JEE ADVANCED EXAMINATION 2015 QUESTIONS WITH …€¦ · T T P = eA T 4 P A = e. 4 (400 R)2 × T A 4 P B = e. 4pR2 × T B 4 (400R)2 × T A 4 = 104 × T B 4 16 × 104 × T A 4 = 10

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JEE ADVANCED

EXAMINATION 2015

QUESTIONS WITH SOLUTIONS

PAPER - 1 [CODE - 0]

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PHYSICS

IIT-JEE 2015 Solutions by Motion Edu. Pvt. Ltd. Kota

1. Consider a concave mirror and a convex lens(refractive index = 1.5) of focal length 10cm each, separated by a distacne of 50 cmin air (refractive index = 1) as shown in thefigure. An object is placed at a distacne of15 cm from the mirror. Its erect image formedby this combination has magnification M1.When the set-up is kept in a medium ofrefractive index 7/6, the magnification

becomes M2. The magnitude 2

1

MM is

Sol. 0007

15cm Image for mirror

v1

+ )15(1

= – 101

v1

= 151

– 101

= 30

32 = –

301

v = – 30 cm & m1 = – uv

= – )15()30(

= – 2

Image from lens

20cm

v1

– u1

= f1

v1

+ 201

= 101

v1

= 101

201

v = – 20 cm

m2 = – 1|M1| = |m1 × m2| = 2In air for lens

f1

=

21

R2

R = 10 cm

When dipped

'f1

=

176

23

102

= 72

× 102

= 704

= 235

1v '

= 352

– 101

= 140

78 m2 = 7 |M2| = 14

|M2|/|M1| = 7

CODE-0

2. A young’s double slit interference arrangementwith slits S1 and S2 is immersed in water(refractive index = 4/3) as shown in the figure.The positions of maxima on the surface ofwater are given by x2 = p2m22–d2, where isthe wavelenght of light in air (refractive index= 1), 2d is the separation between the slitsand m is an interger. The value of p is

Sol. 3

S1

S2

d

d 2

2d

x

2

2

dx

air

4/3 water

x2 = p2m22 – d2

134

22 dx = m = 9. m2 2 – d2

p2 = 9 ; p = 3

3. Two identical uniform discs roll without slippingon two different surface AB and CD (see figure)starting at A and C with linear speeds v1 andv2, respectively, and always remain in contactwith the surfaces. If they reach B and D withthe same linear speed and v1 = 3 m/s, then v2

in m/s is (g = 10 m/s2)

Corporate Head Office : Motion Education Pvt. Ltd., 394 - Rajeev Gandhi Nagar, Kota-5 (Raj.)Page 2

Sol. 7

12

mv2 + 12

2MR2

× 2

2

vR

43

m × 32 + m × 30 × 20 = 43

m.v2

43

mu22 + m × 27 × 10 =

43

m2

= 43

m × 3 2 + 300 m = 427

+ 30

43

v22 =

4147

; v2 = 7

4. A bullet is fired vertically upwards withvelocity from the surface of a sphericalplanet. When it reaches its maximum height,its acceleration due to the planet’s gravity is1/4th of its value at the surface of the planet.If the escape velocity from the planet is

escv v N , then the value of N is (ignore

energy loss due to atmosphere)Sol. 2

2

eRh

1

g

=

84

; eR

h1 = 2

Reh

= 1

h = Re

eRm.GMe

+ 12

Mv2 = e

e

GM m2R

12

m2 = e

e

GM m2R ; =

e

e

GMR

ve = 0v2

ve = egR2

ev 2 ; v N v

= egR

N = 2

5. Two spherical stars A and B emit blackbodyradiation. The radius of A is 400 times thatof B and A emits 104 times the power emitted

from B. The ratio A

B

of their wavelengths

A and B at which the peaks occur in theirrespective radiation curves is

Sol. 2

400RR

B

A

= A

B

TT

P = eA T4

PA = e. 4 (400 R)2 × TA4

PB = e. 4pR2 × TB4

(400R)2 × TA4 = 104 × TB

4

16 × 104 × TA4 = 104 × TB

4

A

B

TT

= 2

6. A nuclear power plant supplying electricalpower to a village uses a radioactive materialof half life T years as the fuel. The amountof fuel at the beginning is such that the totalpower requirement of the village is 12.5% ofthe electrical power available from the plantat that time. If the plant is able to meet thetotal power needs of the village for a maximumperiod of nT years, then the value of n is

Sol. 3T50%25%2T3T

7. An infinitely long uniform line chargedistribution of charge per unit length liesparallel to the y-axis in the y-z plane at

3z

2 a (see figure). If the magnitude of

the flux of the electric field through therectangular surface ABCD lying in the x-y

plane with its centre at the origin is 0

Ln (0

permittivity of free space), then the value ofn is


Sol. 6

360 2nr

;2nr60 60

360

0

R3r 4 3r

n = 6Since angle is 60° hence total flux will be1/6th of total flux.

