Solution to IIT JEE 2018 (Advanced) : Paper -...

0518/IITEQ18/Paper2/QP&Soln/Pg.1

Solution to IIT JEE 2018 (Advanced) : Paper - II

PART I – PHYSICS SECTION 1 (Maximum Marks:24)

This section contains SIX (06) questions.

Each question has FOUR options for correct answer(s). ONE OR MORE THAN

ONE of these four option(s) is (are) correct option(s).

For each question, choose the correct option(s) to answer the question.

Answer to each question will be evaluated according to the following marking

scheme:

Full Marks : +4 If only (all) the correct option(s) is (are) chosen.

Partial Marks : +3 If all the four options are correct but ONLY three options are chosen.

Partial Marks : +2 If three or more options are correct but ONLY two options are chosen,

both of which are correct options.

Partial Marks : +1 If two or more options are correct but ONLY one option is chosen and

it is a correct option.

Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered).

Negative Marks : —2 In all other cases.

For Example: If first, third and fourth are the ONLY three correct options for a question

with second option being an incorrect option; selecting only all the three correct options

will result in +4 marks.

Selecting only two of the three correct options (e.g. the first and fourth options), without

selecting any incorrect option (second option in this case), will result in +2 marks.

Selecting only one of the three correct options (either first or third or fourth option) ,

without selecting any incorrect option (second option in this case), will result in +1

marks. Selecting any incorrect option(s) (second option in this case), with or without

selection of any correct option(s) will result in -2 marks.

1. A particle of mass m is initially at rest at the origin. It is subjected to a force and starts moving

along the x-axis. Its kinetic energy K changes with time as dK/dt = t, where is a positive

constant of appropriate dimensions. Which of the following statements is (are) true?

(A) The force applied on the particle is constant

(B) The speed of the particle is proportional to time

(C) The distance of the particle from the origin increases linearly with time

(D) The force is conservative

1. (A), (B), (D)

dK

tdt

K = 21mv

2

dK 1 dv

m 2v tdt 2 dt

dv t

vdt m

vdv t dtm

2 2v t

2 m 2

(2) Vidyalankar : IIT JEE 2018 Advanced : Question Paper & Solution


v tm

(proportional to time)

a = dv

dt m

F ma m (constant)

2. Consider a thin square plate floating on a viscous liquid in a large tank. The height h of

the liquid in the tank is much less than the width of the tank. The floating plate is pulled

horizontally with a constant velocity u0. Which of the following statements is (are) true?

(A) The resistive force of liquid on the plate is inversely proportional to h

(B) The resistive force of liquid on the plate is independent of the area of the plate

(C) The tangential (shear) stress on the floor of the tank increases with u0

(D) The tangential (shear) stress on the plate varies linearly with the viscosity of the liquid

2. (A), (C), (D)

Viscous force F = Adv

dh

0uF A

h

F 1

h

Shear stress = 0

Fu

A

F

A

3. An infinitely long thin non-conducting wire is parallel to the

z-axis and carries a uniform line charge density . It pierces a thin

non-conducting spherical shell of radius R in such a way that the

arc PQ subtends an angle 120° at the centre O of the spherical

shell, as shown in the figure. The permittivity of free space is 0 .

Which of the following statements is (are) true ?

(A) The electric flux through the shell is 03R /

(B) The z-component of the electric field is zero at

all the points on the surface of the shell

(C) The electric flux through the shell is 02R /

(D) The electric field is normal to the surface of the shell at all points

3. (A), (B)

Gauss law

= 0

QE dA

where Q = L

L = PQ = 2R sin 60

L = R 3

0

R 3

Electric field lines are radially outwards, perpendicular to length of wire. Hence

component of E.F. is zero along z-axis.

u0 F

h

120

Q

P

Z

E

E

IIT JEE 2018 Advanced : Question Paper & Solution (Paper – II) (3)


4. A wire is bent in the shape of a right angled triangle and is placed in front of a concave

mirror of focal length f , as shown in the figure. Which of the figures shown in the four

options qualitatively represent(s) the shape of the image of the bent wire? (These figures

are not to scale.)

(A) (B)

(C) (D)

4. (D)

If |u| = | f |

v

If | u | < | f |

Image is virtual, erect and magnified.

Let a point ‘P’ at distance ‘x’ from f

u = (f x)

1 1 1

V f x f

1 1 1 x

V f x f f (f x)

f (f x)

Vx

Magnification M = I

0

h v f

h u x

where h0 = x Ih f (independent of x)

5. In a radioactive decay chain, 23290 Th nucleus decays to 212

82 Pb nucleus. Let N and N be

the number of and particles, respectively, emitted in this decay process. Which of the

following statements is (are) true?

(A) N = 5 (B) N = 6 (C) N = 2 (D) N = 4

5. (A), (C)

Z = 90 82 = 8

A = 232 212 = 20

Number of -particle N = 20

54

Z = 5 2 = 10

Number of -particles N = 10 8 = 2

90Th232

82Pb212

f

2

f

45

x P



6. In an experiment to measure the speed of sound by a resonating air column, a tuning fork

of frequency 500 Hz is used. The length of the air column is varied by changing the level

of water in the resonance tube. Two successive resonances are heard at air columns of

length 50.7 cm and 83.9 cm. Which of the following statements is (are) true?

(A) The speed of sound determined from this experiment is 332 m s1

(B) The end correction in this experiment is 0.9 cm

(C) The wavelength of the sound wave is 66.4 cm

(D) The resonance at 50.7 cm corresponds to the fundamental harmonic

6. (A), (B), (C)

For 1st resonance

0

1

3Vf

4 L e

……….. (1)

For 2nd

resonance

0

2

5Vf

4 L e

……….. (2)

1 2

3V 5V

4 L e 4 L e

2 13 L e 5 L e

2e = 3L2 5L1

e = 3 83.9 5 50.7 251.7 253.5 1.8

0.92 2 2

0

1

3Vf

4 L e

V = 500 4 49.8

20 16.6 332m / s3 100

= 0

V 332100 66.4cm

f 500

SECTION II (Maximum Marks:24)

This section contains EIGHT (08) questions. The answer to each question is a

NUMERICAL VALUE.

For each question, enter the correct numerical value (in decimal notation,

truncated/rounded-off to the second decimal place; e.g. 6.25, 7.00, -0.33, -.30,

30.27, -127.30) using the mouse and the onscreen virtual numeric keypad in the

place designated to enter the answer.

Answer to each question will be evaluated according to the following marking

scheme:

Full Marks : +3 If ONLY the correct numerical value is entered as answer.

Zero Marks : 0 In all other cases.

7. A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass

m = 0.4 kg is at rest on this surface. An impulse of 1.0 N s is applied to the block at time

t = 0 so that it starts moving along the x-axis with a velocity v(t) = t /e

0v

, where v0 is a

constant and = 4 s . The displacement of the block, in metres, at t = is ________.

Take e1

= 0.37.

7. [6.3]

t /

0v v e F = 1 N.S.



t /

0

dSv v e

dt

S

t/

0

0 0

dS v e dt

S = t /

0 0v e

S = 0

1v 1

e

Impulse = P = mv

v = 1

2.50.4

s = 4 2.5 1 0.37 10 0.63 6.3m/ s

8. A ball is projected from the ground at an angle of 45° with the horizontal surface. It

reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground

for the first time, it loses half of its kinetic energy. Immediately after the bounce, the

velocity of the ball makes an angle of 30° with the horizontal surface. The maximum

height it reaches after the bounce, in metres, is ________ .

8. [30 m]

H = 2 2u sin 45

2g

h = 2 2v sin 30

2g

22

22

1v

h 1 v 141H 2 u 4

u2

9. A particle, of mass 103

kg and charge 1.0 C, is initially at rest. At time t = 0, the particle

comes under the influence of an electric field E (t) = E0 sin t i , where E0 = 1.0 N C1

and = 103 rad s

1 . Consider the effect of only the electrical force on the particle. Then

the maximum speed, in m s1

, attained by the particle at subsequent times is ______.

