1

SOLUTIONS TO IRP FULL TEST-2 ONLINE (HELD ON 12TH & 14TH MAY 2018) PHYSICS (PAPER-1)

SECTION 2:

1. In the

figure two

cylindrical conductor

s with same cross-sections and different

resistivity’s A and B are put end to end. A

current I is flowing from A to B. C is the

boundary of conductor A and B. VA and VB

are potential difference across the conductor

respectively. Select the correct option/s

(A) EA < EB if A B

(B) EA = EB if A B

(C) At the common boundary some charge will

be developed

(D) VA = VB if A B and lA = lB

Solution (ABCD):

E

J E

Also if A B A B A B A B& R R V V

2. Two capacitors C1

and C2 are charged

to q1 and q2 and

connected with

capacitor C as shown in figure. As switch S is

closed:

(A) C gets extra charge when 21

1 2

CC

q q

(B) C gets extra charge when 21

1 2

CC

q q

(C) C gets extra charge when 21

1 2

CC

q q

(D) C does not gets extra charge for any value

of q1 and q2.

Solution (AB): C will not get any extra charge only if voltage

across C1 and C2 is same

3. The diagram shows a

circuit with two identical

resistors. The battery has a negligible internal

resistance. When switch

S is closed, (A) Equivalent resistance of the circuit

decreases (B) Ammeter reading will increase

(C) Voltmeter reading will increase

(D) Power dissipated across R in right branch

will become zero.

Solution (ABCD):

One of the resistor gets shunted if key is

closed so current in circuit will increase.

4. A thin conducting rod of length l is moved

such that its end B

moves along the X-axis

while end A moves

along the Y-axis. A

uniform magnetic field 0ˆB B k exists in the

region. At some instant, velocity of end B is v

and the rod makes an angle of 60 with

the x-axis as shown in the figure. Then, at

this instant

(A) angular speed of rod AB is 2v

3

(B) angular speed of rod AB is 3v

2

(C) e.m.f. induced in rod AB is B v 3

(D) e.m.f. induced in rod AB is B v

2 3

Solution (AD):

u vcot

v 2v

sin 3

Consider a imaginary conducting loop ABC.

e.m.f. in this loop = motional e.m.f. in the rod

B d(xy)

2 dt

B dy B vcos2[yv xu] u

2 dt 2sin

5. The system shown in figure is in

equilibrium well above the ground. When the string is cut at t = 0, the

time at which the spring is at its

natural length will be (Take 2 10)

(A) 1

s6

(B) 1

s2

(C) 1

s3

(D) 2

s3

Solution (AB):

T 2k

Now,

T 3T 5Tt , , etc

4 4 4

6. Choose CORRECT statement(s) regarding X-

ray: (A) If the target material in a Coolidge tube is

changed keeping the acceleration voltage

2

constant the minimum wavelength of the

x-rays produced will change.

(B) X-ray do not get deflected on application of electric or magnetic fields

(C) X-ray can ionize a gas

(D) X-ray can be reflected, diffracted and

polarized-under suitable condition.

Solution (BCD): Conceptual

7. A hypothetical experiment conducted to

determine Young’s modulus formula x

3

.TY .

If the errors found in , ,T and Y

are 1%, 2%, 1% and 5% respectively

then the possible value(s) of x can be (A) x=0 (B) x=4

(C) x=6 (D) x=10

Solution (ABCD):

Y T 3 T

xY T

5 x 2 3 x 0,4,6,10

8. A circular conducting loop of

radius r0 and having

resistance per unit length

as shown in the figure is

placed in a magnetic field B which is constant

in space and time. The ends of the loop are

crossed and pulled in opposite directions with

a velocity v such that the loop always remains circular and the radius of the loop goes on

decreasing, then (A) Radius of the loop changes with r as r = r0

–vt/

(B) EMF induced in the loop as a function of

time is 0e 2Bv[r vt/ ]

(C) Current induced in the loop is Bv

I2

(D) Current induced in the loop is Bv

I

Solution (ABD):

If x is radius of loop, then d(2 r)

2vdt

dr v

(constant)dt

0

r t

0 0

r 0

v vt vtdx dt r r r r

Emf 2d d dr

B r 2B rdt dt dt

0vt v

2B r

2B rv Bv

current2 r

9. In the diagram shown, a

light ray is incident on

the lower medium boundary at an angle

45° with the normal.

Which of the following

statement is/are true?

(A) If 2 2 then angle of deviation is 45°

(B) if 2 2 then angle of deviation is 90°

(C) If 2 2 then angle of deviation is 135°

(D) If 2 2 then angle of deviation is 0°

Solution (AB):

For ci i , TIR will occur and then

180 2 45

For ci i ray will finally graze on upper

surface

10. A point object is kept at (1, 0, 0). A circular

plane mirror of radius 1m is kept in yz-plane such that its centre is at the origin. The

reflecting side faces positive x-axis. At which

of the following points can the image of the

object be seen? (A) ( ( 0.5, 0, 0.5) (B) (2, 2, 2)

(C) (1,1.5,1.5) (D) (1, 1,1.5)

Solution (BD):

Image cannot be seen in the region with sign

SECTION 3:

1. Match the following: COLUMN I

(A) Vd – Vc (in volt)

(B) Va – Vb (in volt)

(C) work done by 12V cell is (in J)

(D) work done by 24V cell is (in J)

COLUMN II (P) –8

(Q) +4

(R) –192

(S) 384

Solution (A-Q; B-P; C-R; D-S):

In the loop q q

24 12 04 2

3q

12 q 16 C4

1m

1m

B

vv

3 2

2

1=245°

3

(A) d cq 16

V V 4voltC 4

(B) A b A Bq q

V V V V 8volt2 2

(C) work done qE 16 12 192

(D) work done 24 16 384J

2. In a string a standing wave is set up whose

equation is given as y = 2A sin kx cos t. The

mass per unit length of the string is .

COLUMN I

(A) at t = 0

(B) at T

t8

(C) at T

t4

(D) at T

t2

COLUMN II

(P) Total energy per unit length at x=0 is 2 22 A .

(Q) Total energy per unit length at x /4 is

2 22 A .

(R) Total energy per unit at x is 2 22 A .

(S) Power transmitted through a point at

x is 0.

(T) Power transmitted through a point at

x /4 is 0.

Solution (A-PRST; B-ST; C-QST; D-PRST):

Total energy per unit length is 2 21

(2A) .2

For (A): At t=0

Power F. (at extreme, =0 for all points)

For (B): At T

t8

Power F. ( = 0 at node)

For (C): At T

t8

Power F. (F for all points)

For (D): At T

t2

Power F. (at extreme, 0 for all points)

SECTION 4:

1. A 50 kg man is standing at one end of

a 3m long plank of

mass 100 kg. The

plank is sliding towards the right with

uniform velocity of 2 m/s on the frictionless

horizontal ground. The man walks to the other end of the plank in 5s. Find distance

traveled by the man relative to the ground in

this 5 s interval. Solution (8):

11(50 100)2 50(v 3/5) 100v v

5

M/g P/g M/p M/g

M/g

11 3 8; ;

5 5 5

S 5 8m

2. A light of wavelength 3540Å falls on a metal

having work function of 2.5 eV. If ejected

electron collides with another target metal

inelastically and its total kinetic energy is utilized to raise the temperature of target

metal. The mass of target metal is 10–3 kg

and its specific heat is 160 J/kg/°C. If 1018

electrons are ejected per second, then find the

rate of raise of temperature (in °C/s) of the metal [Assume there is no loss of energy of

ejected electron by any other process, all the

electron are reaching the target metal with

max kinetic energy and take hc = 12400 ev-Å] Solution (1):

Energy of 1 photon hc 12400

eV 3.5eV3540

Thus K.E.max of e ejected = 1 eV

Thus energy lost per electron = 1 eV

= 191 1.6 10 J

Total energy gained by target metal per

second 19 181.6 10 10 J 0.16J

Thus 3

dT dT 0.160.16 ms °C/sec

dt dt 10 160

3. An electron and a proton are separated by a large distance and the electron approaches

the proton with a kinetic energy of 4.11 eV. If

the electron is captured by the proton to form

hydrogen atom in the ground state, the

wavelength of photon given off is 210 Å?

Fill the value of in your OMR sheet. Solution (7):

Since initially electron and proton are

separated by large distance potential

energy = 0

Thus when electron and proton combine to

form a hydrogen atom E (13.6 4.11)eV

hc 12400E Å 700Å

17.71

4

2m m

l

2020

W E

N

S

1045

4. An open organ pipe containing air resonates

in fundamental mode due to a tuning fork.

The measured values of length (in cm) of the

pipe and radius r (in cm) of the pipe are

94 0.1, r 5 0.05. The velocity of the

sound in air is accurately known. The

maximum percentage error in the

measurement of the frequency of that tuning

fork by this experiment is given by 2% . Find

the value of 10 .

Solution (4):

v

f2( 2e)

where e = end correction = 0.6r

v v

f2( 2 0.6r) 2( 1.2r)

f v ( 1.2r)

f v 1.2r

v 1.2 r

v 1.2r

Here v

v

= 0 (given)

f100

f

1.2 r

1001.2r

For maximum % error : = 0.1, r = 0.05

max

f 0.1 1.2 0.5100 100

f 94 1.2 5

0.16%

5. A horizontal plane with the coefficient of

friction µ supports two bodies: a bar and an electric motor with a battery on a block. A

thread attached to the bar is wound on the

shaft of the electric motor. The distance

between the bar and the electric motor is

equal to l. When the motor is switched on, the bar, whose mass is twice as great as that of

the other body, starts moving with a constant

acceleration w. The bodies collide at

t(3w µg)

nl , n here is?

