Jee Advanced 2013 Paper 1 Solutions English

7/28/2019 Jee Advanced 2013 Paper 1 Solutions English

1/13

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IIT-JEE 2013 PAPER-I WITH SOLUTION [CODE - 3]

PART - I [PHYSICS]1. A particle of mass m is projected from the groundwith an intial speed u

0at an angle with the horizontal.

At the highest point of its trajectory, it makes acompletely inelastic collision with another identicalparticle, which was thrown vertically upward from theground with the same intial speed u

0. The angle that

the composite system makes with the horizontalimmediately after the collision is

(A)

4

(B)

4

(C)

2

(D)

2

Ans. A

u0

final kinetic energy of both the particles issame because the vertical displacement is alsosame.

mgh = 12

mv2

(for both the particles) The final angle which the composite system

makes with the horizontal is 45.

2. The image of an object, formed by a plano-convex lens at a distance of 8 m, behind the lens, isreal and is one-third the size of the object. The

wavelength of light inside the lens is3

2times the

wavelength in free space. The radius of the curvedsurface of the lens is(A) 1 m (B) 2 m (C) 3 m (D) 6 mAns. C

h

O

24m

8mh/3

from wavelength ratio = 1.5 (from

1/

2)

1

8+

1

24= (1.5 1)

1

R

R = 3m Ans.

3. The diameter of a cylinder is measured using a Verniercallipers with no zero error. It is found that the zero ofthe Vernier scale lies between 5.10 cm and 5.15 cm ofthe main scale. The Vernier scale has 50 divisionsequivalent to 2.45 cm. The 24th division of the Vernierscale exactly coincides with one of the main scaledivisions. The diameter of the cylinder is(A) 5.112 cm (B) 5.124 cm(C) 5.136 cm (D) 5.148 cmAns. B

L.C.=

32.450.05 cm=10 cm.50

Reading = 5.10 cm + 24 103 cm.= 5.124 cm.

4. The work done on a particle of mass m by a

force,

j

yx

yi

yx

xK

2/3222/322 (K being a constant

of appropriate dimensions), when the particle is takenfrom the point (a,0) to the point (0,a) along a circular

path of radius a about the origin in the x-y plane is

(A)a

K2 (B)

a

K

(C)a2

K(D) 0

Ans. Dx2+y2=a2

dw =

a 0

3 30 a

x ydx+ dy

a a

w =

a 02 2

3

0 a

1 x y

2 2a

w = 0 Ans.

5. One end of a horizontal thick copper wire oflength 2L and radius 2R is welded to an end of anotherhorizontal thin copper wire of length L and radius R.When the arrangement is stretched by applying forcesat two ends, the ratio of the elongation in the thin wireto that in the thick wire is(A) 0.25 (B) 0.50

(C) 2.00 (D) 4.00


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Ans. C

F2R

F

R

FF

y =

22 F / 4R

y = 2

1

F

2 R

y = 22

F

R

2

1

= 2 Ans

6. A ray of light travelling in the direction j3i2

1

is incident on a plane mirror. After reflection, it travels

along the direction j3i2

1 . The angle of incidence is

(A) 30 (B) 45(C) 60 (D) 75Ans. A

3

2

3

2

-

1

2

i r

1

2

Mirror in xz planeSince y component is reversed

tan=(1/2)

( 3/2)=

1

3

=30

7. Two rectangular blocks, having identicaldimensions, can be arranged either in configuration I orin configuration II as shown in the figure. One of theblocks has thermal conductivity K and the other 2K.The temperature difference between the ends alongthe x-axis is the same in both the configurations. ittakes 9 s to transport a certain amount of heat fromthe hot end to the cold end in the configuration I. Thetime to transport the same amout of heat in theconfiguration II is

2K

K

Configuration I Configuration IIfoU;kl I foU;kl II

2KK

(A) 2.0 s (B) 3.0 s (C) 4.5 s (D) 6.0 sAns. A

Equivalent circuit diagram

R R/2

T1 T2

R

R/2T1 T2

Q =1 2T T 93R

2

..... (i) Q = 1 2

T Tt

R

3

..... (ii)

Solving (i) & (ii)t = 2 sec. Ans.

