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JEE ADVANCED

EXAMINATION 2014

QUESTIONS WITH SOLUTIONS

PAPER - 2 [CODE - 4]

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PHYSICS

IIT-JEE 2014 Solutions by Motion Edu. Pvt. Ltd. Kota

1. A metal surface is illuminated by light of twodifferent wavelengths 248 nm and 310 nm. The maxi-mum speeds of the photoelectrons corresponding tothese wavelengths are u1 and u2, respectively. Ifthe ratio u1 : u2 = 2 : 1 and and hc= 1240 eV nm,the work function of the metal is nearly

(A) 3.7 eV (B) 3.2 eV(C) 2.8 eV (D) 2.5 eV

Sol. A

E1 = 248012400

=5eV

E2 = 310012400

= 4eV

5 = + 4k .....(1)4 = + k .....(2)On solving

= 311

= 3.66 = 3.7 eV

2. If Cu is the wavelength of K X-ray line ofcopper (atomic number 29) and M0 is the wave-length iof the K X-ray line of molybdenum (atomicnumber 42), then the ratio Cu/Mo is close to

(A) 1.99 (B) 2.14(C) 0.50 (D) 0.48

Sol. BBy Mosley’s Law :

f = a (z – 1)

As v = f f = c

Inverse relation and hence :

2)1Z(1

2

1

= 2

1

2

1Z1Z

1

0

= 2

2

)124()142(

= 2.14

3. Parallel rays of light of intensity I=912Wm-2 are incident on a spherical black body kept insurroundings of temperature 300 K. Take Stefan-Boltzmann constant =5.7×10-8Wm-2K-4 and assumethat the energy exchange with the surroundings isonly through radiation. The final steady statetemeprature of the black body is close to

(A) 330 K (B) 660 K(C) 990 K (D) 1550 K

Sol. A912×R2 = 5.7×10–18×1×[T4–(300)4] × 4 R2

108 (121) = T4

T = 330 K

4. During an experiment with a metre bridge,the galvanometer shows a null point when the jockeyis pressed at 40.0 cm using a standard resistance of90 , as shown in the figure. The least count of thescale used in the metre bridge is 1 mm. The un-known resistance is

(A) 60 0.15(B) 135 0.56(C) 60 0.25(D) 135 0.23

Sol. CR × 60 = 90 × 40

R = 90 × 1

2

n R = n 90 + n 2 + n 1

RdR

= 2

2d

+

1

1d

dR60

= 60

1.0 +

401.0

dR = 0.1 + 0.15= 0.25 R = (60 ± 0.25)

5. A planet of radius 1R

10 ×(radius of Earth)

has the same mass density as Earth. Scientists dig

a well of depth R5

on it and lower a wire of the same

length and of linear mass density 10-3 kgm-1 into it.If the wire is not touching anywhere, the force ap-plied at the top of the wire by a person holding it inplace is (take the radius oif Earth=6×106m and theacceleration dur to gravity on Earth is 10 ms-2)

(A) 96 N (B) 108 N(C) 120 N (D) 150 N

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Sol. B

R = 101

Re

dF =

10/Rx1

109.dx.

e

= 10–3 × 1

2

ee

50R

50R

× 2R10

e

= 10–3 × 6 × 106

9001

501

= 108

6. A wire, which passes through the hole in a smallbead, is bent in the form of quarter of a circle. Thewire is fixed vertically on ground as shown in thefigure. The bead is released from near the top ofthe wire and it slides along the wire without friction.As the bead moves from A to B, the force it applieson the wire is

(A) always radially outwards.(B) always radially inwards.(C) radially outwards initially and radially inwards later.(D) radially inwards intially and radially outwards later.

Sol. DBy theory

7. A glass capillary tube is of the shape of a trun-cated cone with an apex angle so that its twoends have cross sections of different radii. whendipped in water vertically, water rises in it to a heighth, where the radius of its cross section is b. If thesurface tension of water is S, its density is , andits contact angle with glass is , the value of h willbe (g is the acceleration due to gravity)

(A) 2s cosb g

(B) 2s cos

b g

(C) 2s cos /2b g

(D) 2s cos /2

b g

Sol. D

h = T cos ( + /2)

h = 2sb g cos ( + /2)

8. Charges Q, 2Q and 4Q are uniformly distributed inthree dielectric solid spheres 1,2 and 3 of radii R/2,R and 2R respectively, as shown in figure. If magni-tudes of the electric fields at point P at a distance Rfrom the centre of spheres 1,2 and 3 are E1, E2 andE3 respectively, the

(A) E1>E2>E3 (B) E3>E1>E2(C) E2>E1>E3 (D) E3>E2>E1

Sol. C

E1 = 2RKQ

E2 = 2RKQ2

E3 = 3)R2(R.KQ4

= 2R2KQ

E2 > E1 > E3

9. A tennis ball is dropped on a horizontal smoothsurface. It bounces back to its original position af-ter hitting the surface. The force on the ball duringthe collision is proportional to the length of com-pression of the ball. Which one of the followingsketches describes the variation of its kinetic en-ergy K with time t most appropriately? The figuresare only illustrative and not to the scale.

