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IIT-JEE 2012 / Actual Paper –2Questions and Solutions

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Analysis IIT-JEE

Actual Paper –22012


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PART– 1 : PHYSICS

SECTION I : Single Correct Answer Type

This section contains 8 multiple choice question. Each question has four choices (A),(B),(C) and (D) outof which ONLY ONE is correct.

1. A thin uniform cylindrical shell, closed at both ends, is partially filled with water. It is floating verti-

cally in water in half-submerged state. If cP is the relative density of the material of the shell with

respect to water, then the correct statement is that the shell is

(A) more than half-filled if cP is less than 0.5

(B) more than half-filled if cP is more than 1.0.

(C) half-filled if cP is more than 0.5

(D) less than half-filled if cP is less than 0.5

Sol.1. A

2. In the given circuit, a charge of + 80 Cµ is given to the upper plate of the 4 Fµ capacitor. Then in the

steady state, the charge on the upper plate of the 3 Fµ capacitor is

(A) + 32 Cµ (B) 40 C+ µ (C) 48 C+ µ (D) 80 C+ µ


... 3

Sol.2. C

1 2a a

2 3= (Parallel)

1 2a a 80C+ =

2 22

a a 803

⇒ + =

33

a 80 48C5

⇒ = × =

3. Two moles of ideal helium gas are in a rubber balloon at 30ºC. The balloon is fully expandable andcan be assumed to require no energy in its expansion. The amount of heat required in raising thetemperature is nearly (take R = 8.31 J/mol.K)(A) 62 J (B) 104 J (C) 124 J (D) 208 J

Sol.3. D

4. Consider a disc rotating in the horizontal plane with a constant angular speed ω about its centre O.The disc has a shaded region on one side of the diameter and an unshaded region on the other sideas shown in the figure. When the disc is in the orientation as shown, two pebbles P and Q aresimultaneously projected at an angle towards R. The velocity of projection is in the y-z plane and issame for both pebbles with respect to the disc. Assume that (i) they land back on the disc before

the disc has completed 18

rotation, (ii) their range is less than half the disc radius, and (iii) ω re-

mains constant throughout. Then

(A) P lands in the shaded region and Q in the unshaded region.(B) P lands in the unshaded region and Q in the shaded region.(C) Both P and Q land in the unshaded region.(D) Both P and Q land in the shaded region.

Sol.4. C or D


... 4

5. A student is performing the experiment of Resonance Column. The diameter of the column tube is4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is 38ºC in which the speed ofsound is 336 m/s. The zero of the meter scale coincides with the top end of the Resonance Columntube. When the first resonance occurs, the reading of the water level in the column is(A) 14.0 cm (B) 15.2 cm (C) 16.4 cm (D) 17.6 cm

Sol.5. B

6. Two identical discs of same radius R are rotating about their axes in opposite directions with thesame constant angular speed ω . The discs are in the same horizontal plane. At time t = 0, thepoints P and Q are facing each other as shown in the figure. The relative speed between the two

points P and Q is rv . In one time period (T) of rotation of the discs, v, as a function of time is best

represented by

(A)

(B)


... 5

(C)

(D)

Sol.6. A

rV 0 at t 0= =V

r is max at point A again V

r = 0 at point B and maximum at point C.

Hence, graph (A).

B

C

A

P Q

C

A

B


... 6

7. A loop carrying current l lies in the x-y plane as shown in the figure. The unit vector k is coming outof the plane of the paper. The magnetic moment of the current loop is

(A) 2a Ik (B) 2I a Ik

2π +

(C) 2I a Ik

2π − +

(D) 2(2 I)a Ikπ +

Sol.7. B

22(a / i) ˆM iA i 4 a k

2

π= = × +

2 Îa 1 k2π = +


... 7

8. An infinitely long hollow conducting cylinder with inner radius R/2 and outer radius R carries a

uniform current density along its length. The magnitude of the magnetic field, | B |

as a function of

the radial distance r from the axis is best represented by

(A)

(B)

(C)

(D)


... 8

Sol.8. D

From r > 0 and R

r ,2

< B = 0.

From R

r2

< to r < R. B increases from zero to max value.

1B

r∝ when r > R. Hence (D).


... 9

SECTION II : Paragraph Type

This section contains 6 multiple choice questions relating to three paragraphs with the two questionsone each paragraph. Each question has four choice (A), (B) (C) and (D) out of which ONLY ONE iscorrect.