9. [2]

Velocity is maximum when acceleration is zero.

F = 0

E = E0 sin t = 0

sin t = 0

t = 0 or

t

E = E0 sin t

F = qE = qE0 sin t

a = 0qEqEsin t

m m

a = 0qEdvsin t

dt m

v t

0

0 0

qEdv sin t

m

45

30

v

u



v = t

0 0

0

qE qEcos t1 cos t

m m

replacing t =

v = 02qE

m

max 3 3

2 1 1v 2

10 10

10. A moving coil galvanometer has 50 turns and each turn has an area 2 104

m2. The

magnetic field produced by the magnet inside the galvanometer is 0.02 T. The torsional

constant of the suspension wire is 104

N m rad1

. When a current flows through the

galvanometer, a full scale deflection occurs if the coil rotates by 0.2 rad. The resistance of

the coil of the galvanometer is 50 . This galvanometer is to be converted into an

ammeter capable of measuring current in the range 0 1.0 A. For this purpose, a shunt

resistance is to be added in parallel to the galvanometer. The value of this shunt

resistance, in ohms, is __________.

10. [5.56]

Number of turns = 50

Area = 2 104

m2

B = 0.02 T

Torsional constant C = 104

At full scale deflection = 0.2 rad

Resistance R = 50

Torque = C = niAB

Full scale current i = 4

4

C 0.2 10

nAB 50 2 10 0.02

i 0.1amp

0.9 S = 0.1 R

S = R 50

9 9

S 5.56

11. A steel wire of diameter 0.5 mm and Young's modulus 2 1011

N m2

carries a load of

mass M. The length of the wire with the load is 1.0 m. A vernier scale with 10 divisions is

attached to the end of this wire. Next to the steel wire is a reference wire to which a main

scale, of least count 1.0 mm, is attached. The 10 divisions of the vernier scale correspond

to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero

of main scale. If the load on the steel wire is increased by 1.2 kg, the vernier scale

division which coincides with a main scale division is _______ .

Take g = 10 m s2

and = 3.2.

11. [3]

Diameter d = 0.5 mm

Y = 2 1011

N/m2

Strain = Stress

Y

L F

L AY

G

S 0.9 A

0.1 A 1.0 A



L =

2

3

11

FL 1.2 10 1

AY 0.5 103.2 2 10

4

L = 5

5

1230 10 0.3mm

1.6 0.25 10

L.C. of Vernier scale = 0.1 mm

Number of division which coincide with main scale = 3.

12. One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume

becomes eight times its initial value. If the initial temperature of the gas is 100 K and the

universal gas constant R = 8.0 Jmol1

K1

, the decrease in its internal energy, in Joule,

is _________ .

12. [900 J]

Adiabatic Expansion

PV = constant

T.V1

= constant

T2 = T1

1

1

2

V 5where

V 3

T2 = T1

2/3

1T125K

8 4

U = f 3

nR T 1 8 75 900J2 2

13. In a photoelectric experiment a parallel beam of monochromatic light with power of

200 W is incident on a perfectly absorbing cathode of work function 6.25 eV. The

frequency of light is just above the threshold frequency so that the photoelectrons are

emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency

is 100%. A potential difference of 500 V is applied between the cathode and the anode.

All the emitted electrons are incident normally on the anode and are absorbed. The anode

experiences a force F = n 104

N due to the impact of the electrons. The value of

n is _______ .

Mass of the electron me = 9 1031

kg and 1.0 eV = 1.6 1019

J .

13. [24]

P = 200 w

KEmax = h = 0

Energy of 1 photon E1 = h = = 6.25 eV

Number of photons incident per unit time

n = 19

19

1

P 20020 10

E 6.25 1.6 10

= 2 10

20

Change in momentum of 1 photon

P1 = 2mk 2me( v)

Force F = n(P1)

F = 2 1020

31 192 9 10 1.6 10 500

= 2 1020

489 1.6 10

= 2 12 1020

1024

= 24 104



14. Consider a hydrogen-like ionized atom with atomic number Z with a single electron. In

the emission spectrum of this atom, the photon emitted in the n = 2 to n = 1 transition has

energy 74.8 eV higher than the photon emitted in the n = 3 to n = 2 transition. The

ionization energy of the hydrogen atom is 13.6 eV. The value of Z is _______ .

14. [3]

E = 13.6 z2

2 2

1 2

1 1

n n

E1 = 13.6z2 21 3

1 13.6z4 4

E2 = 13.6z2 21 1 5

13.6z4 9 36

E1 E2 = 13.6 z2

3 5

4 36

13.6z2

2274.8

36

z2 =

74.8 189

13.6 11

z = 3

SECTION III (Maximum Marks:12) This section contains FOUR (04) questions.

Each question has TWO (02) matching lists: LIST-I and LIST-II.

FOUR options are given representing matching of elements from LIST-I and

LIST-II. ONLY ONE of these four options corresponds to a correct matching.

For each question, choose the option corresponding to the correct matching.

For each question, marks will be awarded according to the following marking

scheme :

Full Marks : +3 If ONLY the option corresponding to the correct matching is chosen.


Negative Marks : 1 In all other cases.

15. The electric field E is measured at a point P(0,0, d) generated due to various charge

distributions and the dependence of E on d is found to be different for different charge

distributions. List-I contains different relations between E and d. List-II describes

different electric charge distributions, along with their locations. Match the functions in

List-I with the related charge distributions in List-II.

LIST-I LIST-II

P. E is independent of d 1. A point charge Q at the origin

Q. E

1

d

2. A small dipole with point charges Q at (0, 0,l) and

Q at (0, 0, l). Take 2l << d.

R. E

2

1

d

3. An infinite line charge coincident with the x-axis,

with uniform linear charge density

S. E

3

1

d

4. Two infinite wires carrying uniform linear charge

density parallel to the x- axis. The one along (y = 0,

z = l) has a charge density + and the one along

(y = 0, z = l) has a charge density . Take 2l << d

5. Infinite plane charge coincident with the xy-plane

with uniform surface charge density

(A) P 5; Q 3, 4; R 1; S 2 (B) P 5; Q 3; R1, 4; S 2

(C) P 5; Q 3; R 1, 2; S 4 (D) P 4; Q 2, 3; R 1; S 5



15. (B)

(P) (5), (Q) (3), (R) (1), (4); (S) (2)

(1) E.F. due to a point charge at origin.

2 2

kq 1E E

d d

(2) E.F. at any point on axis of dipole

E = 3 3

2KP 4KQL

d d

3

1E

d

(3) E.F. due to an infinite long charge

E = 2K

d

1

Ed

(4) E.F. due to two infinite long wires

1 2E E E

2 22 2

2K 2K 2k (2L) 4K LE

d L d L d Ld L

If d >> L E = 2

4K L

d

2

1E

d

(5) E.F. due to infinite plane charge

0

E

(independent of d)

16. A planet of mass M, has two natural satellites with masses m1 and m2. The radii of their

circular orbits are R1 and R2 respectively. Ignore the gravitational force between the

satellites. Define v1, L1, K1 and T1 to be, respectively, the orbital speed, angular

momentum, kinetic energy and time period of revolution of satellite 1; and v2, L2, K2 and

T2 to be the corresponding quantities of satellite 2. Given m1 / m2 = 2 and R1/R2 = 1/4,

match the ratios in List-I to the numbers in List-II.

