paper – 2 section 1: (integer value correct type) this...

Paper – 2 Section 1: (Integer value correct type)

This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.

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

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

massless rod of length l and this assembly is free to move along the line connecting them. All three

masses interact only through their mutual gravitational interaction. When the point mass nearer to M is at

a distance r = 3l from M, the tension in the rod is zero for .288

Mm k

æ öç ÷= ç ÷è ø

The value of k is

Solution

(7)

Since tension in the rod is zero, for the first mass m FBD gives

2

22

mGM Gmma

gl l = (1)

for the second mass m, FBD gives

2

2 216mGM Gm

mal l

+ = (2)

From Eqs. (1) and (2)

2

mGM

g l

2

2

Gm

l

216

mGM

l=

2

2

Gm

l+

29 16

M Mm =

72

144

Mm=

7288 288

M Mm K

æ ö æ öç ÷ ç ÷= =ç ÷ ç ÷è ø è ø

Therefore,

K = 7

2. The energy of a system as a function of time t is given as E(l) = A2 exp (l), where = 0.2 s1 . The

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

percentage error in the value of E(l) at l = 5 e is

Solution

(4)

E = A2et

2 ( ) 2 ( )t tdE A e dl A dA ea aa = +

Relative error 2dE dA

dtE A

aæ öç ÷= + ç ÷è ø

2dE dt dA

tE t A

aæ ö æ öç ÷ ç ÷= +ç ÷ ç ÷è ø è ø

Therefore, % error (at t = 5 s) = (0.2 5 1.5) + (2 1.25)

Considering the worst case, all errors are added

% error = 1.5 + 2.5 = 4

3. The densities of two solid spheres A and B of the same radii R vary with radial distance r as

( )A

rr k

Rr


and 5

( ) ,B

rr k

Rr


respectively, where k is a constant. The moments of inertia of the

individual spheres about axes passing through their centres are IA and IB, respectively. If ,10

B

A

I n

I= the

value of n is

Solution

(6) For 1st shell

21

2( )

3dI dm n=

where dm is the mass of the elemental shell shown shaded 21 (4 )dm r drr p=

Therefore,

2 21

24

3

KrdI r dr r

Rp

é ùæ öê úç ÷= ç ÷ê úè øë û

5 51

0

8 8

3 18

RK KI r dr R

R

p pæ öç ÷= = ç ÷è ø

ò

For 2nd shell

5

2 22

24

3

rdI K r r

Rp

é ùæ öê úç ÷= ç ÷ê úè øë û

58

30

KR

pæ öç ÷= ç ÷è ø

Therefore,

2

1

18 66

30 10 10

I nn

I= = = Þ =

4. Four harmonic waves of equal frequencies and equal intensities I0 have phase angle 0, x/3, 2/3 and .

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

Solution

(3) Let y1 = a sin t

y2 = a sin(t + /3)

y3 = a sin(t + 2/3)

y4 = a sin(t + )

y = y1 + y2 + y3 + y4 = R sin (t f )

where R is the amplitude of the resultant wave using phasor method

2 2 22 cos /3 3R a a a ap= + + =

I = R2 = 3a2 = 3I0 = nI0

n = 3

5. For a radioactive material, its activity A and rate of change of its activity R are defined as dN

Adt

=

and ,dA

Rdt

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

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

RQ , respectively. If P

Q

,R n

R e= then the value of n is

Solution

(2) Law of radioactivity N = N0et

where = decay constant

Activity 0| | tdNA N e

dtll

= =

Rate of activity 20

td AR N e

dlll = =

At t = 0, A1 = A2

NOP P = NOQ Q

At t = 2J;

2P(25)

( )25OPP P P

Q(25)Q Q OQ Q

Q PNR e

eR N e

ll l

l

l l

l l

æ ö æ öç ÷ ç ÷= =ç ÷ ç ÷è ø è ø

Mean life 1

Jl

=

Therefore, 25

1 1

25 5P P

Q Q

Re

R

l

l

æ öç ÷ç ÷è ø= 1P

Q

el

l=

P P

Q Q

1R n

R e e

l

l= =

P

Q

22

Jn

J

l

l= = =

6.A monochromatic beam of light is incident at 60 on the face of an equilateral prism of refractive index

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

3n = , the value of is 60 and .d

mdn

q= The value of m is

Solution

(2)

Using Snell’s law

sin 60 = n sin r1 (1)

1

3 1sin

22 3r = =

´

r1 = 30

Also

n sin r2 = 1 sin

Also

r1 + r2 = A = 60

Therefore,

n sin (60 r1 ) = 1 sin (2)

Differentiating on both sides, we get

11 1sin(60 ) cos(60 ) cos

dr dr n r

dw dw

qq° ° =

Differentiating Eq. (1) on both sides, we get

11 10 sin cos

drr n r

dw= +

11 30 3

2 2

dr

dn= + ×

Therefore,

1 1

3

dr

dn

=

Hence putting r1 = 30

1 1

3

dr

dt

=

1sin 30 3 cos30 cos60

3

d

dn

qæ öç ÷° ° = °ç ÷è ø

1 3 1

2 2 3 2

d

dn

q+ =

´

2d

dn

q=

7. In the following circuit, the current through the resistor R (=2 ) is I amperes. The value of I is

Solution

(1)

ACGA constitutes a Wheatstone bridge, hence 8 is redundant and hence can be removed

Therefore,

AG

3 62

9R

AGDFA again constitutes a Wheatstone bridge 10 which is redundant and hence can be removed.

AB

6 184.5

24R

6.51 A

6.5I = =

8. An electron in an excited state of Li2, ion has angular momentum 3h/2. The de Broglie wavelength of

the electron in this state is pa0 (where a0 is the Bohr radius). The value of p is

Solution

(2) Angular momentum 2

nhmvr

p= where r = 3a0 where n = 3, that is, electron in Li2+ is in second excited

state

0

hp a

mvl p= =

0

3( )

3 3 2 2

mvr p h phn p mva p

pp p

p

æ öæ öç ÷ç ÷Þ = = = =ç ÷ ç ÷

è ø è ø

Therefore, p = 2

Section 2: (One or More Options Correct Type) This section contains EIGHT questions. Each has four options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct.

9. Two spheres P and Q of equal radii have densities 1 and 2, respectively. The spheres are connected

by a massless string and placed in liquids L1 and L2 of densities 1 and 2 and viscosities 1 and 2,

respectively. They float in equilibrium with the sphere P in L 1 and sphere Q in L 2 and the string being

taut (see the below figure). If sphere P alone in L 2 has terminal velocity ,QV

then

(A) P 1

2Q

V

V

h

h=

(B) P 2

1Q

V

V

h

h=

(C) P Q 0V V

(D) P Q 0V V

Solution

(A, D)

where s = density of solid

l = density of liquid

For sphere in L2 Q: T + Fbf = mg

3 22 2

4 4

3 3T r g r gp s p r= =

32 2

4( )

3T r gp r s= , that is 2 > 2

For sphere in L1 P: bfT m g F¢ ¢+ =

31 2

4( )