Paragraph for Questions 9 and 10

Most materials have the refractive index, n> I. So, when a light ray from air enters a naturally occurring

material, then by Snell’s law, 1 2

2 1

sin n,

sin n

θ=

θ it is understood that the refracted ray bends towards the

normal. But it never emerges on the same side of the normal as the incident ray. According to electromag-

netism, the refractive index of the medium is given by the relation, r rc

n ,v

= = ± ε µ where c is the speed

of electromagnetic waves in vacuum, v its speed in the medium, rε and rµ are the relative permittivity and

permeability of the medium respectively.

In normal materials, both rε and rµ are the relative permittivity and permeability of the medium respec-

tively. In normal materials, both rε and rµ are negative, one must choose the negative root of n. Such

negative refractive index materials can now be artificially prepared and are called meta-materials. Theyexhibit significantly different optical behavior, without violating any physical laws. Since n is negative, itresults in a change in the direction of propagation of the refracted light. However, similar to normal, mate-rials, the frequency of light remains unchanged upon refraction even in meta-materials.

9. Choose the correct statement.(A) The speed of light in the meta-material is v = c|n|

(B) The speed of light in the meta-material is c

v| n |

=

(C) The speed of light in the meta-material is v = c

(D) The wavelength of the light in the meta-material ( )mλ is given by m air | n |,λ = λ where airλ is

the wavelength of the light in air

Sol.9. B

Speed C

| n |= [Speed > 0]


... 10

10. For light incident from a on a meta-material, the appropriate ray diagram is

(A) (B)

(C) (D)

Sol.10. C

1 2

2 1

sin n

sin n

θ=

θ

n < 0 for meta-material,

2 2sin 0 0.θ < ⇒ θ <


... 11


The general motion of a rigid body can be considered to be a combination of (i) a motion of its centre ofmass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass.These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) hori-zontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated aboutthe origin on a horizontal frictionless plane with angular speed ω, the motion at any instant can be taken ascombination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of the discthrough an instantaneous vertical axis passing through its centre of mass (as is seen form the changedorientation of points P and Q) Both these motions have the same angular speed ω in the case.

Now consider two similar systems as shown in the figure: Case (a) the disc with its face vertical andparallel to z-z plane, Case (b) the disc with its face making an angle 45° with x-y plane and its horizontaldiameter parallel to x-axis, In both the cases, the disc is welded at point P, and the systems are rotatedwith constant angular speed ω about the z-axis.

11. Which of the following statements regarding the angular speed about the instantaneous axis(passing through the centre of mass) is correct?

(A) It is 2ω for both the cases (B) It is ω for cases (a); and 2

ωfor case (b)

(C) It is w for case (a); and 2ω for case (b) (D) It is ω for both the cases


... 12

Sol.11. D

Orientation of the rign doest not affect the angular velocity.

12. Which of the following statements about the instantaneous axis (passing through the centre ofmass) is correct?(A) It is vertical for both the cases (a) and (b).(B) It is vertical for case (a); and is at 45º to the x-z plane and lies in the plane of the disc for cases

(b).(C) It is horizontal for case (a); and is at 45º to the x-z plane and is normal to the plane of the disc

for case (b).(D) It is vertical for case (a); and is at 45º to the x-z plane and is normal to the plane of the disc for

case (b).

Sol.12. A


The β -decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory, aproton (p) and an electron (e–) are observed as the decay products of the neutron. Therefore, consideringthe decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energyof the electron should be a constant. But experimentally, it was observed that the electron kinetic energy

has a continuous spectrum. Considering a three-body decay process, i.e. en p e e v ,−→ + + + around

1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino ( )ev to be

massless and possessing negligible energy and the neutron to be at rest, momentum and energy conser-vation principles are applied. From this calculation, the maximum kinetic energy of the electron is 0.8 × 106

eV. The kinetic energy carried by the proton is only the recoil energy.

13. If the anti-neutrino had a mass of 3 eV/c2 .(where c is speed of light) instead of zero mass, whatshould be the range of the kinetic energy, K, of the electron?

(A) 60 K 0.8 10 eV≤ ≤ × (B) 63.0 eV K 0.8 10 eV≤ ≤ ×

(C) 63.0 eV K 0.8 10 eV≤ < × (D) 60 K 0.8 10 eV≤ < ×

Sol.13. D

14. What is the maximum energy of the anti-neutrino?(A) Zero. (B) Much less than 0.8 × 106 eV.(C) Nearly 0.8 ×106 eV. (D) Much larger than 0.8 × 106 eV.

Sol.14. C


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SECTION III : Multiple Correct Answer(s) Type

This section contains 6 multiple choice questions. Each question has four choices (A), (B),(C) and (D)out of which ONE or MORE are correct.