8. A pulse of light of duration 100 ns is absorbedcompletely by a small object initially at rest. Power of thepulse is 30 mW and the speed of light is 3 108 ms-1. Thefinal momentum of the object is(A) 0.3 10-17 kg ms-1 (B) 1.0 10-17 kg ms-1

(C) 3.0 10-17 kg ms-1 (D) 9.0 10-17 kg ms-1

Ans. BEnergy given to particle= 100 109 30 103 = 3000 1012J

EP=C

12

8300010= 310= 1 1017 kg m s1

9. In the young's double slit experiment using amonochromatic light of wavelength , the path difference(in terms of an integer n) corresponding to any pointhaving half the peak intensity is

(A) 2

1n2

(B) 4

1n2

(C) 2n 18

(D)

161n2

Ans. B

I=4I0cos2

2=2I

0 cos2

2=

1

2

=

2 x

4or (2n+1)

4

10. Two non-reactive monoatomic ideal gases have theiratomic masses in the ratio 2 : 3. The ratio of theirpartial pressures, when enclosed in a vessel kept at aconstant temperature, is 4 : 3. The ratio of their densitiesis(A) 1 : 4 (B) 1 : 2

(C) 6 : 9 (D) 8 : 9Ans. D

2rms

1P= v

3where v

rms=

3RT

M

P =1 3RT

3 M

P

M

1

2

4 3=

3 2

1

2

=8

9 Ans.


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11. Two non-conducting solid spheres of radii R and2R, having uniform volume charge densitites

1and

2

respectively, touch each other. The net electric field ata distance 2R from the centre of the smaller sphere,

along the line joining the centres of the spheres, is

zero. The ratio2

1

can be

(A) -4 (B)25

32 (C)

25

32(D) 4

Ans. B,D

E1

=

31

2

4K R

3

2R

R P

2R

E2

=2

0

R

3

as field is zero at PR

2RP

2R

31

202

0

1 4R

R4 3

32R

1

2

= 4

and 1

/2

= 32/25

12. A horizontal stretched string, fixed at two ends,is vibrating in its fifth harmonic according to the equation,

y(x,t) = (0.01 m) sin [(62.8 m-1

)x] cos 1

[ 628s t]

Assuming =3.14, the correct statement (s) is (are)(A) The number of nodes is 5.(B) The length of the string is 0.25 m.(C) The maximum displacement of the midpoint of thestring, from its equilibrium position is 0.01 m.(D) The fundamental frequency is 100 Hz.Ans. BC

K =2

= 10m

P

L =5

2

= 25 cm A = 0.01 m

13. In the circuit shown in the figure, there are twoparallel plate capacitors each of capacitance C. Theswitch S

1is pressed first fo fully charge the capacitor

C1and then released. The switch S

2is then pressed to

charge the capacitor C2. After some time, S

2is released

and then S3is pressed. After some time,

2V0 V0C2C1

S1 S2 S3

(A) the charge on the upper plate C1is 2CV

0.

(B) the charge on the upper plate of C1is CV

0.

(C) the charge on the upper plate of C2is 0.

(D) the charge on the upper plate of C2is -CV

0.

Ans. BD

Hint.V'+

2CV0 0

2CV0=2CV

V = V0 Final situation

CV0

CV0

CV0

CV0

C1 C2

14. A particle of mass M and positive charge Q,

moving with a constant velocity 11

msi4u

, enters a

region of uniform static magnetic field normal to the x-yplane. The region of the magnetic field extends from x =0 to x = L for all values of y. After passing through thisregion, the particle emerges on the other side after 10

milliseconds with a velocity ji32u2

ms-1. The

correct statement (s) is (are)(A) The direction of the magnetic field is -z direction.(B) The direction of the magnetic field is +z direction.

(C) The magnitude of the magnetic fieldQ3

M50units.

(D) The magnitude of the magnetic field is

Q3

M100units.

Ans. AC

tan =1

3, = 30

Also T = 12 103 =2 M

qB

y

x

q,m

L

2 3 i 2j+

B = 32 M 500 M

q 12 10 3q


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15. A solid sphere of radius R and density isattached to one end of a mass-less spring of forceconstant k. The other end of the spring is connected toanother solid sphere of radius R and density 3. The

complete arrangement is placed in a liquid of density 2and is allowed to reach equilibrium. The correctstatement(s) is (are)

(A) the net elongation of the spring isk3

gR4 3

(B) the net elongation of the spring isk3

gR8 3

(C) the light sphere is partially submerged.(D) the light sphere is completely submerged.Ans. AD

Hint :(A) For upper block

kx + 43

R3 = 43

R3 2

x =4

3

3R

k

(D) Cannot provide upward force if partially submerged.