3T r gp s r= , that is 1 1r s<

Given liquids are immiscible, 2 > 1

From above we get 2 > 2 > 1 > 1

We know viscous force FV = 6 rv

Terminal velocity, 1

Vm

µ , for a given viscous force. If P is put in L 2, 2

1| | ,V

rm

µ 1 2 ,r s< vfF ¯

, that is

PV

If Q is put in L 1 , 1

1| | ,QV

mµ 2 1 ,r s> vfF

, that is QV ¯

Therefore, 1

2

P

Q

V

V

m

m=

P QV V

× cos180P Q P QV V V V° =

that is, 0P QV V

× <

10. In terms of potential difference V, electric current I, permittivity 0 , permeability 0 and speed of

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

(A) 2 20 0I Vm e= (B) 0 0I Ve m=

(C) 0I cVe= (D) 0 0cI Vm e

Solution

(A, C) Dimensions of [V] = [M 1 L2T3A−1 ]

[I] = [A]

[E0] = [M −1 L3T4A2]

[0] = [M 1 L1 T2A2]

[C] = [L1 T1 ]

(A) 0I2 = E0V2

LHS → [M 1 L1 T2];

RHS → [M 1 L1 T2] dimensionally correct

(B) E0I = 0V

LHS → [M 1 L3T4 A3]

RHS → [M 2L3T5 A3] dimensionally incorrect

(C) I = E0CV

LHS → [A];

RHS → [A]

(D) 0CI = E0V

LHS → [M 1 L2T3A1 ],

RHS → [L1 T1 A1 ] dimensionally incorrect

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

distribution, a spherical cavity of radius R2, centred at P with distance OP = a = R1 R2 (see the below

figure) is made. If the electric field inside the cavity at position r

is ( ),E r

then the correct statement(s)

is(are)

(A) E

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

(B) E

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

(C) E

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

(D) E

is uniform and both its magnitude and direction depend on a

Solution

(D)

For a uniformly charged sphere, at r < R1 , 03

erE

E=

where r

is the position vector of the point from the

center.

Considering a point P ¢ distant r1 from O and r2 from P: 1 2 1 2

0 0

( ) const3 3

aE E E r r

E E

r r= = = =

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

on the yaxis and stress on the xaxis as shown in the figure. Then the correct statement(s) is(are)

(A) P has more tensile strength than Q

(B) P is more ductile than Q

(C) P is more brittle than Q

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

Solution

(A, B) Young’s modulus of elasticity stress 1

strain slopey = =

Slope of Q < slope of P

yQ > yP

For a given stress, strain P > strain Q

Therefore, is more ductile.

Breakpoint of P > breakpoint of Q

Therefore, P has more tensile strength.

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

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

(A) P(r = 0) = 0

(B) ( 3 /4) 63

( 2 /3) 80

P r R

P r R

==

=

(C) ( 3 /5) 16

( 2 /5) 21

P r R

P r R

==

=

(D) ( /2) 20

( /3) 7

P r R

P r R

==

=

Solution

(B,C)

34

3

M

R

r

p

=

dm (of elemental strip or spherical shell of radius r and thickness dr) = 4 r2 dr

Under equilibrium,

2

( )( )

Gm dmdp A

r=

Therefore,

2

0

4

3p G r

dp

r p

r = =ò

24 rr pæ öç ÷ç ÷è ø

2

dr

r 2(4 )rp

R

r

ò

2 22 24 4

3 3 2

R

r

R rG r dr G

pr r

æ öç ÷= = ç ÷è ø

ò

Therefore, P = (R2 r2) where 2 4

6G

pb r= ×

P(r = 0) = R2 0

2 323 9 7

4 16 16

R R RP r Rb b

æ öæ öç ÷ç ÷= = =ç ÷ ç ÷

è ø è ø

2 222 4 5

3 9 9

R R RP r Rb b

æ öæ öç ÷ç ÷= = =ç ÷ ç ÷

è ø è ø

Therefore,

3

634

2 80

3

RP r

RP r

æ öç ÷=ç ÷è ø =æ öç ÷=ç ÷è ø

Choice (2)

2 223 9 16

5 25 25

R R RP r Rb b

æ öæ öç ÷ç ÷= = =ç ÷ ç ÷

è ø è ø

2 222 4 21

5 25 25

R R RP r Rb b

æ öæ öç ÷ç ÷= = =ç ÷ ç ÷

è ø è ø

Therefore,

3

165

2 21

5

RP r

RP r


Choice (3)

2 22 3

2 4 4

R R RP r Rb b

æ öæ öç ÷ç ÷= = =ç ÷ ç ÷

è ø è ø

2 22 8

3 9 9

R R RP r Rb b

æ öæ öç ÷ç ÷= = =ç ÷ ç ÷

è ø è ø

Therefore,

272

32

3

RP r

RP r


1 4. A parallel plate capacitor having plates of area S and plate separation d, has capacitance C1 in air.

When two dielectrics of different relative permittivities (1 = 2 and 2 = 4) are introduced between the

two plates as shown in the below figure, the capacitance becomes C2. The ratio 2

1

C

C is

(A) 6/5 (B) 5/3

(C) 7/5 (D) 7/3

Solution

(D)

Originally 1

GsC

d=

Let the given arrangement be divided into three capacitances

1 00

1 1 1 12 2

/2p

SK E E S

C K K C Cd d

= = = =

2 00

2 2 1 12 4

/2Q

SK E

E SC K K C C

d d= = = =

1 001 1

1 12

2 2R

SK E E SK K

C C Cd d

= = = =

Equivalent arrangement would be

12 1 1

74

3 3AB

CC C C C= = + =

Therefore, 2

1

7

3

C

C=

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

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

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

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

cylinder, the correct statement(s) is(are)

(A) If V2 = 2V1 and T2 = 3T1 , then the energy stored in the spring is 1 1

1

4PV

(B) If V2 = 2V1 and T2 = 3T1 , then the change in internal energy is 3P1 V1

(C) If V2 = 3V1 and T2 = 4 T1 , then the work done by the gas is 1 1

7

3PV

(D) If V2 = 3V1 and T2 = 4 T1 , then the heat supplied to the gas is 1 1

17

6PV

Solution

(A, B, C)

1

21 2 1

1( )

2

kxP P

A

W P dV P V V kx

2 2 1 1

3 3( ) ( )

2 2VU nC T nR T PV PV

For case 1: If V2 = 2V1 , T2 = 3T1

1 11 1

1 1 1 1spring

53 ,

4

17,

4 4

PVU PV W

PV PVQ U W U

For case 2: If V2 = 3V1 and T2 = 4 T1

1 1 1 1

1 1 1 1spring

7,

2 3

41,

6 3

PV PVU W

PV PVQ U W U

Using 1 1 2 2

1 2

PV PV

T T=

If V2 = 2V1 , T2 = 3T1

2 1

3

2P P=

2 2 1 13

2V

PV PVRU nC T n

nR nR

æ öæ öç ÷ç ÷D = D = ç ÷ç ÷

è øè ø

( )2 2 1 1

3

2PV PV=

= 3P1 V1

1

21 2 1

2 2 1 1

1( )

2

3 3( ) ( )

2 2V

kxP P

A

W PdV P V V kx

U nC T nR T PV PV

For case 1:

1 11 1

1 1 1 1spring

53

4

17

4 4

PVU PV W

PV PVQ U W U

For case 2:

