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IIT JEE 2011 Paper 1 (Code 4)

Part - I: Chemistry

SECTION - I (Total Marks: 21)

(Single Correct Answer Type)

This section contains 7 multiple choice questions. Each question has 4 choices (A), (B),

(C) and (D) out of which ONLY ONE is correct.

1. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density

1.15 g/mL. The molarity of the solution is

(A) 1.78 M (B) 2.00 M

(C) 2.05 M (D) 2.22 M

Sol. (C)

Mass of solution = 1120 g

So, volume of solution =Mass 1120

973.9mlDensity 1.15

wt. of solute 1000Molarity

mol. wt. of solute volume of solution120 1000

Molarity 2.05M60 973.9

2. AgNO3 (aq.) was added to an aqueous KCl solution gradually and the conductivity of

the solution was measured. The plot of conductance ( ) versus the volume of AgNO3is

(A) (P) (B) (Q)

(C) (R) (D) (S)

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Sol. (D)

Ag+ and K+ have nearly same ionic mobility so first part of the curve does not show

increase in conductivity. After the equivalence point i.e. in second part of the curve,

conductivity increase because of increase in concentration of Ag+ and NO3- ions.

3 3AgNO + KCl AgCl + KNO

3. Among the following compounds, the most acidic is

(A) p-nitrophenol (B) p-hydroxybenzoic acid

(C) o-hydroxybenzoic acid (D) p-toluic acid

Sol. (C)

o-hydroxybenzoic acid is most acidic because of stabilisation of its conjugate base by

intramolecular hydrogen bonding and ortho effect.

4. The major product of the following reaction is

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Sol. (A)

5. Extra pure N2 can be obtained by heating

3 4 3

4 2 2 7 3 2

(A) NH with CuO (B) NH NO

(C) (NH ) Cr O (D) Ba(N )

Sol. (D)

Extra pure N2 can be obtained by heating Ba(N3)2 since there is no gaseous by

product formed in the reaction whereas other reactants involve formation of

gaseous by products.

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3 2 2Ba(N ) Ba(s) + 3N

6. Geometrical shapes of the complexes formed by the reaction of Ni2+

with Cl-

, CN-

and

H2O, respectively are:

(A) Octahedral, tetrahedral and square planar

(B) Tetrahedral, square planar and octahedral

(C) Square planar, tetrahedral and octahedral

(D) Octahedral, square planar and octahedral

Sol. (B)

+2 - -24

+2 - 24

+2 +22 2 6

Ni + 4Cl [NiCl ] (tetrahedral)

Ni + 4CN Ni(CN) (square planar due to strong field ligand)

Ni + H O Ni(H O) (octahedral)

7. Bombardment of aluminium by - particle leads to its artificial disintegration in twoways, (i) and (ii) as shown. Products X, Y and Z respectively are:

(A) Proton, neutron, positron (B) Neutron, positron, proton

(C) Proton, positron, neutron (D) Positron, proton, neutron

Sol. (A)

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SECTION - II (Total Marks: 16)

(Multiple Correct Answers Type)

This section contains 4 multiple choice questions. Each question has four choices

(A), (B), (C) and (D) out of which ONE OR MORE may be correct.

8. Amongst the given options, the compound(s) in which all the atoms are in one plane

in all the possible conformations (if any), is (are)

Sol. (B, C)

A. Only two conformers cis- and trans- will have all atoms in same plane. Two

hydrogens will lie in different plane due to rotation along C2- C3 bond.

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B. All atoms lie in one plane hence it is planar.

C. All atoms lie in one plane hence it is planar.

D. It has terminal hydrogens lying perpendicular to each other and hence don't lie in

same plane.

9. According to kinetic theory of gases

(A) Collisions are always elastic

(B) Heavier molecules transfer more momentum to the wall of the container

(C) Only a small number of molecules have very high velocity

(D) Between collisions the molecules move in straight lines with constant velocities

Sol. (A, B, C)

According to Kinetic theory of gases

(i) Collisions are always elastic

(ii) Momentum transferred on wall by one collision along x component = 2 mvx

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(iv) During each collision, the speed and the direction of motion of molecules

undergo a change.