(A) (B)

(C) (D)


Sol. BV = gt

KE = 21

m (gt)2

Relation is parabolic

10. A point source S is placed at the bottom of atransparent block of height 10 mm and refractiveindex 2.72. It is immersed in a lower refractive indexliquid as swhon in the figure. It is found that thelight emerging from the block to the liquid forms acircular bright spot of diameter 11.54 mm on the topof the block. The refractive index of the liquid is

(A) 1.21 (B) 1.30(C) 1.36 (D) 1.42

Sol. C

sin C = 100254.11

254.11

2

Also sin C = 72.2n

3.13377.5

= M

2.72n = 1.36

Paragraph 11 & 12The figure shows a circular loop of radius a with twolong parallel wires (numbered 1 and 2) all in theplane of the paper. The distance of each wire fromthe centre of the loop is d. The loop and the wiresare carrying the same current I. The current in theloop is in the counterclockwise direction if seen fromabove.

11. When d a but wires are not touching the loop,it is found that the net magnetic field on the axis ofthe loop is zero at a height h above the loop. Inthat case(A) current in wire 1 and wire 2 is the direction PQand RS, respectively and h a(B) current in wire 1 and wire 2 is the direction PQand SR, respective and h a(C) current in wire 1 and wire 2 is the direction PQand SR, respectively and h 1.2a(D) current in wire 1 and wire 2 is the direction PQand RS, respectively and h 1.2a

Sol. C

222322

20

22 ha

i

ha

ia o/

cos

02 2

i

2 a h

cos

222322

2haha

a/

22122 /

ha

a

222

2 4

haa

a2 = 0.4a2 + 0.4h2

0.6a2 = 0.4h2

h2 = 23

a2

h 1.2 a

12. consider d>>a, and the loop is rotated about itsdiameter parallel to the wires by 30° from the posi-tion shown inthe figure. If the currents in the wiresare in the opposite directions, the torque on theloop at its new position will be (assume that the netfield due to the wires is constant over the loop)

(A) 2 2

0l ad

(B)

2 20l a2d

(C) 2 2

03 l ad

(D) 2 2

03 l a2d

Sol. B

= MB sin

= ia2× dio

22

sin 300

= dai

dai oo

221 2222


Paragraph 13 & 14In the figure a containner is shown to have a movalble(without friction) piston on top. The container andthe piston are all made of perfectly insulating mate-rial allowing no heat transfer between outside andinside the container. The container is divided intotwo compartments by a rigid partition made of athermally conducting material that allows slow trans-fer of heat. The lower compartment of the con-tainer is filled with 2 moles of an ideal monatomicgas at 700 K and the upper compartment is filledwith 2 moles of an ideal diatomic gas at 400 K. Theheat capacities per mole of an ideal monatomic gas

are V P3 5C R,C R2 2

, and those for an ideal diatomis

gas are V P5 7C R,C R2 2

.

13. Consider the partition to be rigidly fixed so thatit does not move. When equilibrium is achieved, thefinal temperature to the gases will be

(A) 550 K (B) 525 K(C) 513 K (D) 490 K

Sol. D

2 × 23

R (700 – T)

= 2R (T – 400) + 25

R (T – 400) × 2

T = 490 K

14. Now consider the partition to be free to movewithout friction so that the pressure of gases inboth compartments is the same. Then total workdone by the gases till the time they achieve equilib-rium will be

(A) 250 R (B) 200 R(C) 100 R (D) -100 R

Sol. D|Q1| = |Q2|

52

× 2 × R (700–T) = 27

× 2 × R (T – 400)

T = 525 KW.D. by lower gas = 2 × R × 175 = 350 RW.D. upper gas = 2 × R × 125 = 250 R net W.D. = –100 R

Paragraph 15 & 16A spray gun is shown in the figure where a pistonpushes air out of a nozzle. A thin tube of uniformcross section is connected to the nozzle. The otherend of the tube is in a small liquid container. As thepiston pushes air through the nozzle, the liquid fromthe containwer rises into the nozzle and is sdprayedout. For the spray gun shown, the radii of the pistonand the nozzle are 20 mm and 1 mm respectively.The upper end of the container is open to theatmosphere.

15. If the piston is pushed at a speed of 5 mms-1, theair comes out of the nozzle with a speed of

(A) 0.1 ms-1 (B) 1 ms-1

(C) 2 ms-1 (D) 8 ms-1

Sol. CFrom equation of continuityA1V1 = A2V2(20)2 × 5 × t = (1)2 × v × tv = 400 × 5 = 2000 mm/s= 2 m/s

16. If the density of air is a and that of the liquidl, then for a given piston speed the rate (volumeper unit time) at which the liquid is sprayed will beproportional to

(A) a

l

(B) a l

(C) a

l (D) l

Sol. A

Pressure at point in nozzle PP = P0 – 1 a V1

2 .....(1)

P = P0 – 21

l V22 – lgh ......(2)

equation (1) and (2)

21

a V12 =

21

lV22 + lgh

Neglecting the term regh

V2 = a

1l

V

Rate of liquid flow = AV2

= AV1 a

l

Rate of liquid flow

a

l


17. A person in a lift is holding a water jar, whichhas a small hole at the lower end of its side. Whenthe lift is at rest, the water jet coming out of thehole hits the floor of the lift at a distance d of 1.2 mfrom the person. In the following, state of the lift’smotion is given in list I and the distance where thewater jet hits the floor of the lift is given in List II.Match the statements from List I with those is ListII and select the correct answer using the codegiven belsow the lists.