LIST-I LIST-II

P. 1

2

v

v

1. 1

8

Q. 1

2

L

L

2. 1

R. 1

2

K

K

3. 2

S. 1

2

T

T

4. 8

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

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

16. (B)

(P) (3), (Q) (2), (R) (4), (S) (1)

1

2

m2

m



1

2

R 1

R 4

(P) For orbital speed

F = 2

2

mv GMm

R R

v = GM

R

1 2

2 1

V R4 2

V R

(Q) Angular momentum

L = mVR

1 1 1 1

2 2 2 2

L m V R 12 2 1

L m V R 4

(R) Kinetic energy

K = 21mv

2

2

21 1 1

2 2 2

K m V2 (2) 8

K m V

(S) Time period of revolution

T = 2 R

V

1 1 2

2 2 1

T R V 1 1 1

T R V 4 2 8

17. One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown

schematically in the PV-diagram below. Among these four processes, one is isobaric, one

is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in

List-I with the corresponding statements in List-II.

LIST-I LIST-II

P. In process I 1. Work done by the gas is zero

Q. In process II 2. Temperature of the gas remains unchanged

R. In process III 3. No heat is exchanged between the gas and its

surroundings

S. In process IV 4. Work done by the gas is 6P0V0

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

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

M m1 m2



17. (C)

(P) (3), (Q) (4), (R) (1), (S) (2)

(P) Process I is adiabatic

Hence No heat is exchanged between gas and surrounding.

(Q) Process II is isobaric.

w = PV = 3P0 (3V0 V0) = 6P0V0

(R) Process is isochoric

w = 0

(S) Process is isothermal

T = constant

18. In the List-I below, four different paths of a particle are given as functions of time. In

these functions, and are positive constants of appropriate dimensions and . In

each case, the force acting on the particle is either zero or conservative. In List-II, five

physical quantities of the particle are mentioned: p is the linear momentum, L is the

angular momentum about the origin, K is the kinetic energy, U is the potential energy and

E is the total energy. Match each path in List-I with those quantities in List-II, which are

conserved for that path.

LIST-I LIST-II

P. r(t) = ˆˆt t t j 1. p

Q. r(t) = cos t i + sin t j 2. L

R. r(t) = (cos t i + sin t j ) 3. K

S. r(t) = 2ˆ ˆt i t j

2

4. U

5. E

(A) P 1, 2, 3, 4, 5; Q 2, 5; R 2, 3, 4, 5; S 5

(B) P 1, 2, 3, 4, 5; Q 3, 5; R 2, 3, 4, 5; S 2, 5

(C) P 2, 3, 4; Q 5; R 1, 2, 4; S 2, 5

(D) P 1, 2, 3, 5; Q 2, 5; R 2, 3, 4, 5; S 2, 5

18. (A)

(P) (1), (2), (3), (4), (5); (Q) (2), (5); (R) (2), (3), (4), (5); (S) (5)

(P) ˆ ˆr t i t j

dr ˆ ˆv i jdt

(constant)

a 0

F 0

Angular momentum L m r v

ˆ ˆL m t k t k 0(conserved)

I adiabatic

II isobaric

III isochoric

IV isothermal



(Q) ˆ ˆr a cos t i sin t j

Particle is moving on an elliptical path.

Angular momentum about origin and total energy remains conserved.

(R) ˆ ˆr a cos t i sin t j

Particle moves on a circular path

radius = a

angular velocity =

speed v = a

but v constant

hence P constant

Angular momentum 2 ˆL m a kisconstant

Kinetic energy E = 21mv

2 (constant)

P.E. U = constant

E = constant

(S) 2tˆ ˆr t i j

2

dr ˆ ˆv i t jdt

(depends on t)

hence P mv also depends on t.

dv ˆa jdt

ˆF ma m j (constant)

2

2 tˆ ˆL m r v m t k ( k)2

2m t ˆL k

2

(depends on t)

Only total energy remains conserved.



PART II-CHEMISTRY

SECTION 1 (Maximum Marks: 24)


Each question has FOUR options for correct answer(s). ONE OR MORE THAN ONE

of these four option(s) is (are) correct option(s).


Answer to each question will be evaluated according to the following marking scheme:


Partial Marks : +3 If all the four options are correct but ONLY three

options are chosen.

Partial Marks : +2 If three or more options are correct but ONLY two options are

chosen, both of which are correct options.

Partial Marks : +1 If two or more options are correct but ONLY one option is chosen

and it is a correct option.



For Example: If first, third and fourth are the ONLY three correct options for a question

with second option being an incorrect option; selecting only all the three correct options

will result in +4 marks. Selecting only two of the three correct options (e.g. the first and

fourth options), without selecting any incorrect option (second option in this case), will

result in +2 marks. Selecting only one of the three correct options (either first or third or

fourth option) ,without selecting any incorrect option (second option in this case), will

result in +1 marks. Selecting any incorrect option(s) (second option in this case), with or

without selection of any correct option(s) will result in ‐2 marks.

1. The correct option(s) regarding the complex [Co(en)(NH3)3(H2O)]3+

(en = H2NCH2CH2NH2) is (are)

(A) It has two geometrical isomers

(B) It will have three geometrical isomers if bidentate ‘en’ is replaced by two cyanide

ligands

(C) It is paramagnetic

(D) It absorbs light at longer wavelength as compared to [Co(en)(NH3)4]3+

1. (A, B, D)

Complex [Co(en) (NH3)3 (H2O)]+3

has two geometrical isomers.

[Co(NH3)3 (H2O) (CN)2]+ has three geometrical isomers.

Co

NH3

NH3

NH3

H2O

Co

NH3

H2O

NH3

NH3

(Fac) (Mer)

en en



Co

NH3

NH3

NH3

H2O

NC

CN

Co

NH3

NH3

NH3

NC

OH2

NC

Co

NH3

H2O

NH3

NC

NH3

CN

(Fac) (Fac) (Mer)

[Co(en) (NH3)3 (H2O)]+3

is diamagnetic

Co = [Ar]18 4s2 3d

7

Co+3

= [Ar]18 3d6

en is strong field ligand.

_ _ t2g

[Co(en) (NH3)3 (H2O)]+3

[Co (en) (NH3)4]+4

0 = hc

0 1

[Co (en) (NH3)3 (H2O)]+3

absorbs light at longer wavelength as compared to [Co(en)

(NH3)4]+3

2. The correct option(s) to distinguish nitrate salts of Mn2+

and Cu2+

taken separately is (are)

(A) Mn2+

shows the characteristic green colour in the flame test

(B) Only Cu2+

shows the formation of precipitate by passing H2S in acidic medium

(C) Only Mn2+

shows the formation of precipitate by passing H2S in faintly basic medium

(D) Cu2+/

Cu has higher reduction potential than Mn2+

/Mn (measured under similar

conditions)

2. (B), (D)

Both Cu+2

and Mn+2

gives green colour in the flame test.

Ksp (CuS) < Ksp (MnS)

Only Cu+2

shows the formation of precipitate by passing H2S in acidic medium.

2 20 0

Cu /Cu Mn /CuE E

Less electro positive metals have higher reduction potential.

3. Aniline reacts with mixed acid (conc.HNO3 and conc.H2SO4) at 288K to give

P(51 %),

Q (47%) and R (2%). The major product(s) of the following reaction sequence is (are)

1) Ac2O, pyridine 1) Sn/HCl

2) Br2, CH3CO2H 2)Br2/H2O (excess)

R S major product(s)

3) H3O+

3) NaNO2, HCl/273278 K

4) NaNO2, HCl/273 278 K 4) H3PO2

5) EtOH,

eg

0

0



(A) (B)

(C) (D)

3. (D)

NH2

NO2

2Ac O, pyridine

NH

NO2

C CH3

O

2

3

Br

CH COOH

NH

NO2

AC

Br

3H O

NH2

NO2

Br

2NaNO , HCl

273 278 k

Et OH

NO2

Br

Sn/HCl

NH2

Br

2 2Br /H O

excess

NH2

Br

Br

BrBr

2NaNO , HCl

273 278

3 2H PO

Br

Br

BrBr

(R)

4. The Fischer presentation of D-glucose is given below.

D-glucose

The correct structure(s) of β-L-glucopyranose is (are)

(A) (B) (C) (D)



4. (D)

CH2OH

CHO

H OH

OH H

H OH

H OH

CH2OH

CHO

HOH

OHH

HOH

HOH

D-Glucose L-Glucose

C

HCH2OH

H

OH

H

H

O C

HOH

H

OH

OH

H

H

CH2OH

H

O

O

H

OH

OH

H

H

CH2OH

H

H

OH

HO

HO

HO

L-Glucose

HO ....