Solution (2): For Bar

T 2mg 2mw

T mg mw

mg mw 2mw

w g 2w

So, 21 2(w w )t t

2 w w

2

tg 3w

6. In a car race, car A takes time t less than car

B and passes the finishing point with a

velocity v more than the velocity with which

car B passes the point. Assuming that the

cars start from rest and travel with constant

accelerations 9 m/s2 and 4 m/s2

respectively. Then the value of v/t is n. Find n Solution (6):

B Bv v 0 9(t t)

2 2

B B1 1

9(t t) 4t2 2

B B B3(t t) 2t t 3t

B Bv 0 4t

B Bv v 9(t t)

B Bv 4t 9t 9t

Bv 5t 9t

v 15t 9t 6t

v/t 6

7. Two ships A and B are 10 km apart on a line running south to north. Ship A farther north

is streaming west at 20 km/hr and ship B is

streaming north at 20 km/hr. Their distance

of closest approach is n 2 km. Find n ?

Solution (5):

Velocity of B w.r.t. A = ˆ ˆ20j 20i

Closest distance

10sin45

5 2

8. A charged particle is moving

with constant speed v in x-y

plane in a straight line as

shown in figure. Suddenly a

uniform magnetic field is switched on along y direction when particle is

at origin. The value of (in degree) so that

the particle passes through point P(0, ,

2 3 ) in the minimum possible time is .n

Find n?

Solution (3): Since x is zero and z coordinates is non zero

so particle has completed half circle, So

T m

vcos vcos2 qB

… (i)

mvsin

2 3 2R 2qB

… (ii) From (i) and (ii)

3

cos . tan 3sin 3

5

5

CHEMISTRY (PAPER-1)

SECTION 2:

1. Select the incorrect statement(s):

(A) 1IE of nitrogen atom is less than 1IE , of oxygen atom

(B) Electron gain enthalpy of oxygen is less negative than selenium

(C) Electronegativity an Pauling scale is 2.8 times the electronegativity an Mulliken scale.

(D) 6Cr is smaller then

3Cr .

Solution (AC):

2. For 3Mn

pairing energy is 28000 1cm

, o for 3

6[Mn(CN) ] is 38500

1cm then which of the

following is/are correct:

(A) Complex will be coloured (B) Complex will be low spin complex

(C) Net 1CFSE 33600cm (D) Complex will be colourless

Solution (BCD):

7cb

1259 10 em 259 nm

38500

(u.V region) and 2,1,1 0,02gt , eg

3. Select the correct statement with respect to the p d dative bond:

(A) In 3 2(Ph Si) O, the Si O Si group is nearly linear.

(B) Silanols such as ( 3 3(CH ) SiOH are stronger protonic acids then their carbon analogs

(C) Trisilysl phosphine 3 3(H Si) P is planar

(D) 4ClO does not polymerise at all

Solution (ABD):

(A) steric repulsions of bulkier groups and p d dative bonding favour for linear Si O Si

groups

(B) Due to stabilization of the conjugate base only by O(p ) Si(d ) bonding motion.

(C) It is pyramidal because p d bonding is not effective due to bigger size of phosphorous

atom

(D) Mont effective, p d overlapping due to small size of chlorine

4. The reaction 22NO Br 2NOBr follows the mechanism;

1. fast

2 2No Br NOBr

2. slow

2NOBr NO 2NOBr

Which of the following is/are true regarding this:

(A) The order of reaction with respect to NO is two (B) The molecularity of the steps (1) and (2) are two each.

(C) The moleuclarity of the overall reaction is three.

(D) The overall order of reaction is three.

Solution (ABD):

5. 10 moles of a liquid are 50% converted into its vapour at its boiling point o(273 C) and of a pressure of

1 atm. If the value of latent heat of vaporization of liquid is 273 L atom/mole then which of the

following statement is/are. Correct: Assume volume of liquid to be negligible and vapour of the liquid to

behave ideally:

(A) work done by the system in the above process is 224 L atm

(B) the enthalpy change ( H) for the above process is 1365 L atm (magnitude only)

(C) the entropy of the system increases by 2.5 L atm in the above process

(D) the value of U for the above process is 1589 L atm

Solution (ABC):

fV 5 R 546 224L

ext.W P ( V) 1atm(224L) 224 L atm

Work done by system 224L atm

6

OH

OH

O

OHOH

OOO

COOHOH

HOOC

OH

2CH I

OHOOC

O

Enthalpy change ( H) Q 273 5 1365 L atm

vap.H 1365S 2.5L atm/k

T 546

U q w 1365 224 1141L atm

6. Select the correct statement(s) related to hexagonal close packing of identical sphere in three

dimensions: (A) In one unit cell there are 12 octahedral voids and all are completely inside the unit cell

(B) In one unit cell there are six octahedral voids and all are completely inside the unit cell

(C) In one unit cell there are six octahedral voids and out of which three are completely inside the unit

cell and other three are from contribution of octahedral voids which are partially inside the unit cell

(D) Co-ordination number of every sphere is 12 in hcp lattice Solution (BD):

hcp AB AB AB.......... pattern repeat

For calculating voids between two layers A and B octahedral void=3

So total octahedral voids=6=all are completely inside.

7. Which of the following statements are correct:

(A) Average velocity of molecules of a gas in a container is zero

(B) All molecules in gas are moving with the same speed

(C) If an open container is heated from 300 K to 400 K the fraction of air which goes out with

respect to originally present is 1/4

(D) It compressibility factor of gas at STP is less than unity then its molar volume is less than 22.4l at STP

Solution (ACD):

C- 1 1 2 2n T n T Escaped out

3n n

14n 4

2n 300 n 400

23

n n4

D- PV

1 Z 1RT

Molar volume for Z 1 is 22.4 littre so for Z 1 molar volume is less than 22.4 litre at STP.

8. The products of the following reaction can be:

(2) CH I(i) CHCl /NaOH (3) (i) AgNO NH OH2 23 3 4

(ii) HP

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

Solution (AC):

OHOH

3(1) CHCl /NaOH

Re imer Tiemann reaction

OHC

ONa

ONa

+

CHO

ONa

ONa

CHO

+2 2CH I

O

O

O

O

CHO(3)

COOH

+

O

O

O

O

COOH

7

7

2H C14

2C H

O

182H /H O

9.

Which cannot be the product:

(A)

14

2 218

H C CH| |

HO HO

(B)

14

2 2

18

CH CH| |

HO OH

(C)

2 218 18

CH CH| |OH OH

(D)

14

2 2CH CH| |OH OH

Solution (CD):

Cis-(1R-3S)-disec-butylcyclobutane

10. What is/are true about the following compound:

(A) it has centre of symmetry (inversion centre) (B) It has plane of symmetry

(C) it does not have two fold axis 2(C ) of symmetry (D) It is an chiral molecule

Solution (BCD):

It is a meso achiral compound and it has no. centre of symmetry and no. axis of symmetry but plane of

symmetry.

SECTION 3:

1. Match the following column I & column II.

Given o

2Cu /CuE 0.34V,

o

Cl /Cl2

E 1.36V,

o

Br /Br2

E 1.08V ,

o

I /I2

E 0.54V

COLUMN I COLUMN II

(A) 2

2Cu 2Cl Cu Cl (P) Can produce electricity in the galvanic cell

(B) 2

2Cl Cu Cu 2Cl (Q) Can be made to occur in electrolytic cell

(C) 2I starch solution + chlorine water (R) appearance of brown colour

(D) 42Br CCl chlorine water (S) Appearance of violet colour

Solution (A-Q, B-P, C-S, D-R): o ored red 2Cl /Cl Cu /Cu2

(E ) (E )

A Q So reduction of 2Cl to Cl will occur at cathode and oxidation of Cu to 2Cu

will occur at

anode. So reverse reaction is possible it means such revere reaction can be made to occur in an

electrolysis cell.

B P as mentioned in (a) the reaction is feasible and hence can produce electricity in the galvanic

cell.

C S 2 22I Cl I (violet) 2Cl

182H O14

2 2H C CH

O

H14 14

2 2 2 2CH CH CH CH

18HO OH HO 18OH

+

a

b

3H C3CHH

5 2H C

Cis-(1R-3S)-disec-butylcyclobutane

H

2 5C H

8

NH NH+

2CH

2NO 2CH

O

+

o ored redCl /Cl Br /Br2 2

(E ) (E )

D R 2 22Br Cl Br (brown) 2Cl

o ored redCl /Cl Br /Br2 2

(E ) (E )

2. Match the following column I & column II:

COLUMN I COLUMN II

(A) (P) Stabilised by hyperconjugation

(B) (Q) Stabilised by -I effect

(C) (R) Stabilised by +M effect

(D) (S) Stabilised by M effect

Solution (A-PR; B-QS; C-P; D-R)

SECTION 4:

1. Electrons in a sample of H-atoms make. Transitions from state n=x to some lower excited state. The emission spectrum from the sample is found to contain only the lines belonging to a particular series. If

one of the photons had on energy of 0.6375eV. Then find the value of x 3

(take 0.6375eV 0.85eV)4

Solution (8):

We have 3

E 0.85eV4

as energy 0.6375 the photon will belong to Brackett series (as for

Brackett 0.31 E 0.85)

2 2

1 1 10.85 1 13.6

4 4 n

21 13.6 4

0.85 1 14 16 n

n 8

2. At 2 atm. pressure the value of x/m will be where x is the mass gas adsorbed per unit mass of

adsorbent according to given graph:

Solution (4):

x 1log logK logP

m n

|aK |0/2 K 2 nx

K(P)m

11

n n 1

xx2 2 4

m

log 2

log P

o45log x/m

9

9

O

O

H O

Br Br

Br

2 5OC H

3. 3NH is heated initially of 15 atm from o27 C to

o127 C of constant volume. At o127 C equilibrium is

established. The new pressure at equilibrium at o127 C becomes 30 atm for the reaction

3 2 22NH (g) N (g) 3H (g). Then find the fraction of moles of 3NH actually decomposed.