List I List IIP. Lift is acceleration vertically up. 1. d = 1.2 mQ. Lift is accelerating vertically 2. d> 1.2 m with an acceleration less than the gravitational acceleration.R. Lift is moving vertically up with 3. d < 1.2 m constant speed.S. Lift is falling freely. 4. No water

leaks out of thejar

Code :(A) P-2, Q-3, R-2, S-4(B) P-2, Q-3, R-1, S-4(C) P-1, Q-1, R-1, S-4(D) P-2, Q-3, R-1, S-1

Sol. C

d = vt = gh2 gh2

= 2h which is independent

of g.But when the lift falls freely no water leaksout of the jar as geff = 0.

18. For charges Q1, Q2, Q3 and Q4 of same magni-tude are fixed along the x axis at x = -2a, -a, +aand +2a, respectively. A positive charge q is placedon the positive y axis at a distance b > 0. Fourpotions of the signs of these charges are given inList I. The direction of the forces on the charge q isgiven in List II. Match List I with List II and selectthe correct answer using the code given below thelists.

List I List IIP. Q1, Q2, Q3, Q4 all positive 1. +xQ. Q1, Q2 positive; Q3, Q4 negative 2. -xR. Q1, Q4 positive; Q2, Q3 negative 3. +yS. Q1, Q3 positive; Q2, Q4 negative 4. -yCode :

(A) P-3, Q-1, R-4, S-2(B) P-4, Q-2, R-3, S-1(C) P-3, Q-1, R-2, S-4(D) P-4, Q-2, R-1, S-3

Sol. A

(P) All +ve

(Q) Q1, Q2 +ve

Q3, Q4 –ve

(R) Q1, Q3 –ve

Q2, Q3 –ve

(S) Q1, Q3 +ve

Q2, Q4 –ve


19. Four combinations of two thin lenses aregiven in List I. The radius of curvature of all curvedsurface is r and the refractive index of all the lensesis 1.5. Match lens combinations in List I with theirfocal length in List II and select the correct answerusing the code given below the lists.

List I List II

P. 1. 2r

Q. 2. r/2

R. 3. -r

S. 4. r

Code :(A) P-1, Q-2, R-3, S-4(B) P-2, Q-4, R-3, S-1(C) P-4, Q-1, R-2, S-3(D) P-2, Q-1, R-3, S-4

Sol. D

f1

=

1

m

L

21 R1

R1

rrf111

231

1 1

1 1 2f 2 r

rf11

1

eq

1 1 1f r r

2rfeq

1

1 3 11f 2 r

rf 211

1

21

111fffeq

eq

1 1f r

feq = r

rf 211

1

2

1 3 1 11f 2 r

rrfeq 21

211

feq = -r

1f1

=

123

r1

r1

= r1

2

1f =

3 12

1 1

r

= r21

eq

1 1 1f r 2r

feq = 2r

P - 2 Q - 4 R - 3 S - 1

20. A block of mass m1 = 1 kg another massm2=2kg are placed together (see figure) on aninclined plane with angle of inclination . Variousvalues of are given in list I. The coefficient ofstatic and dinamic friction between the block m2and the plane are equal to =0.3. In List II expres-sions for the friction on block m2 are given. Matchthe correct expression of the friction in List II withthe angles given in List I, and choose the correctoption. The acceleration due to gravity is denoted by g.

List I List IIP. = 5° 1. m2g sin Q. = 10° 2. (m1+m2) g sin R. = 15° 3. m2 g cos S. = 20° 4. (m1+m2) g cos qCode :

(A) P-1, Q-1, R-1, S-3(B) P-2, Q-2, R-2, S-3(C) P-2, Q-2, R-2, S-4(D) P-2, Q-2, R-3, S-3

Sol. DThere will be no slipping if frictionbalances the net force acting downwardsalong the incline

maxsf = m2 g cos Force in the downward direction is (m1+ m 2) g s in thus angle at whichsl ipping starts(m1 + m2) g sin = m2 g cos

tan = 3

23.0 = 0.2

given tan 11.5° = 0.2Thus = 11.5°Thus for angles less than 11.5° there won’tbe any slipping hence friction is static andequal to (M1 + M2) g sin for greater than11.5° the friction is dynamic and is equal to m2 g cos .


