: aakash tower, plot no.-4, sec-11, mlu, dwarka, new delhi ...€¦ · (7) when s2 is pressed and...

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Answers & Solutionsforforforforfor

JEE (Advanced)-2013

Time : 3 hrs. Max. Marks: 180

INSTRUCTIONS

Question Paper FormatThe question paper consists of three parts (Physics, Chemistry and Mathematics). Each part

consists of three sections.

Section 1 contains 10 multiple choice questions. Each question has four choices (A), (B), (C) and

(D) out of which ONLY ONE is correct.

Section 2 contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and

(D) out of which ONE or MORE are correct.

Section 3 contains 5 questions. The answer to each question is a single-digit integer, ranging from

0 to 9 (both inclusive).

Marking SchemeFor each question in Section 1, you will be awarded 2 marks if you darken the bubble corresponding

to the correct answer and zero mark if no bubbles are darkened. No negative marks will be awarded

for incorrect answers in this section.

For each question in Section 2, you will be awarded 4 marks if you darken all the bubble(s)

corresponding to only the correct answer(s) and zero mark if no bubbles are darkened. In all other

cases, minus one (–1) mark will be awarded.

For each question in Section 3, you will be awarded 4 marks if you darken the bubble corresponding

to only the correct answer and zero mark if no bubbles are darkened. In all other cases, minus

one (–1) mark will be awarded.

DATE : 02/06/2013

PAPER - 1 (Code - 4)

CODE

4

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075Ph.: 011-47623456 Fax : 011-47623472

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PART–I : PHYSICS

SECTION - 1 : (Only One Option Correct Type)This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D) out of whichONLY ONE is correct.

1. In the Young's double slit experiment using a monochromatic light of wavelength , the path difference (interms of an integer n) corresponding to any point having half the peak intensity is

(A) (2 1)

2n (B)

(2 1)4

n

(C)

(2 1)8

n (D)

(2 1)16

n

Answer (B)

Hint :

2max cos

2I I

21 cos2 2

cos = 0

3 5 7, , ,

2 2 2 2

3 5, ,

4 4 4x

(2 1)4

x n

2. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulseis 30 mW and the speed of light is 3 × 108 ms–1. The final 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

Answer (B)

Hint :

–3 –9–17 –1

830 10 100 10 10 kg ms

3 10E P tpC C

3. One end of a horizontal thick copper wire of length 2L and radius 2R is welded to an end of another horizontalthin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends,the ratio of the elongation in the thin wire to that in the thick wire is(A) 0.25 (B) 0.50(C) 2.00 (D) 4.00

Answer (C)Hint : Force will be same

Now, FLAY

21

22

(2 ) 22

L RLR

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4. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is

one-third the size of the object. The wavelength of light inside the lens is 23

times the wavelength in free

space. The radius of the curved surface of the lens is(A) 1 m (B) 2 m(C) 3 m (D) 6 m

Answer (C)

Hint :

32

air

med

fcf

Now, = +8 m,

1 24 m3

m uu

1 1 1f u

1 1 1 4

8 24 24ff = 6 m

1

Rf

6 3 m0.5R R

5. A particle of mass m is projected from the ground with an initial speed u0 at an angle with the horizontal.At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle,which was thrown vertically upward from the ground with the same initial speed u0. The angle that thecomposite system makes with the horizontal immediately after the collision is

(A) 4 (B)

4

(C) –

2 (D) 2

Answer (A)Hint : Speed of first particle at highest point = u0cos

u0cosu0

Hu0

Speed of second particle at highest point = 20 2u gH

Now,

2 20 sin

2uH

g Speed of 2nd particle = u0cos

Final momentum = 0 0ˆ ˆcos cosmu i mu j

Angle = 4

6. The work done on a particle of mass m by a force,

2 2 3/2 2 2 3/2

ˆ ˆ( ) ( )

x yK i jx y x y

(K being a constant of

appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circularpath of radius a about the origin in the x – y plane is

(A)2K

a (B)K

a

(C)

2K

a (D) 0

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Answer (D)

Hint :

2 2 3/2 2 2 3/2

ˆ ˆ( ) ( )

xi yjF kx y x y

.dW F dx

2 2 3/2( )

xdx ydydW kx y

let x2 + y2 = r2

xdx + ydy = r2

3 2krdr kdW drr r

2

1

r

r

kWr

Now, r1 = a, r2 = a W = 07. The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero

of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisionsequivalent to 2.45 cm. The 24th division of the Vernier scale 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 cm