(represent an … 1x 10 )

Solution (5):

3 2 2a 0 0

a 2x x 3x

2NH N 3H

2 212

1 2

P PP 15P 20 atm

T T 300 400

Now; a 2x 30 1

x aa 20 4

Fraction of 32x 1

NH 0.5a 2

4. Among the following complexes how many have ‘spin only’ magnetic moment d 2.83 BM? 2 2 2 4

4 4 2 3 2 4 6 6[Ni(CO) ], [Ni(CN) ] , [NiCl (PPh ) ], [NiCl ] ,[NiF ] , [NiF ] ,

2 2 23 6 3 2 6[Ni(NH ) ] ,[Ni(en) ] ,[Ni(H O) ]

Solution (6): 2 4 2 2 2

2 3 2 4 6 3 6 3 2 6[NiCl (PPh ) ],[NiCl ] ,[NiF ] ,[Ni(NH ) ] [Ni(en) ] ,[Ni(H O) ]

5. (i) CH MgBr(3equivalent)3

(ii) H /H O2

product

How many bromine atoms are present in final major product?

Solution (2):

6. Glucose is subjected to bond cleavage by 4HIO . The number of formic acid unit(s) formed per unit of

glucose is…..

Solution (5):

O

O

H O

Br Br

2 5C OC H(i) CH MgBr(3equivalent)3

(ii) H /H O2

OH

Br

2 5COOC H

CH3CH

Br Br OH

2CH OH

HO

H

H

OH

OH

H

O

H OH

O O

C H

5H C OH 1H C H

10

3CH3CH

OH OH

Ph Ph

7. In the stander dization of 2 2 3Na S O using 2 2 7K Cr O by iodometry the equivalent weight of 2 2 7K Cr O is

molecular weight

x

Solution (6):

2 32 27

2

2 32 2 27

Cr O 14H 6e 2Cr 7H O

(2I I 2e ) 3

Cr O 14H 6I 2Cr 3I 7H O

So eq. weightmoleuclar weight

6

8. Conc. H SO .2 4 Products

How many number of different type of carbonyl products (only structural isomers) can be formed (major

or minor) in this reaction, (considering all type of possible migrations): Solution (4):

Pincol pinacolone rearrangement.

11

MATHEMATICS (PAPER 1) SECTION 2:

1. If the roots of the equation

2 2c x c a b x ab 0 are sinA, sinB where

A, B and C are the angles and a, b, c are the

opposite sides of the triangle ABC, then

(A) the triangle is right angled

(B) the triangle is acute angled

(C) the triangle is obtuse angled

(D) sinA cos A a b /c

Solution (AD):

2

c a b sinA sinBsinA sinB

sinCc

sinC 1 oC 90 oA B 90

cos A sinB

Hence sinA cosA sinA sinB a b

c

.

2. The vectors a,b & c are of the same length

and taken pair-wise, they form equal angles.

If ˆ ˆa i j and ˆ ˆb j k the coordinates of c

are

(A) 1,0,1 (B) 1,2,3

(C) 1,1,2 (D) 1 4 1

, ,3 3 3

Solution (AD):

ˆ ˆa i j and ˆ ˆb j k a b 2 .

Hence c 2 .

Let 1 2 3ˆ ˆ ˆc c i c j c k .

2 2 21 2 3c c c 2 2 2 2

1 2 3c c c 2

Also ˆ ˆ ˆ ˆ ˆ ˆa.b i j 0k . 0i j k

a b cos 0 1 0 2. 2 cos 1

1

cos2

1 2 3ˆ ˆ ˆ ˆ ˆa.c i j . c i c j c k

1 2 3ˆ ˆ ˆ ˆ ˆa . c cos i j . c i c j c k

1 21

2. 2. c c2 1 2c c 1

Similarly 2 3c c 1

1 2 3 2c 1 c ,c 1 c

2 2 21 2 3c c c 2

2 222 2 21 c c 1 c 2

22 23c 4c 2 2 2

4c 0 or

3

For 2c 0 : 1 3c 1, c 1 ˆ ˆc i k

c 1,0,1

For 24

c3

: 1 31 1

c , c3 3

1 4 1ˆ ˆ ˆc i j k3 3 3

1 4 1

, ,3 3 3

3. If 1 2 nA ,A ,....,A are n independent events

such that i1

P A ,i 1,2,...,ni 1

. The

probability that none of 1 2 nA ,A ,...,A occurs

is

(A) n

n 1 (B)

1

n 1

(C) less than1

n (D) greater than

1

n

Solution (BC):

Since 1 2 nA ,A ,...,A are independent events

therefore

/ / /2 nP A A ... A / / /

n1 2P A P A ....P A

1 1 1

1 1 ... 12 3 n 1

1 2 3 n 1 n 1 1

... .2 3 4 n n 1 n 1 n

4. Number of integral values of ‘b’ for which

inequality 2 2a 1 x 4 a b x 2 0 is true

for atleast one ‘x’ & a R , is

(A) 0 (B) 1

(C) 2 (D) 3

Solution (A):

Since 2 2a 1 x 4 a b x 2 0 , for at

least one x, hence discriminant must be

positive.

2 24 a b 4 a 1 2 0 a R

2 2 216 a b 2ab 8 a 1 0 a R

2 2 22 a b 2ab a 1 0 a R

2 2a 4ab 2b 1 0 a R

2 2a a 4b 2b 1 0 a R

Since above quadratic in ‘a’ must be positive a R , hence its discriminant must be

negative.

2 24b 4 1 2b 1 0

2 216b 8b 4 0 28b 4 0 , which is

not possible. Hence number of possible

integral values of b is 0.

5. Consider a square on the complex plane. The

complex numbers corresponding to its four vertices are the four distinct roots of the

equation with integer coefficients

12

4 3 2x px qx rx s 0 , then the minimum

area of the square is

(A) 1

2 (B) 1

(C) 2 (D) 2

Solution (C):

Let origin be the centre of the square and edge

length be 2a units.

Area 22a 2a 4a

As per question:

a ai, a ai, a ai & ia a are root of the

equation: 4 3 2x px qx rx s 0

Product of the roots 4 s

P 11

a ai a ai a ai a ai

s a 1 i a 1 i a 1 i a 1 i

4s a 1 i 1 i 1 i 1 i

4s a 2 2 4s 4a

Since s is an integer, therefore minimum

value of 4a for which 44a is an integer is

4 1a

4 .

Area 2 4 14a 4 a 4

4

14 2

2 sq.units.

6. If is the imaginary cube root of unity such

that

n r n r

P 1 P 1

r 1 p 1 r 1 p 1

r. 155 r. 155

, where n is multiple of 3, then possible

values of n can be

(A) 27 (B) 30

(C) 33 (D) 36 Solution (B):

n r

p 1

r 1 p 1

r. 155

n

o 1 2 r 1

r 1

r .... 155

1 0 1 0 1 21 2 3

0 1 2 3 0 1 2 n 14 ... n ...

155

n1 4 7 .... terms

3

2 2n2 5 8 ... terms 155 1

3

2n n n2 1 1 3

3 2 3 3 2

2n2 2 1 3 155 155

3

2 2n n2 n 3 4 n 3 155 155

6 6

2 2n nn 1 n 1 155 155

6 6

2n nn 1 155 155 n 1

6 6

As per question:

2n nn 1 155 155 n 1

6 6

2n nn 1 155 155 n 1

6 6

2n nn 1 155 155 n 1

6 6

is purely real

n

155 n 1 06

155 6 n n 1

2n n 155 6 0 n 30 n 31 0

n 30, 31

From the given options only option (B) is

correct.

7. If the grid shown below consists of n rows

then number of triangles orienting the same way as ABC are

(A) n 12C

(B) n 23C

(C) n 13C

(D) n 22C

Solution (C):

There is one triangle of height n, and 1 2

triangles of heights n 1 (these are AQR,

PBN, LMC). Similarly there are 1 2 3

triangles of height n 2 , and 1 2 ...n

triangles of height 1. Hence the number of

triangles in the grid the same way as ABC are

1 1 2 1 2 3 ... 1 2 ... n .

n 1 n 1 2 n 2 3 ... 1 n

n

r 1

n r 1 r

13

n n n n

2 2

r 1 r 1 r 1 r 1

n 1 r r n 1 r r

n n 1 n n 1 2n 1

n 12 6

n n 1 2n 1n 1

2 3

n n 1 3n 3 2n 1

2 3

n n 1 n n 1 n 2n 2

2 3 6

n 23C

8.

n

nn

3n 1lim

4n 1

is equal to

(A) 3

4 if n is even (B) 0 if n is even

(C) 3

4 if n is odd (D) none of these

Solution (AC):

CASE I: If n is even, say n 2k. Then the

limit is:

2k

2kk k

6k 1 6k 1lim lim

8k 18k 1

k

16

6 3klim1 8 4

8k

CASE II: If n is odd, say n 2k 1. Then the

limit is:

2k 1

2k 1k k

3 2k 1 1 6k 3 1lim lim

8k 4 14 2k 1 1

k

46

3klim5 4

8k

n

nk

3n 1 3lim

44n 1 1

, n even or odd.