CHEMISTRY

21. For the identification of -naphthol using dyetest, it is necessary to use(A) dichloromethane solution of -napthol(B) acidic solution of -naphthol(C) neutral solution of -naphthol(D) alkaline solution of -naphthol

Sol. D

22. For the elementary reaction M N, the rateof disappearance of M increases by a factorof 8 upon doubling the concentration of M.The order of the reaction with respect to Mis :(A) 4 (B) 3(C) 2 (D) 1

Sol. (B)r = K [M]x

8r = K [2M]x

x = 3

23. For the processH2O (l) H2O (g)

at T=100ºC and 1 atomsphere pressure, thecorrect choice is :(A) Ssystem > 0 and Ssurroundings > 0(B) Ssystem > 0 and Ssurroundings < 0(C) Ssystem < 0 and Ssurroundings > 0(D) Ssystem > 0 and Ssurroundings < 0

Sol. (B)H2O (l) + Heat H2O (g) Ssystem > 0& Ssurroundings < 0

24. Isomers of hexane, based on their branch-ing, can be divided into three distinct classesas shown in the figure.

(I)

(II)

(III)

The correct order of their boiling point is :(A) I > II > III (B) III > II > I(C) II > III > I (D) III > I > II

Sol. B

Boiling point surface area branching1

25. Assuming 2s-2p mixing is NOT operative, theparamagnetic species among the following is:(A) Be2 (B) B2(C) C2 (D) N2

Sol. (C)if 2s-2p mixing is not allowed then MOD willbe like O2.

1s * 1s 2s *2s 2px ||

z

y2p

2p

*

||

*z

y2p

2p

*2px.

26. The product formed in the reaction of SOCl2with white phosphorous is :(A) PCl3 (B) SO2Cl2(D) SCl2 (D) POCl3

Sol. (A)P4 + SOCl2 PCl3 + SO2 + S2Cl2white

27. The major product in the following reactionis

(A)

(B)

(C) (D)

Sol. D


28. Hydrogen peroxide in its reaction with KlO4

and NH2OH respectively, is acting as a(A) reducing agent, oxidising agent(B) reducing agent, reducing agent(C) oxidising agent, oxidising agent(D) oxidising agent, reducing agent

Ans. (A)

29. The acidic hydrolysis of ether (X), shownbelow is fastest when

(A) one phenyl group is replaced by a methylgroup.(B) one phenyl group is replaced by a para-methoxyphenyl group(C) two phenyl groups are replaced by twopara-methoxyphenyl groups(D) no structural change is made to X

Ans. CRate of SN1 reaction stability of carbocation. If two phenyl group is replaced by two p-methoxy phenyl group then it gives moststable carbocation so fastest reaction.

30. Under ambient conditions, the total numberof gases released as products in the finalstep of the reaction scheme shown below is

XeF6

CompleteHydrolysis

P + other product

Q

OH /H O-2

Products

slow disproportionation in OH /H O-2

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

Ans. (C)

XeF6 + H2O CompleteHydrolysis

XeO3 + HF

XeO3 + OH– (basic medium) HXeO4–

HXeO4– + OH– Slow

Disproportionation XeO6

4– +

Xe + O2 + H2O

Paragraph For Question 31 and 32X and Y are two volatile liquids with molar weightsof 10g mol–1 and 40g mol–1 respectively. Two cottonplugs, one soaked in X and the other soaked in Y,are simultaneously placed at the ends of a tube oflength L = 24cm, as shown in the figure. The tube isfilled with an inert gas at 1 atmosphere pressureand a temperature of 300 k. Vapurs of X and Y reactto form a product which is first observed at a dis-tance d cm from the plug soaked in X. Take X and Yto have equal molecular diameters and assume idealbehaviour for the inert gas and the two vapous.

31. The value of d in cm (shown in the figure),as estimated from Graham's law, is(A) 8 (B) 12(C) 16 (D) 20

Sol. (C)

L = 24 cm

Cotton woolsoaked in Y

Cotton woolsoaked in X

Initial formation of

the product

d

x

y

y

x

MMMM

raterate

x

y

y

x

MMMM

tdtd

//

1040

24

dd

d = 48 – 2d d = 16

32. The experimental value of d is found to besmaller than the estimate obtained usingGraham's law. This is due to(A) larger mean free path for X as comparedto that of Y.(B) larger mean free path for Y as comparedto that of X.(C) increased collision frequency of Y withthe inert gas as compared to that of X withthe inert gas.(D) increased collision frequency of X withthe inert gas as compared to that of Y withthe inert gas.

Sol. (B, D)


Paragraph For Questions 33 and 34Schemes 1 and 2 describe sequential transforma-tion of alkynes M and N. Consider only the majorproducts formed in each step for both the schemes.

33. The product X is

(A)

(B)

(C)

(D)

Sol. AHO–CH2–CH2–CCH NaNH2 HO–CH2–CH2–CH

C–Na+ CH –CH I3 2 HO–CH2–CH2CC–CH2–CH3

CH –I3 CH3–O–CH2–CH2–CC–CH2–CH3

H /Lindlar's catalst

2

34. The correct statement with respect to prod-uct Y is(A) It gives a positive Tollens test and is afunctional isomer of X.(B) It gives a positive Tollens test and is ageometrical isomer of X.(C) It gives a positive iodoform test and is afunctional isomer of X.(D) It gives a positive iodoform test and is ageometrical isomer of X.

Sol. C

It gives + Ve iodoform test and it is func-tional isomer if X .