15. Six point charges are kept at the vertices of a regular hexagon of side l. and centre O, as shown in

the figure. Given that 20

1 qK ,

4 L=

πε which of the following statement(s) is (are) correct?

(A) The electric field at O is 6K along OD(B) The potential at O is zero.(C) The potential at all points on the line PR is same(D) The potential at all points on the line ST is same

Sol.15. A, B, C

OE = OO = OC = L

14 ε0π

qL22E =2

14 ε0π

qL22E =3

2q4 ε0π L22E =1

1 2 3E E E E 6k∴ = + + =

Potential at 0 os cancelled by symmetric +q and –q charges.


... 14

16. Two solid cylinders P and Q of same mass and same radius start rolling down a fixed inclined planefrom the same height at the same time. Cylinder P has most of its mass concentrated near itssurface, while Q has most of its mass concentrated near the axis. Which statement(s) is (are)correct?(A) Both cylinders P and Q reach the ground at the same time.(B) Cylinder P has larger linear acceleration than cylinder Q.(C) Both cylinders reach the ground with same translational kinetic energy(D) Cylinder Q reaches the ground with larger angular speed.

Sol.

16. D P QI I>

mgsin f ma

fR 1∝

θ − = =

solving we get

2

mgsina

Im

R

θ= +

P QI I>

P Qa a⇒ <

2V 2as=

T TP Q PV V KE KEθ⇒ < ⇒ <

Pw wθ⇒ <

17. Two spherical planets P and Q have the same uniform density P, masses MP and M

Q and surface

areas A and 4A, respectively. A spherical planet R also has uniform density P and its mass is

( )P QM M+ . The escape velocities from the planets P, Q and R, and VP, VQ and VR respectively.

Then(A) VQ > VR > VP (B) VR > VQ > VP

(C) VR / VP= 3 (D) P Q1

V / V2

=

Sol.17. B, D

3P P

4M R

3= π∫ and

30

4M R

3 θ= π∫2P4 R Aπ = and 24 R 4Aθπ =

PR 2Rθ⇒ = ...(1)


... 15

Also, 3

P C4

M M P R3

+ = π

3 3 3PR R Rθ⇒ + = Also, 3 3

P9R R=

...(ii) ...(iii)

From (i), (ii) and (iii),

2R

2

R P 2/ 3

2

P 2 / 3

8V GP R

3

8GP 4RV V V as V

3 9

8GP RV

3 9

θ θ

= π

π> > =

π =

P

O

V 1V 2

=


... 16

18. The figure shows a system consisting of (i) a ring of outer radius 3R rolling clockwise withoutslipping on a horizontal surface with angular speed / 2.ω The ring and disc are separated by friction-less ball bearings. The system is in the x-z plane. The point P on the inner disc is at a distance Rfrom the origin, where OP makes an angle of 30º with the horizontal. Then with respect to thehorizontal surface.

(A) the point O has a linear velocity 3R iω

(B) the point P has a linear velocity 11 3R i R k

4 4ω + ω

(C) the point P has a linear velocity 13 3R i R k

4 4ω − ω

(D) the point P has a linear velocity 3 1

3 R i R k4 4

− ω + ω

Sol.18. A, B


... 17

19. In the given circuit, the AC source has ω = 100 rad/s. Considering the inductor and capacity to beideal, the correct choice (s) is (are)

(A) The current through the circuit, I is 0.3 A. (B) The current through the circuit, I is 0.3 2 A.

(C) The voltage across 100Ω resistor 10 2= V

(D) The voltage across 50Ω resistor = 10 V.

Sol.19. C or A, C

20. A current carrying infinitely long wire is kept along the diameter of a circular wire loop, withouttouching it. The correct statement(s) is (are)(A) The emf induced in the loop is zero if the current is constant(B) The emf induced in the loop is finite if the current is constant(C) The emf induced in the loop is zero if the current decreases at a steady rate.(D) The emf induced in the loop is finite if the current decreases at a steady rate.

Sol.20. A, C

Net flox passing through the coil is zero.


... 18

PART II : CHEMISTRY


This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)out of which ONLY ONE is correct.

21. The reaction of white phosphorus with aqueous NaOH gives phosphine along with another phosphoruscontaining compound. The reaction type; the oxidation states of phosphorus in phosphine and theother product are respectively(A) redox reaction; – 3 and –5 (B) redox reaction; + 3 and + 5(C) disproportionation reaction; –3 and +5 (D) disproportionation ; –3 and +3

Sol.21. Zero marks to all.