10. The correct statement(s) pertaining to the adsorption of a gas on a solid surface is

(are)

(A) Adsorption is always exothermic

(B) Physisorption may transform into chemisorption at high temperature

(C) Physisorption increases with increasing temperature but chemisorption decreases

with increasing temperature

(D) Chemisorption is more exothermic that physisorption, however it is very slow

due to higher energy of activation

Sol. (B, D)

A. Adsorption is not always exothermic; chemisorption of H2 on glass is an

endothermic process.

B. At high temperature, sufficient activation energy for chemical adsorption is

provided so physisorption may transform into chemisorption.

C. Physisorption occurs at low temperatures and increases with increasing

temperature but chemisorption occurs at high temperatures and decreases with

increasing temperature.

D. Chemisorption is more exothermic than physisorption. Enthalpy of adsorption for

chemisorptions is larger than that for physical adsorption.

11. Extraction of metal from the ore cassiterite involves

(A) Carbon reduction of an oxide ore (B) Self reduction of a sulphide ore

(C) Removal of copper impurity (D) Removal of iron impurity

Sol. (A, C, D)

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Cassiterite ore contains 0.5-10% of meal as SnO2, the rest being impurities of Fe,

Cu.

o1200-1300 C2SnO + C Sn CO

SECTION - III (Total Marks: 15)

(Paragraph Type)

This section contains 2 paragraphs. Based upon one of the paragraphs, 2 multiple

choice questions and based upon the second paragraph, 3 multiple choicequestions have to be answered. Each of these questions has four choices (A), (B),


Paragraph for Questions Nos. 12 and 13

An acyclic hydrocarbon P, having molecular formula C6H10, gave acetone as the only

organic product through the following sequence of reactions, in which Q is an

intermediate organic compound.

12. The structure of compound P is

Sol. (D)

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13. The structure of the compound Q is

Sol. (B)

Paragraph for Questions Nos. 14 to 16

When a metal rod M is dipped into an aqueous colourless concentrated solution of

compound N, the solution turns light blue. Addition of aqueous NaCl to the blue

solution gives a white precipitate O. Addition of aqueous NH3 dissolves O and gives

an intense blue solution.

14. The metal rod M is

(A) Fe (B) Cu

(C) Ni (D) Co

Sol. (B)

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3 3 2Cu + AgNO Cu(NO ) + Ag

(M) (N) Blue solution

The metal rod dipped in the solution is of Copper.

15. The compound N is

3 3 2

3 3 3 2

(A) AgNO (B) Zn(NO )

(C) Al(NO ) (D) Pb(NO )

Sol. (A)

N is silver nitrate solution in which the metal rod is dipped.

+3 3 2Cu + AgNO Cu(NO ) + Ag

(M) (N) Blue solution

16. The final solution contains

2+ 2- 3+ 2+3 4 4 3 4 3 4

+ 2+ + 2+3 2 3 4 3 2 3 6

(A) [Pb(NH ) ] and [CoCl ] (B) [Al(NH ) ] and [Cu(NH ) ]

(C) [Ag(NH ) ] and [Cu(NH ) ] (D) [Ag(NH ) ] and [Ni(NH ) ]

Sol. (C)

+

White ppt (O)

+3 3 2

2+ 23 3 4

Ag + NaCl AgCl

AgCl + 2NH [Ag(NH ) ] Cl

Cu 4NH [Cu(NH ) ]

(Bluesolution)

SECTION - IV (Total Marks: 28)

(Integer Answer Type)

This section contains 7 questions. The answer to each of the questions is a single-

digit integer, ranging from 0 to 9. The bubble corresponding to the correct answer

is to be darkened in the ORS.

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17. The work function ( ) of some metals is listed below. The number of metals which

will show photoelectric effect when light of 300 nm wavelength falls on the metal is

Sol. (4)

Energy of incident photon =34 8

9 19hc 6.62 10 3 10

4.14 eV300 10 1.6 10

Kinetic Energy; E = h - work function

Energy of incident photon should be higher than work function to slow photoelectric

effect. Li, Na, K and Mg have work function less than 4.14 eV so they'll show

photoelectric effect.

18. To an evacuated vessel with movable piston under external pressure of 1 atm., 0.1

mol of He and 1.0 mol of an unknown compound (vapour pressure 0.68 atm. at 0C)

are introduced. Considering the ideal gas behavior, the total volume (in litre) of the

gases at 0C is close to

Sol. (7)

Partial pressure of He = 1 - 0.68 = 0.32 atm.