HO

-L-Glucopyranose

5. For a first order reaction A(g) → 2B(g) + C(g) at constant volume and 300 K, the total

pressure at the beginning (t = 0) and at time t are P0 and Pt, respectively. Initially, only A

is present with concentration [A]0, and t1/3 is the time required for the partial pressure of

A to reach 1/3rd

of its initial value. The correct option(s) is (are)

(Assume that all these gases behave as ideal gases)

(A) (B)

(C) (D)



5. (A, D)

A(g) 2 B(g) + C(g)

oe

o t

2P1k n

t 3P P

Kt = 0 o tn(2P ) n(3P P )

o t on(3P P ) n(2P ) kt

At t = t1/2

1/3

n3t

k

t1/3 is independent of o[A]

For Ist

order Rxn, rate constant is independent of initial concentration of reaction.

6. For a reaction, A ⇌ P, the plots of [A] and [P] with time at temperatures T1 and T2 are

given below.

If T2 > T1, the correct statement(s) is (are)

(Assume ΔHƟ and ΔS

Ɵ are independent of temperature and ratio of lnK at T1 to lnK at

T2 is greater than T2/T1. Here H, S, G and K are enthalpy, entropy, Gibbs energy and

equillibrium constant, respectively.)

(A) ΔHƟ < 0, ΔS

Ɵ < 0 (B) ΔG

Ɵ < 0, ΔH

Ɵ > 0

(C) ΔGƟ < 0, ΔS

Ɵ < 0 (D) ΔG

Ɵ < 0, ΔS

Ɵ > 0

Slope = k

t

o tn(3P P )

R

ate

const

ant

o[A]

o[A]

t1/3



6. (A), (C)

From the graph (1) and (2), we get H = ve i.e. exothermic.

Given : 1 2

2 1

n K T

nK T

1 2

1

2

S h 1

R R T T

TS h 1

R R T

2 1

S(T T ) 0

R

S 0



NUMERICAL VALUE.


truncated/rounded‐off to the second decimal place; e.g. 6.25, 7.00, ‐0.33, ‐.30, 30.27,

‐127.30) using the mouse and the onscreen

virtual numeric keypad in the place designated to enter the answer.




7. The total number of compounds having at least one bridging oxo group among the

molecules given below is ____.

N2O3, N2O5, P4O6, P4O7, H4P2O5, H5P3O10, H2S2O3, H2S2O5

7. [5.00 or 6.00]

2 3N O : NO

N OO N N OO

O

Symmetrical

or

Unsymmetrical

2 5N O : N O NO O

O

4 6P O :O

OP

OPO

P

O

P

O

7 7P O :O

O

P

OP

O

P

O

P

OO



4 2H P : P O PO OH H

H H

O O

2 3 10H P : P O PO OH P

OH OH

O H

OH

O O O

2 2 3H S : S O HOH

O

S

2 2 5H S : S S OHOH

O

O

O

8. Galena (an ore) is partially oxidized by passing air through it at high temperature. After

some time, the passage of air is stopped, but the heating is continued in a closed furnace

such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of

O2 consumed is ____.

(Atomic weights in g mol1

: O = 16, S = 32, Pb = 207)

8. [6.47]

PbS + O2 Pb + SO2

2 bO Pn n [Equal moles]

b b2 2

2 b

p po o

O p

w ww w

M M 32 207

Let, 2ow = 1 kg

bPw =

207

32 = 6.47 kg

9. To measure the quantity of MnCl2 dissolved in an aqueous solution, it was completely

converted to KMnO4 using the reaction,

MnCl2 + K2S2O8 + H2O KMnO4 + H2SO4 + HCl (equation not balanced).

Few drops of concentrated HCl were added to this solution and gently warmed. Further,

oxalic acid (225 mg) was added in portions till the colour of the permanganate ion

disappeared. The quantity of MnCl2 (in mg) present in the initial solution is ____.

(Atomic weights in g mol−1

: Mn = 55, Cl = 35.5)

9. [126.00]

2 2 2 8 2 4 2 4MnCl K S O H O KMnO H SO HCl

(ON) 5

(ON) 2

24 2 2 4 2KMnO H C O Mn CO

(ON) 5

(ON) 2



No. of equivalents of MnCl2 = No. of eq. of KMnO4 = No. of eq. of H2C2O4

5 2MnCln = 2

2 2 4H C On

5 w

126 = 2

225

90

w = 2 225 126

90 5

= 126 mg

10. For the given compound X, the total number of optically active stereoisomers is ____.

10. [7.00]

Total number of stereomers of compound X = 8

X has 7 optically active stereisomers

X has 1 optically inactive

11. In the following reaction sequence, the amount of D (in g) formed from 10 moles of

acetophenone is ____.

(Atomic weights in g mol1

: H = 1, C = 12, N = 14, O = 16, Br = 80. The yield (%)

corresponding to the product in each step is given in the parenthesis)

11. [495.00]

NaOBr

H3O+

COOH

NH3, Br/KOH

CONH2

NH2

Br2

CH3COOH

NH2

BrBr

Br

A (60%)

6 moles

B (50%)

3 moles

D

C (50%)

1.5 moles

(3 eq)

D (100%)

1.5 moles

10 moles

No. of moles of D = 1.5

(C6H4NBr3) = 1.5 330 = 495 gm



12. The surface of copper gets tarnished by the formation of copper oxide. N2 gas was passed

to prevent the oxide formation during heating of copper at 1250 K. However, the N2 gas

contains 1 mole % of water vapour as impurity. The water vapour oxidises copper as per

the reaction given below:

2Cu(s) + H2O(g) Cu2O(s) + H2(g)

2HP is the minimum partial pressure of H2 (in bar) needed to prevent the oxidation at

1250 K. The value of ln2H(p ) is ____.

(Given: total pressure = 1 bar, R (universal gas constant) = 8 JK−1

mol−1

, ln(10) = 2.3.

Cu(s) and Cu2O(s) are mutually immiscible.

At 1250 K: 2Cu(s) + ½ O2(g) Cu2O(s); ΔGƟ = − 78,000 J mol

−1

H2(g) + ½ O2(g) H2O(g); ΔGƟ = − 1,78,000 J mol

−1; G is the Gibbs energy)

12. [14.60]

(s) 2 (g) 2 (s) 2 (g)2Cu H O Cu O

At 1250 K (s) 2 (g) 2 (s)

12Cu O Cu O G 78000

2 J/mol = 78 kJ/mol

2 (g) 2 (g) 2 (g)

1H O H O

2 G° = 178000 J/mol = 178 kJ/mol

(s) 2 (g) 2 (s) 2 (g)2Cu H Cu O G 100 kJ/mol

G = G° + RT n K

0 = 100 + RT 2

2

(pH )n

(pH O)

100 = 2

2

pH81250 n

1000 pH O

Given : pH2O = 1 % of total pressure (1 bar)

= 1

1100

100 = 22

pH81250 n

1000 10

10 = 22n pH n10

10 = 2n pH 2 n10

10 = 2n pH 2(2.3)

10 = 2n pH 4.6

2n pH 14.60 bar

13. Consider the following reversible reaction,

A(g) + B(g) AB(g).