16. A uniform circular disc of mass 50 kg and radius0.4 m is rotating with an angular velocity of 10 rad s-1

about its own axis, which is vertical. Two uniform circularrings, each of mass 6.25 kg and radius 0.2 m, are gentlyplaced symmetrically on the disc in such a manner thatthey are touching each other along the axis of the disc

and are horizontal. Assume that the friction is largeenough such that the rings are at rest relative to thedisc and the system rotates about the original axis. Thenew angular velocity (in rad s-1) of the system isAns. 8

Angular momentum conservationI

1

1= I

2

2

2 22 2

1 2

MR MR2(mr mr )

2 2

250 0.4

2 10 =

2

2 250 0.4

2 6.25 0.2 0.22

2

40 = [4 + 1] 2

2= 8 rad/s

17. The work functions of Silver and Sodium are 4.6and 2.3 eV, respectively. The ratio of the slope of thestopping potential versus frequency plot for Silver tothat of Sodium is

Ans. 1

V

f

V=hf

e e

Slope is planks constant so 1.

18. A bob of mass m, suspended by a string of lengthl1, is given a minimum velocity required to complete a

full circle in the vertical plane. At the highest point, itcollides elastically with another bob of mass m

suspended by a string of length l2, which is initially atrest. Both the strings are mass-less and inextensible. Ifthe second bob, after collision acquired the minimumspeed required to complete a full circle in the vertical

plane, the ratio2

1

l

lis

Ans. 5

Vbottom

=2

5g , At top, Vtop

=1

g

Thus Vtop

= Vbottom

1g = 25g

1

2

= 5 Ans.

19. A particle of mass 0.2 kg is moving in onedimension under a force that delivers a constant power0.5 W to the particle. If the initial speed (in ms-1) of theparticle is zero, the speed (in ms-1) after 5 s isAns. 5

2 2m(v u )

2= pt

20. A freshly prepared sample of a radioisotope ofhalf-life 1386 s has activity 103 distintegrations persecond. Given that ln 2 = 0.693, the fraction of theinitial number of nuclei (expressed in nearest integerpercentage) that will decay in the first 80 s afterpreparation of the sample isAns. 4

Fraction in % = t

0

0

N (1 e )

N

= t(1 e )

= .04(1 e ) 4%


5/13



PART - II [CHEMISTRY]

SECTION ASingle Correct

21. The standard enthalpies of formation of CO2(g),H2O() and glucose(s) at 25C are 400 kJ/mol,300 kJ/mol and 1300 kJ/mol, respectively. Thestandard enthalpy of combustion per gram ofglucose at 25C is(A) +2900 kJ (B) 2900 kJ(C) 16.11 kJ (D) +16.11 kJ

Sol. CC

6H

12O

6(s) + 6O

2(g) 6CO

2(g) + 6H

2O ()

fH 1300 0 400 300

rH = (

fH)

P (

fH)

R

= [6(400) + 6(300)] [(1300)]= [2400 1800] + 1300

= 2900mole

kJ

mole

gm.wtmolecular

mole

kJH

gm

kJH

c

c

= 180

2900

= 16.11 kJ/gm

22. KI in acetone, undergoes SN2 reaction witheach of P,Q,R and S. The rates of the reactionvary as. H

3CCl

O

ClClH CCl2

(A) P>Q>R>S (B) S>P>R>Q(C) P>R>Q>S (D) R>P>S>Q

Sol. B

S > P > R > QTransition state of compound S is highly stabi-

lize by Ph C =O

group so it has more rate of reactiontowards SN2.

order of stability of T.S. S > P > R > Q.In compund R, T.S is stabilize by p-orbital

attached to carbon having leaving group.

23. The compound that does NOT liberate CO2, ontreatment with aqueous solium bicarbonatesolution is(A) Benzoic acid (B) Benzenesulphonic acid(C) Salicylic acid (D) Carbolic acid (Phenol)

Sol. D

Benzoic acid, Benzene sulphonic acid & salicylicacid are more acidic than H

2CO

3so reaction

proceed in forward direction and release CO2

but not phenol.