22. The shape of XeO2F2 molecule is(A) trigonal bipyramidal (B) square planar (C) tetrahedral (D) see-saw

Sol.22. D

23. For dilute solution containing 2.5 g of a non-volatile non-electrolyte solute in 100 g of water, theelevation in boiling point at 1 atm pressure is 2°C. Assuming concentration of solute is much lowerthan the concentration of solvent, the vapour pressure (mm of Hg) of the solution is (take Kb = 0.76K kg mol–1)(A) 724 (B) 740 (C) 736 (D) 718

Sol.

23. A b bT K .m∆ =

or 2 3

2.5M0.76

100 10−=

×

or, 2.5 2M 7.6

=

solutionP

xP

∆ =

or,

2.5P M

2.5 100760M 18

∆ =+

or,

2P 7.6

2 1007607.6 18

∆ =+

or, P∆ = 36 mm of HgV.P = 760 – 36 = 724 mm of Hg


... 19

24. The compound that undergoes decarboxylation most readily under mild condition is

(A)

COO HCH COO H2

(B)

COO HO

(C)

COO HCOO H

(D)

CH CO OH2

O

Sol.

24. B β − keto acid undergoes decarboxylation most readily.

25. Using the data provided, calculate the multiple bond energy (kJ mol–1) of a C C≡ bond in C2H2. Thatenergy is (take the bond energy of a C-H bond as 350 kJ mol–1.2C(s) + H2(g) → C2H2(g) ∆H = 225 kJ mol–1

2C(s) → 2C(g) ∆H = 1410 kJ mol–1

H2(g) → 2H(g) ∆H = 330 kJ mol–1

(A) 1165 (B) 837 (C) 865 (D) 815

Sol.25. D

2C (S) + H2 (g) H C C H→ − ≡ −

( ) ( )atom (s) b.d H2H 2 H C H ∆ = ∆ + ∆ – ( ) ( )bd b.dC C C H

H H≡ − ∆ + ∆

( ) ( )bd C C1410 330 H 2 350≡

= + − ∆ + ×

Finaly, ( )bd C CH 815≡

∆ =

26. The major product H of the given reaction sequence is

2 495%H SOCN3 2 3 Heat

CH CH CO CHΘ

− − − → →G H

(A) 3

3

CH CH C COOH|

CH

− = − (B) 3

3

CH CH C CN|CH

− = −

(C) 3 2

3

OH|

CH CH C COOH|

CH

− = − (D) 3 2

3

CH CH C CO NH|

CH

− = − −

Sol.26. A


... 20

27. NiCl2P(C2H5)2(C6H5)2 exhibits temperature dependent magnetic behaviour (paramagnetic /diamagnetic). The coordination geometries of Ni2+ in the paramagnetic and diamagnetic states arerespectively(A) tetrahedral and tetrahedral (B) square planar and square planar(C) tetrahedral and square planar (D) square planar and tetrahedral

Sol.

27. C In the case of 2fNC The outer most configuration is3dB 45º

There are two unpaired electrons due to this. It shows paramagnetic properties and the shape willbe tetrahedral with sp3 hybridization.Similarly, if we consider all electrons are paired then it will be diamagnetic and shape will be squareplanaz with dsp2 hybridization.

28. In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agentsused are(A) O2 and CO respectively (B) O2 and Zn dust respectively(C) HNO3 and Zn dust respectively (D) HNO3 and CO respectively

Sol.28. B


... 21


This section contains 6 multiple choice questions relating to three paragraphs with two questions oneeach paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


Bleaching powder and bleach solution are produced on a large scale and used in several house holdproducts. The effectiveness of bleach solution is often measured by iodometry.

29. 25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 41 acetic acid.In the titration of the liberated iodine, 48 mL of 0.25 N Na2S2O3 was used to react the end point. Themolarity of the household bleach solution is(A) 0.48 M (B) 0.96 M (C) 0.24 M (D) 0.024 M

Sol.

29. C1 1

Ca(OCl)Cl+ −

1 1 1

22Cl 2 I 2Cl I+ − −

+ → +

2 2 2 3 2 4 6I Na S O 2NaI Na S O+ → +

meg. meg. 2 meg. 2 2 3n of Cl n of I n of Na S O±

= =

= 48 × 0.25= 12

n factor of OCl = 2

nm-mole = 12

62

=

Mobility = moles

solution

n

v (inL)

=

36 1025

1000

−× = 0.24M

30. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of thatoxoacid is(A) Cl2O (B) Cl2O7 (C) ClO2 (D) Cl2O6

Sol.

30. A 02 22HOCl Cl O H O→ +


... 22


3 2 2

3 2

3

(CH CO) O (i) H ,Pd/ CCH COONa (ii) SOCl

(iii) anhyd.AlCl

→ →I J

J(C9H

8O

2) gives effervescence on treatment with NaHCO

3 and a positive Baeyer’s test.