Answer (B)Hint : 1 MSD = 5.15 – 5.10 = 0.05 cm

1 VSD = 2.4550 = 0.049 cm

LC = 1 MSD – 1 VSD = 0.001 cm

Reading = 5.10 + L.C. × 24 = 5.10 + 0.024 = 5.124 cm

8. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partialpressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is(A) 1 : 4 (B) 1 : 2(C) 6 : 9 (D) 8 : 9

Answer (D)

Hint :

RTPM

1 1 2

2 1 2

P MP M

1

2

4 33 2

1

2

89

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9. Two rectangular blocks, having identical dimensions, can be arranged either in configuration-I or inconfiguration-II as shown in the figure. One of the blocks has thermal conductivity and the other 2. Thetemperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 sto transport a certain amount of heat from the hot end to the cold end in the configuration-I. The time totransport the same amount of heat in the configuration-II is

Configuration-I Configuration-II

2x

2

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

Answer (A)

Hint :

1 2

2

T TQl ltA A

[in case (i)]

1 2

2( )'

Q A AT Tt l l

' 2

9tt

t' = 2 s

10. A ray of light travelling in the direction 1 ˆ ˆ( 3 )2

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

along the direction 1 ˆ ˆ( – 3 )2

i j . The angle of incidence is

(A) 30° (B) 45°(C) 60° (D) 75°

Answer (A)

Hint :

ˆ ˆ3 – 3.2 2

cos(180 2 ) ˆ ˆ3 – 32 2

i j i j

i j i j

i 32

j –i 32

180° – 2

^ j

(1 3)4cos21

1cos22

1cos22

2 60 = 30°

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SECTION - 2 : (One or More Options Correct Type)This section contains 5 multiple choice questions. Each question has four choices (A), (B), (C) and (D) out ofwhich ONE OR MORE are correct.

11. A solid sphere of radius R and density is attached to one end of a mass-less spring of force constant k. Theother end of the spring is connected to another solid sphere of radius R and density 3. The completearrangement is placed in a liquid of density 2 and is allowed to reach equilibrium. The correct statement(s)is (are)

(A) The net elongation of the spring is 343R g

k

(B) The net elongation of the spring is 383R g

k(C) The light sphere is partially submerged(D) The light sphere is completely submerged

Answer (A, D)Hint : At equilibrium, for upper sphere R

R 3

2

3 34 4 (2 )3 3

kx R g R g

34

3R gx

k

The system is completely submerged as total weight = Total Buoyant force.

12. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switchS1 is pressed first to fully charge the capacitor C1 and then released. The switch S2 is then pressed to chargethe capacitor C2. After some time, S2 is released and then S3 is pressed. After some time,(A) The charge on the upper plate of C1 is 2CV0

(B) The charge on the upper plate of C1 is CV0

(C) The charge on the upper plate of C2 is 0

S1

2V0

C1V0

S2 S3

C2

(D) The charge on the upper plate of C2 is –CV0

Answer (B, D)Hint : When S1 is pressed and released

2V0 C1 V0

S2 S3

C2+–

+2CV0

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When S2 is pressed and released

2V0 V0

S2 S3

C2+–

CV0

S1

+CV0

When S3 is pressed

2V0 C1 V0C2+–

CV0 CV0

CV0

–+

13. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities 1 and 2respectively, touch each other. The net electric field at a distance 2R from the centre of the smaller sphere,

along the line joining the centres of the spheres, is zero. The ratio

1

2 can be

(A) –4 (B) 3225

(C)3225 (D) 4

Answer (B, D)Hint : EP = 0

31 2

200

4( )3

34 (2 )

R RR P' P

R 2R

2R

1

24

' 0PE

3 31 2

2 20 0

4 4 (2 )3 3

4 (2 ) 4 (5 )

R R

R R

1

2

3225

14. 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[(628 s–1)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 the string, from its equilibrium position is 0.01 m(D) The fundamental frequency is 100 Hz

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Answer (B, C)

Hint : No. of nodes = 6,

2k =

2 3.14 0.1 m62.8

Length =

5 0.25 m2

The mid-point is antinode. Its maximum displacement = 0.01 m

f =

20 Hz2 2vl k l

15. A particle of mass M and positive charge Q, moving with a constant velocity 11

ˆ4 msu i , enters a regionof uniform static magnetic field, normal to the x-y plane. The region of the magnetic field extends from x = 0to x = L for all values of y. After passing through this region, the particle emerges on the other side after

10 milliseconds with a velocity 1

2ˆ ˆ2( 3 )msu i j . 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 field 50

3M

Q units

(D) The magnitude of the magnetic field is 100

3M

Q units

Answer (A, C)

Hint :

MtqB

Clearly = 30° = 6

C

F

x L=

ˆ2 j v

ˆ2 3i

ˆ4ix = 0

3100 50

6 36 10 10M M MB

q QqB must be in –z direction.