24. Consider the following complex ion P, Q and RP = [FeF6]

3, Q = [V(H2O)6]2+ and

R = [Fe(H2O)6]2+

The correct order of the complex ions, accordingto their spin-only magnetic moment value (in B.M.) is(A) R < Q < P (B) Q < R < P(C) R < P < Q (D) Q < P < R

Sol. BIn all complex ligands are WFL(P) [Fe F

6]3

Fe+3 3d5 4s 4p

sp d3 2

M.M = 5(5 2)

= 35 BM

(Q) [V(H2O)

6]2

V+2 3d2 4s 4p

d sp2 3

M.M = 3(3 2)

= 15 BM

(R) [Fe(H2O)

6]2

Fe+2 3d6 4s 4p

sp3 d2

M.M = 4(4 2)

= 24 BM

Order of M.M Q < R < P


6/13



25. In the reaction,P + Q R + Sthe time taken for 75%reaction of P is twicethe time taken for 50%reaction of P. Theconcentration of Qvaries with reactiontime as shown infigure. the overall orderof the reaction is

[Q]o[Q]

Time

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

Linear variation ofconc. with time suggestthat it is zero orderreaction w.r.t. 'Q'

Q

Timet75%

= 2t50%

Ist ordes w.r.t. 'P'

Hence rate law expressionr = k[P]1 [Q]0

over all order = 1 + 0 = 1

26. Concentrated nitric acid, upon long standing,turns yellow-brown due to the formation of(A) NO (B) NO

2(C) N

2O (D) N

2O

4

Sol. BOn long standing nitric acid decomposes as

HNO3 H2O + NO2 + O2 brown& hence due to dissolved NO2 it turns brown.

27. The arrangement of X ioins around A+ ion insolid AX is given in the figure (not drawn toscale). If the radius of Xis 250 pm, the radius ofA+ is(A) 104 pm

X

A+(B) 125 pm(C) 183 pm(D) 57 pm

Sol. AThis is ocathedral viod. with coordinationno. = 6

r

r= 0.414 (In ideal octahedral void)

r+

= 0.414 250 pm = 103.5 pm

28. Upon treatment with ammoniacal H2S, the metalion that precipitates as a sulfide is(A) Fe(III) (B) Al(III) (C) Mg(II) (D) Zn(II)

Sol. DUpon treatment with ammionical H

2S

ZnS ppt as NH4OH + H2S in group reagent forIVth group radicals.

29. Methylene blue, from its aqueous solution, isadsorbed on activated charcoal at 25C. For thisprocess the correct statement is(A) The adsorption requires activation at 25C.

(B) The adsorption is accompanied by adecreases in enthalpy(C) The adsorption increases with increase of

temperature.(D) The adsorption is irreversible.

Sol. BAdsorption is always exothermic.

30. Sulfide ores are common ofr the metals(A) Ag, Cu and Pb (B) Ag, Cu and Sn(C) Ag, Mg and Pb (D) Al, Cu and Pb

Sol. AAg

2

S Argentite/silver glanceCu

2S Chalcocite

PbS Galena

31. Benzene and naphthalene form an ideal solutionat room temeprature. For this process the truestatement(s) is (are)(A) G is positive (B) Ssystem is positive(C) Ssurrounding = 0 (D) H = 0

Sol. B, C, D(A) G < 0 for mixing(B) S

sys> 0 because disorder increases

(C) Ssurr

= 0 No heat is exchanged in case of

ideal Solution(D) H

mix= 0 for ideal solution

32. The pair(s) of coordination complexes/ionsexhibiting the same kind of isomerism is(are)(A) [Cr(NH3)5Cl2]

+ and [Cr(NH3)4Cl2]Cl(B) [Co(NH3)4Cl2]

+ and [Pt(NH3)2(H2O)Cl]+

(C) [CoBr2Cl2]2 and [PtBR2Cl2]

2

(D) [Pt(NH3)3(NO3)]Cl and [Pt(NH3)3Cl]BrSol. B,D(A) [Cr(NH

3)

5Cl]Cl

2no isomerism

[Cr(NH3)

4Cl

2]Cl only geometrical isomersism

(B) [Co(NH3)4]Cl2]

only geometrical isomersism[Pt(NH3)

2(H

2O)Cl] only geometrical isomersism

(C) [CoBr2Cl

2]2 no isomerism because complex

is tetrahedral[PtBr

2Cl

2]2 only geometrical isomersism

(D) [Pt(NH3)