31. The compound K is

(A) (B) O O

(C) O

(D) O

Sol.

31. C

O H

I

O

CH C3

O

CH C2

H

+CH CO O3

–

solvent

|C — CH 2 — C |H

OH O ||

OH C3 C —

CH CH –CO OH i) H , PD/C2 2 2CH = CH – CO OH2

ii) SOCl2

CH – C H – C – C l2 2

O

Intram olecularelectrophilicsubstitu tion reaction

anh. A | C |3

J(gives effervescence w ithNaHCO and positiveBaeyer ’s test)

3

OK

32. The compound I is

(A)

O H

(B)

OH

(C)

O CH 3

(D)

HH

H

Sol.32. A


... 23


The electrochemical cell shown below is a concentration cell.

M | M2+ (saturated solution of a sparingly soluble salt, MX2) | | M2+ (0.001 mol dm–3) | M

The emf of the cell depends on the difference in concentrations of M2+ ions at the two electrode. Theemf of the cell at 298 K is 0.059 V.

33. The value of ∆G (kJ mol–1) for the given cell is (take 1F = 96500 C mol–1)(A) –5.7 (B) 5.7 (C) 11.4 (D) –11.4

Sol.33. D ∆G = –nFEcell

= – 2 × 96500 × 0.059 = – 11.4 KJ mol–1.

34. The solubility product (Ksp; mol3 dm–9) of MX2 at 298 K based on the information available the givenconcentration cell is (take 2.303 × R × 298/F = 0.059 V)(A) 1 × 10–15 (B) 4 × 10–15 (C) 1 × 10–12 (D) 4 × 10–12

Sol.34. B M | M2+ (10–3 M)M

22 S 25

MX M 2X−+ +

2 2spK [M ][X ]+ −=

= 453

on [M2+] = S =

1/ 3

SpK

4

E°cell = 0

1/ 3

Sp2K

M(S) M 2e4

+ +

2 3

2eM (10 M) M+ −

+

1/ 3

Sp2 3 2K

M M (10 M) M M4

+ − + + +

1/ 3

sp

3

K

40.059Ecell E cell log

2 10−

= ° −


... 24

or 0.059 =

1/ 3

sp

3

K

40.059log

2 10−

−

or –2 =

1/ 3

sp

3

K

4log

10−

or, log 10–2 =

1/ 3

sp

3

K

4log

10−

or

1/ 3

spK

4

= 10–5

or Ksp = 4 × 10–15


... 25

SECTION III : Multiple Correct Answer(s) Type

This section contains 6 multiple choice questions.. Each question has four choices (A), (B), (C) and (D)out of which ONE or MORE are correct.

35. For the given aqueous reactions, which of the statement(s) is (are) true?

excess K I + K [Fe(CN) ]3 6

dilute H HO2 4 brown ish-ye llow solution

ZzSO 4

white precipitate + brown ish-yellow filtrate

Na S O2 2 3

colourless solu tion

(A) The first reaction is a redox reaction.(B) White precipitate is Zn3[Fe(CN)6]2.(C) Addition of filtrate to starch solution gives blue colour.(D) White precipitate is soluble in NaOH solution.

Sol.35. A,C,D

KI + 0.53Fe(CN)6 K4[Fe(CN)6] + I2

↓(Brownish-yellow)

Zn [Fe(CN) ] + K i3white ppt. B rownish - ye llow

Na2S2O3Na S O (colourless sol )

2 6

2 4 6n

36. With respect to graphite and diamond, which of the statement(s) given below is (are) correct(A) Graphite is harder than diamond.(B) Graphite has higher electrical conductivity than diamond.(C) Graphite has higher thermal conductivity than diamond.(D) Graphite has higher C-C bond order than diamond.

Sol.36. B, D


... 26

37. With reference to the scheme given, which of the given statement(s) about T, U, V and W is (are)correct?

O

H C3 T

LiA lH 4

U

excess(CH CO) O3 2V W

CrO / H3⊕

(A) T is soluble in hot aqueous NaOH(B) U is optically active(C) Molecular formula of W is C10H18O4

(D) V gives effervescence on treatment with aqueous NaHCO3

Sol.37. A, C, D

O

T

O

H C3

LiA lH 4 H C3 OH

(not opticallyactive)

U

CrO|H3

+excess (CH

CO) O

3

2

CO OH

CO OH

(v)

gives effervescence w ith NaHC O 3

O – C – CH3

O

O – C – CH3

O

(M olecular formula

C H O )10

184


... 27

38. Which of the given statement(s) about N, O, P and Q with respect to M is (are) correct?

(A) M and N are non-mirror image stereoisomers(B) M and O are identical(C) M and P are enantiomers(D) M and Q are identical

Sol.38. A,B,C

39. The reversible expansion of an ideal gas under adiabatic and isothermal conditions is show in thefigure. Which of the following statement(s) is (are) correct?

