616016

−=⎟⎠

⎞⎜⎝

⎛−−

⎟⎠

⎞⎜⎝

⎛+−

∴ AB is diameter of circle i.e. centre C(4, 0)

CAREER POINT


[CODE – A]

34

416016mPC =

−−

=

22416016mPB −=

−−

=

tanθ = 2)2(

341

)2(34

=−⎟

⎠⎞

⎜⎝⎛+

−−

Q.82 Tangents are drawn to the hyperbola 4x2 – y2 = 36 at the points P and Q. If these tangents intersect at

the point T(0, 3) then the area (in sq. units) of ΔPTQ is :

(1) 545

(2) 354

(3) 360

(4) 536 Ans. [1]

Sol. Hyperbola 1369

22

=−yx

PQ is a chord of contact w.r.t T (0, 3) Eq. of PQ T = 0

⇒ 136

)3(9

)0(=−

yx

y = –12 y = –12 intersect the hyperbola at P & Q

(0, 3)

QP

T

O

∴ P( 12,45 −− ) Q )12,45( −

Area of ΔPTQ = 1124513011245

21

−

−− = 545

CAREER POINT


[CODE – A]

Q.83 If L1 is the line of intersection of the planes 2x – 2y + 3z – 2 =0, x – y + z + 1 = 0 and L2 is the line of intersection of the planes x + 2y – z – 3 = 0, 3x – y + 2z – 1 = 0, then the distance of the origin from the plane, containing the lines L1 and L2, is:

(1) 24

1

(2) 23

1

(3) 22

1

(4) 2

1

Ans. [2] Sol. Eq. of line L1 is

0

41

01

5 −=

−=

+ zyx

Eq. of line L2 is

70

578

375

−−

=−

−=

− zyx

Eq. of plane contain line L1 & L2 given by –7(x + 5) +7(y – 0) –8(z – 4) = 0 Where (–7, 7, –8) are Dr's of normal of plane obtained by

753

011

ˆˆˆ

−−

kji

So eq. of plane is –7x + 7y – 8z– 3 = 0

Perpendicular distance from origin is 23

1

Q.84 The length of the projection of the line segment joining the points (5, –1, 4) and (4, –1, 3) on the plane,

x + y + z = 7 is:

(1) 3

2

(2) 32

(3) 31

(4) 32

Ans. [4]

CAREER POINT


[CODE – A]

Sol.

θ (4, –1, 3)

(5, –1, 4)

1, 0, 1 → dr's of line

cosθ = 32

32101

=++

AB = 211 =+

Projection = AB sinθ =3212 −

= 32

312 =

Q.85 Let ur be a vector coplanar with the vectors ar kji ˆˆ3ˆ2 −+= and kjb ˆˆ +=r

. If ur is perpendicular to

ar and burr

⋅ = 24, then 2uv is equal to:

(1) 336 (2) 315 (3) 256 (4) 84

Ans. [1]

Sol. Let kcjbiu ˆˆˆa 111 ++=r

ur is coplanar with vectors ar & br

∴ 110132

a 111

−cb

= 0

⇒ a1(4) – b1(2) + c1(2)= 0

⇒ 4a1 – 2b1 + 2c1 = 0 ….(1) ur is perpendicular to ar ∴ arr

⋅u = 0 ⇒ 2a1 + 3b1 – c1 = 0 ….(2)

and burr

⋅ = 24 ⇒ b1 + c1 = 24 . …(3) Solving (1) (2) & (3) we get a1 = –4, b1 = 8, c1 = 16

CAREER POINT


[CODE – A]

ur = kji ˆ16ˆ8ˆ4 ++−

25664162++=ur

3362=ur

Q.86 A bag contains 4 red and 6 black balls. A ball is drawn at random from the bag, its colour is observed

and this ball along with two additional balls of the same colour are returned to the bag. If now a ball is drawn at random from the bag, then the probability that this drawn ball is red, is :