3(NO

3)]Cl only ionization isomerism

[Pt(NH3)

3Cl]Br only ionization isomerism

33. The initial rate of hydrolysis of methyl acetate(1M) by a weak acid (HA, 1M) is 1/100th of astrong acid (HX, 1M) at 25C. The Ka of HA is(A) 1 104 (B) 1 105

(C) 1 106 (D) 1 103

Sol. A

r1 = k[Acetate][H+]HAr

2= k[Acetate][H+]

HX


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1 HA

2 Hx

r [H ] 1

r 100[H ]

(given)

acK 1

1M 100

a1

cK

100

cKa= 104

,K

a=

410

c

= 1 104

34. The hyperconjugative stabilities of tert-butylcation and 2-butene, respectively, are due to(A) p (empty) and electron

delocalisations(B) and electron delocalisations(C) p (fil led) and electron

delocalisations(D) p (fil led) and electron

delocalisationsSol. A

C C H

H

H

CH3

CH3

vacantp-orbital

It is p(empty) electron delocalisation.

C C

H

H CH3

H

H

H

C

It is electron delocalisation.

35. Among P,Q,R and S, the aromatic compound(s)is/are Cl

AlCl3 P

QNaH

OO

(NH ) CO4 2 3

100-115 C

HCl

R

S

O

(A) P (B) Q (C) R (D) SSol. ABCD

Cl

AlCl3 P

QNaH

OO

(NH ) CO4 2 3

100-115 C

HCl

R

S

+

Aromatic

Aromatic

Non aromaticaldol condensation

Aromatic

OHO

CH3CH3 N

36. The total number of lone-pairs of electrons in

melamine isSol. 0006

NN

NH2H N2 N

NH2

Malamine

Lone pairs of electrons in malamine is 6.

37. The total number of carboxylic acid group in theproduct P is

Sol. 0002

O

O

O

O

OH O3

2CO2Hydrolysis followed by

decarboxylation

O

O

O3H2O2

oxidative ozonolysis

O

HOC

HOC

O O

O

38. The atomic masses of He and Ne are 4 and 20a.m.u. respectively. The value of the de Brogliewavelength of He gas at 73 C is "M" timesthat of the de Broglie wavelength of Ne at 727C. M is

Sol. 0005

=mkT3

h

Ne

He

=HeHe

NeNe

T.M

T.M

=)27373(4

)273727(20

=

55 = 5


8/13



PART - III [MATHEMATICS]

39. EDTA4 is ethylenediaminetertraacetate ion. Thetotal number of NCoO bond angles in[Co(EDTA)]1 complex ion is

Sol. 0006

Co

ON

N O

O

O

1

6

4

32

5

octahedral compelxtotal N CO O angle is 6

40. A tetrapeptide has COOH group on alanine. Thisproduces glycine (Gly), valine (Val), phenylalanine (Phe) and alanine (Ala), on completehydrolysis. For this tetrapeptite, the number ofpossible sequencess (primary structures) with

NH2 group attached to a chiral center isSol. 0004

Phe Gly Val AlaPhe Val Gly AlaVal Gly Phe AlaVal Phe Gly Ala(Glycile has no chiral centre)

SECTION ASingle Correct

41. The value of cot

23

1n

n

1k

1 k21cot is

(A)25

23(B)

23

25

(C)24

23 (D)23

24

Sol. B

cot

23

1n

n

1k

1 k21cot

= cot

231

2n 1

n 1 ntan

1 n n)

=

23

1n

1

)1n(n1

n)1n(tancot

=

23

1n

11 ntan)1n(tancot

= cot (tan1 24 tan1 (1))

= cot

1241

124tan 1 =

23

25

42. Let k2ji3PR

and

SQ = k4j3i

determine diagonals of a parallelogram PQRS and

k3j2iPT

be another vector. Then the

volume of the parallelepiped determined by the

vectors PQ,PT and PS is

(A) 5 (B) 20(C) 10 (D) 30

Sol. CRequired volume

321

431

213

2

1

= 10

43. Let complex numbers and1

lie on circles

(x x0)2 + (y y0)

2 = r2 and (x x0)2 +

(y y0)2 = 4r2, respectively. If z0 = x0 + iy0

satisfies the equation 2 |z0|2 = r2 + 2, then

||=(A)

2

1(B)