SECTION - 3 : (Integer Value Correct Type)This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9(both inclusive).

16. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Giventhat ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) thatwill decay in the first 80 s after preparation of the sample is

Answer (4)

Hint : 0tN N e

0

tN eN

(9)

ln 2 801386

0

N eN

0.693 801386

0

N eN

0.04

0

N eN

0.04

0

1NN e

Fraction of nuclei decayed =

0.04

0

11 1NN e = 0.04 = 4%

17. A bob of mass m, suspended by a string of length l1, is given a minimum velocity required to complete a fullcircle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspendedby a string of length l2, which is initially at rest. Both the strings are mass-less and inextensible. If the secondbob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the

ratio 1

2

ll is

Answer (5)

Hint : Speed of first bob at highest point = 1gl

For elastic collision between objects of same mass, velocities are exchanged speed of 2nd 1bob = gl

but 1 25gl gl

1

2 5l

l

18. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s–1

about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m,are gently placed symmetrically on the disc in such a manner that they are touching each other along theaxis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at restrelative to the disc and the system rotates about the original axis. The new angular velocity (in rad s–1) ofthe system is

Answer (8)Hint : By conservation of angular momentum,

2 2 21 150 (0.4) 10 50 (0.4) 2 2 6.25 (0.2)

2 2

40 8 rad/s4 1

19. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stoppingpotential versus frequency plot for Silver to that of Sodium is

Answer (1)

Hint :

hfVe e

Slope = .he

It is same for both.

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20. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W tothe particle. If the initial speed (in ms–1) of the particle is zero, the speed (in ms–1) after 5 s is

Answer (5)

Hint : As power is constant, P × t = kE

210.5 5 (0.2) 02

v

2.5 = 0.1 v2

v = 5 m/s

PART–II : CHEMISTRY


21. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25°C. For this process, thecorrect statement is

(A) The adsorption requires activation at 25°C

(B) The adsorption is accompanied by a decrease in enthalpy

(C) The adsorption increases with increase of temperature

(D) The adsorption is irreversible

Answer (B)

Hint : The adsorption of methylene blue on activated charcoal is physiosorption which is exothermic, multilayerand does not have energy barrier.

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

Answer (D)Hint : Upon treatment with ammoniacal H2S, Zn2+ ion gets precipitated as ZnS. Fe3+ ion and Al3+ ions also get

precipitated as hydroxides but not as sulphide.23. In the reaction,

P + Q R + SThe time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Qvaries with reaction time as shown in the figure. The overall order of the reaction is

[Q]0

[Q]

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

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Answer (D)Hint : The order of reaction with respect to P is one since t3/4 is twice of t1/2. From the given graph the order

of reaction with respect to Q is zero. Therefore, overall order of reaction is one.24. KI in acetone, undergoes SN2 reaction with each of P, Q, R and S. The rates of the reaction vary as

H C – Cl3

P

ClO

ClCl

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

Answer (B)Hint :

P

ClO

Cl

Q

Cl

RS

Compounds : CH –Cl3 : :

Relative reactivities towards S reactionN2

1,00,000 200 79 0.02: : :

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

Answer (C)Hint : C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)

H° = 6 (–400) + 6(–300) – (–1300)H° = –2900 kJ/mol

2900H 16.11 kJ / gm180

26. Consider the following complex ions, P, Q and R.P = [FeF6]3–, Q = [V(H2O)6]2+ and R = [Fe(H2O)6]2+

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

Answer (B)Hint : The electronic configuration of central metal ion in complex ions P, Q and R are

3 36P [FeF ] ; Fe :

3d

Q = [V(H2O)6]2+; V2+:3d

R = [Fe(H2O)6]2+; Fe2+

3d

The correct order of spin only magnetic moment is Q < R < P

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27. The compound that does NOT liberate CO2, on treatment with aqueous sodium bicarbonate solution, is

(A) Benzoic acid (B) Benzenesulphonic acid

(C) Salicylic acid (D) Carbolic acid (Phenol)

Answer (D)

Hint : Carbolic acid (Phenol) is weaker acid than carbonic acid and hence does not liberate CO2 on treatmentwith aq. NaHCO3 solution. Benzoic acid, benzenesulphonic acid and salicylic are more acidic than carbonicacid and hence will liberate CO2 with aq. NaHCO3 solution.