Paragraph For Questions 35 and 36An aqueous solution of metal ion M1 reacts sepa-rately with reagents Q and R in excess to give tet-rahedral and square planar complexes, respectively.An aqueous solution of another metal ion M2 alwaysforms tetrahedral complexes with these reagents.Aqueous solution of M2 on reaction with reagent Sgives white precipitate which dissolves in excess ofS. The reactions are summarized in the scheme givenbelow:SCHEME:

Tetrahedral Square planarM1RQ

excess excess

Tetrahedral M2RQ

excess excessTetrahedral

S, stoichiometric amount

White precipitate White precipitateS

excess

35. M1, Q and R, respectively are(A) Zn2+, kCN and HCl (B) Ni2+, HCl and KCN(C) Cd2+, KCN and HCl (D) Co2+, HCl and KCN

Ans. (B)

Ni2+ KCN [Ni(CN) ]42–

dsp sq. planar

2

Ni2+ HCl [NiCl ]42–

sp (tetrahedral)3

M1


Zn + KCN 2+ [Zn(CN) ]42–

(Tetrahedral sp )3

Zn + HCl2+ [Zn(Cl ] + H42– +

(Tetrahedral sp )3

M2

Zn2+ + KOH Zn(OH)2 OH–

(white ppt.)[Zn(OH)4]

2–

(color less solution)

36. Reagent S is(A) K4[Fe(CN)6] (B) Na2HPO4(C) K2CrO4 (D) KOH

Ans. (D)

37. Match each coordination compound in List-Iwith an appropriate pair of characteristicsfrom List-II and select the correct answerusing the code given below the lists.{en = H2NCH2CH2NH2; atomic numbers : Ti =22, Cr = 24, Co = 27; Pt = 78}List-I(P) [Cr(NH3)4Cl2]Cl(Q)[Ti(H2O)5Cl (NO3)2](R) [Pt(en)(NH3)Cl]NO3](S) [Co(NH3)4(NO3)2]NO3

List-II(1) Paramagnetic and exhibits ionisation isom-erism(2) Diamagnetic and exhibits cis-trans isom-erism(3) Paramagnetic and exhibits cis-trans isom-erism(4) Diamagnetic and exhibits ionisation isom-erism

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

Ans. B38. Match the orbital overlap figure shown in List-

I with the description given in List-II andselect the correct answer using the codegiven below the lists.List-I

(P)

(Q)

(R)

(S)

List-II(1) p–d antibonding(2) d–d bonding(3) p–d bonding(4) d–d antibonding

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

Ans. C39. Different possible thermal decomposition

pathways for perosysters are shown below.match each opathway form List with an ap-propriate structure from List II and selectthe correct answer using the code given be-low the lists.

List-I

(P) Pathway P

(Q) Pathway Q

(R) Pathway R

(S) Pathway S

List-II

(1)

(2)

(3)

(4)


Code :

P Q R S

(A) 1 3 4 2

(B) 2 4 3 1

(C) 4 1 2 3

(D) 3 2 1 4

Sol. A

Compound (I) follow the pathway (P) andform formaldehyde as carbonyl. It remove CO2

in I step

Compound (II) follow the pathway (S) andform formaldehyde as carbonyl. It remove CO2

in II step

Compound (III) follow the pathway (Q) anddoes not form carbonyl. It remove CO2 in Istep

Compound (IV) follow the pathway (R) anddoes not form carbonyl. It remove CO2 in IIstep

40. Match the four starting materials (P, Q, R, S)given in List I with the corresponding reac-tion schemes (I, II, III, IV) provided in List-IIand select the correct answer using the codegiven below the lists.

List-I(P)

(Q)

(R)

(S)

List-II1. Scheme I(i) KMnO4, HO–, heat (ii) H+, H2O(iii) SoCl2 (iv) NH3

\2. Scheme II(i) Sn/HCl (ii) CH3COCl (iii) Conc. H2SO4(iv) HNO3 (v) dil. H2SO4, heat (vi) HO–

3. Scheme III(i) red hot iron, 873 K(ii) fuming HNO3, H2SO4 heat(iii) H2S. NH3 (iv) NaNO2, H2SO4(v) hydrolysis

4. Scheme IV(i) conc. H2SO4, 60ºC(ii) conc. HNO3, conc. H2SO4(iii)dil. H2SO4, heatCode :

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

Sol. C(P) Acetylene Benzene

Dinitrobenzene

m-nitro aniline

m-nitro benzene diazonium ion

m-nitro phenol(C6H5NO3)

(Q)


(R)

(S)


MATHEMATICS

SECTION – ASingle Correct

41. In a triangle the sum of two sides is x andthe product of the same two sides is y.If x2 – c2 = y, where c is the third side of thetriangle, then the ratio of the in-radius tothe circum-radius of the triangle is

(A) 3y

2x(x c) (B) 3y

2c(x c)

(C) 3y

4x(x c) (D) 3y

4c(x c)

Sol. Ba + b = x, ab = y(a + b)2 – c2 = ab

cos C = 12

C = 120°

also x – C = y

x c

Now r (s c) tan60cR

2sin120

= (s c) 3 · 3

c

= y 13· ·2(x c) c

42. The common tangents to the circle x2 + y2 =2 and the parabola y2 = 8x touch the circleat the points P, Q and the parabola at thepoints R, S. Then the area of the quadrilat-eral PQRS is(A) 3 (B) 6(C) 9 (D) 15