9. Consider a circle with its centre lying on the

focus of the parabola 2y 2px such that it

touches the directrix of the parabola. Then

point(s) of intersection of the circle and the

parabola is/are:

(A) p

,p2

(B) p

, p2

(C)p

,p2

(D)p

, p2

Solution (AB):

2y 2px 2 py 4 x

2

…(i)

Hence focus is at p

,02

& equation of

directrix is p

x2

2x p 0 . Since the

circle touches the directrix therefore the

radius r of circle is:

2 2

p2 p

2p2r p

22 0

Hence equation of the circle is:

22 2p

x y p2

2

2 2 3px y px 0

4

…(ii)

Eliminating y from (i) and (ii):

2

2 3px 2px px 0

4 2 23

x px p 04

2 2p p 3p p 2px

2 2

p 3p

x ,2 2

p

x2

, (x cannot be negative)

2 py 2p

2

y p .

Hence the points of intersection are

p p,p , , p

2 2

.

10. A variable point P on the ellipse 2 2

2 2

x y1

a b is

joined to its foci S & /S . Let e be the

eccentricity of this ellipse. The correct

statements are (A) The locus of the incentre of the triangle

/PSS is the ellipse

2 22

2 2 2 2

e 1 yx1

a e b e

.

(B) The locus of the incentre of the triangle /PSS is the hyperbola

2 22

2 2 2 2

e 1 yx1

a e b e

.

(C) The eccentricity of the locus of the

incentre of the triangle /PSS is

1/21 e

2e

.

(D) The eccentricity of the locus of the

incentre of the triangle /PSS is

1/22e

1 e

.

14

Solution (AD):

Let the point P be acos ,bsin .

Also S ae,0 and /S ae,0

PS a ex a aecos /PS a ex a aecos /SS 2ae

Hence coordinates h,k of incenter of the

triangle /PSS are given by:

1 2 3ax bx cxh

a b c

2ae a cos a aecos ae

a aecos ae

2ae a aecos a aecos

2 22ae a cos 2a e cos

2ae 2a

2aecos ae cos

e 1

aecos

1 2 3ay by cyk

a b c

2ae bsin a aecos 0

a aecos 0

2ae a aecos a aecos

2ae bsin besin

2ae 2a e 1

k e 1h

cos & sinae be

Eliminating :

22h e 1

k 1ae be

222

2 2 2 2

y e 1x1

a e b e

.

Hence option (A) is correct. Also eccentricity /e of the above ellipse is given by:

2 2

2/

2 2

b e

e 1e 1

a e

2

22

b1

a e 1

2

2

1 e1

e 1

,

22

2

b1 e

a

2

2

2e 2e 2e

e 1e 1

Hence option (D) is correct.

SECTION 3:

1. Match the following:

‘n’ whole numbers are randomly chosen and multiplied.

COLUMN-I COLUMN- II

(A) The probability that the (P)n n

n

8 4

10

last digit is 1, 3, 7 or 9 is

(B) The probability that the (Q)n n

n

5 4

10

last digit is 2, 4, 6, 8 is

(C) The probability that the (R)

n4

10

last digit is 5 is (D) The probability that the

last digit is zero is

(S) n n n n

n

10 8 5 4

10

Solution (A R, B P, CQ, D S):

(A) The required event will occur if last digit

in all chosen numbers is 1, 5, 7 or 9.

Required probability =

n4

10

(B) The required probability is equal to the

probability that last digit is 2, 4, 6, 8

n n

n

8 4P 1,2,3,4,5,6,7,8,9 P 1,3,7,9

10

(C) n n

n

5 4P 1,3,5,7,9 P 1,3,7,9

10

(D) n n n nP 0,5 P 5 10 8 5 4

n n n n

n

10 8 5 4

10

2. Match the following:

COLUMN I COLUMN II

(A) If x 1 & 1 2sin cot x 1 dx

(P) 0

f x k , where f 1 0 , then

the value of 2f e is equal to

(B) If

2

2012

sec xdx g x C

tan x tanx

, (Q) 1

where n2

g4 2011

, then

x

2

lim g x

is equal to

(C) If the function x

f xx

(R) 2

is self inverse, then

can be

15

(D) Given x 0,1 & (S) even number

x

20

dtf x

1 t

,

then the value of

22f x f 1 x

is equal to

Solution (ARS, B PS, C PQR, DQ):

(A) 1 2sin cot x 1 dx f x k

1

dx f x kx

lnx C f x k

f x lnx & k C

2 2f 1 0 & f e lne 2

(B)

2

2012

sec xdx g x C

tan x tanx

Let tanx t 2sec x dx dt

2012

dtg x C

t t

2012

2011

t dtg x C

1 t

Let 20111 t y

20122011 t dt dy

2012 1t dt dy

2011

dy

g x C2011 y

1

ln y C g x C2011

20111ln 1 t C g x C

2011

20111ln 1 tan x C g x C

2011

20111g x ln 1 tan x

2011

2011

2011

1 cos xln 1

2011 sin x

x

2

1 0lim g x ln 1

2011 1

1 1

ln1 0 02011 2011

.

(C) x

f xx

Let x

yx

xy y x xy x y

y

xy 1

1 xf x

x 1

Since f x is self in verse.

x x

, x Rx x 1

2 2 2x x x x x x

x R

2 2x 1 x 1 1 0

x R

21 0 & 1 0 & 1 0

1 & R

(D) x

20

dtf x

1 t

x

1

0

f x sin t

1f x sin x

2 1 2f 1 x sin 1 x

2 1f 1 x cos x

2 1 12 2f x f 1 x sin x cos x

2

12

.

SECTION 4:

1. Given that the vectors a and b are non-

collinear. The value of x y for which the

vector equality 2u v w holds, where

u xa 2yb, v 2ya 3xb, w 4a 2b , is

Solution (2):

2u v w

2 xa 2yb 2ya 3xb 4a 2b

2x 2y 4 a 3x 4y 2 b 0

2x 2y 4 0 and 3x 4y 2 0

a,b are linearly independent

10 4

x ,y7 7

x y 2 .

2. Let N be the number of natural numbers

which are smaller than 82 10 and which can

be written by means of the digits 1 and 2. The

value of N 760 is

Solution (6): 82 10 200000000, which is nine digit

number. Hence all the numbers of 1 digit, 2

16

digit, …, 8 digits formed by 1 and 2 are clearly

smaller that 82 10 . Also all the 9 digit

numbers beginning with 1 are smaller than 82 10 .

Number of one digit numbers 12 2

Number of two digit numbers 22 2 2

Number of three digit numbers 32 2 2 2

Number of eight digit numbers 82

Number of nine digit numbers beginning with one (the first place is to be filled with 1 & remaining 8 places can be filled in 2ways viz:

1 & 2) 8 81 2 2 .

Hence total number of required numbers

8

2 3 8 8 82 2 1

N 2 2 2 ... 2 2 22 1

9 82 2 2 512 2 256 766 .

N 760 6 .

3. The value of 11 12sin 11 sin 126 6

19 20... sin 19 sin 206 6

is

Solution (0): 11 13 15 17 1911 ,13 ,15 ,17 ,19 are odd and

12 1412 ,14 , 16 18 2016 ,18 ,20 are even. Hence

For n 11,13,15,17,19

n 1sin n sin

6 6 2

& for

n 12,14,16,18,20. n 1sin n sin

6 6 2

Hence the required sum

1 1 1 1 1 1 1 1 1 10

2 2 2 2 2 2 2 2 2 2 .

4. The value of k for which the function

4x

2 2

2

e 1f x

x xsin log 1

2k

, x 0 , f 0 8

may be a continuous function at x 0 is

(given that k 0 ):

Solution (2):

4

x

2 2x 0

2

e 1lim

x xsin log 1

2k

2 24

x 22

2 2x 0

2

x x

e 1 2klim 2kx x x

sin log 12k

4 2 2 21 1 1 2k 2k

For continuity at x 0 :

x 0

f 0 lim f x

28 2k k 2

5. If the product of the perpendiculars drawn

from the point 1,2 to the pair of lines

2 2x 4xy y 0 is P then 13

4P is equal to

Solution (1):

We know that the product of perpendiculars

drawn from 1 1x ,y to the lines represented by

2 2ax 2hxy by 0 is

2 21 1 1 1

2 2

ax 2hx y by

a b 4h

.

Here 1a 1,2h 4,b 1,x 1 , 1y 2 .

Required product

2 2

2 2

1 1 4 1 2 1 2 13P

41 1 4 2

13

14P

6. If g x

30

dtf x

1 t

, where

cos x

2

0

g x 1 sint dt then the value of

/1 f2

is equal to

Solution (0):

g x

30

dtf x

1 t

//

3

g xf x

1 g x

…(i)

Also cos x

2

0

g x 1 sint dt

/ 2g x 1 sin cos x sinx …(ii)

Hence from (i) and (ii):

2

/

2

1 sin cos x sinxf x

1 g x

…(iii)

Also cos

22

0

g 1 sint dt2

0

2

0

1 sint dt 0

…(iv)

17

2

/

2

1 sin cos sin2 2

f2

1 g2

2

1 sin01

1 0

/1 f 02

7. The number of zeros in the end in the product

of 6 7 8 9 515 6 7 8 ... 50 must be

Solution (356):

Since, one pair of 2 and 5 makes 10 (i.e.,

2 5 10 ) which gives one zero at the end.