If V is the total volume of gases at 0oC, then

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23. The maximum number of electrons that can have principal quantum number, n = 3,

and spin quantum number, s1

m , is2

Sol. (9)

Number of orbitals for n = 3 is n2 = (3)2 = 9

So, total no. of electrons in n = 3 is = 2n2 = 18

The number of electrons with ms=12

will be 9.

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PART-II : Physics

SECTION - I (Total Marks : 21)


This section contains 7 multiple choice questions. Each question has 4 choices

(A), (B), (C) and (D) out of which ONLY ONE is correct

24. The wavelength of the first spectral line in the Balmer series of hydrogen atom is

6561oA . The wavelength of the second spectral line in the Balmer series of singly-

ionized helium atom is

(A) 1215oA (B) 1640

oA (C) 2430

oA (D) 4687

oA

Sol. (A)

2

2 21 2

1 1 1Rz

n n

22 2

1

22 2

2

21

2

1 1 1R 1

2 31 1 1

R 22 4

527

56561 1215

27

25. A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m.

The ball is rotated on a horizontal circular path about vertical axis. The maximum

tension that the string can bear is 324 N. The maximum possible value of angular

velocity of ball (in radian/s) is

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(A) 9 (B) 18 (C) 27 (D) 36

Sol. (D)

2

2

maxmax

T cos mg

T sin mlsin

T ml

T 32436rad/s

ml 0.5 0.5

26. A meter bridge is set-up as shown, to determine an unknown resistance X using a

standard 10 ohm resistor. The galvanometer shows null point when tapping key is at

52 cm mark. The end corrections are 1 cm and 2 cm respectively for the ends A and

B. The determined value of X is

(A) 10.2 ohm (B) 10.6 ohm (C) 10.8 ohm (D) 11.1 ohm

Sol. (B)

Condition of Wheatstone bridge

X 10

52 1 48 2

10 53X 10.6

50

27. A 2 F capacitor is charged as shown in the figure. The percentage of its stored

energy dissipated after the switch S is turned to position 2 is

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29. 5.6 liter of helium gas at STP is adiabatically compressed to 0.7 liter. Taking the

initial temperature to be T1, the work done in the process is

1 1 1 19 3 15 9

(A) RT (B) RT C RT D RT8 2 8 2

Sol. (A)

1 11 1 2 2

11

2 12

2 / 31 1

2 1ext

1

T V T V

{ 5 / 3}

VT T

V

T 8 4T

5.6 1no. of moles

22.4 4R T T

19

RT8

30. Consider an electric field 0 0E E x, where E

is a constant. The flux through the

shaded area (as shown in the figure) due to this field is

2

2 2 2 00 0 0

E a(A) 2E a B 2 E a C E a D

2

Sol. (C)

Projected area on y-z plane = a2

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Flux = Eoa2

SECTION - II (Total Marks : 16)

(Multiple Correct Answer Type)

This section contains 4 multiple choice questions. Each question has four choices

(A), (B), (C) and (D) out of which ONE OR MORE may be correct.

31. A metal rod of length 'L' and mass 'm' is pivoted at one end. A thin disk of mass 'M'

and radius 'R' (< L) is attached at its center to the free end of the rod. Consider two

ways the disc is attached. (Case A): The disc is not free to rotate about its center

and (case B) the disc is free to rotate about its center. The rod-disc system performs

SHM in vertical plane after being released from the same displaced position. Which of

the following statement(s) is (are) true?

(A) Restoring torque in case A = Restoring torque in case B

(B) Restoring torque in case A < Restoring torque in case B

(C) Angular frequency for case A > Angular frequency for case B

(D) Angular frequency for case A < Angular frequency for case B

Sol. (A, D)

A B2

2B

2 22

A

B A B A

since FBD is same.

mlI Ml

3

ml MRI Ml

3 2I I , so

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QA = QE = Q1 + Q2 + Q3. So, option (A) is correct.

QlT QR is least for E.

KA So, option (C) is correct.

Option (B) can be correct or incorrect.

33. An electron and a proton are moving on straight parallel paths with same velocity.

They enter a semi-infinite region of uniform magnetic field perpendicular to the

velocity. Which of the following statement(s) is/are true?

(A) They will never come out of the magnetic field region

(B) They will come out travelling along parallel paths

(C) They will come out at the same time

(D) They will come out at different times

Sol. (B, D)

By diagram, option (B) is true.