(A) T1 = T2 (A) T3 > T1

(C) wisothermal > wadiabatic (D) ∆Uisothermal > ∆Uadiabatic

Sol.39. A, D


... 28

40. The given graphs / data I, II, III and IV represent general trends observed for different physisorptionand chemisorption processes under mild conditions of temperature and pressure. Which of thefollowing choice(s) about I, II, III and IV is (are) correct?

(A) I is physisorption and II is chemisorption(B) I is physisorption and III is chemisorption(C) IV is chemisorption and II is chemisorption(D) IV is chemisorption and III is chemisorption

Sol.40. A, C


... 29

Part III : Mathematics


This section contains 8 multiple choice questions. Each question has four choices (A), (B), (C) and (D)out of which ONLY ONE is correct.

41. If a and b

are vectors such that a b 29+ =

and ( ) ( ) a 2i 3j 4k 2i 3j 4k b,× + + = − − ×

then a

possible value of ( ) ( )a b 7i 2j 3k+ ⋅ − + +

is

(A) 0 (B) 3 (C) 4 (D) 8

Sol.

41. C ˆ ˆ ˆ ˆ ˆ âx(2i 3 j 4k) (2i 3 j 4k) b+ + = + + +

ˆ ˆ ˆ(a b) (2i 3 j 4k) b⇒ + × + + =

ˆ ˆ â b (2i 3 j 4k)⇒ + = λ + +

given | a b | 2a 1⇒ + = ⇒ λ = ±

ˆ ˆ ˆ(a b).( 7i 2 j 3k)∴⇒ + − + +

( ) ( )2i 3j 4k 7i 2j 3k 4= ± + + − + + = ±

42. If P is a 3 × 3 matrix such that PT = 2P + I, where PT is the transpose of P and I is the 3 × 3 identity

matrix, then there exists a column matrix

x 0

x y 0

z 0

= ≠

such that

(A)

0

PX 0

0

=

(B) PX = X (C) PX = 2X (D) PX = – X

Sol.

42. D

a b c

Let P d e f

g h i

=

given PT = 2P + I


... 30

a d g 2a 1 b c

b e h d 2c 1 f

c f i g h 2i 1

+ ⇒ = + +

1 0 0

P 0 1 0

0 0 1

− ⇒ = − −

PX X∴ = −

43. Let (a) and (a)α β be the roots of the equation

( ) ( ) ( )23 1 a 1 x 1 a 1 x 6 1 a 1 0+ − + + − + + − = where a > – 1

Then a 0 a 0

lim (a) and lim (a)+ +→ →α β are

(A) 5

and 12

− (B) 1

and 12

− − (C) 7

and 22

− (D) 9

and 32

−

Sol.

43. B ( ) ( ) ( )2 63 1 a 1 x 1 a 1 x 1 a 1 0+ − + + − + + − =

( ) ( ) ( )3 6

2

a 0t

1 a 1 1 a 1 1 a 1lim x x 0

1 a 1 1 a 1 1 a 1→

+ − + − + −+ + =+ − + − + −

21 1 1x x 0

3 2 6⇒ + + =

22x 3x 1 0⇒ + + =

∴Roots are 12

− and – 1

44. Four fair dice D1, D2, D3 and D4, each having six faces numbered 1, 2, 3, 4, 5 and 6, are rolledsimultaneously. The probability that D4 shows a number appearing on one of D1, D2 and D3 is

(A) 91

216(B)

108216

(C) 125216

(D) 127216

Sol.44. A Total number of ways for all the four dice to show forces in 64.

Number of favourable cases = 6 3 6 3 2 2 6 13 1 2 2 1 1 1 1P C C C C C C C× + × × × + ×

∴ Required Probability 91

216=


... 31

45. The value of the integral

/ 22

/ 2

xx ln cos x dx

x

π

−π

π + + π − ∫ is

(A) 0 (B) 2

42

π − (C) 2

42

π + (D) 2

2π

Sol.

