Paper 1 Code 0 Solution Aakash

8/13/2019 Paper 1 Code 0 Solution Aakash

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Answers & Solutionsforforforforfor

JEE (Advanced)-2013

Time : 3 hrs. Max. Marks: 180

INSTRUCTIONS

Question Paper Format

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

consists of three sections.

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

(D) out of which ONLY ONEis correct.

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

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

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

0 to 9 (both inclusive).

Marking Scheme

For each question in Section 1, you will be awarded 2 marksif you darken the bubble corresponding

to the correct answer and zero markif no bubbles are darkened. No negativemarks 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 markif no bubbles are darkened. In all other

cases, minus one (1) markwill be awarded.

For each question in Section 3, you will be awarded 4 marksif 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) markwill be awarded.

DATE : 02/06/2013

PAPER - 1 (Code - 0)

CODE

0

Regd. Off ice : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075

Ph.: 011-47623456 Fax : 011-47623472


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PARTI : 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. The work done on a particle of mass mby a force,

2 2 3/2 2 2 3/2

( ) ( )

x yK i j

x y x y(Kbeing a constant of

appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular

path of radius aabout the origin in the x yplane is

(A)2K

a(B)

Ka

(C)

2

K

a (D) 0

Answer (D)

Hint :

2 2 3/2 2 2 3/2

( ) ( )

xi yjF k

x y x y

.dW F dx

2 2 3/2( )

xdx ydydW k

x y

let x2+y2= r2

xdx+ ydy= r2

3 2

krdr kdW dr

r r

2

1

r

r

kW

r

Now, r1= a, r

2= aW= 0

2. Two rectangular blocks, having identical dimensions, can be arranged either in configuration-I or in

configuration-II as shown in the figure. One of the blocks has thermal conductivity and the other 2. The

temperature difference between the ends along the x-axis is the same in both the configurations. It takes 9 s

to transport a certain amount of heat from the hot end to the cold end in the configuration-I. The time to

transport 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)


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Hint :

1 2

2

T TQ

l lt

A A

[in case (i)]

1 2

2( )

'

Q A AT T

t l l

' 2

9

t

t

t' = 2 s

3. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial

pressures, 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 :

RT

PM

1 1 2

2 1 2

P M

P M

1

2

4 3

3 2

1

2

8

9

4. A particle of mass mis projected from the ground with an initial speed u0at 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 the

composite 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

2

uH

g

Speed of 2ndparticle = u0cos

Final momentum = 0 0 cos cosmu i mu j

Angle = 4


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5. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse

is 30 mW and the speed of light is 3 108ms1. The final momentum of the object is

(A) 0.3 1017kg ms1 (B) 1.0 1017kg ms1

(C) 3.0 1017kg ms1 (D) 9.0 1017kg ms1

Answer (B)

Hint :

3 917 1

8

30 10 100 1010 kg ms

3 10

E P tp

C C

6. In the Young's double slit experiment using a monochromatic light of wavelength , the path difference (in

terms of an integer n) corresponding to any point having half the peak intensity is

(A)

(2 1)2

n (B)

(2 1)4

n

(C)

(2 1)8

n (D)

(2 1)16

n

Answer (B)

Hint :

2max cos

2I I

21 cos

2 2

cos = 0

3 5 7

, , ,2 2 2 2

3 5

, ,4 4 4

x

(2 1)4

x n

7. 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 is2

3times 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 :

3

2

air

med

fc

f

Now, = +8 m,

1

24 m3

m uu

1 1 1

f u

1 1 1 4

8 24 24f

f= 6 m

1

Rf

6 3 m0.5

RR


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8. One end of a horizontal thick copper wire of length 2Land radius 2Ris welded to an end of another horizontal

thin copper wire of length Land 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, FL

AY

21

22

(2 )2

2

L R

LR

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

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

along the direction1 ( 3 )2

i j . The angle of incidence is

(A) 30 (B) 45

(C) 60 (D) 75

Answer (A)

Hint :

3 3.