28. Sulfide ores are common for the metals

(A) Ag, Cu and Pb (B) Ag, Cu and Sn

(C) Ag, Mg and Pb (D) Al, Cu and Pb

Answer (A)

Hint : Silver, copper and lead are commonly found in earth's crust as Ag2S (silver glance), CuFeS2 (Copperpyrites) and PbS (Galena)

29. The arrangement of X– ions around A+ ion in solid AX is given in the figure (not drawn to scale). If the radiusof X– is 250 pm, the radius of A+ is

X–

A+

(A) 104 pm (B) 125 pm(C) 183 pm (D) 57 pm

Answer (A)

Hint : Cation A+ occupies octahedral void formed by anions X–. The maximum radius ratio for a cation toaccommodate a octahedral void without distortion is 0.414. Radius of anion X– is 250 pm.

A

X

R0.414

R

AR 0.414 250 103.50 104 pm

30. Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of(A) NO (B) NO2

(C) N2O (D) N2O4

Answer (B)Hint : Conc. HNO3 slowly decomposes as

4HNO3 4NO2 + 2H2O + O2

It acquires yellow-brown colour due to the formation of NO2.

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31. Among P, Q, R and S, the aromatic compound(s) is/areCl

P

Q

AlCl3

NaH

R(NH ) CO4 2 3

100-115 °C

O O

SHCl

O

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

Answer (A, B, C, D)

Hint : (i) ClAlCl3

AlCl4 (P)

(AROMATIC)

(ii)NaH

(Q) + H2

(AROMATIC)

Na

(iii)O O

(NH ) CO4 2 3

100–115 CºR

(NH ) CO4 2 3

2NH + CO + H O3 2 2

O O+ NH3

O O NH2 O NH2OH

IMPE

N

H

–2H O2

N

H

HO OH

IMPEN

H

O OH

H

(R) AROMATIC

H

(iv) O HCl OH Cl (S)

(AROMATIC)

32. The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100th of that of a strongacid (HX, 1M), at 25°C. The Ka of HA is(A) 1 × 10–4 (B) 1 × 10–5

(C) 1 × 10–6 (D) 1 × 10–3

Answer (A)

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Hint : Rate with respect to weak acidR1 = K[H+]WA[ester]

and rate with respect to strong acidR2 = K[H+]SA[ester]

WA1

2 SA

[H ]R 1R 100[H ]

SA

WA

[H ] 1H 0.01 M C100 100

= 0.01Ka for weak acid = C2

= 1(0.01)2

= 1 × 10–4

33. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s)is(are)(A) G is positive (B) Ssystem is positive(C) Ssurroundings = 0 (D) H = 0

Answer (B, C, D)Hint : Benzene and naphthalene form an ideal solution. For an ideal solution, H = 0, Ssystem > 0 and

Ssurroundings = 0 because there is no exchange of heat energy between system and surroundings.34. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are)

(A) [Cr(NH3)5Cl]Cl2 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]BrAnswer (B, D)Hint : The pair of complex ions [Co(NH3)4Cl2]+ and [Pt(NH3)2(H2O)Cl]+ show geometrical isomerism. The pair of

complexes [Pt(NH3)3(NO3)]Cl and [Pt(NH3)3Cl]Br show ionisation isomerism. The other pairs given do nothave same type of isomerism.

35. The hyperconjugative stabilities of tert-butyl cation and 2-butene, respectively, are due to(A) p (empty) and electron delocalisations(B) * and electron delocalisations(C) p (filled) and electron delocalisations(D) p (filled) and electron delocalisations

Answer (A)Hint : In hyperconjugation p (empty) electron delocalization for tert-butyl carbocation and * electron

delocalization for 2-butene will take place.

SECTION - 3 : (Integer Value Correct Type)This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9(both inclusive).