45. B/ 2 2/ 2

xI x ln cosx dx

x

π−π

π + = + π − ∫/ 2

2

/ 2

xI x in cos x dx

x

π

−π

π − = + π + ∫

/ 22

/ 2

2I x cos x dxπ

−π⇒ = ∫

/ 22

0

I 2 x cos x dxπ

⇒ = ∫

( ) / 22

02 x sinx 2xcosx 2sinx

π= + −

2 22 2 4

4 2

π π= − = −

46. Let a1, a2, a3, ... be in harmonic progression with a1 = 5 and a20 = 25. The least positive integer n forwhich an < 0 is(A) 22 (B) 23 (C) 24 (D) 25

Sol.46. D Let ‘d’ be the common difference of the Reciprocals of the terms of the H.P

20 1

1 1 419d d

a a 19.25−= + ⇒ =

( )n

1 1 4n 1

a 5 19.25− = + −

for an < 0 we get 99

n4

>


... 32

47. The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and

x – y + z = 3 and at a distance 2

3 from the point (3, 1, – 1) is

(A) 5x – 11y + z = 17 (B) 2x y 3 2 1+ = −

(C) x y z 3+ + = (D) x 2y 1 2− = −

Sol.47. A Let the equation of plane is

(x 2y 3z 2) (x y z 3) 0+ + − + λ − + − =distance from (3, 1, –1) to the plane is

( ) ( ) ( )2 2 2

2 2d

31 2 3

λ= =

+ λ + λ − + λ +

72

−⇒ λ =

48. Let PQR be a triangle of area ∆ with a = 2 7 5

b and c ,2 2

= = where a, b and c are the lengths of the

sides of the triangle opposite to the angles at P, Q and R respectively.

Then 2sin P sin2 P

equals2sin P sin2 P

−+

(A) 3

4∆(B)

454∆

(C) 23

4 ∆

(D) 245

4 ∆

Sol.

48. C

2 227 5

258 292 2

cosP7 3 70 3522 2

+ − = = =⋅ ⋅

2012sinP sin2P 1 cosP 6 735

292sinP sin2P 1 cosP 64 32135

−− −= = = =+ + + +

1 35(5 a)(5 b)(5 c) 4 2 6

2 2∆ = − − − = × × × =

23 9 340 16 6 32

= = ×


... 33


This section contains 6 multiple choice questions relating to three paragraphs with two questions oneach paragraph. Each question has four choices (A), (B), (C) and (D) out of which ONLY ONE is correct.


Let an denote the number of all n-digit positive integers formed by the digits 0.1 or both such that noconsecutive digits in them are 0. Let bn = the number of such n-digit integers ending with digit 1 and cn =the number of such n-digit integers ending with digit 0.

49. Which of the following is correct?

(A) a17 = a16 + a15 (B) 17 16 15c c c≠ + (C) 17 16 16b b c≠ + (D) 17 17 16a c b= +

Sol.

49. A 17 16 15 14 13 12 15 10 917 0 1 2 3 4 5 6 7 8a C C C C C C C C C= + + + + + + + +

16 15 14 13 12 11 10 9 816 0 1 2 3 4 5 6 7 8a C C C C C C C C C= + + + + + + + +

15 14 13 12 11 10 9 815 0 1 2 3 4 5 6 7a C C C C C C C C= + + + + + + +

16 15 17a C a∴ + =

50. The value of b6 is(A) 7 (B) 8 (C) 9 (D) 11

Sol.

50. B

Number of ways to fill above four places will be 4 4 40 1 2C C C 3 8+ + − =


A tangent PT is drawn to the circle x2 + y2 = 4 at the point ( )P 3,1 . A straight line L, perpendicular to PT

is a tangent to the circle (x – 3)2 + y2 = 1.

51. A common tangent of the two circles is

(A) x = 4 (B) y = 2 (C) x 3y 4+ = (D) x 2 2y 6+ =


... 34

Sol.51. D Let slope of common tangent be ‘n’ equation of tangents will be

2y mx 2 1 m= + +

( ) 2y m n 3 1 m= − + +

Comparing the coefficients we get 1

n2 2

= ±

52. A possible equation of L is

(A) x 3y 1− = (B) x 3y 1+ = (C) x 3y 1− = − (D) x 3y 5+ =

Sol.

52. A Slope of tangent at P is 3−

⇒ Slope of line parpendicular to tangent is 1

.3

Equation of tangent is 1 1

y (x 3) 123

= − ± +


... 35


Let f(x) = (1 – x)2 sin2 x + x2 for all x IR,∈ and let

x

l

2(t 1)g(x) ln t f(t) dt

t 1− = − + ∫ for all x (1, ).∈ ∞

53. Consider the statements :

P : There exists some x IR∈ such that f(x) + 2x = 2(1 + x2)

Q : There exists some x IR∈ such that 2f(x) + 1 = 2x(1 + x)Then(A) both P and Q are true (B) P is true and Q is false(C) P is false and Q is the (D) both P and Q are false