Here,

6 11 16 21 26 315 ,10 ,15 ,20 ,25 ,30 ,

41 46 5140 ,45 ,50

Total numbers of fives

6 11 16 21 26 31 36 41 46 51 26 51

10

6 51 772

356

Since, number of s2 is greater than that of 5.

Thus, number of zeros at the end 356 .

8. In a ABC , the maximum value of

2 Aa cos

21000

a b c

must be

Solution (750):

2 2 2A B Ca cos bcos ccos

2 2 2

a b c

a 1 cos A b 1 cosB c 1 cosC

2 a b c

a b c a cos A bcosB ccosC

2 a b c

1 a cos A bcosB ccosC

2 4s

1 R

sin2A sin2B sin2C2 4s

a b c2R

sinA sinB sinC

1 R

4sinAsinBsinC2 4s

2

1 R a b c 1 abc4. . .

2 4s 2R 2R 2R 2 9R s

2

1 4R 1 r 1 r 1 11 1

2 2 2R 2 R 2 28R s

r 1

R 2

2 2 2A B Ca cos bcos ccos

32 2 2

a b c 4

2 2 2A B Ca cos bcos ccos

2 2 21000

a b c

31000 750

4

**************

18

IO

Principle

axis

IO

Principle

axis

PHYSICS (PAPER-2) SECTION 1:

PARAGRAPH FOR NEXT TWO QUESTIONS:

A light of wavelength is incident on a metal

sheet of work function 2eV. The wavelength

varies with time as = 3000 + 40t, where is

in Å and t is in second. The power incident on metal sheet is constant at 100 W. This signal is

switched on and off for time intervals of 2

minutes and 1 minutes respectively. Each time

the signal is switched on, the start from initial

value of 3000Å. The metal plate is grounded and

electron clouding is negligible. The efficiency of

photoemission is 1% (hc = 12400 eVÅ, 1eV = 1.6

× 10–19 J) 1. The time after which photoemission will stop

is

(A) 120 s (B) 80 s

(C) 60 s (D) 180 s Solution (B):

wf 00

hc6200 Å

for emission

0 3000 40t 6200 t 80sec

2. The variation of rate of emission vs time is

(A) (B)

(C) (D)

Solution (B):

dN P 100

(3000 40t)dt hc hc

PARAGRAPH FOR NEXT TWO QUESTIONS In the shown circuit the key is shifted from position 1 to position 2

3. Final charge on capacitor B will be

(A) 2 coulomb (B) 4 coulomb (C) 1 coulomb (D) zero

Solution (A): Charge on B will not charge due to shifting of

key because the circuit becomes open

4. Extra work done by 2 volt cell due to shifting

of key is

(A) 8 J (B) zero

(C) 12 J (D) 4 J

Solution (C): Charge through cell = 6 coulomb

Work done = qE = 12 joule

SECTION-2:

1. The figure shows

positions of object O

and its diminished

image I. This is

possible if (A) a concave mirror is placed to the right of I

(B) a convex mirror is placed between O and I

(C) a concave lens is placed to the right of I

(D) a concave lens is placed between O and I

Solution (BC): Conceptual

2. Two pulses on the same string are described

by the following wave equations:

1 22 2

5 5y and y

(3x 4t) 2 (3x 4t 6) 2

where x and t are in m and second

respectively. Choose the correct statement.

(A) Pulse y1 and pulse y2 travel along positive

and negative x-axis respectively. (B) At t=3/4s, displacement at all points on

the string is zero.

(C) At x=1m displacement is zero for all time

(D) Energy of string is zero at t = 3/4s

Solution (ABC):

For (A): 1 1 2 2y f (x vt) and y f (x vt)

For (B): At 1 2

3t s, y y y

4

2 2

5 50

(3x 3) 2 (3x 3) 2

For (C): At x = 1m

2 2

5 5y 0

(3 4t) 2 (4t 3) 2

For (D): At 3

t s4

potential energy of string is

minimum and kinetic energy is maximum.

3. Two identical incandescent light bulbs are connected as shown in the figure. When an

AC voltage source of frequency f is applied in

circuit, which of the following observations

will be wrong

2V

A

2F 2F

B

1

2

4V

22

2V2

initial

24

2V

final

19

O I

(A) Maximum brightness

of both does not

depend on frequency

f, if 1

f 1/LC2

(B) Both bulbs will glow

with same

brightness provided frequency

1

f 1/LC2

(C) Bulb b1 will light up initially and goes off, bulb b2 will be ON constantly

(D) Bulb b1 will blink and bulb b2 will be ON

constantly

Solution (ACD): At resonance both bulbs will glow with same

brightness. At resonance,

L CX X or 1

2 fL2 fC

or 1

f2 LC

4. The diagram shows a

convex lens and an object

O placed at a distance

greater than 2f on its

optical axis. Which of the

following statements is/are true?

(A) if O is moved towards lens I will move

away from lens

(B) if O is moved away from lens I will move

towards lens (C) If lens is rotated clockwise about its pole, I

will move downwards

(D) If lens is rotated clockwise about its pole, I

will move upwards

Solution (A): In convex lens image moves in same direction as that of object. Even if less will rotate, image

will move along principle axis

5. Two concentric metallic shells of radii R and

2R are placed in vacuum. The inner shell is having charge Q and outer shell is uncharged.

If they are connected with a conducting wire.

Then

(A) Q amount of charge will flow from inner to

outer shell.

(B) Q/e number of electrons will flow from outer to inner shell, where ‘e’ is charge on

an electron.

(C)

2KQ

4Ramount of heat is produced in the

wire.

(D)

2KQ

2Ramount of heat is produced in the

wire.

Solution (ABC):

(After connection entire charge moves towards

outer shell) Q

Q ne ne

Amount of heat produced = loss in electric

potential energy 2 2 2KQ KQ KQ

2R 4R 4R

6. A hollow cylinder, a spherical shell, a solid

cylinder and a solid sphere are allowed to roll

on an inclined rough surface of coefficient of

friction and inclination . The correct

statement(s) is/ are

(A) Frictional force will be equal for all the rolling objects, if having same mass.

(B) If all the objects are rolling and have same

mass, the K.E. of all the objects will be

same at the bottom of inclined plane.

(C) Work done by the frictional force will be

zero, if objects are rolling. (D) If cylindrical shell can roll on inclined

plane, all other objects will also roll.

Solution (BCD):

mgsin f ma … (i)

2fR I mK … (ii)

K is radius of gyration

From (i) and (ii), 2 2 2

2

gRsin gsin;

(R K ) K1

R

7. A standing wave of time period T is set up in a string clamped between two rigid supports. At

t = 0 antinode is at its maximum is placement

2A.

(A) The energy density of a node is equal to

energy density of an antinode for the first time at t = T/4.

(B) The energy density of node and antinode

becomes equal after T/2 second.

(C) The displacement of the particle at

antinode at T

t is 2 A8

(D) The displacement of the particle at node is zero

Solution (CD):

20

60°

60°

30°30

°

3m

3m

3O

A

B

A

B

C

Equation of SHM of particle which is at

antinode is 2

y 2Asin tT

at time T

t8

y 2Asin 2A4

Displacement of particle at node is always

zero.

8. In a photoelectric effect experiment, if f is the

frequency of radiations incident on the metal surface and I is the intensity of the incident

radiations, then choose the correct

statement(s).

(A) If f is increased keeping I and work

function constant then maximum kinetic

energy of photoelectron increases. (B) If distance between cathode and anode is

increased stopping potential increases

(C) If I is increased keeping f and work

function constant then stopping potential

remains same and saturation current increases.

(D) If work function is decreased keeping f

and I constant then stopping potential

increases

Solution (ACD):

max wfKE hf

Saturation current intensity (I):

maxS

KEV

e

SECTION-4: 1. A non conducting ring

(of mass m, radius r,

having charge Q) is

placed on a rough

horizontal surface (in

a region with transverse magnetic field). The field is increasing with time at the rate R and

coefficient of friction between the surface and

the ring is . For ring to remain in

equilibrium should be greater than QrR

xmgx

is …

Solution (2): For ring to be in equilibrium

2r R

mgr Q .r2 r

2. Figure shows a

square loop 10

cm on each side

in the x-y plane with its centre at

the origin. An

infinite wire is at z = 12 cm above y-axis. Then

magnetic flux through ABCD loop due to

straight infinite wire is x Weber. Then value of

x is

Solution (0): Net flux through rectangular plane is zero.

3. Figure shows circular

region of radius

R 3m in which

upper half has

uniform magnetic field

ˆB 0.2 k T and

lower half has uniform

magnetic field ˆB 0.2kT . A very thin parallel

beam of point charges each having mass

m 2gm, speed v 0.3m/sec and charge

q 1mC are projected along the diameter as

shown in figure. A screen is placed

perpendicular to inclined velocity of charges

as shown. If the distance between the points on screen where positive and negative charges

will strikes is 4X meters, then calculate X.

Solution (3): When positive charge enters

mv

RqB

= 3

3

AB 3 sin602

B C BB tan60

(OA OA)tan60

3

(2 3 ) 32

3

32

4. Full scale

deflection current

for galvanometer is

1Amp. What

should be the

value of shunt resistance in so that

galvanometer shows half scale deflection.