Sol. D

R

ma2,

ma2 ,

m

a2,maS 2

y = mx + m2

A

Q

S

B

RP

21m

m200

2

2)1m(m

422

m4 + m2 – 2 = 0 (m2 – 1) (m2 + 2) = 0 m = ±1R(2, 4) & S(2, –4)P(–1, 1) Q(–1, –1)

Area = 21

(PQ + RS) · AB = 21

(2 + 8)· 3

43. The quadratic equation p(x) = 0 with realcoefficients has purely imaginary roots.Then the equationp(p(x)) = 0 has(A) only purely imaginary roots(B) all real roots(C) two real and two purely imaginary roots(D) neither real nor purely imaginary roots

Sol. Ap(p(x)) = 0 p(x) = purely imaginary roots No real value of x will satisfy For real x, LHS is purely real

44. Six cards and six envelopes are numbered 1,2, 3, 4, 5, 6 and cards are to be placed inenvelopes so that each envelope containsexactly one card and no card is placed in theenvelope bearing the same number andmoreover the card numbered 1 is alwaysplaced in envelope numbered 2. Then thenumber of ways it can be done is(A) 264 (B) 265(C) 53 (D) 67

Sol. CRequired ways

= 5

!61

!51

!41

!31

!21!6

= 5

1630120360 =

5265

= 53

45. Three boys and two girls stand in a queue.The probability, that the number of boysahead of every girl is at least one more thanthe number of girls ahead of her, is

(A) 12

(B) 13

(C) 23 (D)

34

Sol. ABGBGB BGBBG BBGGB BBGBG BBBGG

Probability = 21

!5)!2!3(5


46. Let f : [0, 2] R be a function which iscontinuous on [0, 2] and is differentiable on

(0, 2) with f(0) = 1. Let F(x) = 2x

0f( t) dt for

x [0, 2]. If F'(x) = f'(x) for all x (0, 2),then F(2) equals(A) e2 – 1 (B) e4 – 1(C) e – 1 (D) e4

Sol. B

F(x) = 2x

0dt)t(f

differentiatingf'(x) = f(x) . 2x

x2)x(f)x('f

n f(x) = x2 + kf(0) = 1 k = 0

f(x) = 2xe

F(x) = 2x

0

t dte

F(x) = [2xe – 1]

F(2) = )1e(Lim2x

2x

= e4 – 1

47. The function y = f(x) is the solution of the

differential equation dydx + 2

xyx –1

= 4

2

x 2x

1– x

in (–1, 1) satisfying f(0) = 0. Then

32

3–2

f(x) dx

is

(A) 3–

3 2

(B) 3–

3 4

(C) 3–

6 4

(D) 3–

6 2

Sol. B

dxdy

–

2x1x

y = 2

4

x1

x2x

21 2x dx2 1 xI.F. e

I.F. = )x1(ln

21 2

e

= 2x1

y 2x1 =

2

4

x1

x2x 2x1 dx

y 2x1 = 5x5

+ x2

y = 2

25

x15

x5x

y = 2

25

x15

x5x

y = 3 /2 5

23 /2

x dx5 1 x

+ dxx15

x52/3

2/32

2

= 2

2/3

02

2

x15

x5dx

= 2

2/3

02

2

x1

11x dx

= 2

2/3

02

2/3

0

2 dxx1

1dx)x1(

= 2

3

dx)x1(2/3

0

2

= 2 3 /2

2 1

0

x 11 x sin x2 2 3

= 2 3 14 2 6 3

= – 43

– 3

+ 32

3

– 43


48. For x (0, ), the equation sinx + 2sin 2x –sin 3x = 3 has(A) infinitely many solutions(B) three solutions(C) one solution(D) no solution

Sol. Dsin x + 4 sin x cos x – 3 sin x + 4 sin3 x – 3 = 04 sin x (cos x + 1 – cos2 x) – 2 sin x – 3 = 0

2

21xcos

+ 43

(cosecx – 1) = 0

Hence No solution

49. The following integral 2

17

4

(2 cosec x)

dx is

equal to

(A) log(1 2) u –u 160 2(e e ) du

(B) log(1 2) u –u 170 (e e ) du

(C) log(1 2) u –u 170 (e – e ) du

(D) log(1 2) u –u 160 2(e – e ) du

Sol. A

cot 2x

= eu

x = 2 cot–1 eu

dx = u

2u2e

1 e

du

cosec x =

2xtan2

2xtan1 2

= 2u

u

11e

2e

= u

u2

e2e1

I =

0

)12(ln

17

u

u2

ee1

. u

2u2e du1 e

= log( 2 1)

u u 16

0

2(e e ) du

50. Coefficient of x11 in the expansion of

(1 + x2)4 (1 + x3)7(1 + x4)12 is(A) 1051 (B) 1106(C) 1113 (D) 1120

Sol. CGiven expression is(1 + 4x2 + 6x4 + 4x6 + x8).

(1 + 7C1 x3 + 7C2

x6 + 7C3x9 +.....).

(1 + 12C1 x4 + 12C2

x8 + .......)Coefficient ofx11 = 1 . 7C1 .