1 1 1 1 1 1

1 1spring

7 41

2 3 6

3

PV PV PVU W Q

PVU

1 6. A fission reaction is given by 236 160 9492 34 38U Xe Sr ,x y® + + + where x and y are two particles. Considering

23692 U to be at rest, the kinetic energies of the products are denoted by KXe, KSr Kx(2 MeV) and Ky

(2 MeV), respectively. Let the binding energies per nucleon of 23692 U, 140

84 Xe and 9438Sr be 7.2 MeV,

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

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

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

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

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

Solution

(A) 236 140 9492 54 38U Xe Sr x y® + + +

For the given fission reaction

Q value = (140 8.5) + (94 8.5) (236 7.5) = 219 MeV

Therefore, KE of Xe and Sr combined = Q value KEx KEy

= 219 2 2 = 215 MeV

Conservation of charge implies x and y to be neutral

Conservation of mass implies x and y are neutrons 10 r

0conservation of momentum

0i

f

P

P

ü= ïý

= ïþ

Xe Sr| | | | | |P P P

= = say

2

KE2

p

m=

Therefore, 1

KEm

µ

Section 3: (Paragraph Type) This section contains TWO paragraphs. Based on each paragraph, there will be TWO questions. Each question has four options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct.

Paragraph for Questions 17 and 18: In a thin rectangular metallic strip a constant current I flows along

the positive xdirection, as shown in the below figure. The length, width and thickness of the strip are l,

w and d, respectively.

A uniform magnetic field B

is applied on the strip along the positive ydirection. Due to this,

the charge carriers experience a net deflection along the zdirection. This results in accumulation of

charge carriers on the surface PQRS and appearance of equal and opposite charges on the face opposite

to PQRS. A potential difference along the zdirection is thus developed. Charge accumulation continues

until the magnetic force is balanced by the electric force. The current is assumed to be uniformly

distributed on the crosssection of the strip and carried by electrons.

1 7. Consider two dif ferent metallic strips (1 and 2) of the same material. Their lengths are the same,

widths are 1 and 2 and thickness of d1 and d2, respectively. Two points K and M are symmetrically

located on the opposite faces parallel to the xy plane (see the above figure). V1 and V2 are the potential

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

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

(A) If 1 = 2 and d1 = 2d2, then V2 = 2V1

(B) If 1 = 2 and d1 = 2d1 , then V2 = V1

(C) If 1 = 22 and d1 = d2, then V2 = 2V1

(D) If 1 = 22 and d1 = d2, then V2 = V1

Solution

(A, D) Given I1 = I2

ne11 dA v ne=

22 dA v where vd = drift velocity

1 21 2d dA v A v=

1 21 1 2( ) ( )d x dd v d vw w=

Potential difference V = Bvd

Therefore,

1

2

11 2

2 2 1

d

d

VV d

V V d

w

w= =

For choice (1) if 1 = 2 and d1 = 2d2; 2

1 2

d

d

VV =

Therefore,

12 1

2

12

2

VV V

V= Þ =

For choice (2), if 1 = 22 and d1 = d2

2 12d dV V=

Therefore,

1

2

1V

V=

1 8. Consider two different metallic strips (1 and 2) of same dimensions (length l, with and thickness d)

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

magnetic field B2 both along positive ydirections. Then V1 and V2 are the potential differences

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

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

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

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

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

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

Solution

(A, C) Given I1 = I2

1 21 1 2 2d dn eAV n eA V=

1 1 1(n d w1 2 2 2) (dV n d w=

2) dV

1 21 2d dn V n V=

Also V = BVd

11 11

2

dB VV

V

w=

22 2dB V w1 2

2 1

B n

B n=

If B1 = B2, n1 = 2n2, 1

2

1

2

V

V=

If B1 = 2B2, n1 = n2, 1

2

2V

V=

Paragraphs for Questions 19 and 20: Light guidance in an optical fibre can be understood by

considering a structure comprising of thin solid glass cylinder of refractive index n1 surrounded by a

medium of lower refractive index n2. The light guidance in the structure takes place due to successive

total internal reflections at the interface of the media n1 and n2 as shown in the below figure. All rays

with the angle of incidence i less than a particular value im are confined in the medium of refractive

index n1 . The numerical aperture (NA) of the structure is defined as sin im.

1 9. For two structures namely S1 with 1 45/4n = and n2 = 3/2, and S2 with n1 = 8/5 and n2 = 7/5 and

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

(A) NA of S1 immersed in water is the same as that of S2 immersed in a liquid of refractive index 16

3 15

(B) NA of S1 immersed in liquid of refractive index 6

15 is the same as that of S2 immersed in water

(C) NA of S1 placed in air is the same as that of S2 immersed in liquid of refractive index 4

15

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

Solution

(A, C)

TIR occurs at the core cladding interface if ≥ iC where iC is the critical angle

≥ iC Also 1sin

sin m

ni

r n= (Snell’s law)

(90 r) ≥ iC

sin (90 r) ≥ sin iC

cos r ≥ sin iC = 2

1

n

n

As i↑, r↑, = (90 r)↓

Thus there is a maximum value of i (say im) beyond which no TIR occurs at the interface

im = maximum angle of acceptance

Now n1 sin rm = n1 sin (90 iC) = n1 cos iC

21 1 sin Cn i=

22

1 21

1n

nn

=

2 21 2n n=

2 22 2 1 2

2(NA) sin m

m

n ni

n

= =

If 1

45,

4n = 2

3,

2n =

4

3mn =

Therefore, 9

NA sin16

mi= =

If 1

8,

5n = 2

7,

5n =

16;

3 15mn =

9NA sin

16mi= =

If 1

45,

4n = 2

3,

2n = nm = 1;

3NA sin

4mi= =

If 1

8,

5n = 2

7,

5n =

4

15mn = ;

3NA sin

4mi= =

20. If two structures of same crosssectional area, but different numerical apertures NA1 and NA2

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

(A) 1 2

1 2

NA NA

NA NA+ (B) NA1 NA2

(C) NA 1 (D) NA2

Solution

(D) For TIR to occur in both, NA should be the least of the two constituting the combined structure.

Paper – 2 Section 1: (Integer value correct type)

This section contains EIGHT questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.

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

Solution (4)

22. Among the following the number of reaction(s) that produce(s) benzaldehyde is

Solution (4)

23. In the complex acetylbromidodicarbonylbis(triethylphosphine)iron(II), the number of Fe – C

bond(s) is Solution (3) The structure is

From the above structure, we can conclude the number of Fe – C bond(s) is 3. 24. Among the complex ions, [Co(NH 2CH 2CH 2NH2)2Cl 2]+ , [CrCl 2(C 2O 4 )2]3−, [Fe(H2O) 4 )OH) 2]+ ,

[Fe(NH3)2(CN) 4 ]−, [Co(NH 2CH 2CH 2NH2)2(NH3)Cl] 2+ and [Co(NH 3)4 (H2O)Cl] 2+ , the number of complex ion(s) that show(s) cis-trans isomerism is

Solution (6) The complex with MA

2B

4 (or) MA

4B

2 (or) M(AA)

2 B

2 (or) M(AA)

2 BC exhibits cistrans

isomerism. 25. Three moles of B

2H

6 are completely reacted with methanol. The number of moles of boron

containing product formed is ______________. Solution (6) The reaction is

2 6 3 3 3 2B H 6CH OH 2B(OCH ) 6 H

1 mol of B 2H6 reacts with 6 mol of CH 3OH to produce 2 mol of B(OCH3)