Solution (2):

For half deflection 1

i Amp.2

s s1 38

(20 18) 9.5 R R 22 2 9.5

5. An electron is shot into one end of a solenoid.

As it enters the uniform magnetic field within

the solenoid, its speed is 800m/s and it

velocity vector makes an angle of o30 with the

central axis of the solenoid. The solenoid carries 4.0 A current and has 8000 turns

along its length. Find number of revolutions

made by the electron within the solenoid by

++++

++++++++++++++++

++++ B

Top view

21

the time it emerges from the solenoid’s

opposite end. (Use charge to mass ratio e

m

for electron 113 10 C/kg ). Fill your

answer of y if total number of revolution is

6y10 .

5

Solution (8):

70

3000B ni 4 10 4

t800 cos30

No. of revolutions 7t e 4 10 32000

2 m 800cos30 2 m

qB

11 1013 10 10 32

3800

2

3208

400

6. A potential difference of 500V is applied

across a parallel plate capacitor. The

separation between the plates is 2×10–3m.

The plates of the capacitor are vertical. An electron is projected vertically upwards

between the plates with a velocity of 105 m/s

and it moves un-deflected between the plates.

If the magnetic field acting perpendicular to

the electric field has magnitude of

2n 5Wb/m

10

, then n is: (neglect gravity)

Solution (5): qE qvB

5

3

500 50010 B B 2.5 n 5

2002 10

7. 90% of a radioactive sample is left undecayed

after time t has elapsed. The percentage of the initial sample will decay in a total time 2t is

n1.9 10 .the value of n is

Solution (1): In time ‘t’ sec 10% has decayed so in next t

sec 10% of removing i.e. 90%. i.e. 9% more %

will decay so total decay percentage is 19%

8. The K, L and M energy levels of platinum lie

roughly at 78, 12 and 3 keV respectively. If

the ratio of wavelength of K line to that of

K line in X ray spectrum is

2n

22 find n.

Solution (5):

211 2

2 1

E 25E 66,E 75,

E 22

22

22

CHEMISTRY (PAPER-2)

SECTION 1:

PARAGRAPH FOR NEXT TWO QUESTIONS:

The dissolution of ammonia gas in water does not obey Henry’s law. On dissolving, a major portion of

ammonia molecules unite with 2H O to form 4NH OH molecules. 4NH OH again dissociates into

4NH and OH ions. In solution therefore, we have 3NH molecules, 4NH OH molecules and 4NH

ions

and the following equilibrium exist:

3NH (g) (pressure P and concentration c) Initially 3 2 4 4NH ( ) H O NH OH NH OH l

Let 1 3c mol/L of NH pass in liquid state which on dissolution in water forms 2c mol/L of 4NH OH . The

solution contains 3c mol/L of 4NH ions:

1. Total concentration of ammonia, which can be determined by volumetric analysis is equal to:

(A) 1 2c c (B) 1c (C) 1 3c c (D) 2 3c c

Solution (B):

The intermediate solution of acid will react with all the 3NH present in solution

2. Degree of dissociation of ammonium hydroxide is:

(A) 1c (B) 3 1c /c (C) 3c /c (D) 3 2c /c

Solution (D):

34

2

cMole of NH OH dissociated

Total mole c

PARAGRAPH FOR NEXT TWO QUESTIONS:

3B PBr C

C NaCN D

2 4 6 10 2D H SO E(C H O )

2E SOCl F

3F AlCl G

H SOLiAlH 2 44G H I

3. Find out structure of I:

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

4. The compounds D and F must be:

(A) (B)

(C) (D)

Solution (Q.3-C, Q.4-A):

23

23

N

M

Time

[A]0/2

[A]0

H SOPBr NaCN 2 43

2 2 2 2(C) (D)

Ph CH CH Br Ph CH CH CN

SECTION 2:

1. If n nr ,U and nE are radius of orbit speed of electron and energy of electron in H-like atom shell then,

which of the following are proportional to number of shell ‘n’?

(A) n nr E (B) n nr U (C) n

n

U

E (D) n

n

r

E

Solution (BC): 2

4 0n

r r .Z

n 0Z

U U .n

2

n 0 2

ZE E .

n

Thus; terms which are proportional to ‘n’ are

(i) n nr .U (ii) n

n

U

E

2. Select the correct statement:

(A) A process of expansion of an ideal gas which is isothermal as well as adiabatic must be a free-

expansion work (B) An adiabatic process in which no work is performed is an example of a constant energy process

(C) A reversible isothermal process can also be adiabatic only at T 0

(D) An irreversible isothermal process can also be adiabatic at T 0

Solution (ABCD):

(A) An isothermal free expansion is also adiabatic and vice versa.

(B) U Q W; for an adiabatic process in which no work is performed; i.e Q=0; W=0; thus;

U 0

(C) For rev isothermal process; 2

1

VU 0; W nRTln .

V

Thus; 2

1

VQ W nRTln

V

thus; at T 0 W 0; thus Q 0

(D) For are irreversible isothermal process extQ W P dV for it to become adiabatic also;

extP 0 ; thus at any T.

3. Which statements are correct about the two first order reactions if made with same initial

concentrations?

M Product

N product

(A) 1/2t for M is more than 1/2t of N

(B) 1/2t for M is less than 1/2t of N

(C) Rate constant for M is more than rate constant for N

(D) Rate constant for M is less than rate constant for N

Br + Mg Ether BrMg O

3(ii) H O

(i)2 2Ph CH CH OH

(A) (B)

SOCl AlCl2 32 2 2 2 (EAS)

Ph CH CH COOH Ph CH CH C Cl

O

(E) (F)

O

4LiAlH

OH

2 4H SO /

(G) (H) (I)

24

24

Solution (AD):

It is clear from graph that; 1/2 1/2t of M t of N

Again; 1/20.693

tK

thus; K for M<K for N.

4. Select species in which d-orbitals are used in hybridization and also involved in bond formation.

(A) 4OsF (B) 2 2XeO F (C) 3XeO (D) 6PCl

Solution (AB): 3

4OsF sp d hybridization and one p bond

33 2XeO F sp d hybridization and three p d bonds

5. Compound P: 2 4 2 2[Cr(H O) Cl ]Br 2H O

Q: 2 4 2 2[Cr(H O) Br ]Cl 2H O

R: 2 5 2 2[Cr(H O) Cl]Br 1H O

S: 2 6 3[Cr(H O) ]Cl

Incorrect statement(s) are:

(A) P and Q are ionization isomers

(B) P and R are hydrated isomers

(C) Given volume of solution of most conducting solution will require rd1/3 volume of equimolar

3AgNO solution for complete precipitation

(D) 2.0 grams of R prepared in 2L will show higher conduction compared with 2.0 gm of Q in 2L

solution

Solution (ABC):

Given vol. of must conducting solution requires thrice vol. of 3AgNO

6. Which of the following statements are not correct?

(A) Copper is extracted by self reduction method

(B) Cast iron is the purest form of iron

(C) The composition of malachite ore is 2 3Cu(OH) .2CuCO

(D) Cupellation process is used for the refining of Ag and Au

Solution (BC):

Conceptual

7. 2 4 2 42MX H SO (conc.) 2HX M SO (M=metal)

Above reaction is correct if ‘X’ is:

(A) F (B) Cl (C) 3NO

(D) I

Solution (ABC):

F and Cl salts produce HF and HCl respectively

I and Br salts produce 2I and 2Br respectively.

8. Which of the following reactions give diastereomeric products?

(A) (B)

(C) (D)

25

25

3CH

3CH Et

m CPBA

3CH

3CH Et

O

+

3CH

3CH Et

O

3CH

2 4Br /CCl

BrBr

3CH

+

BrBr

3CH

Solution (BC):

(B)

(C)

SECTION 4:

1. The mass of molecule A is twice the mass of molecule B. The rms speed of A is twice the rms speed of

B. If two samples of A and B contain same of molecule, the ratio of pressure of gas samples of A and B

in separate containers of equal volume is ……

Solution (8):

A BM 2M

rms rmsU (A) 2U (B)

A B A B A2

A B A B B

3RT 3RT T T T4 8

M M M M T

A AA

n .RTP

V B B A A

BB B

n .RT P TP 8

V P T

2. FeO crystallizes in NaCl type lattice. The crystal is however non-stoichiometric as 0.96Fe O and

deficient in iron. Some cation sites are vacant and some contain 3Fe

so that it becomes electrically

neutral. The % of cation sites vacant are……….

Solution (4):

For 1 mole Fe 4 octahedral voids

Thus for 0.96 mole of Fe 4 0.96 OV are occupied.

thus; % vacant sites4 0.96 4

100 4%4

3. If o o o1 2 3E ,E and E are standard oxidation potentials for

2 2 3 3Fe|Fe ,Fe |Fe and Fe|Fe , then

o oo 2 13

E 2EE

n

. The value of n is……

Solution (3):

2Fe Fe 2e ; o o1 1G 2FE …….(i)

2 3Fe Fe e ; o o2 2G 1 FE …….(ii)

3Fe Fe 3e ; o o3 3G 3 F E …….(iii)

Three;

o oo o o o 1 21 2 3 3

2E EG G G E

3

thus; n 3

4. Out of given aq. ion(s), find no. of ions which produce ppt. with dil. HCl + 2H S :

3 2 2 2 2 3 2Cr ,Zn ,Cd ,Hg ,Fe ,Bi ,Pb

Solution (4):

II group cations produce ppt.

26

26

5. In the compound 2 4 5 4 2Na [B O (OH) ] 8H O, if the

(i) number of B O B bonds is ‘x’

(ii) number of B B bonds is ‘y’

(iii) number o 2sp hybridized B atoms is ‘z’

Calculate the value of x y z .