2

e p

e p

e p

mvqvB

r

mvr

qB

m m

r r

t t

34. A spherical metal shell A of radius RA and a solid metal sphere B of radius RB (< RA)

are kept far apart and each is given charge '+Q'. Now they are connected by a thin

metal wire. Then

on surface on surfaceinside A BA BA A BB A

R(A) E 0 B Q Q C D E E

R

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varying electric field needs to be applied that has an angular frequency , where a

part of the energy is absorbed and a part of it is reflected. As approaches P , all

the free electrons are set to resonance together and all the energy is reflected. This

is the explanation of high reflectivity of metals.

35. Taking the electronic charge as 'e' and the permittivity as ' 0 ', use dimensional

analysis to determine the correct expression for P

2

0 0

20 0

m mNe Ne(A) B C D

m Ne m Ne

Sol. (C)

2 22 3

2 4o

3

1A T

Ne 1Lm TA T

ML M

36. Estimate the wavelength at which plasma reflection will occur for a metal having the

density of electrons N 4 1027 m3. Take 0 1011 and m 1030, where these

quantities are in proper SI units

(A) 800 nm (B) 600 nm (C) 300 nm (D) 200nm

Sol. (B)

2

o

o

2

30 118

19 27

2 c Ne2 f

m

m2 c

Ne

10 102 3.14 3 10600nm

1.6 10 4 10


Phase space diagrams are useful tools in analyzing all kinds of dynamical problems.

They are especially useful in studying the changes in motion as initial position and

momentum are changed. Here we consider some simple dynamical systems in one-

dimension. For such systems, phase space is a plane in which position is plotted

along horizontal axis and momentum is plotted along vertical axis. The phase space

diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time

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flow. For example, the phase space diagram for a particle moving with constant

velocity is a straight line as shown in the figure. We use the sign convention in which

position or momentum upwards (or to right) is positive and downwards (or to left) is

negative.

37. The phase space diagram for a ball thrown vertically up from ground is

Sol. (D)

Momentum is initially positive. Then it decreases and later becomes positive.

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38. The phase space diagram for simple harmonic motion is a circle centered at the

origin.

In the figure, the two circles represent the same oscillator but for different initialconditions, and E1 and E2 are the total mechanical energies respectively. Then

1 2 1 2

1 2 1 2

(A) E 2E B E 2E

(C) E 4E D E 16E

Sol. (C)

1 2

22 2

2

1 1

1 2

p 2p

E p 1

E 4p

E 4E

39. Consider the spring-mass system, with the mass submerged in water, as shown in

the figure. The phase space diagram for one cycle of this system is

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Sol. (B)

From extreme positive position, the momentum starts increasing in negative

direction (with energy dissipation) and gradually returns back to less positive

position as pf< pi (due to energy dissipation)

SECTION - IV (Total Marks : 28)


This section contains 7 questions. The answer to each question is a Single-digit integer,

ranging from 0 to 9. The bubble corresponding to the correct answer is to be darkened in

the ORS.

40. Steel wire of length 'L' at 40C is suspended from the ceiling and then a mass 'm' is

hung from its free end. The wire is cooled down from 40C to 30C to regain its

original length 'L'. The coefficient of linear thermal expansion of the steel is 105/C,

Young's modulus of steel is 1011 N/m2 and radius of the wire is 1 mm. Assume that L

>> diameter of the wire. Then the value of 'm' in kg is nearly.

Sol. (3)

2

11 3 5

mg YA T

10 10 10 10YA Tm

g 10m kg 3.14kg

41. The activity of a freshly prepared radioactive sample is 1010 disintegrations per

second, whose mean life is 109s. The mass of an atom of this radioisotope is 1025

kg. The mass (in mg) of the radioactive sample is

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Sol. (1)

10

10

10 10 9 19

total 119 25

-6

dNN 10

dt

10N 10 10 10 10

m N m

10 10

=10 kg 1mg

42. A block is moving on an inclined plane making an angle 45 with the horizontal and

the coefficient of friction is . The force required to just push it up the inclined plane

is 3 times the force required to just prevent it from sliding down. If we define N = 10

, then N is

Sol. (5)

d

u d

F = mg(sin cos )

F = mg(sin + cos )= 3F

13

112

10 5

So, N=5

43. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the

figure. The stick applies a force of 2 N on the ring and rolls it without slipping with an

acceleration of 0.3 m/s2. The coefficient of friction between the ground and the ring

is large enough that rolling always occurs and the coefficient of friction between the

stick and the ring is (P/10). The value of P is

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Sol. (4)