36. A tetrapeptide has —COOH group on alanine. This produces glycine (Gly), valine (Val), phenyl alanine (Phe)and alanine (Ala), on complete hydrolysis. For this tetrapeptide, the number of possible sequences (primarystructures) with —NH2 group attached to a chiral center is

Answer (4)

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Hint : According to question C – Terminal must be alanine and N – Terminal do have chiral carbon means itsshould not be glycine so possible sequence is :

Val Phe Gly AlaVal Gly Phe Ala

Phe Val Gly Ala

Phe Gly Val Ala

So, answer is (4).37. The atomic masses of 'He' and 'Ne' are 4 and 20 a.m.u., respectively. The value of the de Broglie wavelength

of 'He' gas at –73ºC is "M" times that of the de Broglie wavelength of 'Ne' at 727ºC 'M' is

Answer (5)

Hint :

He Ne Ne

Ne He He

M VM V

NeNe

Ne

HeHe

He

3RTMM

3RTMM

NeNe

Ne

HeHe

He

TMMTMM

Ne Ne

He He

M TM T

205 1000

4 200 He Ne5

38. The total number of lone-pairs of electrons in melamine isAnswer (6)Hint : Structure of melamine is

N NH2

N

NH2

H2N

N

So, melamine has six lone pair of electrons.39. The total number of carboxylic acid groups is the product P is

O

O O

O

O

1. H O , +3

2. O33. H O2 2

P

Answer (2)

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Hint : O

O

O

O

O

H O3+

O

O

O

O

OH

O

O

(i) O3

(ii) H O2 2

O

O

HOOCHOOC

(P)

OHOH (–H O, –CO )2 2

So, final product (P) has two carboxylic acid Heating causes evaporation of water and due to which concentration of acid increases and in

concentrated acidic medium decarboxylation as well as dehydration will take place.40. EDTA4– is ethylenediaminetetraacetate ion. The total number of N—Co—O bond angles in [Co(EDTA)]1–

complex ion isAnswer (8)Hint :

Co

O – C = O

O – C = O

OC CH2 CH2

O

CH2

CH2

N

O NC CH2

OCH2

II

III

IV

I

I

II

So, bond angles are

(1) NI CoOI

(2) NI CoOII

(3) NI CoOIII

(4) NI CoOIV

(5) NII CoOI

(6) NII CoOII

(7) NII CoOIII

(8) NII CoOIV

So, total asked bond angles are 8.

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PART–III : MATHEMATICS


41. Let 1: ,1

2f (the set of all real numbers) be a positive, non-constant and differentiable function such

that f(x) < 2f(x) and

1 12

f . Then the value of 1

1/2

( )f x dx lies in the interval

(A) (2e – 1, 2e) (B) (e – 1, 2e – 1)

(C)

1 , 12

e e (D)

10,2

e

Answer (D)

Hint : 2dy ydx 2 22x xdye ye

dx 2( ) 0xd ye

dx ye–2x is decreasing function

As, 112

x

e–1 > ye–2x > y(1)e–2

e2x–1 > y > y(1)e2x–2

1 1 12 1 2 2

1/2 1/2 1/2

(1) 0x xe dx ydx y e

1

1/2

102

eydx

42. A curve passes through the point 1,

6. Let the slope of the curve at each point (x, y) be

sec , 0y y x

x x.

Then the equation of the curve is

(A)

1sin log2

y xx

(B)

cosec log 2y xx

(C)

2sec log 2y xx

(D)

2 1cos log2

y xx

Answer (A)

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Hint : secdy y ydx x x

Put y = vx

secdvv x v vdx

cos dxv dvx

sin v = ln x + c

It passes through 1,

6

12

c

1sin log2

y xx

43. Perpendiculars are drawn from points on the line

2 1

2 1 3x y z to the plane x + y + z = 3. The feet of

perpendiculars lie on the line

(A)

1 25 8 13x y z

(B)

1 22 3 5x y z

(C)

1 24 3 7x y z

(D)

1 22 7 5x y z

Answer (D)

Hint :

2 1 , 32 1 3

x y z x y z

Any point on this line is (2 – 2, – – 1, 3)

( , , )

B

A

C

(–2, –1, 0)

This point will satisfy the plane(2 – 2) + (– – 1) + (3) = 3

4 – 3 = 3

32

So, point of intersection of plane is

5 91, ,2 2

C .

Now, point on line is (–2, –1, 0) and direction ratio of AB (from figure) is

2 11 1 1

k .