The activation energy of the backward reaction exceeds that of the forward reaction by

2RT (in J mol−1

). If the pre-exponential factor of the forward reaction is 4 times that of the

reverse reaction, the absolute value of ΔGƟ (in J mol

−1) for the reaction at 300 K is ____.

(Given; ln(2) = 0.7, RT = 2500 J mol−1

at 300 K and G is the Gibbs energy)

13. [8500.00]

(g) (g) (g)A B AB

Eb Ea = 2RT J/mol ; Af = 4Ab

RT = 2500 J/mol ; T = 300 K



G° = RT n K

a

b

E /RTf f

E /RTb b

K A eK

K A e

=

b aE E

RTF

b

Ae

A

= 4

2RT

2RTe 4e

G° = RT 2n 4e

= 2500 2n 4e

= 2500 2n 4 n e

= 2500 (2 n 2 + 2)

= 2500 2(0.7) 2

= 2500 3.4

= 8500

Absolute value of G° = 8500.00

14. Consider an electrochemical cell: A(s) | An+

(aq, 2 M) || B2n+

(aq, 1 M) | B(s). The value of

ΔHƟ for the cell reaction is twice that of ΔG

Ɵ at 300 K. If the emf of the cell is zero, the

ΔSƟ (in J K

−1 mol

−1) of the cell reaction per mole of B formed at 300 K is ____.

(Given: n (2) = 0.7, R(universal gas constant) = 8.3 J K−1

mol−1

. H, S and G are enthalpy,

entropy and Gibbs energy, respectively.)

14. [11.62]

n 2n+

s aq, 2M aq,1M sA A B B

Oxidation

nA A ne …. (1)

Reduction

B2n

+ + 2ne B …(2)

multiply equation (1) by (2) and add both

2n n2A B B

Given E.M.F. of the cell is zero; the system is in equilibrium , G = 0

G = RT n k n 2

2n

[A ]G RT n k

[B ]

2G RT n(2)

G H S

G 2 G S (Given H = 2G)

G = TS

G

ST

= RT n4

T

= 2R n 2

= 8.3 0.7 2

= 11.62 J k1

mol1




This section contains FOUR (04) questions.

Each question has TWO (02) matching lists: LIST‐I and LIST‐II.

FOUR options are given representing matching of elements from LIST‐I and LIST‐II.

ONLY ONE of these four options corresponds to a correct matching.


For each question, marks will be awarded according to the following marking scheme:

Full Marks : +3 If ONLY the option corresponding to the correct matching is chosen.


Negative Marks : +1 In all other cases.

15. Match each set of hybrid orbitals from LIST–I with complex(es) given in LIST–II.

LIST–I LIST–II

P dsp2 1. [FeF6]

4−

Q sp3 2. [Ti(H2O)3Cl3]

R sp3d

2 3. [Cr(NH3)6]

3+

S d2sp

3 4. [FeCl4]

2−

5. Ni(CO)4

6. [Ni(CN)4]2−

The correct option is

(A) P → 5; Q → 4, 6; R → 2, 3; S → 1

(B) P → 5, 6; Q → 4; R → 3; S → 1, 2

(C) P → 6; Q → 4, 5; R → 1; S → 2, 3

(D) P → 4, 6; Q → 5, 6; R → 1,2; S → 3

15. (C)

P : dsp2

(6)

2

4[Ni (CN) ]

Ni 4S2 3d

8

Ni2+

CN is a strong ligand Pairing takes place.

dsp2 hybridization square planar

Q : sp3 (4), (5)

(4) [FeCl4]2

Fe 4s2 3d

6

2Fe

Cl Weak ligand No pairing takes place.

sp3 hybridization Tetrahedral structure.

4s 4p 3d

4s 4p 3d



(5) Ni(CO)4

Ni 4s2 3d

8

CO strong ligand pairing from electrons of s orbital.

sp3 hybridization Tetrahedral

R ; (1)

(1) [Fe F6]4

Fe 4s2 3d

6 ; Fe

2+ 3d

6

F weak field ligand No pairing

sp3d

2 hybridization octahedral

S; d2sp

3 2, 3

(2) [Ti (H2O)3Cl3]

Ti 4s2 3d

2

Ti3+

3d1

Weak ligands H2O, Cl

d2sp

3 octahedral

(3) 3

3 3 6Cr(NH )

Cr : 4s1 3d

5 ; Cr

3+ 4s

0 3d

3

d2sp

3 octahedral

soln. (C)

16. The desired product X can be prepared by reacting the major product of the reactions in

LIST-I with one or more appropriate reagents in LIST-II.

(given, order of migratory aptitude: aryl > alkyl > hydrogen)

4s 3p 3d

4s 4p 3d

4s 4p 3d 4d

4s 4p 3d



LIST I LIST II

P.

1. I2, NaOH

Q.

2. [Ag(NH3)2]OH

R.

3. Fehling solution

S.

4. HCHO, NaOH

5. NaOBr

The correct option is

(A) P → 1; Q → 2,3; R → 1,4; S → 2,4

(B) P → 1,5; Q → 3,4; R → 4,5; S → 3

(C) P → 1,5; Q → 3,4; R → 5; S → 2,4

(D) P → 1,5; Q → 2,3; R → 1,5; S → 2,3

16. (D)

OH

Ph

Ph

Me

OHMe

2 4H SO 2I , NaOHX

NaOBr

X

(P) :

Iodo form reaction

P (1), (5)

NH2

Ph

Ph

H

OHMe

2HNO 2N

Ph

Ph

H

OHMe

X

(Q) :

(2) Ag (NH3)2 OH

(3) Fehlings solution

Gives Tollens & Fehlings Test

Q (2), (3)



(R)OH

Ph

OHMe

Me

Ph H2SO

4 Me

CH3

O

Ph

Ph

NaOH

X

Me

CH3

O

Ph

Ph

[Ag(NH3)

2]OH

FehlingsX

+

(1) I2 ,

(2) NaOBr

+ AgNO3

(1)

(2)

(S)

Haloform reaction

P 1, 5 ; Q 2, 3 ; R 1, 5 ; S 2, 3

17. LIST-I contains reactions and LIST-II contains major products.

LIST I LIST II

P.

1.

Q.

2.

R.

3.

S.

4.

5.

Match each reaction in LIST-I with one or more products in LIST-II and choose the

correct option.

(A) P → 1,5; Q → 2; R → 3; S → 4 (B) P → 1,4; Q → 2; R → 4; S → 3

(C) P → 1,4; Q → 1,2; R → 3,4; S → 4 (D) P → 4,5; Q → 4; R → 4; S → 3,4

17. (P) (1), (4)

ONa BrElimination

OH+ +

(Q)

OMeSubstitution

O

Me

HBr

Br+ H++

+ MeOH

(2)

(R)

Brbase

NaOMe

(4)

R (4)



(S)

ONaSubstitution

OMeMeBr

-NaBr+

(3)

Soln. (B)

P 1, 4 ; Q 2 ; R 4 ; S 3

18. Dilution processes of different aqueous solutions, with water, are given in LIST-I. The

effects of dilution of the solutions on [H+] are given in LIST-II.

(Note: Degree of dissociation () of weak acid and weak base is << 1; degree of

hydrolysis of salt <<1; [H+] represents the concentration of H

+ ions)

LIST I LIST II

P. (10 mL of 0.1 M NaOH + 20 mL of

0.1M acetic acid) diluted to 60

mL

1. the value of [H+] does not change

on dilution

Q. (20 mL of 0.1 M NaOH + 20 mL of

0.1 M acetic acid) diluted to 80 mL

2. the value of [H+] changes to half of

its initial value on dilution

R. (20 mL of 0.1 M HCl + 20 mL of

0.1 M ammonia solution) diluted to

80 mL

3. the value of [H+] changes to two

times of its initial value on

dilution

S. 10 mL saturated solution of

Ni(OH)2 in equilibrium with excess

solid Ni(OH)2 is diluted to 20 mL

(solid Ni(OH)2 is still present after

dilution).