Taking torque about P,

p

2

2R - 2 R=Ia

2R - 2 R=2mR .R

2 2 2ma2 2 2 2 0.32 2 1.2

40.4

10P 4

44. Four solid spheres each of diameter 5 cm and mass 0.5 kg are placed with their

centers at the corners of a square of side 4 cm. The moment of inertia of the system

about the diagonal of the square is N 104 kg-m2, then N is

Sol. (9)

2 22 2 2 2

2 2

2 4

4 2

2 2 2 a 2 aI mR mR mR m mR m

5 5 5 52 28

I mR ma58 5

0.5 0.5 4 102 4

9 10 kgmN 9

45. A long circular tube of length 10 m and radius 0.3 m carries a current I along its

curved surface as shown. A wire-loop of resistance 0.005 ohm and of radius 0.1 m is

placed inside the tube with its axis coinciding with the axis of the tube. The current

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varies as I = Iocos(300t) where I0 is constant. If the magnetic moment of the loop is

N o Io sin(300t), then 'N' is

Sol. (6)

o 2 o o 2ring

2o o

2

o o

2

2 4

o o

2 4

I I cos300tr r

L L

I rd300sin300t

dt L

r 300i I sin300t

R LR

M i rr 300

I sin300tLR

r 300N 5.91 6

LR

46. Four point charges, each of +q, are rigidly fixed at the four corners of a square

planar soap film of side 'a'. The surface tension of the soap film is . The system of

charges and planar film are in equilibrium, and

1 /N2qa k

where 'k' is a constant.

Then N is

Sol. (3)

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2 2

2 2

23

12 3

kq kq 12 2 a

2a 2a

qa

qa

N 3

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Part III: Mathematics

SECTION - I (Total Marks: 21)


This section contains 7 multiple choice questions. Each question has 4 choices (A), (B),


47. Let P ={ : sin -cos = 2cos } and Q = : sin +cos = 2 sin } be two sets. Then (A) P Q and Q - P (B) Q P

(C) P Q (D) P = Q

Sol. (D)

We have:

sin cos 2 cos

sin 2 1 cos

1sin cos

2 1

2 1 sin cos

2 sin sin cos

2 sin sin cos

This shows that the given equations are identical.

Hence, their solution sets are equal.

P = Q

48. Let the straight line x = b divide the area enclosed by y = (1 x)2, y = 0 and x = 0

into two parts 1 2R (0 x b) and R (b x 1) such that 1 21

R R .4

Then b

equals

3 1 1 1(A) B c D4 2 3 4

Sol. (B)

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Given, 1 21

R R4

b 12 2

0 b

b 13 3

0 b

3 3

3 3

3

3

11 x dx 1 x dx

4

1 x 1 x 1

3 3 4

1 b 1 b1 10

3 3 3 4

1 b 1 b1 1

3 3 3 4

2 1 b 1 1 3 4 1

3 4 3 12 12

11 b

8

11 b

2

1 1b 1

2 2

49. Let and be the roots of 2 n nnx - 6x - 2 = 0, with > . If a = - for n 1,

10 8

9

a 2athen the value of2a

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

Sol. (C)

Given, and are the roots of 2x - 6x - 2 = 0.

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6, 2

n 1 n 1 n n n 2 n 2

n 1 n n 2n n 2

n 1

n n 2

n 1

10 8

9

Now,

6a a 2a

a 2a6

a

a 2a3 , n 2

2a

Putting n = 10, we get,

a 2a3

2a

50. A straight line L through the point (3, 2) is inclined at an angle 60 to the line3x y 1 . If L also intersects the x-axis, then the equation of L is

(A) y+ 3x 2 3 3 0 (B) y- 3x 2 3 3 0

(C) 3y x 3 2 3 0 (D) 3y x 3 2 3 0

Sol (B)

The equation of the line through (3, 2) can be written as

y + 2 = m(x 3)

It makes angle 60 with the line 3x y 1 .

o m 3tan601 3m

m 33

1 3m

m 3 or m 0

Since the line intersects x-axis also, hence m 0.

m = 3

Thus, the required line is:

y 2 = 3(x - 3)

y 2 = 3x - 3 3

y - 3x + 2 + 3 3 =0

51. Let (x0, y0) be the solution of the following equations

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In2 In3

Inx Iny

0

2x 3y

3 2Then x is

1 1 1

A B C D 66 3 2

Sol. (C)