Any general point on line AB is (k – 2, k – 1, k).This will satisfy equation so, (k – 2) + (k – 1) + k = 3. k = 2Therefore, (, , ) (0, 1, 2).So, equation of line passing through BC is

1 2

7 512 2

x y z

1 22 7 5x y z

(19)

44. Let ˆ ˆ ˆ3 2PR i j k and

ˆ ˆ ˆ3 4SQ i j k determine diagonals of a parallelogram PQRS and ˆ ˆ ˆ2 3PT i j k

be another vector. Then the volume of the parallelepiped determined by the vectors

, andPT PQ PS is(A) 5 (B) 20(C) 10 (D) 30

Answer (C)

Hint :

1ˆ ˆ ˆ3 2PR a b i j k d S b( )

P

R

Q a( )

2ˆ ˆ ˆ3 4SQ a b i j k d

ˆ ˆ ˆ2 3a i j k

ˆ ˆ ˆ2b i j k

Volume of the required parallelopiped = 2 1 3

1 2 11 2 3

= |2(6 – 2) + 1(3 – 1) – 3(2 – 2)|= 10 cubic units

OR

Area of parallelogram =

1 21| |2

d d

Volume of parallelepiped =

3 1 2

1 1 3 42 1 2 3

= 10 cubic units.

45. The value of 23

1

1 1cot cot 1 2

n

n kk

is

(A)2325 (B)

2523

(C) 2324 (D)

2423

Answer (B)

Hint : 23

1

1cot 1 ( 1)

nk k

=23

1 1

1tan ( 1) tan ( )

nk k

= tan–1(24) – tan–1(1)

Now, 1 1cot (tan (24) tan (1)

= 1 24 1cot tan1 24 1

=

1 23cot tan

25 = 2523

(20)

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

0bx ay c is less than 2 2 . Then

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

Answer (A)

Hint : 0ax by c ... (1)

0bx ay c ... (2)

Solving, cxa b

Also from (1) & (2)y = x

Point of intersection lies on y = x

cy

a b

Given, 2 2

1 1 2 2c ca b a b

2 1 2 2ca b

2a b ca b

2 2a b c a b

0a b c

47. Let complex numbers and 1 lie on circles 2 2 2

0 0( ) ( )x x y y r and 2 2 20 0( ) ( ) 4x x y y r ,

respectively. If 0 0 0z x iy satisfies the equation 2 2

02 2,z r then

(A)12 (B) 1

2

(C)17 (D)

13

Answer (C)

(21)

Hint : , lies on 2 2 20 0( ) ( )x x y y r ... (1)

( , )x y0 0

r

21

lies on 2 2 2

0 0( ) ( ) 4x x y y r ... (2)

Let be 1 + i1

lies on 2 2 2 2 20 0 0 0( ) ( ) 2( )x y x y xx yy r

2 2 20 1 0 1 02( )z x y r ... (3)

From eq. (2), 2 21 0 1 002 2

( )1 2 4x yz r

2 2 220 1 0 1 01 2( ) 4z x y r ... (4)

From (3) – (4), 2 2 2 2201 (1 ) 1 4z r

2 2 220( 1)(1 ) 1 4z r ...(5)

Now, 2202 1r z

2

20

21

rz

Substituting in (5),2 21 2(1 4 )

2 21 2 8

27 1

17

48. The number of points in (– , ) , for which 2 sin cos 0x x x x , is

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

Answer (C)

Hint : 2( ) sin cosf x x x x x

( ) 2 cos sin sinf x x x x x x

= x(2 – cosx)f(x) is increasing x > 0f(x) is decreasing x < 0f(0) = –1

( )f

(– )f

(22)

49. The area enclosed by the curves sin cosy x x and cos siny x x over the interval

0,

2 is

(A) 4( 2 1) (B) 2 2( 2 1)

(C) 2( 2 1) (D) 2 2( 2 1)

Answer (B)

Hint :/2 /4 /2

0 0 /4(sin cos ) (cos sin ) (sin cos )x x dx x x dx x x dx

=/2 /2 /4 /4 /2 /2

0 0 0 0 /4 /4cos sin sin cos cos sinx x x x x x

=1 1 1 1(0 1) (1 0) 1 0 12 2 2 2

O /4 /2

= 1 12 2 1 12 2

=2 2 2 2

= 4 2 2

=2 2( 2 1)

50. Four persons independently solve a certain problem correctly with probabilities 1 3 1 1, , ,2 4 4 8 . Then the

probability that the problem is solved correctly by at least one of them is

(A)235256 (B)

21256

(C)3

256 (D)253256

Answer (A)