2 2cos(180 2 )

3 3

2 2

i j i j

i j i j

i 3

2

j i 3

2

180 2

^j^

(1 3)

4cos21

1cos2

2

1

cos22

2 60

= 30

10. 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 divisions

equivalent to 2.45 cm. The 24thdivision of the Vernier scale exactly coincides with one of the main scale

divisions. 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 =2.45

50= 0.049 cm

LC = 1 MSD 1 VSD

= 0.001 cm

Reading = 5.10 + L.C. 24

= 5.10 + 0.024

= 5.124 cm


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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 of

which ONE OR MORE are correct.

11. In the circuit shown in the figure, there are two parallel plate capacitors each of capacitance C. The switch

S1is pressed first to fully charge the capacitor C1and then released. The switch S2is then pressed to chargethe capacitor C

2. After some time, S

2is released and then S

3is pressed. After some time,

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

0

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

0

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

S1

2V0C1

V0

S2 S3

C2

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

0

Answer (B, D)

Hint : When S1is pressed and released

2V0 C1 V0

S2 S3

C2+

+2CV0

When S2is pressed and released

2V0 V0

S2 S3

C2+

CV0

S1

+CV0

When S3is pressed

2V0 C1 V0C2+

CV0 CV0CV0

+

12. A particle of mass Mand positive charge Q, moving with a constant velocity 11

4 msu i , 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= Lfor all values of y. After passing through this region, the particle emerges on the other side after

10 milliseconds with a velocity 1

2 2( 3 )msu i j . The correct statement(s) is (are)

(A) The direction of the magnetic field is zdirection

(B) The direction of the magnetic field is +zdirection

(C) The magnitude of the magnetic field50

3

M

Qunits

(D) The magnitude of the magnetic field is100

3

M

Q

units

Answer (A, C)


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Hint :

Mt

qB

Clearly = 30 =6

C

F

x L=

2j

v

2 3i

4ix= 0

3100 50

6 36 10 10

M M M

B q Qq

B must be in zdirection.

13. 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 m1)x] cos[(628 s1)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

Answer (B, C)

Hint : No. of nodes = 6,

2

k=

2 3.140.1m

62.8

Length =

5

0.25m2

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

f=

20Hz

2 2

v

l k l

14. A solid sphere of radius Rand density is attached to one end of a mass-less spring of force constant k. The

other end of the spring is connected to another 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 correct statement(s)

is (are)

(A) The net elongation of the spring is 343

R g

k

(B) The net elongation of the spring is 383

R 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

3 34 4

(2 )3 3

kx R g R g

R

R 3

2

34

3

R gx

k

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


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15. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities 1 and

2

respectively, touch each other. The net electric field at a distance 2Rfrom the centre of the smaller sphere,

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

1

2can be

(A) 4 (B) 3225

(C)32

25(D) 4

Answer (B, D)

Hint : EP

= 0

31

2

200

4( )3

34 (2 )

RR

R P' P

R 2R

2R

1

2

4

' 0PE

3 31 2

2 20 0

4 4(2 )

3 3

4 (2 ) 4 (5 )

R R

R R

1

2

32

25

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 bob of mass m, suspended by a string of length l1, is given a minimum velocity required to complete a full

circle in the vertical plane. At the highest point, it collides elastically with another bob of mass msuspended

by a string of length l2, which is initially at rest. Both the strings are mass-less and inextensible. If the second

bob, after collision acquires the minimum speed required to complete a full circle in the vertical plane, the

ratio1

2

l

lis

Answer (5)

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

For elastic collision between objects of same mass, velocities are exchanged speed of 2nd1

bob = l

but 1 2

5gl gl

1

2 5

l

l


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17. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to

the particle. If the initial speed (in ms1) of the particle is zero, the speed (in ms1) after 5 s is

Answer (5)

Hint : As power is constant, P t= kE

21

0.5 5 (0.2) 02 v

2.5 = 0.1 v2

v = 5 m/s

18. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping

potential versus frequency plot for Silver to that of Sodium is

Answer (1)

Hint :

hf

Ve e

Slope = .h

e

It is same for both.

19. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103disintegrations per second. Given

that ln2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest integer percentage) that

will decay in the first 80 s after preparation of the sample is

Answer (4)

Hint : 0

tN N e

0

tN eN

ln280

1386

0

N

eN

0.693 80

1386

0

Ne

N

0.04

0

Ne

N

0.04

0

1N

N e

Fraction of nuclei decayed =

0.04

0

11 1

N

N e

= 0.04 = 4%

20. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s1

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 the

axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest

relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s1) of

the 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

408 rad/s

4 1


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PARTII : CHEMISTRY

SECTION - 1 : (Only One Option Correct Type)


21. Consider the following complex ions, P, Q and R.

P = [FeF6]3, Q = [V(H

2O)

6]2+and R = [Fe(H

2O)

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 3

6P [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

22. The arrangement of Xions around A+ion in solid AX is given in the figure (not drawn to scale). If the radius

of 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 Xis 250 pm.

A

X

R0.414

R

AR 0.414 250 103.50 104 pm

23. 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), CuFeS

2(Copper

pyrites) and PbS (Galena)


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24. The standard enthalpies of formation of CO2(g), H

2O(l) and glucose(s) at 25C are 400 kJ/mol, 300 kJ/mol

and 1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25C is

(A) +2900 kJ (B) 2900 kJ

(C) 16.11 kJ (D) +16.11 kJ

Answer (C)

Hint : C6H

12O

6(s) + 6O

2(g) 6CO

2(g) + 6H

2O(l)

H = 6 (400) + 6(300) (1300)

H = 2900 kJ/mol

2900

H 16.11 kJ / gm180

25. 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.

26. Methylene blue, from its aqueous solution, is adsorbed on activated charcoal at 25C. For this process, the

correct statement is

(A) The adsorption requires activation at 25C

(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, multilayer

and does not have energy barrier.

27. 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

Cl

O

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

Cl

O

Cl

Q

Cl

RS

Compounds: CH Cl3 : :

Relative reactivitiestowards S reactionN2

1,00,000 200 79 0.02: : :


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28. In the reaction,

P + Q R + S

The time taken for 75% reaction of P is twice the time taken for 50% reaction of P. The concentration of Q

varies 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

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.

29. Concentrated nitric acid, upon long standing, turns yellow-brown due to the formation of

(A) NO (B) NO2

(C) N2O (D) N

2O

4

Answer (B)

Hint : Conc. HNO3slowly decomposes as

4HNO34NO2+ 2H2O + O2

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

30. 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 CO2on treatment

with aq. NaHCO3solution. Benzoic acid, benzenesulphonic acid and salicylic are more acidic than carbonic

acid and hence will liberate CO2with aq. NaHCO3solution.




31. The initial rate of hydrolysis of methyl acetate (1M) by a weak acid (HA, 1M) is 1/100thof that of a strong

acid (HX, 1M), at 25C. The Kaof HA is

(A) 1 104 (B) 1 105

(C) 1 106 (D) 1 103

Answer (A)


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Hint : Rate with respect to weak acid

R1= K[H+]

WA[ester]

and rate with respect to strong acid

R2= K[H+]

SA[ester]

WA1

2 SA

[H ]R 1

R 100[H ]

SA

WA

[H ] 1H 0.01 M C

100 100

= 0.01

Kafor weak acid = C2

= 1(0.01)2

= 1 104

32. 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.