4. the value of [H+] changes to 1/ 2

times of its initial value on dilution

5. the value of [H+] changes to 2

times of its initial value on dilution

Match each process given in LIST-I with one or more effect(s) in LIST-II. The correct

option is

(A) P → 4; Q → 2; R → 3; S → 1

(B) P → 4; Q → 3; R → 2; S → 3

(C) P → 1; Q → 4; R → 5; S → 3

(D) P → 1; Q → 5; R → 4; S → 1

18. [D]

P : 3 3 2CH COOH NaOH CH COONa H O

10 ml of 0.1m20 ml of 0.1m

The value of H+ does not change on dilution for buffer solution P(1).

Q : 3 3 2NaOH CH COOH CH COONa H O

20 ml of 0.1M20ml of 0.1M

C1 = 0.05 M

C2 = 0.025 M on dilution to 80 ml

M1V1 = M2V2

KH

OH



n

a

KOH K C C

K

aK KH

C

1

HC

1 2

12

H C 0.025

C 0.05H

2

1

H 0.052

0.025H

2 1H 2 H

R : 4 4 2NH OH HCl NH Cl H O

20 ml of 0.1M20 ml of 0.1M

C1 = 0.05 M

C2 = 0.025 M

nH K C

H C

1 1

22

H C

CH

2 2

11

H C 0.025

C 0.05H

1

2

2 1

1H H

2

R (4)

S : The value of H does not change on dilution.



PART- III MATHEMATICS



Each question has FOUR options for correct answer(s). ONE OR MORE THAN

ONE of these four option(s) is (are) correct option(s).




Partial Marks : +3 If all the four options are correct but ONLY three options are

chosen.

Partial Marks : +2 If three or more options are correct but ONLY two options are

chosen, both of which are correct options.

Partial Marks : +1 If two or more options are correct but ONLY one option is

chosen and it is a correct option.

Zero Marks : 0 If none of the options is chosen (i.e. the question is

unanswered).


For Example: If first, third and fourth are the ONLY three correct options for a

question with second option being an incorrect option; selecting only all the three

correct options will result in +4 marks. Selecting only two of the three correct options

(e.g. the first and fourth options), without selecting any incorrect option (second option

in this case), will result in +2 marks. Selecting only one of the three correct options

(either first or third or fourth option) ,without selecting any incorrect option (second

option in this case), will result in +1 marks. Selecting any incorrect option(s) (second

option in this case), with or without selection of any correct option(s) will result in 2

marks.

1. For any positive integer n, define fn : (0, ) → as

n 1

n j 1

1f (x) tan

1 (x j)(x j 1)

for all x (0, )

(Here, the inverse trigonometric function tan1

x assumes values in ,2 2

)

Then, which of the following statement(s) is (are) TRUE?

(A) 5 2jj 1

tan f (0) 55

(B) 10 2j jj 1

1 f (0) sec (f (0)) 10

(C) For any fixed positive integer n, nx

1lim tan(f (x))

n

(D) For any fixed positive integer n, 2

nxlim sec (f (x)) 1

1. (A, B, D) n

1

n

j 1

(x j) (x j 1)f (x) tan

1 (x j) (x j 1)

= n

1 1

j 1

tan (x j) tan (x j 1)

= 1 1 1 ntan (x n) tan (x) tan

1 x(x n)

n 2

ntan(f (x))

1 x nx



nxlim tan(f (x)) 0

2

nxlim sec (f (x)) 1

Also, 2 2

jtan (f (0)) j

5 5

2 2

j

j 1 j 1

tan (f (0)) j 55

Also (B) option is correct.

and (A), (D) are correct.

2. Let T be the line passing through the points P(2, 7) and Q(2, 5). Let F1 be the set of all

pairs of circles (S1, S2) such that T is tangent to S1 at P and tangent to S2 at Q, and also

such that S1 and S2 touch each other at a point, say, M. Let E1 be the set representing the

locus of M as the pair (S1, S2) varies in F1. Let the set of all straight line segments joining

a pair of distinct points of E1 and passing through the point R(1, 1) be F2. Let E2 be the

set of the mid-points of the line segments in the set F2. Then, which of the following

statement(s) is (are) TRUE?

(A) The point (2, 7) lies in E1 (B) The point 4 7

,5 5

does NOT lie in E2

(C) The point 1

,12

lies in E2 (D) The point

30,

2

does NOT lie in E1

2. (D)

Let M (h, k).

PMQ = 90 (prove yourself)

k 5 k 7

1h 2 h 2

E1 = Locus of M is

x2 + y

2 2y 39 = 0

Equation of chord of circle whose midpoint (h, k) is

T = S1

xh + yk (y + k) 39 = h2 + k

2 2k 39

Since this locus passes through (1, 1).

E2 h2 + k

2 h 2k + 1 = 0

Options (B) & (C) are incorrect.

(A) is also incorrect as in that case one circle will be a point option (D) is correct.

3. Let S be the set of all column matrices

1

2

3

b

b

b

such that b1, b2, b3 and the system of

equations (in real variables)

x + 2y + 5z = b1

2x 4y + 3z = b2

x 2y + 2z = b3

has at least one solution. Then, which of the following system(s) (in real variables) has

(have) at least one solution for each

1

2

3

b

b

b

S ?

(A) x + 2y + 3z = b1, 4y + 5z = b2 and x + 2y + 6z = b3

(B) x + y + 3z = b1, 5x + 2y + 6z = b2 and 2x y 3z = b3

(C) x + 2y 5z = b1, 2x 4y + 10z = b2 and x 2y + 5z = b3

(D) x + 2y + 5z = b1, 2x + 3z = b2 and x + 4y 5z = b3

P

(2,7)

Q

(2,5)

M



3. (A),(C), (D)

For atleast one solution, either 0 or = 1 = 2 = 3 = 0.

=

1 2 5

2 4 3 0

1 2 2

1 =

1

2

3

b 2 5

b 4 3

b 2 2

= 0 b1 + 7b2 13b3 = 0

2 =

1

2

3

1 b 5

2 b 3 0

1 b 2

b1 + 7b2 13b3 = 0

Also, 3 = 0

For option (A), =

1 2 3

0 4 5 12 0

1 2 6

, so unique solution.

For option (B), =

1 1 3

5 2 6 0

2 1 3

, 1 = 0, 2 =

1

2

3

1 b 3

5 b 6

2 b 3

= 3 (b1 + b2 + 3b3) 0

So no solution.

For option (C), =

1 2 5

2 4 10 0

1 2 5

Also, 1 = 2 = 3 = 0. So, infinitely many solution.

For option (D), =

1 2 5

2 0 3

1 4 5

= 54 0, so unique solution.

Hence (A), (C), (D) are correct.

4. Consider two straight lines, each of which is tangent to both the circle 2 2 1x y

2 and

the parabola y2 = 4x. Let these lines intersect at the point Q. Consider the ellipse whose

center is at the origin O(0, 0) and whose semimajor axis is OQ. If the length of the

minor axis of this ellipse is 2 , then which of the following statement(s) is (are) TRUE?