Hint : 1 3 1 1( ) , ( ) , ( ) , ( )2 4 4 8

P A P B P C P D

P(A B C D) = 1 ( )P A B C D

= 1 ( )P A B C D

= 1 ( ) ( ) ( ) ( )P A P B P C P D

= 1 1 3 712 4 4 8

= 211

256

= 235256

(23)


51. For 3 × 3 matrices M and N, which of the following statement(s) is (are) NOT correct?

(A) NTMN is symmetric or skew symmetric, according as M is symmetric or skew symmetric

(B) MN – NM is skew symmetric for all symmetric matrices M and N

(C) MN is symmetric for all symmetric matrices M and N

(D) (adj M) (adj N) = adj(MN) for all invertible matrices M and N

Answer (C, D)

Hint :

(NTMN )T = – NTMT(NT )T

NTMTN

(A) If M is skew symmetric, (NTMN)T = –NTMN,

Hence skew symmetric.

If M is symmetric, (MTMN)T = NTMN,

Hence symmetric.

Option (A) is correct

(B) (MN – NM)T = (MN)T – (NM)T

= NTMT – MTNT

= –(MTMT – NTMT)= –(MN – NM)

Skew symmetric option (B) is correct.(C) (MN)T = NTMT

Symmetricity and skew symmetricity depend on nature of M and N, option (C) is incorrect.(D) adj(MM) = adj(N) adjM,

Option (D) is incorrect.

52. A line l passing through the origin is perpendicular to the lines

1ˆ ˆ ˆ: (3 ) ( 1 2 ) (4 2 ) , l t i t j t k t

2ˆ ˆ ˆ: (3 2 ) (3 2 ) (2 ) ,l s i s j 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)

7 7 5, ,3 3 3 (B) (–1, –1, 0)

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

7 7 8, ,9 9 9

Answer (B, D)

(24)

Hint : x y za b c (Equation of line l)

Equation of a line l1, 3 1 4

1 2 2x y z t

Equation of a line l2, 3 3 2

2 2 1x y z s

Direction ratio of a line l is given by,

ˆ ˆ ˆ1 2 22 2 1

i j k

= ˆ ˆ ˆ2 3 2i j k

Equation of a line l is 2 3 2x y z

Point of intersection of l and l1,

2 3 t ...(1)

3 2 1t ...(2)Put the value of t, 3 2( 2 3) 1

3 4 6 1

7 7

1 Point of intersection is (2, –3, 2)

So, 2 2 2(3 2 2) (3 2 3) (2 2) 17s s s

2 2 24 4 1 36 24 4 17s s s s s

29 28 20 0s s

102, 9

s

i.e., intersection points are (–1, –1, 0) and 7 7 8, ,9 9 9

53. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted intoan open rectangular box by folding after removing squares of equal area from all four corners. If the totalarea of removed squares is 100, the resulting box has maximum volume. Then the lengths of the sides of therectangular sheet are(A) 24 (B) 32(C) 45 (D) 60

Answer (A, C)Hint :

(8 2 )(15 2 )V x x x x 15

8 = 3 2 24 46 120x x x

2 212 92 120 0dV x xdx

, at x = 5

260 230 150 0

26 23 15 0

(25)

(6 5)( 3) 0

For = 3 lengths of sides are 45, 24

54. Let

( 1)4 2 2

1( 1)

k kn

nk

S k . Then Sn can take value(s)

(A) 1056 (B) 1088

(C) 1120 (D) 1332Answer (A, D)

Hint :

( 1)4 2 2

1( 1)

k kn

nk

S k

2 2 2 2 2 2 2 2 2 2–1 – 2 3 4 5 6 ........ (4 3) (4 2) (4 1) (4 )nS n n n n

Sn = 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2(3 1 ) (4 2 ) (7 5 ) (8 6 ) (11 9 ) (12 10 )...... (4 1) (4 3) (4 ) (4 2)n n n n

Sn = 2(1 3) 2(4 2) 2(7 5) 2(8 6) ...... 2(4 1 4 3) 2(4 4 2)n n n n

Sn = 2 4 (4 1)2[1 2 3 ..... 4 ]

2n nn

From A, 4n(4n + 1) = 10564n2 + n = 2644n2 + n – 264 = 0n = 8

From B, 4n(4n + 1) = 1088 (Not possible)From C, 4n(4n + 1) = 1120 (Not possible)From D, 4n(4n + 1) = 1332

n = 955. Let f(x) = x sin x, x > 0. Then for all natural numbers n, f (x) vanishes at

(A) A unique point in the interval

1,2

n n

(B) A unique point in the interval

1 , 12

n n

(C) A unique point in the interval (n, n + 1)(D) Two points in the interval (n, n + 1)

Answer (B, C)Hint : f(x) = x sinx

f (x) = sin x + x cos x = 0 – tan x = x

(26)

0 1 32

2 52

12

Clear f (x) has one root in 1 , 12

n n

, also f (x) has one root in (n, n + 1).