2

1

(C)7

1(D)

3

1

Sol. C on |Z Z

0| = r

| Z0|2 = r2

oZ Z0 + |Z0|2 = r2 .....(1)

Similarly

1on |Z Z

0

| = 2r

2

02Z

||

= 4r2

4||

2

o

||

Z

2

o

||

Z

+ |Z

0|2 = 4r2

2

2

||

||

0(Z ) 0Z + |Z0|2 ||2 = 4r2 ||2 ...(2)

equation (1) (2)

(||2 1| + |Z0|2 (1 ||2) = r2. (1 4 ||2)(||2 1 (1 |Z

0|2) = r2 (1 4 ||2)


9/13



(||2 1)

2

2r1

2

= r2 (|4 ||2)

(||2 1)2r

2

= r2 (1 4||2)

1 ||2 = 2 8 ||2

7 ||2 = 1

|| =7

1Use (Z

0)2 =

2

r2+ 1

44. For a > b > c > 0, the distance between (1, 1)and the point of intersection of the lines ax + by

+ c = 0 and bx + ay + c = 0 is less than 22 .

Then(A) a + b c > 0 (B) a b + c < 0(C) a b + c > 0 (D) a + b c < 0

Sol. A

Point of intersectionc c

,a b a b

distance between P (1, 1) < 2 2

2 c

1a b

< 2 2

c + a + b < 2(a + b)a + b c > 0

45. Perpendiculars are drawn from points on the line

3

z

1

1y

2

2x

to the plane x + y + z = 3. The

feet of perpendiculars lie on the line

(A)13

2z

8

1y

5

x

(B) 52z

3

1y

2

x

(C)7

2z

3

1y

4

x

(D)5

2z

7

1y

2

x

Sol. D

(2r2,r1), 3r

x+y+z=3

P

Q

Equation of PQ

1

)2r2(x =

y ( r 1)

1

=

1

r3z =

Q ( + (2r 2), r 1, + 3r)lies on x = x + y + z = 3

3 + 4r + 3 = 0

=3

64

Q

36r5,

33r7,

3r2

so equation of image

2

x3=

7

3y3

=5

6z3

2

x=

7

1y

=5

2z

46. Four persons independently solve a certain

problem correctly with probabilities 8

1

,4

1

,4

3

,2

1

.Then the probability that the problem is solvedcorrectly by at least one of them is

(A)256

235(B)

256

21

(C)256

3(D)

256

253

Sol. AP(Problem is solved)=1 1/2 1/4 3/4 7/8

47. The area enclosed by the curves y = sin x + cos x

and y = |cos x sin x| over the interval

2

,0 is

(A) )12(4 (B) )12(22

(C) )12(2 (D) )12(22

Sol. B

1

/4 /20

2

Area =/ 4

0(sinx cosx) (cosx sinx)]dx

= 4 4/

0dxxsin

= )12(22 square unit


10/13



48. A curve passes through the point

6

,1 . Let

the solpe of the curve at each point (x, y) be

x

y+ sec

x

y, x > 0. Then the equation of the

curve is

(A)2

1xlog

x

ysin

(B) 2xlog

x

yeccos

(C) 2xlogx

y2sec

(D)

2

1xlog

x

y2cos

Sol. A

dx

dy=

x

y+ sec (

x

y)

x

y= v

v + xdx

dv= v + sec v

dvvcos = xdx

sin (y/x) = ln x + cpasses through (1, /6)

2

1= c

sin (y/x) = log x + 1/2

49. Let f :

1,

2

1R (the set of all real numbers) be

a positive, non-constant and differentiable

function such that f'(x) < 2 f (x) and f

2

1= 1.

Then the value of 1

2/1

)x(f dx lies in the interval

(A) (2 e 1, 2e) (B) (e 1, 2e 1)

(C)

1e,

2

1e(D)

2

1e,0

Sol. D

2)x(f

)x('f so

y

dy< 2dx integrating

i.e. lny < 2x + c

f(x) < e2x+c

also f(1/2) = 1 < e1+c

Now 1

2/1

dx)x(f < 1

2/1

cx2 dxe = e1+c

2

1e

Now, 1

2/1

dx)x(f will be less than minimum value

of e1+c

2

1ewhich is

2

1e

so 1

2/1

dx)x(f <

2

1e

also f(x) > 0

so 1

2/1

dx)x(f > 0 so option D.