Solution (7):

6. Identify number of compound from following. Which liberate 2CO on reaction with 3NaHCO .

Solution (5):

7. Identify number of carbohydrate that have L-configuration, from following:

Solution (5):

8. Identify number of substrate those can give NS 1 and NS 2 reaction both:

Solution (6):

27

27

MATHEMATICS (PAPER 2) SECTION 1:

PARAGRAPH FOR NEXT TWO QUESTIONS:

Let f x be non-positive continuous function and x

0

f x f t dt x 0 & f x f x where c 0 and

let g : 0, R be a function such that

dg x

g x x 0dx

and g 0 0

1. The total number of roots of the equation f x g x is

(A) (B) 1 (C) 2 (D) 0

Solution (B):

f x 0 and F x f x f x cF x

F x cF x 0 cx cxe F x ce F x 0

cxde F x 0

dx

cxe F x is an increasing function.

c 0cxe F x e F 0 cxe F x 0

f x 0 , (As f x cF x and c is positive) f x 0

Also,

cx dg xe g x x 0

dx

x xd g xe e g x 0

dx

xe g x is a decreasing function

0xe g x e g 0

g x 0 as g 0 0

Thus, f x g x has one solution x 0

2. The number of solutions of the equation 2x x 6 f x g x is/are

(A) 2 (B) 1 (C) 0 (D) 3

Solution (C):

2x x 6 f x g x 2x x 6 g x no solution

PARAGRAPH FOR NEXT TWO QUESTIONS:

The figure shown below, indicates positions of the ships A & B at t 0 . The ship A sails at 2km/hr due

east and ship B sails at 1km/hr along a course that forms an angle of o60 with the course of ship A.

3. The distance ‘d’ of separation of the ships as a function of time t is

(A) 23 t t 1 (B) 23 t 3t 5 (C) 23 t t 1 (D) 2t t 1

Solution (C):

4. What was the distance between the ships 2 hours before the initial moment shown in figure? (A) 1 (B) 3 (C) 4 (D) 6

Solution (B)

2d 3 2 2 1 3km.

28

28

Solution to question number (3 & 4):

Let at time t hrs the ship A is at /A and the ship B is at /B . Hence /PA 1 2t km and

/PB 2 t km. Let us take the line /PA as x-axis and P as origin. Hence coordinates of /A are

1 2t ,0 and that of /B are o o2 t cos60 , 2 t sin60 t t 3

1 , 32 2

.

22

t t 3d 1 1 2t 3 0

2 2

2 23t t

3 12 2

2 29t t3 1 t

4 4

2

212t 12t 123 t t 1

4

…(i)

2 22 1 1 1

3 t 2. .t 12 2 2

21 3

3 t2 4

21 9

3 t2 4

…(ii)

Clearly d is minimum if 1

t 02

, that is 1

t2

. Hence 1

hrs2

before the initial moment the ships were

minimum distance apart.

SECTION 2:

1. The perpendicular bisector x y 2 0 and x y 1 0 of sides AB & AC of a triangle ABC intersect,

them at 1, 1 & 2,1 respectively. If the middle points of side BC is P, then distance of this point

from the orthocenter of the triangle ABC is

(A) 85 (B) 41 (C) 13 (D) 7

Solution (C):

Since perpendicular bisector of AB, 4 : x y 2 0 and perpendicular bisector of AC, 2L : x y 1 0

are perpendicular to each other, therefore oA 90 .

A is the orthocenter & P will be the circumcenter. Hence the correct figure will be as shown below:

29

29

Solving 1 2L & L :

1 3P ,

2 2

Since ADPE is a rectangle, therefore PA DE

2 2PA DE PD PE

2 2 2

2

1 3 11 1 2

2 2 2PA

31

2

1 1 25 25

PA4 4 4 4

52

134

2. Let

0 1 2 n

n 23 4L 2C C C ... .C

2 3 n 1

(where n

r rC C ). If nn

LS lim

2 , then which of the following

is/are incorrect?

(A) S is prime (B) S is composite (C) S is even (D) S is odd

Solution (ABC): r n

nr

r 0

r 2L C

r 1

r nn

r

r 0

1 n 1L C r 2

n 1 r 1

n

n 1r 1

r 0

1L C r 2

n 1

n nn 1 n 1

r 1 r 1

r 0 r 0

1L r 1 C C

n 1

n n

n 1 r n 1r 1 r 1

r 1 r 0

1L r 1 C .1 C

n 1

n n 11

L n 1 2 2 1n 1

n nn n

L 1 1lim lim n 1 2

n 12 2

n nn n

L 2 1S lim lim 1 1

n 12 2 n 1

3. If the roots of the equation 4 3 2z z 36 15i z mz 0 are the vertices of a square then m

can be equal

(A) 35 45i (B) 35 45i (C) 35 45i (D) 35 45i

Solution (AB):

4 3 2z z 36 15i z mz 0

Clearly one of the roots of above equation is z 0 .

Since roots are vertices of square, therefore, let other three vertices be 1 1 1 1z ,iz ,z iz , (rotation of

complex number)

1 1 1 10 z iz z iz

12z 1 i …(i)

1 1 1 1 1 1 10 z 0 z iz 0 iz z z iz 1 1 1 1 1z iz iz z iz 36 15i

2 2 2 2 21 1 1 1 1z iz iz z iz 36 15i

213iz 36 15i 2

1z 5 12i 113 5 13 5

z i2 2

1z 9 i 4 1z 3 2i

30

30

CASE I: 1z 3 2i 1z 3 2i

From (i):

2 3 2i 1 i

2 3 2 i 2 3 2 1 5i 2 10i

1 1 1 1z z iz iz m 31z 1 i m

3

m 3 2i 1 i m 37 55i m 35 45i

CASE II: 1z 3 2i 1z 3 2i

From (i)

2 3 2i 1 i 2 10i

1 1 1 1z z iz iz m 31z 1 i m

3

m 3 2i 1 i m 37 55i m 35 45i m 35 45i

4. Let area of triangle formed by any line through point 1,3 and coordinate axes in xy-plane is A, then

(A) If A 8 , then number of such lines is 4 (B) If A 4 , then number of such lines is 4

(C) If A 6 , then number of such lines is 3 (D) If A 0 , then number of such lines is 1

Solution (AC):

Equation of line passing through 1,3 & having slope m is:

y 3 m x 1

For x 0 : y 3 m

For y 0 : 3

x 1m

Therefore area of triangle formed by the above line & coordinate axis is:

1 3

A 3 m 12 m

1 3 m

A 3 m2 m

3 m 1 3 m1

A2 m

21 1

A 3 m2 m

23 m1

A2 m

Verifying option (A):

A 8

23 m1

82 m

2

3 m16

m

2

3 m16

m

29 m 6m 16m 2m 6m 16m 9 0

2 2m 22m 9 0 or m 10m 9 0

Hence 4 values of ‘m’ are possible, since both equations have real & distinct roots. Hence option (A) is

correct.

Verifying option (B):

A 4

23 m1

42 m

29 m 6m 8m 2m 6m 8m 9 0

2 2m 14m 9 0 or m 2m 9 0

Hence 2 values of m are possible, since second quadratic have complex roots. Hence option (B) is not correct.

Verifying option (C):

A 6

23 m1

62 m

29 m 6m 12m

2m 6m 12m 9 0 2 2m 18m 9 0 or m 6m 9 0

22m 18m 9 0 or m 3 0

31

31

Hence 3 values of m are possible, since second quadratic have equal roots. Hence option (C) is correct.

Verifying option (D):

A 0

23 m1

02 m

m 3

Also for m 0 or m area of is 0.

Hence number of such lines is 3. Hence option (D) is not correct.

5. Two vertices of an equilateral triangle are 0,0 & 0,2 and the third vertex lies in 2nd quadrant. Then

(A) The equation of circumcircle of the triangle is 2

21 4x y 1

33

(B) The equation of incircle of the triangle is 2

21 1x y 1

33

(C) The area of the square that circumscribe the circum circle of the triangle is 16

3.

(D) The area of the square inscribed in the circumcircle of the triangle is 3

16.

Solution (ABC):

Coordinates of point B are:

o2cos30,2sin30 3,1

Hence centroid G of the triangle OAB is:

0 0 3 0 2 1 1G , ,1

3 3 3

Since the triangle is equilateral therefore all its centres coincide.

Circumradius 2

21 2R GA 2 1

3 3

.

Hence equation of circum circle is:

2

21 4x y 1

33

.

Inradius 2

21 1r GM 1 1

3 3

Hence equation of incircle is:

2

21 1x y 1

33

.

If a square circumscribes a circle of radius R then area of the square is 2

2 2 162R 2

33

.

32

32

6. Which of the following statements are true?

(A) If A is an invertible 3 3 matrix and B is a 3 4 matrix, then 1A B is defined.

(B) It is never true that A B,A B , and AB are all defined.

(C) Every matrix none of whose entries are zero is invertible.

(D) Every invertible matrix is square and has no two rows the same.

Solution (AB):

7. Lines 1L ax by c 0 and 2L x my n 0 intersect at the point P and make an angle with

each other. Consider the line L, different from 2L and passing through the point P and making the

same angle with 1L . Then

(A) The equation of line L is 2 22 a bm ax by c a b x my n 0

(B) The equation of line L is 2 22 a bm ax by c a b x my n 0

(C) The line L is reflection of the line 2L about the line 1L

(D) The line L is reflection of the line 1L about the line 2L

Solution (AC):

Since L & 2L are equally inclined to 1L and L, 1 2L ,L are concurrent therefore 1L is one of the angle

bisectors of L and 2L . Hence L is reflection of 2L about the line 1L . Also, if 1 1A x ,y is any arbitrary

point on 1L then perpendicular distance of the line 2L from A is equal to perpendicular distance of the

line L from A.