8. Consider a hydrogen atom with its electronin the nth orbital. An electromagnetic radiationof wavelength 90 nm is used to ionize theatom. If the kinetic energy of the ejectedelectron is 10.4 eV, then the value of n is(hc = 1242 eV nm)

Sol. 212420

E 13.6 eV900

KE = 10.2 eV Diff. = 13.6 – 10.2 = 3.4 eVEnergy of electron in nth orbital

|E| = 13.6 2

2

n

z eV = 3.4 eV

z = 1n2 = 13.6/3.4 = 4

n = 2

9. The figures below depict two situtions inwhich two infinitely long static line chargesof constant positive line charge density are kept parallel to each other. In theirresulting electric field, point charges q and–q are kept in equilibrium between them. Thepoint charges are confined to move in the xdirection only. If they are given a smalldisplacement about their equilibrium positions,then the correct statement(s) is (are)

x+q

x–q

(A) Both charges execute simple harmonicmotion.

(B) Both charges will continue moving in thedirection of their displacement.

(C) Charge +q executes simple harmonic mo-tion while charge –q continues moving inthe direction of its displacement.

(D) Charges –q executes simple harmonicmotion while charge +q continues movingin the direction of its displacement.

Sol. C

d

+q

(2)(1)

dk2

(2)(1)

d

–q

1F 2F

Right d Right d force due to (2) force due to (2) & due to (1) while due to (1) .Thus Fnet is leftwards. This Fnet is rightwards SHM No SHM

10. Two identical glass rods S1 and S2 (refractiveindex = 1.5) have one convex end of radius ofcurvature 10 cm. They are placed with thecurved surfaces at a distance d as shown inthe figure, with their axes (shown by thedashed line) aligned. When a point source oflight P is placed inside rod S1 on its axis at adistance of 50 cm from the curved face, thelight rays emanating from it are found to beparallel to the axis inside S2. The distance d is

S2

d50 cm

PS1

(A) 60 cm (B) 70 cm(C) 80 cm (D) 90 cm

Sol. B

50cm

x

S1 S2

2 1 2 1

v u R

1 1.5 1 1.5v 50 10

1 15 5v 500 100

1 1 3v 20 100

5 3100

1 2v 100

100v 50

2

now for S2


2 1 2 1

v u R

u = – xv

5.1

– x1

= 10

15.1

x = +20d = 50 + 20 = 70 cm

11. A concutor (shown in the figure) carryingconstant current I is kept in the x-y plane in

a uniform magnetic field B

. If F is the

magnitude of the total magnetic force act-ing on the conductor, then the correctstatement(s) is (are)

Y

x

(A) If B is along z , F (L+R)

(B) If B is along x , F= 0

(C) If B is along y , F (L+R)

(D) If B is along z , F= 0

Sol. A,B,CIf magnetic field is uniform, then we can

define L (length vector) for whole of the wire.

L

L

= 2 (L + R)

Net force on wire will be F = i )BL(

if L B

, then F

=i |L|

|B|

= 2i (L + R) B

if B||L

, then F = 0

12. A container of fixed volume has a mixture ofone mole of hydrogen and one mole of heliumin equilibrium at temperature T. Assuming thegases are ideal, the correct statement(s) is(are)(A) The average energy per mole of the gasmixture is 2RT.(B) The ratio of speed of sound in the gas

mixture to that in helium gas is 6 /5 .

(C) The ratio of the rms speed of helium atomsto that of hydrogen molecules is 1/2.(D) The ratio of the rms speed of helium atomsto that of hydrogen moleucles is 1/ 2 .