50. The number of points in ( , ), for which x2 x sin x cos x = 0, is

(A) 6 (B) 4 (C) 2 (D) 0Sol. C

f() ; f() = f(0) = 1 f'(x) = x (2 cos x)

SECTION BMultiple Correct

51. A rectangular sheet of fixed perimeter with sideshaving their lengths in the ratio 8 : 15 isconverted into an open rectangular box by foldingafter removing squares of equal area from all

four corners. If the total area of removed squares

is 100, the resulting box has maximum volume.Then the lengths of the sides of the rectangularsheet are

(A) 24 (B) 32 (C) 45 (D) 60Sol. A,C

(8x 2y) (15x 2y). y(120x2 46xy + 4y2).y4y3 46xy2 + 120x2ydifferentiate with respective y

12y2 92xy + 120x2 = 0at y = 5

12.25 92x.5 + 120x2 = 06x2 23x + 15 = 0x = 5/6, x = 3

maximum will be at x = 3Ans. is 24 & 45


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52. Let Sn =

n4

1k

2

)1k(k

)1( k2. Then Sn can take

value(s)(A) 1056 (B) 1088(C) 1120 (D) 1332

Sol. A,DS

n= (1)2 + (1)3 22 + (1)6 (3)2 + ......

Sn= (12 22 + 32 + 42) + (52 62 + 72 + 82 ) +.....

....... + (4n 3)2 (4n 2)2 + (4n 1)2 + (4n)2]S

n= 4n2 + (4n 1)2 (4n 3)2 (4n 2)2

Sn= 32 n 12

Sn= 16 n2 + 4n

Sn= 4n (4n + 1)

53. A line l passing through the origin is perpendicularto the lines

l1 : (3 + t) i + ( 1 + 2t)j + (4 + 2t) k , < t <

l2 : (3 + 2s) i + (3 + 2s)j + (2 + s) k , < s < Then, the coordinate(s) of the point(s) on l2 at

a distance of 17 from the point of intersection

of l and l1 is (are)

(A)

3

5,

3

7,

3

7(B) ( 1, 1, 0)

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

9

8,

9

7,

9

7

Sol. B,D

122221kji

= 2 i + 3j 2k

so equation of l is

2

x=

3

y

,

2

z

to find point of intersection

2t3 =

3t21 9 + 3t = 2 + 4t

7t = 7 t = 1So A(2, 3, 2)A.T.Q.

(2s + 1)2 + (2s + 6)2 + s2 = 17, 9s2 + 28s + 20 = 09s2 + 18s + 10s + 20 = 0(9s + 10) (s + 2) = 0

s =9

10, 2

so required points are

(1, 1, 0) &

98,

97,

97

so ans. B and D

54. Let f(x) = x sin x, x > 0. Then for all naturalnumbers n, f' (x) vanishes at

(A) a unique point in the interval

2

1n,n

(B) a unique point in the interval

1n,

2

1n

(C) a unique point in the interval (n, n + 1)(D) two points in the interval (n, n + 1)

Sol. Bf(x) = x sin x x > 0then for all x x n f'(x)f''(x) = x cos x n + sin x 1 = 0

= tan x = x

period of tan x = 1so graph of tan x

0 1/2 1 3/2 2

y=

y=tan x

x

55. For 3 3 matrices M and N, which of the followingstatement(s) is (are) NOT correct ?(A) NTM N is symmetric or skew symmetric,according as M is symmetric or skew symmetric(B) M N N M is skew symmetric for all symmetricmatrices M and N(C) M N is symmetric for all symmetric matricesM and N(D) (adj M) (adj N) = adj (M N) for all invertible

matrices M and NSol. C,Dfrom the standard result


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SECTION CInteger Answer Type