1 1 1 11 1

2 2 2 2

ax by c x my nx my n

m a b m

But 1 1ax by c 0 .

1 11 1

2 2 2 2

x my nx my n

m a b m

2 2 2 2

1

m a b m

2 2 2 2 2a b m m

2 2 2 2 2 2 2 2a b m 2 a bm m

2 2a b

2 a bm

.

2 2a b

L ax by c x my n 02 a bm

2 22 a bm ax by c a b x my n 0

8. Identify the statements which are True.

(A) the equation of the director circle of the ellipse, 2 25x 9y 45 is 2 2x y 14.

(B) the sum of the focal distances of the point 0,6 on the ellipse 2 2x y

125 36

is 10.

(C) The point of intersection of any tangent to a parabola and the perpendicular to it from the focus lies

on the tangent at the vertex.

(D) P and Q are the points with eccentric angles & on the ellipse 2 2

2 2

x y1

a b , then the area of

the triangle OPQ is independent of , where O is origin.

33

33

Solution (ACD):

Let us verify each option individually:

(A) 2 25x 9y 45 2 2x y

19 5

Hence equation of director circle is: 2 2x y 9 5 2 2x y 14.

Hence option (A) is correct.

(B) Sum of focal distances of any point on any ellipse is equal to length of major axis of the ellipse. For

the given ellipse length of major axis is 2 6 12. Hence option (B) is false.

(C) Option (C) is correct (property of parabola).

NOTE: If SM be perpendicular to tangent at any point P, then M lies on the tangent at the vertex

and 2SM OS.SP , where S is the focus and O is the vertex of the parabola

Proof:

Let 2y 4ax be the given parabola. Tangent at 1 1P x ,y is S a,0 …(i)

The above tangent intersects y-axis at i.e.

21

1

1

y2a

4a y0, 0,

y 2

. The equation of the line passing

through S a,0 and perpendicular to the tangent is:

1yy 0 x a

2a

1y

y x a2a

…(ii)

The line (ii) intersects y-axis at 1y0,

2

, which is same as co-ordinates of M. Hence (i) and (ii)

intersect on y-axis i.e. the tangent at vertex of the parabola.

Also in triangle PMS SM

sinSP

In triangle SOM OS

sinSM

OS SM

SM SP 2SM OS.SP

(D) P acos ,bsin

Q acos ,bsin

O 0,0

Hence area of triangle OPQ is given by:

a cos bsin 11

a cos bsin 12

0 0 1

34

34

1

abcos .sin abcos .sin2

1 1

ab sin ab sin2 2

Hence option (D) is correct.

SECTION 4:

1. If /f x g x and /g x f x for all x and /f 2 4 f 2 then the value of 2 2f 19 g 19 27 is

Solution (5):

/ /f x g x ;g x f x

/f 2 f 2 4 g 2 4

Let h(x) 2 2f x g x 27

/ / /h x 2f x .f x 2g x .g x

/ /2f x .f x 2f x . f x 0

h x constant

2 2h 19 h 2 f 2 g 2 27 32 27 5

2. If 6 xf x 7 x and 1g x f x then the value of 7 /ln 7 e g 2 is

Solution (1):

1g x f x

1f g x f f x x

df g x

1dx

/ /f g x .g x 1

/

/

1g x

f g x

/

/

1g 2

f g 2

Let g 2 t

1f 2 t 1f f 2 f t

2 f t f t 2 6 t7 t 2 6 t7 t 2

By trial t 5 satisfies the above equation. Also 6 ty 7 is a decreasing function of t while y t 2 is

an increasing function of t. Hence the graph of 6 ty 7 and y t 2 can intersect at most once. Hence

t 5 is the only solution.

g 2 5

/

/ /

1 1g 2

f g 2 f 5

6 x

x 5

1

7 ln7 1

7

1 1

7ln7 1 ln7 lne 7

1

ln 7 e

7 /ln 7 e g 2 1

3. A variable plane at a distance of 1unit from the origin cuts the co-ordinate axes at A, B and C. If the

centroid G x,y,z of the triangle ABC satisfies the relation 2 2 2

1 1 1k

x y z , then the value of k is

35

35

Solution (9):

Let the plane meets the coordinate axes at A a,0,0 ,B 0,b,0 ,C 0,0,c . Hence the equation of the

plane ABC is x y z

1a b c . Also centroid G of the triangle ABC is

a b c, ,

3 3 3

.

a b c

x , y , z3 3 3

a 3x, b 3y, c 3z .

As per question the distance of the plane x y z

1a b c from origin is 1.

2 2

11

1 1 1

a b c

2 2 2

1 1 11

a b c

2 2 2

1 1 19

x y z .

4. Let 2A,G and H be the roots of cubic equation 3 2x px qx r 0 which are in GP, where p, q are

integers and A, G, H are respectively AM, GM, HM of two positive numbers. If p, q 100, 100 , then

let number of all possible ordered triplets p,q,r is N. Value of N 90 is

Solution (7):

A, 2G ,H are roots of the equation 3 2x px qx r 0 and roots are also in GP.

2

2G A H …(i)

4G A H 4 2 2G G , G AH

2 2G G 1 0 2 2G 0 or G 1 G 1, 0, 1

Since G is geometric mean of two positive numbers.

G 1

AH 1 , (using (i))

1

HA

Hence roots of the equation are:

1A,1,

A

31 r

A 1 1A 1

1 r r 1

1A 1 p

A …(ii)

1

p A 1 2 1, AM GMA

p 3 p 3 p 99, 98,..., 4, 3

1 1A 1 1 A q

A A

1

A 1 qA

…(iii)

1

q A 1A

q 2 1, AM GM q 3 q 3,4,5,...,99

From (ii) & (iii): p q 0

Now since p q 0 , for every value of p, value of q is fixed, possible ordered triplets (p, q, r) are:

99,99, 1 , 98,98,1 ,..., 3,3,1 .

Number of possible ordered triplets are 97.

N 97 N 90 7

36

36

5. If & are roots of the equation

5 5

10 21 1

0 01 1 x 5x 202 2

1 25 401 2 0 1 x 2

1 12 2

then find the

value of 2 2 is

Solution (6):

Let

10

2A

11

2

& 1 1

B2 0

1 10 1 2 0 1 0

2 2AB

1 11 1 2 1 1 0

2 2

1 0I

0 1

AB BA I 5 5

101 1

0 01 12 2

1 2 0 11 1

2 2

5 10 5A B A

10 times

AAAAA BBBB.......BAAAAA

8 times

AAAAI BBB.......B IAAAA , AB BA I

8 times

AAAA BBB.......B AAAA , AI IA A

6 times

AAAI BBB.......B IAAA , AB BA I

6 times

AAA BBB.......B AAA , AI IA A

2 times

A BB A I I I

Given

5 5

10 21 1

0 01 1 x 5x 202 2

1 25 401 2 0 1 x 2

1 12 2

2x 5x 20

1 25 I 40x 2

2x 5x 20

1 25 40x 2

, AI IA A

21 x 5x 20 25 x 2 40

2x 5x 20 25x 50 40

2x 20x 30 0

As , root of the above equation:

2x 20x 30 1 x x

37

37

For x 2 :

2

2 20 2 30 2 2

4 40 30 2 2

6 2 2 6 2 2

6. A function f x satisfies f 1 3600 and 2f 1 f 2 ... f n n f n for all positive integers n 1 .

Find the value of f 9

10 is equal to

Solution (8):

2f 1 f 2 f 3 ... f n 1 f n n f n …(i)

Replacing n by n 1 :

2

f 1 f 2 ... f n 1 n 1 f n 1 …(ii)

Subtracting (ii) from (i):

22f n n f n n 1 f n 1 , n 2

221 n f n n 1 f n 1

2

2

n 1 f n 1f n

n 1

n 1f n f n 1

n 1

…(iii)

Replacing n by n 1 in (iii):

n 2f n 1 f n 2

n

Replacing n by n 2 in (iii):

n 3f n 2 f n 3

n 1

etc.

n 1f n f n 1

n 1

n 1 n 2

. f n 2n 1 n

n 1 n 2 n 3

. . f n 3n 1 n n 1

……………………….

……………………….

n 1 n 2 n 3 3 2 1

. . .... . . .f 1n 1 n n 1 5 4 3

2 1

f 1n 1 n

2

3600n 1 n

2 3600

f 99 1 9

2 360080

10 9

f 98

10 .

7. The least possible value of a for which

3 2

3 2

x 6x 11x 6 a0

30x x 10x 8

does not have a real solution is

Solution (5):

Let a

f x 030

38

38

Where 3 2

3 2

x 6x 11x 6f x

x x 10x 8

x 1 x 2 x 3

x 1 x 2 x 4

x 3

,x 1,2, 4x 4

Range of 2 1

f x R 1, ,5 6

So, the equation does not have solution if

a 2 11, ,

30 5 6

a 30,12,5

8. Let

sin xdF x e

dx x

, x 0 if

4 3sin x

1

3e dx

x

F k F 1 , then the possible value of k is

Solution (8):

In

4 3sin x

1

3e dx

x, put

3x t , so that 33x dx dt .

Also, when x 1,t 1 and when x 4,t 64 .

Therefore,

4 643sinx dx sint2

1 1

3 3 dte e

x x 3x

64 64

sint

1 1

1 de dt F t dt F 64 F 1

t dt

Hence, k 64