Sol. A,B,D

Emix. = 3RT 5RT1 12 2

1 1

= 2RT

CPmix = 3R

mix = 23

Mmix = 11

2141

= 3

Vs = w

RTM

mix

He

VV =

Hemix

He mx

MM

=

34

53

23

= 56

Vrms VM

1

2H

He

VV

= e

2

H

H

M

M =

42

= 2

1

13. In an aluminum (Al) bar of square crosssection, a square hole is drilled and is filledwith iron (Fe) as shown in the figure. Theelectrical resistivities of Al and Fe are 2.7 ×10–8 m and 1.0 × 10–7 m, respectively.The electrical resistance between the twofaces P and Q of the composite bar is

(A) 247564

(B) 187564

(C) 187549

(D) 2475132


Sol. BThe connection is in parallel

RA = A A

AA

=

A6 6(49 10 4 10 )

= 6

38

1045

1050107.2

= 3 × 10–5

RFe = Fe Fe

FeA

= F

6(4 10 )

= 6

37

104

105010

= 225

× 10–7–3+6 = 225

× 10–4

R = FeA

FeA

FRRR

= 45

45

10225

103

10225

103

= 45

45

105.12103

105.12103

= 5

9

10128

1035.12

128375

× 10–10+5 128375

× 10–5

1283750

× 10–6 64

1875

14. For photo-electric effect with incident photonwavelenght , the stopping potential is V0.Identify the correct variation(s) of V0 with and 1/.

(A) (B)

(C) (D)

Sol. A,C

λhc

– = ev0 ; v0 = λe

hc –

e

y = mx – c

15. Consider a Vernier capplipers in which each 1cm on the main scale is divided into 8 equaldivisions and a screw grauge with 100divisions on its circular scale. Inthe Verniercallipers, 5 divisions of the Verner scalecoincide with 4 divisions on the main scaleand in the screw gauge, one complete rotationof the circular scale moves it by two divisionson the linear scale. Then :(A) If the pitch of the screw gauge is twicethe least count of the Vernier callipers, theleast count of the screw gauge is 0.01 mm.(B) If the pitch of the screw gauge is twicethe least count of the Vernier callipers, theleast count of the screw gauge is 0.005 mm.(C) If the least count of the linear scale ofthe screw gauge is twice the least count ofthe Vernier callipers, the least ount of thescrew gauge is 0.01 mm.(D) If the least count of the linear scale ofthe screw gauge is twice the least count ofthe Vernier callipers, the least count of thescrew gauge is 0.005 mm.

Sol. B,CVERNIER1cm = 8MSD 1MSD = 1/8 cm5 VSD = 4 MSD L.C. = 1 MSD – 1 VSD

= 51

MSD = 401

cm = 0.25 mm

SCREW GAUGE1 pitch = 100 div. = 2 LSD = 2 Linear scale div.If 1 pitch = 2 (L.C. of Ver.) = 0.5 mm

L.C. of screw = 100pitch1

= 0.005 mm

1 LSD = 0.5 mm

L.C. of screw = 100pitch1

= 100mm1

= 0.01 mm

16. planck’s constant h, speed of light c andgravitational constant G are used to form aunit of length L and a unit of mass M. Thenthe correct option (s) is (are).

(A) M c (B) M G

(C) L h (D) L GSol. A,C,D

P = h

M1L1T–1 = 1Lh

L2 h

L h


17. Two independent harmonic oscillators of equalmass are oscillating about the origin withangular frequencies 1 and 2 and have totalenergies E1 and E2, respectively. The variationsof their moments p with positions x are shownin the figures. If a/b = n2 and a/R = n, thenthe correct equation(s) is (are)

b

a

p

x

Energy=E1

R

pEnergy=E2

x

(A) E12 = E22 (B) 2

1

2 n

(C) 12 = n2 (D) 2

2

1

1 EE

Sol. B,D

v = 22 xA or v2 = 2 (A2 – x2)or v2 = 2A2 – 2 x2

x=0 x=A

or v2 + 2x2 = 2A2

or 22

2

A

v

+ 2

2

A

x = 1

as A = a m1 a = b

ab

= 1m

1 = n2 (given)

aR = n (given)

E1 = 21

m 22 A2

2

E2 = 21

m 22 R2

m1A1 = b m1a = b

1 = 2mn1

amb

m2A2 = R m2R = R

2 = m1

2

1

= 2mn1

× m = 2n1

1

1E =

2

2E

18. A ring of mass M and radius R is rotating withangular speed about a fixed vertical axispassing through its centre O with two point

masses each of mass M8

at rest at O. These

masses can move radially outwards along twomassless rods fixed on the ring as shown inthe figure. At some instant the angular speed

of the system is 89 and one of the masses

is at a distance of 35

R from O. At this instant

the distance of the other mass from O is.