3

Therefore, 3 mol of B2H6 would react with 18 mol of CH 3OH to produce 6 mol of B(OCH3)

3

26. The molar conductivity of a solution of a weak acid HX (0.01 M) is 10 times smaller than the molar conductivity of a solution of a weak acid HY (0.10 M). If 0 0

X Y, »l l the difference

in their pKa values, pKa(HX) – pKa(HY), is (consider degree of ionization of both acids to be <<1)

Solution (3) For HX and HY since 0 0

X Y, »l l

C HX

C HY

(HX) 1

(HY) 10

(HX) a HXHX

(HY) HY a HY

H ( )( )

H ( ) ( )

K CC

C K C

(1)

(HX)

(HY)

H 0.01 11

H 0.1 10

Taking negative log of both the sides of equation (1), we get

a HX a HY

a a

a a a a

1 1pH(HX) pH(HY) log( ) log( )

2 2

1 10 p (HX) 2 p (HY) 1

2 2

0 p (HX) p (HY) 3 p (HX) p (HY) 3

K C K C

K K

K K K K

27. A closed vessel with rigid walls contains 1 mol of 23892 U and 1 mol of air at 298 K.

Considering complete decay of 23892 U to 206

82 Pb , the ratio of the final pressure to the initial

pressure of the system at 298 K is Solution (9) Let us consider 1 mol of air = X atm

During the reaction of 23892 U 206

82 Pb , 8 particles are released.

That is 8 mol of He particles are released during the reaction. Therefore, 9 mol of gas are present in the vessel.

Thus, total pressure = X 9 = 9 atm that is, 9 times the original pressure. 28. In dilute aqueous H2SO 4 , the complex diaquodioxalatoferrate(II) is oxidized by MnO 4

. For this reaction, the ratio of the rate of change of [H+ ] to the rate of change of [MnO 4

] is Solution (8) Balancing equation for the reaction is

2 3+2 2 2 4 2 4 2 2[Fe(H O) (C O ) ] MnO 8H Mn Fe 4CO 6H O

Therefore, 4

1 d d[H ] [MnO ]

8 d dt t

Thus, the ratio is 8:1

Section 2: (One or More Options Correct Type) This section contains EIGHT questions. Each has four options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct.

29. In the following reactions, the product S is

Solution (A)

30. The major product U in the following reactions is

Solution (B) The reaction is

31. In the following reactions, the major product W is

Solution (A) The reaction is

32. The correct statement(s) regarding, (i) HClO, (ii) HClO

2, (iii) HClO

3 and (iv) HClO

4 is (are)

(A) The number of Cl = O bonds (ii) and (iii) together is two (B) The number of lone pairs of electrons on Cl in (ii) and (iii) toget her is three (C) The hybridization of Cl in (iv) is sp3 (D) Amongst (i) to (iv), the strongest acid is (i) Solution (A, B, C)

In all the oxyacids of chlorine: Cl undergoes sp3 hybridization. HClO 4 is the strongest acid among the given compounds. 33. The pair(s) of ions where BOTH the ions are precipitated upon passing H

2S gas in presence

of dilute HCl, is (are) (A) Ba2+ , Zn2+ (B) Bi3+ , Fe3+

(C) Cu 2+ , Pb 2+ (D) Hg2+ , Bi3+

Solution (C, D) The pairs Cu 2+ , Pb 2+ and Hg2+ , Bi3+ are precipitated as sulphides upon passing H2S in acidic

medium. 34. Under hydrolytic conditions, the compounds used for preparation of linear polymer and for

chain termination, respectively, are (A) CH 3SiCl 3 and Si(CH 3)4

(B) (CH 3)2SiCl 2 and (CH 3)3SiCl

(C) (CH 3)2SiCl 2 and CH 3SiCl 3

(D) SiCl 4 and (CH 3)3SiCl

Solution (B) The reaction involved in the preparation of linear polymer is

35. When O

2 is adsorbed on a metallic surface, electron transfer occurs from the metal to O

2.

The TRUE statement(s) regarding this adsorption is(are) (A) O 2 is physisorbed

(B) Heat is released

(C) Occupancy of π*2p of O 2 is increased

(D) Bond length of O 2 is increased

Solution (B, C, D) During transfer of electrons, adsorption is exothermic. During electronic transfer

metal to π*2p of O2 molecule, bond order of O

2 decreases; hence O – O bond length increases.

36. One mole of a monoatomic real gas satisfies the equation p(V – b) = RT where b is a constant. The relationship of interatomic potential V(r) and interatomic distance r for the gas is given by

Solution (C) As interatomic forces are absent; the interatomic potential does not vary with inter atomic

distance.

Section 3: (Paragraph Type) This section contains TWO paragraphs. Based on each paragraph, there will be TWO questions. Each question has four options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (are) correct.

Paragraph for question 37 and 38: In the following reactions

37.Compound X is

Solution (C)

38. The major compound Y is

Solution (D)

Paragraph for question 39 and 40: When 100 mL of 10 M HCl was mixed with 100 mL of 1.0

M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7°C was measured for the beaker and its contents (Expt. 1). Because the enthalpy of neutralization of a strong acid with a strong base is a constant (−57.0 kJ mol −1 ), this experiment could be used to measure the calorimeter constant. In a second experiment (Expt. 2) 100 mL of 2.0 M acetic acid (Ka = 2.0 × 10 −5 ) was mixed with 100 mL of 1.0 M NaOH (under identical conditions to Expt. 1) where a temperature rise of 5.6°C was measured.

(Consider heat capacity of all solutions as 42 J g −1 K−1 and density of all solutions as 10 g mL −1 ) 39. Enthalpy of dissociation (in kJ mol–1 ) of acetic acid obtained from the Expt.2 is (A) 1.0 (B) 10. 0 (C) 24.5 (D) 51.4 Solution

(A) Energy evolved on neutralization of HCl and NaOH = 0.1 57 = 5.7 kJ = 5700 J Energy utilized to rise the temperature of the solution = msT = 200 1 4.2 5.7 = 4788 J Energy used to increase temperature of calorimeter = 5700 – 4788 = 912 J msT = 912 ms5.7 = 912 ms = 160 J C 1 [Calorimeter constant] Energy evolved by neutralization of CH 3COOH and NaOH = 200 4.2 5.6 + 160 5.6 = 5600 J So, energy used in dissociation of 0.1 mol CH 3COOH = 5700 5600 = 100 J Thus, enthalpy of dissociation = 1 kJmol1 40. The pH of the solution after Expt.2 is (A) 2.8 (B) 4.7 (C) 5.0 (D) 7.0 Solution (B) In Expt. 2, the final solution is a buffer as it contains equimolar amounts of acid and salt.

a

saltpH p log

acidK (1)

5a p log(2 10 )

0.3010 5

4.699 4.7

K

3

3

100[Salt] [CH COONa] 2 1 M

200

200 100 100[Acid] [CH COOH] 2 2 1M

200 200

Substituting the values in equation (1), we get

1pH 4.7+log

1

4.7

Paper 2 (MATHEMATICS)

SECTION 1: (Integer Value Correct Type) This section contains 10 questions. The answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive.