(1) 103

(2) 52

(3) 51

(4) 43

Ans. [2]

Sol. If Red ball is drawn & finally Black ball is drawn 12024

126

104

=×=

If Black ball is drawn & finally Red ball is drawn 12024

124

106

=×=

52

120242

12024

12024)( =×=+=AP

Q.87 If ∑=

=−9

19)5(

iix and ∑

=

=−9

1

2 45)5(i

ix , then the standard deviation of the 9 items x1, x2, ……, x9 is :

(1) 9 (2) 4 (3) 2 (4) 3

Students may find similar question in CP exercise sheet : [JEE Main, Chapter : Statistics, Ex.4, Q. No. 6]

Ans. [3]

Sol. 2

2 11)( ⎟⎠⎞

⎜⎝⎛ ∑−= ∑ iiar d

nd

nxV where di = derivation ; di = (xi – A)

2

991)45(

91

⎟⎠⎞

⎜⎝⎛ ×−=

= 5 – 1 4)( =xVar Stand. dev. (σ) = 2

CAREER POINT


[CODE – A]

Q.88 If sum of all the solutions of the equation 121

6cos

6coscos8 =⎟⎟

⎠

⎞⎜⎜⎝

⎛−⎟

⎠⎞

⎜⎝⎛ −

π⋅⎟

⎠⎞

⎜⎝⎛ +

π⋅ xxx in [0, π] is kπ, then

k is equal to :

(1) 32

(2) 9

13

(3) 98

(4) 9

20

Ans. [2]

Sol. 121

6cos

6coscos8 =

⎭⎬⎫

⎩⎨⎧

−⎟⎠⎞

⎜⎝⎛ −

π⋅⎟

⎠⎞

⎜⎝⎛ +

π xxx

⇒ 116

cos6

cos2cos4 =⎭⎬⎫

⎩⎨⎧

−⎟⎠⎞

⎜⎝⎛ −

π⎟⎠⎞

⎜⎝⎛ +

π xxx

⇒ 113

cos2coscos4 =⎭⎬⎫

⎩⎨⎧ −

π+xx

⇒ 1212coscos4 =

⎭⎬⎫

⎩⎨⎧ −xx

⇒ 1211cos2cos4 2 =

⎭⎬⎫

⎩⎨⎧ −−xx

⇒ 1cos6cos8 3 =− xx

⇒ 1)cos3cos4(2 3 =− xx

cos 3x = 21

cos 3x = cos 60°, cos 30°, cos 420° x = 20°, 100°, 140° 20° + 100° + 140° = kπ.

⇒ 260° = k(180°)

⇒ 9

13180260

=°°

=k

Q.89 PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the

angles of elevation of the top of the tower at P, Q and R are respectively 45°, 30° and 30°, then the height of the tower (in m) is :

(1) 100 (2) 50

(3) 3100

(4) 250

CAREER POINT


[CODE – A]

Ans. [1]

Sol.

Q M Rx k

h 200 m 200 m

P

45°

h

M P

T

30°

h

MQ

T

x

∴ PM = h tan 30° = xh

hx 3=

Now In ΔPOM (200)2 = h2 + 3h2 4h2 = 4 × (100)2 h = 100 meter Q.90 The Boolean expression ~ (p ∨ q) ∨ (~ p ∧ q) is equivalent to :

(1) ~ p (2) p (3) q (4) ~ q

Students may find same question in CP exercise sheet : [JEE Main, Chapter : Mathematical Reasoning, Ex.3, Q. No. 9]

Ans. [1] Sol. By Truth Table

TFTTFFFTTTFTTFFFFFTFTFFFFTTT

qpqpqppqpqpqp )(~)(~~~)(~ ∧∨∨∧∨∨

JEE Main Exam 2018 - dcmep4q5dgnih.cloudfront.net · JEE Main Exam 2018 Paper & Solution | 8th...

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