(A) 2

R3

(B) 1

R3

(C) 3

R5

(D) 4

R5

0

Sol. DLi = Lf

mR2 = 98

(I1 + I2 + I2 + Iring)

mR2 = 98

222

mRx8m

R53

8m

on solving we get

x = 4R5

19. Match the nuclear processes given in col-umn I withthe appropriate option(s) in col-umn II.Column I Column II

(A) Nuclear fusion (P) Absorption of thermal neutrons

by 23592 U(B)

(B) Fission in a (Q) 6027Co nucleus

nuclear reactor(C) –decay (R) Energy production

in stars via hydrogen conversion to helium

(D) y-ray emission (S) Heavy water(T) Neutrino emission


Sol. A (R),(T)

B (P),(Q),(S)

C (Q),(R),(T)

D (P),(Q),(R)/(P),(Q),(R),(T)

20. A particle of unit mass is moving along the x-

axis under the influence of a force and its

total energy is conserved. Four possible forms

of the potential energy of the particle are

given in column I (a and Uo are constants).

Match the potential energies in column I to

the corresponding statement(s) in column II.

Column I Column II

(A)

220

1U x

U (x) 12 a

(P) The

force acting on

the particle is

zero at x=a.

(B)2

02

U xU (x)

2 a

(Q) The force

acting on the

particle is zero

at x=0.

(C)

2 20

3U x x

U (x) exp2 a a

(R) The force

acting on the

particle is zero

at x=–a.

(D)

30

4U x 1 x

U (x)2 a 3 a

(S) The particle

experiences an

attractive force

towards x=0 in

the region x a .

(T) The particle

with total

energy 0U4

can

oscillate about

the point x=–a.

Sol. A(P),(Q),(R),(T)

B(Q),(S)

C(P),(Q),(R),(S)

D(P),(R),(T)

(A) U1 = 2U0

22

ax

1

–a

U

+aX

Umin at 1 – 2

ax

= 0

x = ± a, F = 0 at x = ± a

(B) U2 = 2U0

2

2

AX

x=0x

U

(C) U3 = 2U0

2

ax

2

2

ax

e

(D) U4 = 2U0

3

ax

31

ax

= 3

U

AT x = – a

U4 = 34

2U0 = –

3U0

At x = a,

U4 = 32

× 2U0 =

3U0

a1

– 3

3

a

x = 0

a1

= 3

2

a

x = x = ± a

U=0

U

x

–U /30


CHEMISTRY C-0


SECTION 1 (MAXIMUM MARKS :32)

• This section contains EIGHTS questions• The answer to each question is a SINGLE

DIGIT INTEGER ranging from 0 to 9, bothinclusive

• For each question, darken the bubble corre-sponding to the correct integer in the ORS

• Marking scheme:+4 If the bubble corresponding to the an-swer is darkened0 In all other cases

21. If the freezing point of a 0.01 molal aqueoussolution of a cobalt (III) chloride-ammoniacomplex (which behaves as a strong elec-trolyte) is –0.0558 °C, the number ofchloride(s) in the coordination sphere of thecomplex is[Kf of water = 1.86 K kg mol–1]

Sol. 1

f fT ik m

01.086.10558.0

i

= 3

i = 1 + (n–1) = n = 3n = 3 suggest ‘3’ ions 2Cl– ions as counter ionsHence 1 Cl– ion in coordination sphere.Ans. 1

22. All the energy released from the reactionX Y, r G

° = –193 kJ mol–1 is used for oxi-dizing M+ as M+ M3+ + 2e–, E° = – 0.25V.Under standard conditions, the number ofmoles of M+ oxidized when one mole of X isconverted to Y is[F = 96500 C mol–1]

Sol. 4x yr, G

° –193 kJ mol–1

M+ m+3 + 2eE° = – 0.25 VG = + 2 × F × 1/4

G2

965002F

G = 100048250

kJ/mole

G = 48.25 Kj/mole

Number of mole = 25.48

193 = 4

23. For the octahedral complexes of Fe3+ in SCN–

(thiocyanato-S) and in CN– ligand environ-ments, the difference between the spin-onlymagnetic moments in Bohr magnetons (whenapproximated to the nearest integer) is.[Atomic number of Fe = 26]

Sol. 4

[Fe(SCN)6]3–

n = 5

µ = )2n(n = )25(5 = 35 = 5.916

µ = )2n(n = 1(1 2) = 3 = 1.73Difference of µ = 5.916 – 1.73 = 4.184integer value = 4.