12C2 + 4 . 7C3 . 1 + 6 . 7C1 .

12C1 + 1. 7C1.1= 462 + 140 + 504 + 7= 1113

ParagraphParagraph for Question Nos. 51 to 52

Box 1 contains three cards bearingnumbers 1, 2, 3; box 2 contains five cardsbearing number 1, 2, 3, 4, 5; and box 3 con-tains seven cards bearing numbers 1, 2, 3,4, 5, 6, 7. A card is drawn from each of theboxes. Let xi be the number on the card drawnfrom the ith box, i = 1, 2, 3.

51. The probability that x1 + x2 + x3 are in anarithmetic progression, is

(A) 29

105 (B) 53

105

(C) 57105 (D)

12

Sol. B

)1(Box3,2,1

Box (2)1, 2, 3, 4, 5,

Box (3)1, 2, 3, 4, 5, 6, 7

A card is drawn from each Box OOO + OEE + EOE + EEO

32

.53

.74

+ 32

.52

.73

+ 31

.53

.73

+ 31

.52

.74

= 10553

52. The probability that x1, x2, x3 are in an arith-metic progression, is

(A) 9

105 (B) 10105

(C) 11

105 (D) 7

105Sol. C

The only cases are111, 123, 135, 147, 222, 234, 246, 321, 333,345, 357

Required Probability = 753

11

= 10511


Paragraph for Question Nos. 53 to 54Let a, r, s, t be nonzero real number. LetP(at2, 2at), Q(ar2, 2ar) and S(as2, 2as) bedistinct points on the parabola y2 = 4ax. Sup-pose that PQ is the focal chord and lines QRand PK are parallel, where K is the point (2a, 0).

53. The value of r is

(A) – 1t (B)

2t 1t

(C) 1t (D)

2t –1t

Sol. D

R(ar ,2ar)2

P(at ,2at)2

F(a,0)

k(2a,0)

ta2,t

aQ2

S(as , 2as)2

mQR= r

t12

, mPK = 2t

t2ata2at20

22

22 2t

1 t 2rt

t22t

2

rt1

2

t2tr

t1

r = t – t1

r = t

1t2

54. If st = 1, then the tangent at P and thenormal at S to the parabola meet at a pointwhose ordinate is

(A) 2 2

3(t 1)

2t

(B) 2 2

3a(t 1)

2t

(C) 2 2

3a(t 1)

t

(D) 2 2

3a(t 2)

t

Sol. C

Tangent at P is y = tx

+ at .....(1)

Normal at S is y = – sx + 2as + as3

Now st = 1 so s = t1

we get

y = – tx

+ ta2

+ 3ta

.....(2)

Adding (1) & (2)

y = at + ta2

+ 3ta

= 3

24

taat2at

= 3

24

t)1t2t(a

= 3

22

t)1t(a

Paragraph for Question Nos. 55 to 56Given that for each a (0, 1),

0hLim dt)t1(t

h1

h

1aa

exists. Let this limit be g(a). In addition,it is given that the function g(a) isdifferentiable on (0, 1).

55. The value of g12

is

(A) (B) 2

(C) 2

(D) 4

Sol. A

g(a) = 0hLim

h1

h

1aa dt)t1(t

g(1/2) = 0hLim

h1

h

2/12/1 dt)t1(t


= 0hLim dt

)t1(t1

h1

h

=

1

0 )t1(tdt

t = sin2 dt = 2 sin cos d

=

2/

0

2/

0

d2cossin

dcossin2

= 2/02 =

56. The value of g'12

is

(A) 2

(B)

(C) – 2

(D) 0

Sol. D

g(a) = 0hLim dt)t1(t

h1

h

1aa

= dt)t1(t1

0

1aa

g'(a) = 1

a

0

( t .ln t (1 – t)a–1

+ t–a (1 – t)a–1 n (1 – t)) dt

g'(1/2) = 1

1 /2

0

( t .ln t (1 – t)–1/2

+ t–1/2 (1 – t)–1/2 n (1 – t)) dt

= 1

0

2/1t (1 – t)–1/2 (–ln t + ln (1 – t))dt

........(1)

g'(1/2)= 1

0

2/1)t1( t–1/2 (–ln (1 – t)+ln t ) dt

........(2)Adding (1) and (2)g'(1/2) = 0

Matrix Match TypeMatching List Type (Only One Option Correct)57. List I List II (P) The number of polynomials 1. 8

f(x) with non-negativeinteger coefficients ofdegree 2, satisfying

f(0) = 0 and 10 f(x) dx

= 1, is (Q) The number of points in 2. 2

the interval [– 13, 13 ] at

which f(x) = sin(x2) + cos(x2)attains its maximum value is

(R)2

2–2 x

3x(1 e )

dx equals 3. 4

(S)

121–2120

1+xcos2x log dx1–x

1 xcos2x log dx1– x

equal 4. 0

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

Sol. D (P) f(x) = ax2 + bx

13 2

0

ax bx 13 2

a b 13 2

a b13 2

a 2 b3 2

2a 2 b3

b = 2 – 2a3

N = b 2a

3


a 0 b 2a 3 b 0

(Q) [ 13, 13] : f(x) = sin x2 + cos x2

x2 = 4

, 49

x = ± 4

, ± 49

(R) I = 2 x 22 2

x x2 2

3x 3e x1 e e 1

2I = 2 2

23x dx

I =

2 2

02 3x

I =

23

0

3x3

= 8

(S) numerator is odd function= 0

58. List I List II (P) Let y(x) = cos(3cos–1 x), 1. 1

x [–1,1], x ± 32

.