SECTION - 3 : (Integer Value Correct Type)This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9(both inclusive).56. A pack contains n cards numbered from 1 to n. Two consecutive numbered cards are removed from the pack

and the sum of the numbers on the remaining cards is 1224. If the smaller of the numbers on the removedcards is k, then k – 20 =

Answer (5)Hint : The smallest value of n for which

( 1) 12242

n n

n(n + 1) > 2448 n > 49For n = 50

( 1)2

n n 1275

So, k + (k + 1) = 1275 – 1224 = 51k = 25k – 20 = 5

57. Consider the set of eight vectors V = ˆ ˆ ˆ : , , { 1,1}ai bj ck a b c . Three non-coplanar vectors can be chosenfrom V in 2p ways. Then p is

Answer (5)Hint : Total 8 vectors are shown in the figure.

Total number of vectors = 8C3 = 56

Number of coplaners = 2 × (6 × 2) = 2456 – 24 = 32 25

(27)

58. A vertical line passing through the point (h, 0) intersects the ellipse 2 2

14 3x y at the points P and Q. Let

the tangents to the ellipse at P and Q meet at the point R. If (h) = area of the triangle PQR,

1 =

1/2 1max ( )

hh and 2 =

1/2 1min ( )

hh , then 1 2

8 85

=

Answer (9)Hint :

2 2

+ = 14 3x yS

Let P and Q be (h, ) and (h, –)

So, R is

4 , 0h

Now, =

1 4 × 2 × –2

hh

R

Q(h, – )

h, ( 0)

P h, ( )

=

2 43 1 – × –4h h

h

= 2 3/24 –32

hh

So, d

dh < 0

i.e. is decreasing

i.e. 1 =

12 = 15 45

8

and 2 = (1) = 92

Now 1 –5 2

8 8 9

59. The coefficients of three consecutive terms of (1 + x)n+5 are in the ratio 5 : 10 : 14. Then n =Answer (6)Hint : Let consecutive terms be tr+2, tr + 1, tr

So, 1 105

r

r

tt

( 5) ( 1) 1 2

1n r

r

n – 3r + 3 = 0 …(i)

also, 2

1

1410

r

r

tt

5n – 12r + 6 = 0 …(ii)

Solving, 6n

(28)

60. Of the three independent events E1, E2 and E3, the probability that only E1 occurs is , only E2 occurs is and only E3 occurs is . Let the probability p that none of events E1, E2 or E3 occurs satisfy the equations( – 2)p = and ( – 3)p = 2. All the given probabilities are assumed to lie in the interval (0, 1).

Then 1

3

Probability of occurrence of Probability of occurrence of

EE

Answer (6)

Hint : 1 2 3( ) ( ) ( )P E P E P E ...(i)

1 2 3( ) ( ) ( )P E P E P E ...(ii)

1 2 3( ) ( ) ( )P E P E P E ...(iii)

1 2 3( ) ( ) ( )P E P E P E ...(iv)

Divide (i) by (iv),

1

1

( ) 2( )

2

P EP E

1

1

( )( ) 2

P EP E

1

1

1 ( )( ) 2P E

P E

1

1 1( ) 2P E

1

1( ) 2P E

12( )P E

…(v)

Now, 2

2 3

2

2 3

– 3 = 2 – 4 = 5– 4

= 5 4 …(vi)

Divide (iii) by (iv)

3

3

( ) ( 2 )( )

2

P EP E

(29)

3

3

( ) 25 4( )

P EP E

3

3

( ) 5 4( ) 2

P EP E

3

3

1 ( ) 5 4( ) 2P E

P E

3

1 6 6 6( )( ) 2 2P E

P(E3) = 2

6( )

1

3

2( )

2( )6( )

P EP E

= 6

: aakash tower, plot no.-4, sec-11, mlu, dwarka, new delhi ...€¦ · (7) when s2 is pressed and...

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