24. The total number of lone pairs of electronsin N2O3 is.

Sol. 8

25. Among the triatomic molecules/ions, BeCl2 ,

–3N , N2O, 2NO

, O3, SCl2, –2ICl , –

3I , and XeFeF2,the total number of linear molecule(s)/ion(s)where the hybridization of the central atomdoes not have contribution from thed-orbital(s) is[Atomic number: S = 16, Cl = 17, I = 53and Xe = 54]

Sol. 4


26. Not considering the electronic spin, thedegeneracy of the second excited state (n=3)of H atom is 9, while the degeneracy of thesecond excited state of H– is

Sol. 3

27. The total number of stereoisomers that can

exist for M is

OH3C

CH3H3C

MSol. 2

OH3C

CH3H3C

M

*

28. The number of resonance structures for N isOH

NaOHN

Sol. 9

(1)

O

(2)O

(3)

O

(4)

O

(5)

O

(6)

O

(7)

O

(3)

O

(8)

O

(9)

O

SECTION 2 (MAXIMUM MARKS :40)

• This section contains Ten Questions.• Each question has Four Options (A), (B), (C)

and (D) one or more than one of these forOption(s) is (are) correct.

• For each question darken the Bubbles corre-sponding to all the correct option(s)is theORS.

• Marking Scheme:+4 If only the bubble(s) corresponding to allthe correct option(s) is(are) darkned0 If none of the bubbles is darkned–2 In all other cases

29. The major product of the reaction is

H C3 CO H2

NH2CH3

(A)

H C3 NH2

OHCH3

(B)

H C3 CO H2

OHCH3

(C)

H C3 CO H2

OHCH3

(D)

H C3 NH2

OHCH3

Sol. C


30. The correct statement (s) about Cr2+ and Mn3+

is (are) [Atomic no. of Cr = 24 and Mn = 25]

(A) Cr2+ is reducing agent

(B) Mn3+ is an oxidising agent

(C) Both Cr2+ and Mn3+ exhibit d4 electronic

configuration

(D) when Cr2+ is used as a reducing agent,

the chromium ion attains d5 electronic

configuration

Sol. (A,B,C)

It will take one electron so act as oxidizing

agent.

It will released one electron so act as

reducing agent.

31. Copper is purified by electrolytic refining of

blister copper. The correct statement(s) about

this process is (are)

(A) Impure Cu strip is used as cathode

(B) Acidified aqueous CuSO4 is used as elec-

trolyte

(C) Pure Cu deposits at cathode

(D) Impurities settle as anode-mud

Sol. (B,C,D)

32. Fe3+ is reduced to Fe2+ by using

(A) H2O2 in presence of NaOH

(B) Na2O2 in water

(C) H2O2 in presence of H2SO4

(D) Na2O2 in presence of H2SO4

Sol. (A,B)

Na2O2 + H2O NaOH + H2O2

33. The % yield of ammonia as a function of time in

the reaction

N2(g) + 3H2(g) 2NH3(g), H < 0

at (P, T1) is given below

T1

time

If this reaction is conducted at (P, T2), with

T2 > T1, the % yield of ammonia as a func-

tion of time is represented by

(A)

T1

time

T2


(B)

T1

time

T2

(C)

T1

time

T2

(D)

T1

time

T2

Sol. BInitially on increasing the temperature rateconstant increases there for % yield increasesbut after, equilibrium keq decereses onincresing the temperature there for % yielddecreases.

34. If the unit cell of a mineral has cubic closepacked (ccp) array of oxygen atoms with mfraction of octahedral holes occupied by alu-minium ions and n fraction of tetrahedralholes occupied by magnesium ions, m andn, respectively, are

(A) 1 1

,2 8

(B) 1

1,4

(C) 1 1

,2 2

(D) 1 1

,4 8

Sol. ANo. of octahedral voids - 44m ×(+3) = by ‘Al’No. of tetrahedral voids = 8

)2(n8 = by Mg

12 m + 16 n = 83m + 4n = 2m = 1/2n = 1/8

35. Compound(s) that on hydrogenationproduce(s) optically inactive compound(s)is(are)

(A)

(B)

(C)

(D)

Sol. (B, D)

Achiral molecule

Achiral molecule36. The major product of the following reaction

is

(A) (B)

(C) (D)

Sol. (A)


37. In the following reaction, the major productis :

(A) (B)

(C) (D)

Sol. D

38. The structure of D-(+)- glucose is

HHO

OHH

CHO

CH OH2

H OHH OH

The structure of L-(–)glucose is

(A)

HOH

HOH

CHO

CH OH2

HO HHO H

(B)

HHO

OHH

CHO

CH OH2

H OHHO H

(c)

HOHO

HH

CHO

CH OH2

H OHHO H

(D)

HOHO

HH

CHO

CH OH2

HO HH OH

Sol. AComplete mirror image of ‘D’ glucose is L–Glucose.