(A) For the ellipse, the eccentricity is 1

2and the length of the latus rectum is 1

(B) For the ellipse, the eccentricity is 1

2and the length of the latus rectum is

1

2

(C) The area of the region bounded by the ellipse between the lines 1

x2

and x = 1 is

1( 2)

4 2

(D) The area of the region bounded by the ellipse between the lines 1

x2

and x = 1 is

1( 2)

16



4. (A, C)

Eq. of tangent to y2 = 4x is

y = mx + 1

m …(1)

Eq. of tangent to x2 + y

2 =

21

2

is

y = mx 21 m

2

…(2)

Comparing (1) and (2), we get m2 = 1 m = 1

Q (1, 0)

Eq. of ellipse is 2 2

2

x y

11

2

= 1

Eccentricity = 1

12

= 1

2

Length of latus rectum = 1

22

= 1

Area = 2

1 2

1

2

1 xdx

2

=

12 1

1

2

x 12 1 x sin (x)

2 2

= ( 2)

4 2

5 Let s, t, r be non-zero complex numbers and L be the set of solutions

z = x + iy (x, y , i = 1 ) of the equation sz tz r 0 and z x iy. Then,

which of the following statement(s) is (are) TRUE?

(A) If L has exactly one element, then | s | | t |

(B) If | s | = | t |, then L has infinitely many elements

(C) The number of elements in L ∩ {z : | z 1 + i | = 5 } is at most 2

(D) If L has more than one element, then L has infinitely many elements

5. (A), (B), (C) & (D)

Let, s = 1 2s is

t = 1 2t it (s1, s2, t1, t2, r1, r2 R)

r = r1 + ir2

sz tz r 0

1 1 2 2 1 2 2 1 1 2(s t )x (t s )y r i (s t )x s t y r 0

1 1 2 2 1s t x t s y r 0

And 2 2 1 1 2s t x s t y r 0

i) If 1 1 2 2

2 2 1 1

s t t s,

s t s t

then locus is two coincident lines.

ii) If 1 1 2 2

2 2 1 1

s t t s,

s t s t

then liens are intersecting and hence locus is a point.

If |s| |t|, L has exactly one point.

And if |s| \ |t|, L has infinitely many elements as it’s a straight line.

Q



6. Let f : (0, ) be a twice differentiable function such that

2

t x

f (x) sin t f (t)sin xlim sin x

t x

for all x (0, )

If f6 12

, then which of the following statement(s) is (are) TRUE?

(A) f4 4 2

(B) 4

2xf (x) x

6 for all x (0, )

(C) There exists (0, ) such that f() = 0

(D) f f 02 2

6. (B, C, D)

2

t x

f (x) sin(t) f (t) sin(x)lim sin (x)

t x

2

t x

f (x) cos(t) f (t) sin(x)lim sin (x)

1

2f (x) cos x f (x)sin(x) sin (x)

d f (x)

1dx sin x

f(x) = x sin(x) ( C 0 )

For f(x) = x sin(x), option (B), (C), (D) are correct.



NUMERICAL VALUE.


truncated/rounded‐off to the second decimal place; e.g. 6.25, 7.00, 0.33, 30, 30.27,

127.30) using the mouse and the on screen virtual numeric keypad in the place

designated to enter the answer.




7 The value of the integral

1

2

10 2 6 4

1 3dx

(x 1) (1 x)

is _____ .

7. [2.00] 1/2

2/4 6/4

0

( 3 ) dx

(x 1) (1 x)

=

1/2

1/2

0 2

( 3 1) dx

x 1(1 x)

1 x

Let x 1

t1 x



= 3

1

( 3 1) dt

2t

2

dx dt

(1 x) 2

= ( 3 1) ( 3 1) 2

8. Let P be a matrix of order 3 3 such that all the entries in P are from the set {1, 0, 1}.

Then, the maximum possible value of the determinant of P is _____ .

8. (4.00)

=

1 1 1

2 2 2

3 3 3

a b c

a b c

a b c

= (a1b2c3 + a3c2b1 + a2c1b3) (a1b3c2 + a2c3b1 + a3c1b2)

{ai , bi, ci} {1, 0, 1} i = 1, 2, 3

Maximum value can be 6, but that’s not possible.

(For = 6, 3 entries should be 1 and 6 entries should be 1 in first bracket and 3 entries

should be 1 and 6 entries should be 1 in second bracket. That’s a contradiction !)

5. (Even if one element is 0)

max = 4 which is possible.

Ex :

1 1 0

1 1 1

1 1 1

9. Let X be a set with exactly 5 elements and Y be a set with exactly 7 elements. If is the

number of one-one functions from X to Y and is the number of onto functions from Y

to X, then the value of 1

5!( ) is _________ .

9. (119.00)

= 7C5 5! = 21 5!

= 7C4 3 2

5C2 3! +

7C3 5 4!

5!

= 119.00

10. Let f : → be a differentiable function with f(0) = 0. If y = f(x) satisfies the

differential equation

dy

(2 5y)(5y 2)dx

then the value of xlim f (x)

is ________ .

10. [0.40]

dydx

(2 5y) (5y 2)

1 (2 5y) (5y 2)dy dx

4 (2 5y) (5y 2)

e

1 5y 2log x c

5 4 5y 2

20x c5y 2e

5y 2

20x2 5y

e2 5y

( f (0) 0)



20x

20x

2 (1 e )y

5 (1 e )

x

2lim f (x)

5

11. Let f : → be a differentiable function with f(0) = 1 and satisfying the equation

f(x + y) = f(x) f(y) + f(x) f(y) for all x, y

Then, the value of loge(f(4)) is _____.

11. (2.00)

f(0) = 1

Put x = 0, y = 0

f(0) = f(0) f (0) + f (0) f (0) f(0) = 1

2

put y = 0

f(x) = f(x) f (0) + f (x) f(0)

f(x) = f(x) 1

2 + f (x)

f (x) = f (x)

2

f '(x)

dxf (x) =

1dx

2

loge (f(x)) = x

2+ c

f(x) =

x

2e (C = 0 since f(0) = 1)

loge f(4) = loge (e

2) = 2

12. Let P be a point in the first octant, whose image Q in the plane x + y = 3 (that is, the

line segment PQ is perpendicular to the plane x + y = 3 and the mid-point of PQ lies in

the plane x + y = 3) lies on the z-axis. Let the distance of P from the x-axis be 5. If R

is the image of P in the xy-plane, then the length of PR is _____ .

12. (8.00)

Let P (x0, y0, z0)

0 0 0x x y y z z

1 1 0

= 0 0

2 2

2(x y 3)

1 1

x = 3 y0

y = 3 x0

Since, mirror images lies on zaxis

3 y0 = 0, 3 x0 = 0

x0 = 3, y0 = 3

Also, 2 20 0z y = 5 z0 = 4

PR = 4 (4) = 8

13. Consider the cube in the first octant with sides OP, OQ and OR of length 1, along the

x-axis, yaxis and z-axis, respectively, where O(0, 0, 0) is the origin. Let S1 1 1

, ,2 2 2

be the centre of the cube and T be the vertex of the cube opposite to the origin O such



that S lies on the diagonal OT. If p SP , q SQ, r SR and t ST, then the value

of (p q) r t) is __________

13. (0.5)

SP = p = 1

i2

1 1ˆ ˆj k2 2

SQ = q = 1

i2

+ 1 1ˆ ˆj k2 2

SR = r = 1

i2

1 1ˆ ˆj k2 2

ST = t = 1

i2

+ 1 1ˆ ˆj k2 2

ˆ ˆ î j k

1 1 1p q

2 2 2

1 1 1

2 2 2

= ˆ î j

2

ˆ ˆ î j k

1 1 1r t

2 2 2

1 1 1

2 2 2

= ˆ î j

2

ˆ 1k

| (p q) (r t ) |22

14. Let 2 2 2 2

10 10 10 101 2 3 10X C 2 C 3 C ......... 10 C

where 10

Cr, r {1, 2, …., 10} denote binomial coefficients. Then, the value of

1X is ____.

1430

14. (646)

X = 10 2

10r

r 0

r C

X = 10 2

1010 r

r 0

10 r C

2X = 10 2

10r

r 0

10 C

X = 20105. C

X646.00

1430

R

z

y Q(0,1,0)

P(0,1,0)

x

S 1 1 1

, ,2 2 2

T

O




This section contains FOUR (04) questions.