In2 In3

3y 2

a a

lnx lny

3

y 3

k k

a k + 1

a

0

We have :

2x 3y

In2 log2x In3 log3y

log 2x log 3

2x 3 , 3y 2 , say

Also, we have:

3 = 2

lnx = lny(log 2)

log x = log 2

x = 2 , y = 3 , say

3 = 2

a = 0 and k + 1 = 0

2x = 3

1

x 2

1Hence, x

2

52. The value of

2In3

2 2In2

xsinxdx is

sinx sin In 6 x

1 3 1 3 3 1 3

(A) In (B) In C In D In4 2 2 2 2 6 2

Sol. (A)

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2In3

2 2In2

In3

In22

In3

In2

In3

In2

xsinxI dx

sinx sin In 6 x

1 sintdt ... (1)

2 sint sin In 6 twhere x t

sin In 6 t1dt ...(2)

2 sin In 6 t sint

Adding (1) and (2), we get:

12I dt

2

1 1 1 3I . In3 In2 In

2 2 4 2

53. Let a i j k,b i j k and c i j k be three vectors. A vector v in the

1plane of a and b, whose projection on c is , is given by

3

A i 3j 3k B 3i 3j k C 3i j 3k D i 3j 3k

Sol (C)

Let r a b be the required vector. Then,r i j k i j k i j k

1Projection of r on c i j k is given to be .

3

(r.c) 1

| c | 3

1

3 3

1

Taking 1 and 2, we get,

r 3i j 3k

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SECTION - II (Total Marks: 16)

(Multiple Correct Answer Type)

This section contains 4 multiple choice questions. Each question has four choices (A),

(B), (C) and (D) out of which ONE OR MORE may be correct.

54. Let f : R R be a function such that f(x + y) = f(x) + f(y), x, y R. If f(x)

is differentiable at x = 0, then

(A) f(x) is differentiable only in a finite interval containing zero

(B) f(x) is continuous x R

(C) f'(x) is constant x R

(D) f(x) is differentiable except at finitely many points

Sol. (B.C)

f(x + y) = f(x) + f(y)

Put x = y = 0 in the above equation. Then,

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

f(0) = 2f(0)

f(0) = 0

Also, we have:

h 0 h 0 h 0

f(x h) f(x) f(x) f(h) f(x) f(h) f(0)f '(x) lim lim lim

h h hf '(x) f '(0) k (say)

Integrating both sides, we get,f(x)=kx+cNow, f(0) = 0, so c = 0

f(x)=kxClearly, f(x) is continuous and has constant de

rivative x R.

55. Let the eccentricity of the hyperbola2 2

2 2x y

1a b

be reciprocal to that of the ellipse x2

+ 4y2 = 4. If the hyperbola passes through a focus of the ellipse, then

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2 2

2 2

x y(A) the equation of the hyperbola is 1

3 2

(B) a focus of the hyperbola is (2, 0)

5

(C) the eccentricity of the hyperbola is 3

(D) the equation of the hyperbola is x 3y 3

Sol (B,D)

2 2

The equation of the given ellipse is

x y1

4 1

1 3Eccentricity of the ellipse = e = 1- =

4 2

Foci of ellipse are 3,0 and 3,0 .

1 2Eccentricity of hyperbola =

3 3

2

2 2

2 2

2 2

2

x ySince given hyperbola 1 passes through 3,0 ,we have:

a b

3 01

a b

a 3

2 2 2

2

Now, b a e 1

4b 3 1 1

3

2 2

2 2

Thus, the equation of the hyperbola is

x y1

3 1

x 3y 3Thus, the foci of the hyperbola are ( 2, 0).

56. Let M and N be two 3 3 non-singular skew-symmetric matrices such that MN =

NM. If PT denotes the transpose of P, then M2N2 (MTN)1(MN-1)T is equal to

(A) M2 (B) N2 (C) M2 (D) MN

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SECTION - III (Total Marks: 15)

(Paragraph Type)

This section contains 2 paragraphs. Based upon one of the paragraph, 2 multiple choice

questions and based upon the second paragraph 3 multiple choice questions have to be

answered. Each of these questions has four choices (A), (B), (C) and (D) out of which ONLY

ONE is correct.


Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains

only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from

U1 and put into U2. However, if tail appears then 2 balls are drawn at random from U1 and

put into U2. Now 1 ball is drawn at random from U2.

58. The probability of the drawn ball from U2 being white is

13 23 19 11

(A) B C D30 30 30 30

Sol. (B)

When a coin is tossed then there are two outcomes.

If we get a head, then the probability of drawing a white ball from U2 is

P1 =1 3 2 1

12 5 5 2

If we get a tail, then the probability of drawing a white ball from U2 is

P2 =3 2 3 2

1 1 2 25 5 5

2 2 2

C C C C1 2 11

2 3 3C C C

Required probability

= P1 + P2

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1 3 1 1 6 2 3 1 1

2 5 5 2 10 3 10 10 3

2 1 12 9 1

5 2 30

2 115 30

12 11 23

30 30

59. Given that the drawn ball from U2 is white, the probability that head appeared on the

coin is

17 11 15 12

(A) B C D23 23 23 23

Sol. (D)

Using Baye's theorem,

1 3 1 2 1 4122 5 2 5 2 10Required probability

23 23 2330 30


Let a, b and c be three real numbers satisfying

1 9 7

a b c 8 2 7 0 0 0 ... (E)

7 3 7

60. If the point P(a, b, c), with reference to (E), lies on the plane 2x + y + z = 1, then

the value of 7a + b + c is

(A) 0 (B) 12 (C) 7 (D) 6

Sol. (D)

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2

n

n 0

For b = 6 a + c = -6

a + 7c = -48

On solving, we get,

a = 1, c = -7

Now, and are roots of the given equation ax + bx + c = 0, hence

-b c+ = = -6, = = -7

a a

1 1 6

7

1 1 6 61

7 7

21

.......... 76

17

SECTION - IV (Total Marks: 28)


This section contains 7 questions. The answer to each of the questions is a single-digit

integer, ranging from 0 to 9. The bubble corresponding to the correct answer is to be

darkened in the ORS.

63. Let f : [1, ) [2, ) be a differentiable function such that f(1) = 2.

x 3

1If 6 f t dt 3xf x x for all x 1, thenthe value of f(2) is

Sol. (6)

Note: If we put x = 1 in the given equation, we get, f(1) = 1/3. The correct equation

which satisfies f(1) = 2, should be x

3

16 f t dt 3xf x x 5

Now, on differentiating with respect to x, we have:

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The point of intersection of (1) and (2) is (1, 3) and of (2) and (3) is (-1, -1).

2

1

2

1 2 01 1

1 1 3 6 32 2

1 1 16

Hence, 23

67. Let 1sin

f sin tan ,where . Then the value of 4 4cos2

df is

d tan

Sol. (1)

1 sinLet k tancos2

22

2

2

2 2

22

2

sintank

cos2

sintan k

1 2sin

sinsec k 1 1 2 sin

cos ec 1sec k

cos ec 2

22

2

22

2

22

2 22

cos ec 2cos k

cosec 1

cos ec 21 cos k 1

cosec 11

sin k

cosec 11sin k tan

cotsink tan

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f sink tan

d fHence, 1

d tan

68. The minimum value of the sum of real numbers-5 -4 -3 8 10a , a , 3a , 1, a and a with a >

0 is

Sol. (8)

The given real numbers are-5 -4 -3 8 10a , a , 3a , 1, a and a .

We know that A.M. G.M.

Using this relation for the numbers, -5 -4 -3 -3 -3 8 10a , a , a ,a ,a , 1, a ,a , we have:

-5 4 3 3 3 8 10-5 4 3 3 3 8 10

-5 4 3 8 10

-5 4 3 8 10

a a a a a 1 a aa .a .a .a .a .1.a .a

8

a a 3a 1 a a 8

min a a 3a 1 a a 8

69. The positive integer value of n > 3 satisfying the equation

1 1 1is

2 3sin sin sin

n n n

Sol. (7)

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1 1 1

2 3sin sin sin

n n n

2 3 3 2sin sin sin sin sin sin

n n n n n n2 3 3

sin sin sin sin sinn n n n n

2 2 3sin 2cos sin sin sin

n n n n n

4 3sin sin sin sin

n n n n

4 3sin sin sin

n n n

0

7sin 2cos sin 0

n 2n 2n

72sin cos sin 0

n 2n 2n

7cos 0 sin 0,sin 0

2n n 2n

n 7