41. For any integer k, let cos sin ,7 7

k

k ka i

p pæ ö æ öç ÷ ç ÷= +ç ÷ ç ÷è ø è ø

where 1.i = The value of the expression

12

11

3

4 1 4 21

k kk

k kk

a a

a a

+=

=

å

å is

Solution

(4 ) /7cos sin7 7

ikk

k ka i e pp pæ ö

ç ÷= + =ç ÷è ø

\

12 12( 1) /7 /7

11 1

3 3(4 1) /7 (4 2) /7

4 1 4 21 1

| |i k ikk k

k k

i k i kk k

k k

a a e e

a a e e

p p

p p

++

= =

= =

=

å å

å å

12/7 /7

1

3(4 2) /7 /7

1

1

1

ik i

k

i k i

k

e e

e e

p p

p p

=

=

=

å

å

12

1

3

1

(1)12

431

k

k

=

=

= = =å

å

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

ratio of the sum of the first seven terms to the sum of the first eleven terms is 6:11 and the

seventh term lies in between 130 and 140, then the common difference of this A.P. is

Solution

(9) 7

11

6

11

S

S= …(1)

7130 140t£ £ …(2)

d = ?

Þ

7[2 6 ]

6211 11[2 10 ]2

a d

a d

+=

+

Þ 3 6

5 7

a d

a d

+=

+ …(3)

Þ 4

6

6

4 7

t=

Let t4 = 6 k, t6 = 7k;

2d = k Þ d = k/2

and a + 3 d = 6 k

Þ a = 6 k – 3k/2 = 9k/2

\ 130 ≤ t7 ≤ 140

Þ 9

130 3 1402

kk£ + £

Þ 15

130 1402

k£ £

Þ 52 56

3 3k£ Î

1K NÎ Þ k = 18

Þ 18

92 2

kd = = =

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

Solution

(8) Given expression is

E = (1 + x)(1 + x2)(1 + x3) …….(1 + x100 )

Coefficient of x9 in E

= coefficient of x9 in (1 + x)(1 + x2)(1 + x3) …. (1 + x9)

Þ Terms containing x9

= (1. x9 + x1 .x8 + x2.x7 + x3.x6 + x4 .x5 + x1 .x2.x6 + x1 .x3.x5 + x1 .x4 .x5 )

Þ Term containing x9 is 8x9 in E

Þ Coefficient of x9 = 8

44. Suppose that the foci of the ellipse 2 2

19 5

x y+ = are (f1 , 0) and (f2, 0) where f1 > 0 and f2 < 0.

Let P1 and P2 be two parabolas with a common vertex at (0, 0) and with foci at (f1 , 0) and (2f2,

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

which passes through (f1 , 0). If m1 is the slope of T1 and m2 is the slope of T2, then the value of

222

1

1m

m

æ öç ÷+ç ÷è ø

is

Solution

(4 ) Given ellipse is

2 2

19 5

x y+ = …(i)

foci are (f1 , 0), (f2, 0), f1 > 0, f2 < 0

foci of parabola 1 1( , 0),P fº

focus of parabola 2 2(2 ,0)P fº

Tangent to P1 pass as through (2f2, 0) and tangent to P2 through (f1 , 0).

To find 222

1

1m

m

æ öç ÷+ç ÷è ø

Clearly f1 = 2, f2 = 2

Focus of 1 (2,0)P º

Þ Equation of P1 is y = 8x

Focus of 2 ( 4,0)P º

Þ Equation of P2 is y2 = −16 x

Equation of 1 1

1

2: ,T y m x

m= +

It passes through (−4, 0)

Þ 1

1

24 0m

m + = …(i)

Þ 21

1

2m = …(ii)

||rly equation of T2 :

2

2

( 4),y m x

m

= + which passes through (2, 0)

Þ 2

2

42 0m

m = …(iii)

Þ 22 2m = …(iv)

Þ 222

1

12 2 4m

m+ = + = (Using (ii) and (iv))

45. Let m and n be two positive integers greater than 1. If

cos( )

0lim

2

na

ma

e e e

a®

æ ö æ öç ÷ ç ÷= ç ÷ç ÷ è øè ø

then the value of m

n is

Solution

(2) , 1m x NÎ and m, n > 1

cos( ) cos 1)

0 0

( 1lim lim

n na a

n ma a

e e e eL

a a

® ®

é ù é ù ê ú ê ú= =ê ú ê úë û ë û

cos 1

0

( 1) (cos 1)lim .

(cos 1)

na n

n ma

e e a

a a

®

é ù ê ú=ê úë û

0

cos 1.lim

n

rna

ae

a®


1

10

sin ..lim

.

n n

ma

a nae

m a

®

é ùê ú=ê úë û

2 1

10

sin. .lim

n n

n ma

n a ae

m a a

®

é ùæ öæ ö ê úç ÷ç ÷= ç ÷ ç ÷ê úè ø è øë û

2

0. lim( )

2n m

a

n ee a

m

®

= = (given)

Þ 2

0lim

2n m

a

ma

n

®=

Þ m = 2n

Þ 2m

n=

46. If

121

9 3tan

20

12 9( )

1x x x

a e dxx

+æ ö+ç ÷= ç ÷+è ø

ò

where 1tan x takes only principal values, then the value of 3

log 14

e

pa

æ öç ÷+ ç ÷è ø

is

Solution

(9) 1

219 3tan

20

12 9( ). ,

1x x x

e dxx

a+

æ ö+ç ÷= ç ÷+è øò

Here 1tan , ;2 2

p pæ öç ÷Î ç ÷è ø

3

log 1 ?4

e

pa

æ öç ÷+ =ç ÷è ø

Put 1

2

1tan

1x t dx dt

x = Þ =

+

Þ / 4

9 tan 3 2

0

.(12 9 tan ).t te t dtp

a += +ò

Again put 9 tant + 3 t = z

Þ (9 (sec2t) + 3) dt = dz

Þ (9(tan2t) + 12) dt = dz

Þ 39 3 /4 94

0

1ze dz epp

a

æ ö+ ç ÷+ç ÷è ø= = ò

Þ 9 3 /41 e pa ++ =

Þ ln |1 + α| = 9 + 3π/4

Þ 3

ln (1 ) 94

pa+ =

47. Let f : R → R be a continuous odd function, which vanishes exactly at one point and

1(1) .

2f = Suppose that

1

( ) ( )x

F x f t dt

=ò for all [ 1,2]x Î and 1

( ) ( ( ))x

G x t f f t dt

=ò for all [ 1,2].x Î

If 1

( ) 1lim ,

( ) 14x

F x

G x®= then the value of

1

2f

æ öç ÷ç ÷è ø

is

Solution

(7) F:1R → 1 R is a continuous odd function having a single root,

1

(1) ,2

f = 1

( ) ( )x

F x f t dt

=ò

[ 1,2];x" Î

1

( ) ( ( ))x

G x t f f t dt

=ò [1,2]x" Î

1

( ) 1 1lim , ?