Each question has TWO (02) matching lists: LIST‐I and LIST‐II.

FOUR options are given representing matching of elements from LIST‐I and

LIST‐II. ONLY ONE of these four options corresponds to a correct matching.


For each question, marks will be awarded according to the following marking

scheme:

Full Marks : +3 If ONLY the option corresponding to the correct matching is

chosen.



15. Let 1

xE {x : x 1and 0}

x 1

and 12 1 e

xE {x E :sin log is a real number}

x 1

.

(Here, the inverse trigonometric function sin1

x assumes values in ,

2 2

)

Let f : E1 be the function defined by f(x) = loge

x

x 1

and g : E2 be the function defined by g(x) = sin1

e

xlog

x 1

LIST I LIST II

P The range of f is 1. 1 e, ,1 e e 1

Q The range of g contains 2. (0, 1)

R The domain of f contains 3. 1 1,

2 2

S The domain of g is 4. (, 0) (0, )

5. e,e 1

6. (, 0)

1 e,

2 e 1

The correct option is:

(A) P → 4; Q → 2; R → 1; S → 1

(B) P → 3; Q → 3; R → 6; S → 5

(C) P → 4; Q → 2; R → 1; S → 6

(D) P → 4; Q → 3; R → 6; S → 5

15. (A)

P 4, Q 2, R 1, S 1

E1 : x

0x 1

x (, 0) (1, )

E2 : x

0x 1

and 1 log x

x 1

1



x (, 0) (1, ) 1 x

e x 1

e

0 x

x 1

1

e

x

x 1 e 0

0 (e 1) x 1

e (x 1)

x(e 1) e

x 1

0

x 1

,1 e

(1, ) x (, 1) e

,e 1

Intersection x 1

,1 e

e

,e 1

So, domain of g is x 1 e

, ,1 e e 1

Range of x

x 1is R

+ {1}

Range of f is R {0} or (, 0) (0, )

Range of g is , {0}2 2

16. In a high school, a committee has to be formed from a group of 6 boys

M1, M2, M3, M4, M5, M6 and 5 girls G1, G2, G3, G4, G5

(i) Let 1 be the total number of ways in which the committee can be formed such

that the committee has 5 members, having exactly 3 boys and 2 girls.

(ii) Let 2 be the total number of ways in which the committee can be formed such

that the committee has at least 2 members, and having an equal number of boys

and girls.

(iii) Let 3 be the total number of ways in which the committee can be formed such

that the committee has 5 members, at least 2 of them being girls.

(iv) Let 4 be the total number of ways in which the committee can be formed such

that the committee has 4 members, having at least 2 girls and such that both M1

and G1 are NOT in the committee together.

LIST I LIST II

P. The value of 1 is 1. 136

Q. The value of 2 is 2. 189

R. The value of 3 is 3. 192

R. The value of 4 is 4. 200

5. 381

6. 461


(A) P → 4; Q → 6; R → 2; S → 1

(B) P → 1; Q → 4; R → 2; S → 3

(C) P → 4; Q → 6; R → 5; S → 2

(D) P → 4; Q → 2; R → 3; S → 1

16. (C)

(i) 6C3

5C2 = 200 (P 4)

(ii) 6C1

5C1 +

6C2

5C2 +

6C3

5C3 +

6C4

5C4 +

6C5

5C5

= 461 (Q 6)

(iii)5C2

6C3 +

5C3

6C2 +

5C4

6C1 +

5C5

6C0 = 381

(R 5)

(iv) (5C2

6C2

4C1

5C1) + (

5C3

6C1

4C2

5C0) +

5C4

6C0

= 189 (S 2)



17. Let H : 2 2

2 2

x y1,

a b where a > b > 0, be a hyperbola in the xy-plane whose conjugate

axis LM subtends an angle of 60° at one of its vertices N. Let the area of the triangle

LMN be 4 3 .

LIST I LIST II

P The length of the conjugate axis of H is 1. 8

Q The eccentricity of H is 2. 4

3

R The distance between the foci of H is 3. 2

3

S The length of the latus rectum of H is 4. 4


(A) P → 4; Q → 2; R → 1; S → 3

(B) P → 4; Q → 3; R → 1; S → 2

(C) P → 4; Q → 1; R → 3; S → 2

(D) P → 3; Q → 4; R → 2; S → 1

17. (B)

LMN = 4 3

1

2ab 4 32

ab = 4 3

a

a . 4 33

2a 12 a 2 3

obtan30

a

b = 2 3

23

Length of conjugate axis 2b = 4.

4 1 2

e 1 112 3 3

SS = 2 ae = 2

2 2 3 83

L.R. = 22b 2 4 4

a 2 3 3

18. Let 1 2f : f : ,2 2

2

3 4, f : 1, e 2 and f :

be functions

defined by

(i) 2x

1f (x) sin 1 e

(ii) 12

| sin x |if x 0

f (x) ,tan x

1 if x 0

where the inverse trigonometric function of tan1

x

30 a N(a, 0)

M(0, b)

(0, 0)

L(0, b)

y

x



assume values in ,2 2

,

(iii) f3(x) = [sin(loge(x + 2))], where, for t , [t] denotes the greatest integer less

than or equal to t,

(iv)

2

4

1x sin if x 0

f (x) x

0 if x 0

LIST I LIST II

P The function f1 is 1. NOT continuous at x = 0

Q The function f2 is 2. continuous at x = 0 and NOT differentiable at x = 0

R The function f3 is 3. differentiable at x = 0 and its derivative is NOT

continuous at x = 0

S The function f4 is 4. differentiable at x = 0 and its derivative is

continuous at x = 0


(A) P → 2; Q → 3; R → 1; S → 4

(B) P → 4; Q → 1; R → 2; S → 3

(C) P → 4; Q → 2; R → 1; S → 3

(D) P → 2; Q → 1; R → 4; S → 3

18. (D)

(P) 1f (x) =

2 2

2

x x

x

x e .cos 1 e

1 e

1f (0) does not exists.

(Q) 2f (x) = 1

| sin x |, x 0

tan x

1, x 0

=

1

1

sin x, x 0

tan x

sin x, x 0

tan x

1, x 0

f2(0+) = 1

f2(0) = 1

Not continuous at x = 0

(R) f3(x) = esin log (x 2 , f3 : (1, 2e 2

) R

Given 1 < x + 2 < e/2

0 < elog (x 2)2

f3 (x) = 0

Always continuous and differentiate.



(S) 4f (x) =

2 1x sin , x 0

x

0, x 0

Obviously, f4(x) continuous and differentiable at x = 0.

4f (x) =

2

2

1 1 12x sin x cos , x 0

x x x

0, x 0

=

1 12x sin cos , x 0

x x

0, x 0

4f (0 ) = h 0

1 12h sin cos 0

h hlim

h

= h 0

1cos

1 hlim 2sinh h

= does not exists.

4f (x) is not differentiate at x = 0.

For continuity 4f (x)

4h 0

1 1f (0 ) lim 2hsin cos

h h

does not exists.

4f (0 ) not continuous at x = 0.

Either n(A)

S1 S3 S5 S2 S4

S1 S4 S2 S5 S3

S2 S4 S1 S3 S5

S2 S4 S1 S5 S3

S2 S5 S3 S1 S4

S2 S5 S4 S1 S3

S3 S5 S1 S4 S2

S3 S5 S2 S4 S1

S3 S1 S4 S2 S5

S3 S1 S5 S2 S4

S4 S1 S3 S2 S5

S4 S2 S5 S1 S3

S5 S1 S3 S2 S4

S5 S2 S4 S1 S3

END OF THE QUESTION PAPER

Solution to IIT JEE 2018 (Advanced) : Paper -...

Documents

Transcript of Solution to IIT JEE 2018 (Advanced) : Paper -...