Sol.53. C P) f(x) + 2x = 2(1 + x2)

(1 – x)2 sin2x + x2 + 2x = 2 + 2x2

2 2 2 2x cos x 2xcos x cos x 1 0⇒ − + + =

( )22cos x x 1 1 0⇒ − + = has no solution

Q) 2f(x) + 1 = 2x(1 + x)

( )2 2 2 22 1 x sin x x 1 2x 2x ⇒ − + + = +

( )2 22 1 x sin x 1 2x⇒ − + = have real solutions

54. Which of the following is true?

(A) g is increasing on (1, )∞

(B) g is decreasing on (1, )∞

(C) g is increasing on (1, 2) and decreasing on (2, )∞

(D) g is decreasing on (1, 2) and increasing on (2, )∞

Sol.

54. B g’(x) = 2 2 22(x 1)

lnx (1 x) sin x xx 1

− − − + +

Let h(x) = 2(x 1) 4

lnx 2 lnxx 1 x 1

− − = − −+ +

h(1) = 0

n’(x) = 2

4 10 x (1, )

x(x 1)− − < ∀ ∈ ∞+

⇒ h(x) is decreasing

⇒ g’(x) < 0 x (1, )∀ ∈ ∞


... 36


This section contains 6 multiple choice questions. Each question has four choices (A), (B), (C) and (D)out of which ONLY or MORE are correct.

55. If the straight lines x 1 y 1 z x 1 y 1 z

and2 k 2 5 2 k− + − −= = = = are coplanar, then the plane(s) containing

these two lines is(are)(A) y + 2z = – 1 (B) y + z = – 1 (C) y – z = – 1 (D) y – 2z = – 1

Sol.55. B, C

As the two straight lines are co-planar

1 1 1 1 0 0

2 k 2 0

5 2 k

+ − + −=

⇒ K = 2±Equation of plane containing the lines is

x y 7

2 k 2 constant

5 2 k

=

56. Let X and Y be two events such that ( )1 1 1P(X | Y) , P(Y | X) and P X Y .

2 3 6= = =

Which of the following is (are) correct?

(A) 2

P(X Y)3

= (B) X and Y are independent

(C) X and Y are not independent (D) c 1

P(X Y)3

=


... 37

Sol.56. A, B

P(X Y) 1P(X / Y)

P(Y) 2∩= =

P(X Y) 1P(Y / X)

P(X) 3∩= =

1P(X Y)

6∩ =

⇒1 1

P(X) , P(Y)2 3

= =

1 1 1 2

P(X Y)2 3 6 3

∪ = + − =

57. If the adjoint of a 3 × 3 matrix P is

1 4 4

2 1 7 ,

1 1 3

then the possible value(s) of the determinant of P is

(are)(A) – 2 (B) – 1 (C) 1 (D) 2

Sol.57. A, D

| Ads (P) | = | P |n – 1

2

1 4 4

2 1 7 | P |

1 1 3

=

⇒ | P |2 = 4

⇒ | P | = 2±

58. Let ( )f : 1,1 IR− → be such that 2

2f(cos4 ) for 0, , .

4 4 22 sec

π π π θ = θ ∈ − θ Then the value(s) of

1f

3

is (are)

(A) 3

12

− (B) 3

12

+ (C) 2

13

− (D) 2

13

+


... 38

Sol.58. Zero marks to all

59. For every integer n, let an and bn be real numbers. Let function f : IR IR→ be given by

[ ]( )

n

n

a sin x, for x 2n, 2n 1f(x) ,

b cos x, for x 2n 1,2n

+ π ∈ += + π ∈ − for all integers n.

If f is continuous, then which of the following hold(s) for all n?(A) an–1 – bn–1 = 0 (B) an – bn = 1 (C) an – bn+1 = 1 (D) an–1 – bn = –1

Sol.59. B, D

As f is continuous for all integers ‘n’ we get an + sin 2nπ = bn + cos 2nπ⇒ wn – bn = 1and also wA–1 + sin π (2n – 1) = bn + cos (2n – 1) n⇒ wA–1 = bn – 1

⇒ wn – 1 – bn = –1

60. If 2x r

0f(x) e (t 2)(t 3) dt= − −∫ for all ( )x 0, ,∈ ∞ then

(A) f has a local maximum at x = 2(B) f is decreasing on (2, 3)

(C) there exists some c (0, )∈ ∞ such that f'' (c) = 0

(D) f has a local minimum at x = 3

Sol.60. A, B, C, D

2xf '(x) e (x 2)(x 3)= ⋅ − −

f '(2) f '(3) & f '(x)= is continuous 2 3

+ – + and differentiable.

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