( ) 14 2x

F xf

G x®

æ öç ÷= =ç ÷è ø

Clearly f(t) and and t|f(f(t))| are odd functions for ( 1, )t xÎ and [ 1,2].x Î

\ 1

( ) 1lim

( ) 14x

F x

G x®= (limit is of

0

0 form)

Þ 1

( ) 1lim

( ( )) 14x

f x

x f f x®=

Þ (1) 1

1 ( (1)) 14

f

f f=

Þ 1/2 1

141

2f

=æ öç ÷ç ÷è ø

Þ 1

72

fæ öç ÷ =ç ÷è ø

Þ 1

7,2

fæ öç ÷= ±ç ÷è ø

but being odd, function and continuous f(0) = 0, thus if 1

72

fæ öç ÷= ç ÷è ø

then

f(x) must have another root in 1 1

,1 72 2

fæ ö æ öç ÷ ç ÷Þ ¹ ç ÷ ç ÷è ø è ø

Þ 1

72

fæ öç ÷=ç ÷è ø

48. Suppose that , andp q r

are three noncoplanar vectors in R3. Let the components of a vector

s

along , andp q r

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

along

( ),p q r + +

( )p q r +

and ( )p q r +

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

Solution

(2) 4 3 5s p q r= + +

Also, ( ) ( ) ( )s x p q r y p q r z p q r

= + + + + + +

Þ ( ) ( ) ( )s x y z p x y z q x y z r= + + + + +

Þ

4

3

5

x y z

x y z

x y z

ü + =ïï

= ýï

+ + = ïþ

…(2)

Þ 7

,2

z

= x = 4, 9

2y = Þ 2x + y + z = 9

Ans 2

Section 2 (One or More Options Correct Type) This section contains EIGHT questions. Each has four options (A), (B), (C) and (D). ONE OR MORE THAN ONE of these four option(s) is (a re) correct.

49. Let S be the set of all nonzero real numbers α such that the quadratic equation ax2 – x + a =

0 has two distinct real roots x1 and x2 satisfying the inequality |x1 – x2| < 1. Which of the

following intervals is(are) a subset(s) of S?

(A) 1 1

,2 5

æ öç ÷ ç ÷è ø

(B) 1

,05


(C) 1

0,5


(D) 1 1

,25


Solution

(A), (D) ax2 – x + a = 0 has two distinct real roots x1 and x2; |x1 – x2| < 1.

D > 0 Þ 1 – 4 a2 > 0 Þ 2 1

4a <

Þ 1 1

,2 2

aæ öç ÷Î ç ÷è ø

…(1)

Also |x1 – x2|2 = (x1 + x2)2 – 4 x1 x2 < 1

Þ 2

11 4(1)

a

æ öç ÷> ç ÷è ø

Þ 2

15

a< Þ 2 1

5a >

Þ 1 1

, ,5 5

aæ ö æ öç ÷ ç ÷Î ¥ È ¥ç ÷ ç ÷è ø è ø …(2)

\ Intersection of (1) and (2) gives,

1 1 1 1

. , ,2 25 5

aæ ö æ öç ÷ ç ÷Î Èç ÷ ç ÷è ø è ø

50. If 1 63sin

11a


and 1 43cos ,

9b


where the inverse trigonometric functions take only the

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

(A) cos β > 0

(B) sin β < 0

(C) cos ( α + β) > 0

(D) cos α < 0

Solution

(B), (C), (D) 1 16 63sin 3sin

11 12 2

pa æ ö æ ö

ç ÷ ç ÷= > =ç ÷ ç ÷è ø è ø

Þ 2

pa p< <

Þ 2

pa p< < …(i)

1 14 13cos 3cos

9 2b p æ ö æ ö

ç ÷ ç ÷= > =ç ÷ ç ÷è ø è ø

Also 1 4 33cos

9 2

pp æ ö

ç ÷< <ç ÷è ø

Þ 3

2

pp b< < …(2)

From (1) and (2)

cos α < 0, cos β < 0, sin β < 0,

Also, 3 5

( )2 2

p pa b< + <

Þ cos (α + β) > 0

51. Let E1 and E2 be two ellipses whose centres are at the origin. The major axes of E1 and E2

lie along the xaxis and the yaxis, respectively. Let S be the circle x2 + ( y – 1) 2 = 2. The straight

line x + y = 3 touches the curves S, E1 and E2 at P, Q and R, respectively.

Suppose that 2 2

.3

PQ PR= = If e1 and e2 are the eccentricities of E1 and E2, respectively then

the correct expression(s) is(are)

(A) 2 21 2

43

40e e+ =

(B) 1 2

7

2 10e e =

(C) 2 21 2

5

8e e =

(D) 1 2

3

4e e =

Solution

(A), (B)

2 2,

3PQ PR= =

Let 2 2

1 2 2: 1

x yE

a b+ = …(1)

and 2 2

2 2 21

x yE

c d= + = …(2)

x + y = 3 is a common tangent to s = 0, E1 = 0 and E2 = 0

Þ (1,2);P º

Also equation of tangent to E1 is

2 2y x a b= + +

Þ 2 2 3a b+ =

llrly 2 2 3c d+ =

Þ a2 + b2 = c2 + d2 = 9 …(3)

Þ

Equation of x + y = 3 is given by

1 2 2 2

31/ 2 1/ 2

x yr

= = =±

+

Þ Co ordinates of Q and R are give

by 2

1 ,3

x = 2

23

y = + i.e., 1 8

,3 3


and 2

1 ,3

x = + 2

23

y = i.e. 5 4

,3 3


Point 1 8

,3 3

Qæ öç ÷ç ÷è ø

lie on 2 2 2 21 64 9E b a a bÞ + =

Þ (9 – a2) + 64 a2 = 9a2(9 – a2) (using (3))

Þ (a2 – 1) 2 = 0

Þ a2 = 1

Þ b2 = 8

Also point 5 4

,3 3

Ræ öç ÷ç ÷è ø

lie on 2 2 2 22 25 16 9E d c c dÞ + =

Þ (d2 – 4) 2 = 0

Þ d2 = 4, C2 = 5

Þ 21

1 71 ,

8 8e = = 2

2

4 11

5 5e = =

Þ 2 21 2

7 1 43,

8 5 40e e+ = + = 2 2

1 2

7.

40e e =

Þ 1 2

7

2 10e e =

52. Consider the hyperbola H:x2 – y2 = 1 and a circle S with centre N(x2 ,0). Suppose that H and

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

P intersects the xaxis at point M. If (l, m) is the centroid of the triangle ΔPMN, then the correct

expression(s) is(are)

(A) 2

1 1

11

3

dl

dx x= for x1 > 1

(B)

( )1

21

13 1

xdm

dx x=

for x1 > 1

(C) 2

1 1

11

3

dl

dx x= + for x1 > 1

(D) 1

1

3

dm

dy= for y1 > 0

Solution

(A), (B), (D)

H: x2 – y2 = 1 …(1)

( , )G l mº …(2)

2 2 2 22 1 2 1( ) ( )S x x y x x yº + = +

or 2 2 2 22 1 1 1 22 2 0x y x x x y x x+ + = …(3)

Equation of tangent to circle at P is

2 21 1 2 1 1 1 1 2( ) 2 0xx yy x x x x y x x+ + + =

or 2 21 2 1 1 2 1 1( ) 0x x x y y x x x y + + =

Also equation of tangent to hyperbola (1) at P(x1 , y1 ) is

xx1 – yy1 = 1

It meets x – ay is, where y = 0 at M

Þ 1

1,0M

x

æ öç ÷º ç ÷è ø

Also slope of 1

PMslope of NP

=

Þ 1

11

1 1 2

0 1

01

y

yx

x x x

=

æ ö æ öç ÷ ç ÷ç ÷ ç ÷è ø è ø

Þ 1 1 1 2

21 1

( )