HOH

HOH

CHO

CH OH2

HO HHO H

L(–)

SECTION 3 (MAXIMUM MARKS :16)• This section contains TWO questions.• Each question contains two columns, Column

I and Column II.• Column I has four entries (A), (B), (C) and

(D)• Column II has five entries (P), (Q), (R), (S)

and (T)• Match the entries in Column I with the

entries in Column II.• One or more entries in Column I may match

with one or more entries in Column II.• The ORS contains a 4 × 5 matrix whose lay-

out will be similar to the one shown below :

(P) (Q) (R) (S) (T)

(P) (Q) (R) (S) (T)

(P) (Q) (R) (S) (T)

(P) (Q) (R) (S) (T)

(A)

(B)

(C)

(D)

• For each entry in Column I, darken thebubbles of all the matching entries. For ex-ample, if entry (A) in Column I match with


entries (Q), (R) and (T), then darken thesethree bubbles in the ORS. Similarly, for en-tries (B), (C) and (D).

• Marking scheme :For each entry in Column I+2 If only the bubble(s) cooresponding toall the correct match(es) is(are) darkned0 If none of the bubbles is darkned–1 In all other cases

39. Match the anionic species given the ColumnI that are present in the ore(s) given inColumn II.Column I Column II(A) Carbonate (P) Siderite(B) Sulphide (Q) Malachite(C) Hydroxide (R) Bauxite(D) Oxide (S) Calamine

(T) ArgentiteSol. ( A)PQS ; (B) T; (C) Q,R; (D) R ;

Siderite FeCO3Malachite Cu(OH)2.CuCO3Bauxite Al2O3.2H2O

AlOx(OH)3–2x0<x<1[AlO0.5(OH)2]xx = 2

Calamine ZnCO3Argentite Ag2S

40. Match the thermodynamic processes givenunder Column I with the expressions givenunder Column II.Column-I(A) Freezing of water at 273 K and 1 atm(B) Expansion of 1 mole of an ideal gas intoa vacuum under isolated conditions(C) Mixing of equal volumes of two idealgases at constant temperature and pressurein an isolated container(D) Reversible heating of H2(g) at 1 atm from300 K to 600 K, followed by reversible cool-ing to 300 K at 1 atm

Column-II(P) q = 0(Q) w = 0(R) Ssys < 0(S) U = 0(T) G = 0

Sol. A–RT, B–PQS, C–PQS, D– ST or PQST(A) H2O(l, 1 atm, 273 K) H2O (s, 1 atm, 273K)

q 0 , w 0, Ssys < 0, U 0, G = 0(B) Free expansion ideal gas.

q 0 , w 0, S > 0, U 0, G < 0(C) q 0 , w 0, Ssys > 0, U 0, G < 0(D) Ssys = 0, U 0, G = 0


MATHEMATICS


Q.41 Let F(x) =

2x6

2

x

2cos t

dt for all x R and

f : 10,2

[0, ) be a continuous function.

For a 10,2

, if F'(a) + 2 is the area of the

region bounded by x = 0, y = 0, y = f(x) andx = a, then f(0) is.

Sol. 3

F(x) =

2x6

2

x

2 cos tdt

F’(x) = 2cos22x

6

. 2x – 2cos2 x .1

F”(x) = – 4cos2 2x .sin x .2x.2x

6 6

+ 2cos2 2x .2

6

+ 4 cosx . sinx

y

1/2x

F”(0) = 0

+ 2 cos2 .26

= 3

0

f(x)dx F '( ) 2

f() = F”()f(0) = F”(0) = 3

Q.42 A cylindrical container is to be made fromcertain solid material with the followingconstraints : It has a fixed inner volume of Vmm3, has a 2 mm thick solid wall and is openat the top. The bottom of the container is asolid circular dise of thickness 2 mm and is ofradius equal to the outer radius of thecontainer.If the volume of the material used to makethe container is minimum when the inner ra-

dius of the container is 10 mm, then the value

of V

250 is.

Sol. 4

2mm

h

2mmr

M = (r + 2)2n – r2n + (r + 2)2.2

M = 2

22 2

r 2 V r v r 2 .2r r

M = 2

22

v r 2V r 2 .2

r

3 2dm 0 – 8 4v 2 r 2 .2dr r r

–8 40v 2 12 01000 1000

–48v 48 01000

v = 1000 pNow

V 1000 4

250 250

Q.43 Let n be the number of ways in which 5 boysand 5 girls can stand in a queue in such away that all the girls stand consecutively inthe queue. Let m be the number of ways inwhich 5 boys and 5 girls can stand in a queuein such a way that exactly four girls standconsecutively in the queue. Then the value

of mn

is.