Then 1

y(x)

22

2d y(x) dy(x)(x –1) x

dxdx

equal (Q) Let A1, A2,...., An (n > 2) 2. 2

be the vertices of a regularpolygon of n sides with itscentre at the origin. Let

ka

be the position vectorof the point Ak, k = 1, 2,...,n.

If n–1k k 1k 1 a a

=

n–1k k 1k 1 a · a

, then the

minimum value of n is (R) If the normal from the 3. 8

point P(h, 1) on the ellipse2x6

+ 2y3

= 1 is perpendicular

to the line x + y = 8, thenthe value of h is

(S) Number of positive 4. 9solutions satisfying theequation

tan–11

2x 1

+ tan–11

4x 1

= tan–122x

is

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

Sol. A

(P) y = sin(3cos–1x) 2

3

1 x

21 x y' = 3 sin(3 cos–1x)

21 x y" – 1

2 2

2x cos(3cos x)y ' 92 1 x 1 x

2

2 2

(1 x )y " xy ' 9y

1 x 1 x

(x2 – 1) y" + y' = 9y

(Q)

na1a n

2

3a

5a

2aO

A2

A1

A3

A5A4

n 1 n 1

k k 1 k k 1k 1 k 1

(a a ) (a ·a )

(n – 1) k k 12| a || a | sinn

= (n – 1) k k 12| a || a | cosn

tan 2n

= 1

2n 4

n 8


(R) y + x = 8slope of Normal = + 1

dxdy

= – y6x3

Normal (h, 1) slope at (h, 1)

is h36

= 1

h = 2

(S) tan–1

1 12x 1 4x 1

1 112x 1 4x 1

2 24x 1 2x 1 2

8x 6x 1 x x

36x 2 2

2x(4x 3) x

3x 1 24x 3 x

3x2 + x = 8x + 6

3x2 – 7x – 6 = 03x2 – 9x + 2x – 6 = 03x(x – 3) + 2(x – 3) = 0

x = 3, x = 23

x = 3

tan–1 1 1 1

1 11 1 13 77 13tan tan tan

1 17 13 9017 13

tan–1 29

R.H.S. tan–1 29

x = – 23 tan–1

11 1tan4 81 13 3

tan–1 13 3tan1 5

– 5

59. Let f1 : R R, f2 : [0, ) R, f3 : R R andf4 : R [0, ) be defined by

f1 (x) = x

| x | if x <0

e if x 0

f2 (x) =x2;

f3 (x) = 2 1

2 1

f (f (x)) if x <0f (f (x)) 1 if x 0

(P) f4 is 1. onto but not one-one (Q) f3 is 2. neither continuous nor one-one (R) f2of1 is 3. differentiable but not one-one (S) f2 is 4. continuous and one-one

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

Sol. Df1 : R R, f2 : [0, ) R, f3 : R Rf4 : R [0, )

f1(x) = x

x x 0

e x 0

(P) f4(x) = 2 1

2 1

f (f (x)) x 0f (f (x)) 1 x 0

= 2

2x

x x 0

e 1 x 0

onto but not one-one


(Q) f3(x) =

y

x

f3 differentiable but not one-one.

(R) f2 0 f1(x) = 2

2x

x x 0

e x 0

neither continuous nor one-one

(S) f2(x) = x2 x 0

continuous and one-one

y

x

y=x2

60. Let zk = cos2k10

+ i sin2k10

; k = 1,2,...,9.

List I List II (P) For each zk there 1. True

exists a zj such thatzk · zj = 1

(Q) There exists a 2. Falsek {1,2,...,9} suchthat z1 · z = zk has nosolution z in the set ofcomplex numbers.

(R) 1 2 9|1– z ||1– z |...|1– z |10 equal 3. 1

(S) 1 – 9k 1

2kcos10

equal 4. 2

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

Sol. C

(P) Zk = 10k2i

e

Zk = 10th roots of unity P is true

(Q) k {1, 2, 3, ...... 9}

Z = )1k(

102i

102i

10k2i

1

k e

e

eZZ

Z obuiously has a solution in the set ofcomplex numbers. False

(R) (x10–1) = (x–1)(x–z1)(x–z2) ...... (x – z9)

(x – z1)(x–z2)(x–z3)......(x–z9) = 1x1x10

= 1 + x + x2 + x3 + ....... + x9

Now put x = 1(1 – z1) (1 – z2) ....... (1 – z9)

= 1 + 1 + 1 ....... + 1 |1 – z1| |1 – z2| ......... |1 – z9| = 10

110

|z1|.......|z1||z1| 921

(S) 1 – 9

k 1

2kcos10

1 – 2 4 6 18cos cos cos ........... cos10 10 10 10

1 –

9sin cos10

sin10

= 1 + 1 = 2

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