1

x y x x

x y

=

Þ 2 21 1 1 2( )( 1)x y x x x=

Þ 2 21 1 1 2 1( 1) ( )( 1)x x x x x = ( x1 > 1)

Þ x1 = −x1 + x2

Þ 2x1 = x2

Þ 1 2 1

1

1 1 1( , ) , ( )

3 3G l m x x y

x

æ öæ öç ÷ç ÷º º + +ç ÷ç ÷

è øè ø

Þ 1

1

1 13 ,

3l x

x

æ öç ÷= +ç ÷è ø

1

3m = 2

1 1

11

3y x=

( y1 > 0)

Þ 2

1 1

11 ,

3

dl

dx x= 1

21 13 1

xdm

dx x=

and 1

1

3

dm

dy=

53. The option(s) with the value of a and L that satisfy the following equation is(are) 4

6 4

0

6 4

0

(sin cos )

?

(sin cos )

t

t

e at at dt

L

e at at dt

p

p

+

=

+

ò

ò

(A) a = 2, 4 1

1

eL

e

p

p

=

(B) a = 2, 4 1

1

eL

e

p

p

+=

+

(C) a = 4, 4 1

1

eL

e

p

p

=

(D) a = 4, 4 1

1

eL

e

p

p

+=

+

Solution

(A), (C) For a = 2,

6 64

6 40

0

(sin 2 cos 2 )

(sin cos 2 )

t

t

e t tL dt

e st t dt

p

p

+=

+ò

ò

Let 4

6 41

0

(sin 2 cos 2 )tL e t t dtp

= +ò

2 3 4

1 2 30 2 2

( ) ( ) ( ) ( ) yf t dt f t dt f t dt f t dt I I I Ip p p p

p p p

= + + + = + + +ò ò ò ò

In 2nd integrated put t = π + x

Þ dt = dx

and t = π Þ x = 0, t = 2π Þ x = π

i.e. 20

( )I f x dxp

p= +ò

In I3, put t = 2π + x

Þ dt – dx, t = 2π Þ x = 0, t = 3π, x = 2π

Þ 30

(2 )I f x dxp

p= +ò

In I4 , put t = 2π + x Þ dt = dx,

t = 3π Þ x = 0, t = 4 π Þ x = π

Þ 40

(3 )x f x dxp

p= +ò

\ 2 2 3 41

1 1

I I I ILL

I I

+ + += =

2 3

1 1 1 1

1

I e I e I e I

I

p p p+ + +=

4

2 2 1(1 )

1

ee e e

e

pp p p

p

+ + + =

Similarly, 4 1

1

eL

e

p

p

=

for a = 4

54. Let f, g:[−2, 2] → R be continuous functions which are twice differentiable on the interval

(−1, 2). Let the values of f and g at the points −1, 0 and 2 be as given in the following table:

x = −1

x = 0 x = 2

f(x) 3 6 0 g(x) 0 1 −1

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

correct statement(s) is(are)

(A) ( ) 3 ( ) 0f x g x¢ ¢ = has exactly three solutions in ( 1,0) (0,2) È

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

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

(D) ( ) 3 ( ) 0f x g x¢ ¢ = has exactly two solutions in (−1, 0) and exactly two solutions in (0,

Solution

(B), (C) f, g: [−1, 2] → R,

f(x) is twice diff on (−1, 2)

f(−1) = 3, g(−1) = 0,

f(0) = 6, g(0) = 1

f(2) = 0, g(2) = −1

( 3 ) 0f g ¢¢ ¹ on (−1, 0) and (0, 2)

No. of sol. of ( ) 3 ( ) 0f x g x¢ ¢ = in ( 1,0) (0,2) ? È =

Let f(x) = f(x) – 3g(x)

f(−1) = f(−1) – 3g(−1) = 3,

f(0) = f(0) – 3g(0) = 6 – 3(1) = 3

\ By Rolle’s theorem, ( )f x¢ i.e., ( ) 3 ( ) 0f x g x¢ ¢ =

has atleast one root in (−1, 0).

Also f(2) = f(2) – 3g(2) = 0 – 3(−1) = 3

\ again by Rolle’s theorem, ( ) 3 ( ) 0f x g x¢ ¢ = has at least one root in (0, 2)

i.e. ( ) 3 ( ) 0f x g x¢ ¢ = has at least 2 roots in (−1, 2).

Since ( 3 ) 0f g ¢¢ ¹ for (−1, 0) and (0, 2)

Þ f(x) has no point of inflexion in (−1, 0) and (0, 2)

Þ ( 3 )( ) 0f g x¢ ¢ ¹ in (−1, 0) and (0, 2)

Þ ( 3 )( ) 0f g x¢ ¢ = exactly once in (−1, 0)

and exactly once in (0, 2)

55. Let f(x) = 7 tan8x + 7 tan 6 x – 3 tan4 x – 3 tan2x for all , .2 2

xp pæ ö

ç ÷Î ç ÷è ø

Then the correct

expression(s) is (are)

(A) /4

0

1( )

12xf x dx

p

=ò

(B) / 4

0

( ) 0f x dxp

=ò

(C) /4

0

1( )

6xf x dx

p

=ò

(D) / 4

0

( ) 1f x dxp

=ò

Solution

(A), (B) f(x) = 7 tan8x + 7 tan 6 x – 3 tan4 x – 3 tan2x

,2 2

xp pæ ö

ç ÷" Î ç ÷è ø

= 7 tan6 x .sec2x – 3tan2x.sec2x

= (7 tan6 x – 3tan2x).sec2x

Þ / 4 / 4

6 2 2

0 0

( ) (7 tan 3tan )secf x dx x x dxp p

= ò ò

1

2 7 3 10

0

(7 603 ) [ ] 0t dt t t= + = =ò

Also / 4

0

( )I xf x dxp

=ò

/4

/46 2 2 6 2 2

00

. (7 tan 3tan )sec 1. (7 tan 3tan )secx x x x dx x x x dxp

p

= ò ò ò

/4

/47 3 7 3

00

.(tan tan ) (tan tan )x x x x x dxp

p

= ò

/4

3 4

0

0 tan (tan 1)x x dxp

= ò

/ 4

3 2 2

0

tan (tan 1)(sec )x x x dxp

= ò

1

5 3

0

( )t t dt= ò

16 4

0

1 1 1

6 4 4 6 12

t té ù é ùê ú= = =ê úê ú ê úë ûë û

56. Let 3

4

192( )

2 sin

xf x

xp¢ =

+ for all x RÎ with

10.

2f

æ öç ÷=ç ÷è ø

If 1

1/2

( ) ,m f x dx M£ £ò then the possible

values of m and M are

(A) m = 13, M = 24

(B) 1 1

,4 2

m M= =

(C) m = −11, M = 0

(D) m = 1, M = 12

Solution

(D) 3

4

192( ) 1 ,

2 sin

xf x x R

xp¢ = " Î

+

1

0.2

fæ öç ÷=ç ÷è ø

If 1

1/2

( ) ,m f x dx M£ £ò to find m and M.