Sol. 5n = 6! × 5!

m = 5! . 6C2 . 5!

4!1! . 2! . 4!


4x [0, 8]Total number of solutions 8

Q.48 Let the curve C be the mirror image of theparabola y2 = 4x with respect to the line x +y + 4 = 0. If A and B are the points of inter-section of C with the line y = –5, then thedistance between A and B is.

Sol. 4y2 = 4x(at2, 2at) (t2, 2t), a =1

2 2x t y 2t 2(t 2t 4)1 1 2

x – t2 = – t2 – 2t – 4x = – 2t – 4

x 4 t2

y – 2t = – t2 – 2t – 4y = – t2 – 4

y = – 2(x 4) 4

4

4y = –x2 – 8x – 16 – 16x2 + 4y + 8x + 32 = 0y = – 5 (given)x2 – 20 + 8x + 32 = 0x2 + 8x + 12 = 0x = – 6, x = – 2A(–6, –5), B(–2, –5)AB = |–6 + 2| = 4

Q.49 Let P and Q be distinct points on theparabola y2 = 2x such that a circle withPQ as diameter passes through the veriex Oof the parabola. if P lies in the first quadrantand the area of the triangle OPQ is 3 2 ,then which of the following is (are) thecoordiantes of P?

(A) 4,2 2 (B) 9,3 2

(C) 1 1,4 2

(D) 1, 2

(A) 4,2 2 (B) 9,3 2

(C) 1 1,4 2

(D) 1, 2

Sol. A, Dy2 = 2x

P : 21 1

1 t t2

mPO . mQO = – 1

1

2

t 01 t 02

. 2

tx 0 11 t 02

4 11/2

t1 t2 = – 4 = 21 1

22

0 0 11 1t t 12 2

1t t 12

12

2 22 2 1

1 1h t t t 3 22 2

|t1 t2 |t1 – t2|| = 12 2

|t1– t2| = 3 2

t1 – t2 = 3 2 , or – 3 2

11

4tt

= 3 2 t12 + 3 2 t1 + t2 = 0

or t12 – 3 2 t1 + 4 = 0

t1 = 2 & t2 = 2 2 P(at1

2, 2at1)

t1 = 2 2 , a = 12

t1 = 2 , a = 21

P 1 1(8), 2 2 22 2

P 1 1(2), 2 2

2 2

4, 2 2 P 1, 2

Q.50 Let y(x) be a solution of the differentialeqution (1 + ex)y’ + yex = 1. If y(0) = 2, thenwhich of the following statements is (are)true?(A) y(–4) = 0 (B) y(–2) = 0(C) y(x) has a critical point in the interval(–1, 0)(D) y(x) has no critical point in the interval (–1, 0)

Sol. A,Cy(x)(1 + ex) y1 + ex y = 1

dydx

+ x

x

e1 e

y = x

11 e


(D) There is an x R such that (gof)(x) = 1Sol. A, B, C

(A) f(x) = sin sin sinx6 2

x R – 1 sin x 1

– sinx2 2 2

– 1 sin sin x 12

–

sin sinx6 6 2 6

– 12

sin 1sin sinx

6 2 2

Range f1 1,2 2

(B) f{g(x)} = sin sin sin sinx6 2 2

again

– sin sinx2 2 2 2

– 1 sin sin sin 12 2

1 1sin sin sin sinx2 6 2 2 2

Hence Range f{g(x)} = 1 1,2 2

(C) x 0

sin sin sinx sin sinx6 2 6 2lim

6sinx sin sinx2 6 2

(D) sin sin sin sin x 12 6 2

– sin sinx6 6 2 6

1 1sin sin sinx2 6 2 2

– sin 1 1sin sin sin sinx sin2 6 2 2

Q.54 Let PQR be a triangle. Let a

= QR

,

b = RP

and c

= PQ

. If |a

| = 12, |b

| =

4 3 and b.c

= 24, then which of the follow-ing is (are) true?

(A) 2| c |

2

–|a| = 12

(B) 2| c |

2

+ |a| = 30

(C) |a b c a

| = 48 3

(D) a.b

= –72

Sol. A,C,D

120°

30° 30°

34b

34c

12a

P

Q R

| a| 12

|b | 4 3

a b c 0

b c a

2 2 2| b| | c | 2b.c | a|

48 + 2| c | 48 144

2| c | 144 – 96 = 48

| c| 4 3

(A)2| c | | a|

2


12 2

11 14

2 2 2 29

2 – + 2 – 2 = ± 62 – + 2 = 8 or 2 – + 2 = – 4

Q.58 In R3, let L be a straight line passing throughthe origin. Suppose that all the points onL are at a constant distance from the twoplanes P1 : x + 2y – z + 1 = 0 and P2 : 2x – y+ z – 1 = 0. Let M be the locus of the feet ofthe perpendiculars drawn from the points onL to the plane P1. Which of the following pointslie(s) on M?