56. A vertical line passing through the point (h, 0)

intersects the ellipse3y

4x 22 = 1 at the points P

and Q. Let the tangents to the ellipse at P andQ meet at the point R. If (h) = area of the

triangle PQR, 1 = 1h2/1max

(h) and 2 = 1h2/1min

(h), then5

81 82 =

Sol. 9

4x

2

+3y

2

= 1 P is

4h

13,h

2

Q is

4

h13,h

2

T at P is4

x2+

2hy 3 1

4

3

= 1 so R is =h

4

so area ofPQR = 2

1

. 2

4

h1.3

2

h

h

4

=2

3

h

)h4( 2/32

Nowdn

dA< 0 so ftn is decreasing

so 2min (h) = (1) = 9/2

2

1 h 1

1= min (h) =

1

2

=8

5.45

21 h 1 so 5

8 , 82 = 9

57. The coefficients of three consecutive terms of(1 + x)n + 5 are in the ratio 5 : 10 : 14. Then n =

Sol. 6(1 + x)n + 5

three consecutive termsn + 5C

r: n + 5C

r + 1: n + 5C

r + 2= 5 : 10 : 14

)1r5n)(r5n(

!)5n(

:)1r5n)(1r(

)5n(

:

)2r)(1r(

)5n(

= 5 : 10 : 14

)1r5n)(r5n(

1

: )1r5n)(1r(1

:

)2r)(1r(

1

= 5 : 10 : 14

take two at a time

(i))1r5n)(r5n(

1

= )1r5n)(1r(1

= 5/10

r5n

1r

=10

5

2

1

r5n

1r

3r = n + 3 ...(1)

(ii))1r5n)(1n(

1

= )2r)(1r(1

= 10 : 14

1410

2r

11r5n

1

24 r = 10n +1212r = 5n + 16 ...(2)solve one (1) & (2)

12

6n5 =

3

3n n = 6

58. Consider the set of eight vectors

V = { kcjaia : a, b, c { 1, 1}}. Three

non-coplanar vectors can be chosen from V in2p ways. Then p isSol. 5

P = 5pair of coplanar vector

1 1 1 1, 1, 11 1 1 1, 1, 11 1 1 1, 1, 1

1 1 1 1 : 1, 1for one pair 6 possibilitiesfor four pair 2p possibilitiesSo No. n-coplaner vector 8C3 24 = 322p= 32P = 5

59. Of the three independent events E1, E2 and E3the probability that only E1 occurs is , only E2occurs is and only E3 occurs is . Let the prob-ability p that none of events E1, E2 or E3 occurssatisfy the equations ( 2) p = and(3)p = 2. All the given probabilities are as-sumed to lie in the interval (0, 1).

Then3

1

EofoccurrenceofobabilityPr

EofoccurrenceofobabilityPr=


13/13


IVRS N 0744 2439051 52 53 ti iitj i f @ ti iitj 13

Sol. 6E

1& E

2& E

3independent

P(E1 E

2' E

3') = = E

1E

2' E

3'

P(E1' E

2 E

3') = = E

1' E

2E

3'

P(E1' E2' E3) = r = E1' E2' E3P(E

1' E

2' E

3') = p = E

1' E

2' E

3'

( 2) p = ( 3r) p = 2r(E

1E

2' E

3' 2E

1' E

2E

3') E

1' E

2' E

3' = E

1E

2' E

3' E

1' E

2

E3'

E1E

2' E

3' 2E

1' E

2E

3' = E

1E

2E

3'

E1E

2' 2E

1' E

2= E

1E

2

2

2

E '

E 1

1

2E '

E = 1

(E1' E

2E

3' 3E

1' E

2' E

3) E

1' E

2' E

3' = 2E

1' E

2E

3'

E1' E

2' E

3

E1' E2 E3' 3E1' E2' E3 = 2 E2 E1' E3E2E

3' 3E

2' E

3= 2E

2E

3

3

3

E

'E

2

2

E

'E3= 2

3

3

E

'E 3

1

1

E

'E21 = 2

b

b1 3

a

)a1(21 = 2

b

b1

3 a

)a1(6

= 2

b

1 1 3 6

1

a

1= 2

b

1 1 3

a

6+ 6 = 2

b

1

a

6= 0

b

1=

a

6

b

a= 6

60. A pack contains n cards numbered from 1 to n.Two consecutive numbered cards are removedfrom the pack and the sum of the numbers on the

remaining cards is 1224. If the smaller of thenumbers on the removed cards is k, then k 20 =Sol. 5

2

)1n(n k k 1 = 1224

2

)1n(n 2k 1 = 1224

2

)1n(n 2k = 1225

n(n + 1) 4k = 2450

n =2

)2450k4(411

=2

k1698011

Now 9801+16k should be perfect square992 + 16k = a2 so 16k = a2 992

16k = (a 99) (a + 99)at k = 25 a = 101k 20 = 5

Jee Advanced 2013 Paper 1 Solutions English

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