Here ( )1 1

1/ 2 1/ 2

( ) ( )f x f x dx¢=ò ò ò …(1)

Here, 3 3 3

4

192 192 192

3 2 sin 2

x x x

xp£ £

+

Þ 3

3 3

41/ 2 1/ 2 1/2

192 192 192

3 2 sin 2

x x xxx dx dx x dx

xp£ £

+ò ò ò

Þ 4 4

1/2

192 1 192 1( )

3 4 64 2 4 64

xx xf x dx

æ ö æ ö¢ç ÷ ç ÷ £ £ ç ÷ ç ÷

è ø è øò

Þ 4 41 1 1

64 ( ) 964 64 2 4 64

x xf x f

é ù é ùæ öê ú ê úç ÷ £ £ ç ÷ê ú ê úè øë û ë û

Þ 1 15 51

1/21/2 1/2

16 24 3( )

5 5 2

x xx f x dx x

é ù é ùê ú ê ú £ £ ê ú ê úë û ë û

ò

Þ 1

1/2

26 39( )

10 10f x dx£ £ò

Þ 4 41 1

64 ( ) 964 64 4 64

x xf x

é ù é ùê ú ê ú £ £ ê ú ê úë û ë û

Þ 1 1 1

4 4

1/ 2 1/2 1/ 2

3(16 1) ( ) 24

2x dx f x dx x dx

æ öç ÷ £ £ ç ÷è ø

ò ò ò

Clearly 26 39

, (1,12)10 10

æ öç ÷Ìç ÷è ø

SECTION 2: Comprehension Type (Only One Option Correct)

This section contains 2 paragraphs, each describing theory, experiments, data etc. Six questions

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

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

Paragraph for Questions 57 and 58: Let n1 and n2 be the number of red and black balls,

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

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

randomly out of this box. The ball was found to be red. If the probability that this red ball was

drawn from box II is 1

,3

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

is(are)

(A) n1 = 3, n2 = 3, n3 = 5, n4 = 15

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

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

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

Solution

(A), (B)

2

22

1 2

1 2

. ( )1

3. ( ) . ( )

AP P E

EEP

A A AP P E P P E

E E

æ öç ÷ç ÷æ ö è øç ÷= =ç ÷ æ ö æ öè ø ç ÷ ç ÷+ç ÷ ç ÷

è ø è ø

Þ

3

3 4

31

1 2 3 4

1.21

3 1 1. .2 2

n

n n

nn

n n n n

æ öç ÷ç ÷+è ø=

æ öæ öç ÷ç ÷ +ç ÷ ç ÷+ +è ø è ø

Þ 3 1 2

1 3 4 3 1 2

( )1

3 ( ) ( )

n n n

n n n n n n

+=

+ + +

Þ 1 3 4

3 1 2

1 1

( )31

( )

n n n

n n xn

=+

++

Þ 1 3 4

3 1 2

( )2

( )

n n n

n n n

+=

+

Which are satisfied by options (A) and (B)

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

a red ball from box I, after this transfer, is 1

,3

then the correct option(s) with the possible value

of n1 and n2 is(are)

(A) n1 = 4 and n2 = 6

(B) n1 = 2 and n2 = 3

(C) n1 = 10 and n2 = 20

(D) n1 = 3 and n2 = 6

Solution

(C), (D) P(red from I after transfer) 1

3= (given)

( )Red from I Red from I 1

(Red transfer from I). Black tranfer from I .Red transfer from I Black transfer from I 3

P P P Pæ ö æ öç ÷ ç ÷+ =ç ÷ ç ÷è ø è ø

Þ 1 1 2 1

1 2 1 2 1 2 1 2

1 1. .

1 1 3

n n n n

n n n n n n n n

æ ö æ ö æ ö æ öç ÷ ç ÷ ç ÷ ç ÷+ =ç ÷ ç ÷ ç ÷ ç ÷+ + + + è ø è ø è ø è ø

Þ 3(n1 (n1 – 1) + n2.n1 ) = (n1 + n2)(n1 + n2 – 1)

2 21 1 2 2 22 2 0m n n n n n + + =

Which are satisfied for n1 = 3, n2 = 6

or n1 = 10, n2 = 20

Paragraph for Questions 59 and 60: Let F : R → R be a three differentiable function. Suppose

that F(1) = 0, F(3) = −4 and ( ) 0F x¢ < for all (1/2, 3).x Î Let f(x) = xF(x) for all .x RÎ

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

(A) (1) 0f ¢ <

(B) f(2) < 0

(C) ( ) 0f x¢ ¹ for any (1,3)x Î

(D) ( ) 0f x¢ = for some (1,3)x Î

Solution

(A), (B), (C) F : R → R, thrice differentiable,

F(1) = 0, F(3) = 4,

( ) 0f x¢ < (1/2, 3)," Î

f(x) = x F(x) 1x R" Î

Since ( ) 0F x¢ < 1

,32

xæ öç ÷" Î ç ÷è ø

Þ F(x) is a decreasing function on 1

,32


Þ F(1) > F(x) > F(3) 1 3x" < <

Þ 0 > F(x) > −4 (1,3)x" Î

\ ( ) ( ) ( ) 0veve

f x xF x F x

¢ ¢= + <

(1,3)x" Î

Also (1) (1) (1) (1) 0f F F F¢ ¢ ¢= + = +

Þ (1) (1) 0f F¢ ¢= < as ( ) 0F x¢ < 1

,32

xæ öç ÷" Î ç ÷è ø

Further (2) 2 (2) 0ve

f F

= <

60. If 3

2

1

( ) 12x F x dx¢ = ò and 3

3

1

( ) 40,x F x dx¢¢ =ò then the correct expression(s) is(are)

(A) 9 (3) (1) 32 0f f¢ ¢+ =

(B) 3

1

( ) 12f x dx =ò

(C) 9 (3) (1) 32 0f f ¢ + =

(D) 3

1

( ) 12f x dx = ò

Solution

(C), (D)

32

1

( ) 12x F x dx¢ = ò and 3

3

1

( ) 40;x F x dx¢¢ =ò

9 (3) (1) 32 ?,f f¢ ¢+ ± = 3

1

( )f x dx P=ò

Here,

9 (3) (1) 9(3 (3) (3) ( (1) (1))f f F F F F¢ ¢ ¢ ¢± = + ± +

9(3 (3) 4) (1) 27 (3) (1) 36F F F F¢ ¢ ¢ ¢= ± = ± ..(1)

Now 3 3

32 2

11 1

( ) ( ) 2 ( )x F x dx x F x xF x dx¢ = ò ò

3

1

[9 (3) (1)] 2 ( )F F f x dx= ò

Þ 3

1

12 36 2 ( )f x dx = ò

Þ 3

1

( ) 12f x dx = ò ..(2)

Also 3 3

33 3 2

11 1

( ) ( ) 3 ( )x F x dx x F x x F x dx¢¢ ¢ ¢= ò ò

Þ 40 27 (3) (1) 3( 12)F F¢ ¢=

Þ 27 (3) (1) 4F F¢ ¢ = …(3)

\ From (1) and (3);

27 (3) (1) 36 4 36 32F F¢ ¢ = =

i.e., 9 (3) (1) 32 0f f¢ ¢= + =

paper – 2 section 1: (integer value correct type) this...

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