(A) 5 20,– ,–6 3

(B) 1 1 1– ,– ,6 3 6

(C) 5 1– ,0,6 6

(D) 1 2– ,0,3 3

Sol. A, B

ˆˆ î j kn 1 2 1

2 1 1

ˆˆ î(2 1) j(1 2) k( 5)

ˆˆ î 3j 5k

x y z1 3 5

then equation of line x y z1 2 1

(, 2, –) + 4 + + 1 = 0 = – 1/6

x 1 / 61

=

y 2 /63

=z 1 / 6

5

A,B, only good

Section 3 (Maximum Marks :16)• This section contains TWO questions.• Each question contains two columns, Col-

umn I and Column II.• Column I has four entries (A), (B), (C) and (D)• Column II has five entries (P),(Q),(R),(S) and (T)

• Match the entries in Column I with the cenriesin Column II.

• One or more entries in Column I may matchwith one or more entries in Column II.

• The ORS contains a 4 × 5 matrix whoselayout will be similar to the one shown below :

• For each entry in Column I, darken the bubblesof al l the matching entries. Forexample, if entry (A) in Column I matc withentries (Q), (R) and (T), then darken thesethree bubbles in the ORS. Similarly, forentries (B), (C) and (D).

• Marking scheme :For each entry in Column I+2 If only the bubble(s) cooresponding to allthe correct match(es) is(are) darkned0 If none of the bubbles is darkned–1 In all other cases

Q.59 Column I(A) In R2, if the magnitude of the projection

vector of the vector ˆ î j on ˆ ˆ3i j is

3 and if a = 2 3 , then possible value(s)of || is (are)

(B) Let a and b be real numbers such that

the function f(x) = 2

2

–3ax – 2, x 1

bx a , x 1

is

differentiable for all x R. Then possiblevalue(s) of a is (are)

(C) Let 1 be a complex cube root ofunity. If (3 – 3 + 22)4 + 3 + (2 + 3 – 32)4

+ 3 + (– 3 + 2 + 32)4 + 3 = 0, then possiblevalue(s) of n is (are)

(D) Let the harmonic mean of two positivereal numbers a and b be 4. If q is positivereal number such that a, 5, q, b is anarighmetic number such that a, 5, q, b is anarithmetic progression, then the value(s) of|q – a| is (are)

Column II

(P) 1 (Q) 2

(R) 3 (S) 4

(T) 5

Sol. (A) P,Q (B) P,Q (C) P,Q,S,T

(D) Q,T


{0, 1}. Then the value(s) of F() +

83 2 , when = 0 and = 1, is (are)

Column II

(P) 1 (Q) 2

(R) 3 (S) 5

(T) 6

Sol. ( A) P,R,S (B) P (C) P,Q (D) S,T

(A)c

ya z

b

x

2(a2 – b2) = c2

2(sin2x – sin2y) = sin2z

2sin(x + y) sin(x – y) = sin2zsin(x y) 1

sinz 2

cos1n2

n = 1, 3, 5

(B)c

ya z

b

x

(1 – cos 2y) + (cos 2x – cos 2y) = 2 sinx siny2sin2y + 1 – 2 sin2x – 1 + 2sin2y = 2 sinx siny4sin2y – 2 sin2x 2sinxsinya2 + ab – 2b2 = 02b2 – a2 = aba2 + 2ab–ab – 2b2 = 0a(a+2b) – b(a+2b)=0

ab = 1

(C) x 3, 1

y 1, 3

z , (1 )

ˆ ˆ(i j) C Q

x – y = 01 32 2

2 – 1 = 32 = 4 = 2

(D)

9 = 4X

9 x4

2

0

(3 2 x)dx =

23/2

0

x3x 232

= 2

0

2/3x34x3

= 23822

346

= 6

Documents

JEE ADVANCED EXAMINATION 2015 QUESTIONS WITH …€¦ · T T P = eA T 4 P A = e. 4 (400 R)2 × T A 4 P B = e. 4pR2 × T B 4 (400R)2 × T A 4 = 104 × T B 4 16 × 104 × T A 4 = 10