33. The pair(s) of coordination complexes/ions exhibiting the same kind of isomerism is(are)(A) [Cr(NH

3)5Cl]Cl

2and [Cr(NH

3)4Cl

2]Cl (B) [Co(NH

3)4Cl

2]+and [Pt(NH

3)2(H

2O)Cl]+

(C) [CoBr2Cl

2]2and [PtBr

2Cl

2]2 (D) [Pt(NH

3)3](NO

3)Cl and [Pt(NH

3)3Cl]Br

Answer (B, D)

Hint : The pair of complex ions [Co(NH3)4Cl

2]+and [Pt(NH

3)2(H

2O)Cl]+show geometrical isomerism. The pair of

complexes [Pt(NH3)3(NO

3)]Cl and [Pt(NH

3)3Cl]Br show ionisation isomerism. The other pairs given do not

have same type of isomerism.

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

Cl

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)


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Hint : (i) ClAlCl3

AlCl4 (P)

(AROMATIC)

(ii)

NaH

(Q) + H2

(AROMATIC)

Na

(iii)O O

(NH ) CO4 2 3

100115 CR

(NH ) CO4 2 3

2NH + CO + H O3 2 2

O O

+ NH3O O NH2 O NH2

OH

IMPE

N

H

2H O2

N

H

HO OH

IMPEN

H

O OH

H

(R) AROMATIC

H

(iv) OHCl

OH Cl (S)

(AROMATIC)

35. Benzene and naphthalene form an ideal solution at room temperature. For this process, the true statement(s)is(are)

(A) G is positive (B) Ssystemis 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.


This section contains 5 questions. The answer to each question is a single digit integer,ranging from 0 to 9(both inclusive).

36. 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 73C is "M" times that of the de Broglie wavelength of 'Ne' at 727C 'M' is

Answer (5)

Hint :

He Ne Ne

Ne He He

M V

M V

NeNe

Ne

HeHe

He

3RTM

M

3RTMM


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NeNe

Ne

HeHe

He

TM

M

TM

M

Ne Ne

He He

M T

M T

205 1000

4 200

He Ne5

37. EDTA 4 is ethylenediaminetetraacetate ion. The total number of NCoO bond angles in [Co(EDTA)]1

complex ion is

Answer (8)

Hint :

Co

O C = O

O C = O

O

C CH2 CH2O

CH2

CH2

N

O N

C CH2

OCH2

II

III

IV

I

I

II

So, bond angles are

(1) NI CoOI

(2) NICoOII

(3) NICoOIII

(4) NICoOIV

(5) NIICoOI

(6) NIICoOII

(7) NIICoOIII

(8) NIICoOIV

So, total asked bond angles are 8.

38. The total number of carboxylic acid groups is the product P is

O

O O

O

O

1. H O ,+

3

2. O3

3. 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

HOOC

HOOC

(P)

OH

OH (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.

39. 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 (primary

structures) with NH2group attached to a chiral center is

Answer (4)

Hint : According to question C Terminal must be alanine and N Terminal do have chiral carbon means its

should not be glycine so possible sequence is :

Val Phe Gly Ala

Val Gly Phe Ala

Phe Val Gly Ala

Phe Gly Val Ala

So, answer is (4).

40. The total number of lone-pairs of electrons in melamine is

Answer (6)

Hint : Structure of melamine is

N NH2

N

NH2

H2N

N

So, melamine has six lone pair of electrons.


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PARTIII : MATHEMATICS

SECTION - 1 : (Only One Option Correct Type)


41. 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,c

xa b

Also from (1) & (2)

y = x

Point of intersection lies ony = x

c

ya b

Given,

2 2

1 1 2 2c c

a b a b

2 1 2 2c

a b

2a b c

a b

2 2a b c a b

0a b c

42. The area enclosed by the curves sin cosy x xand cos siny x x over the interval

0,

2is

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

(C) 2( 2 1) (D) 2 2( 2 1)Answer (B)


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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 1

2 2 1 12 2

= 2 2 2 2

= 4 2 2

= 2 2( 2 1)

43. 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> 0

f(x) is decreasing x< 0

f(0) = 1

( )f

( )f

44. The value of23

1

1 1

cot cot 1 2n

n k

k

is

(A)23

25(B)

25

23

(C)23

24(D)

24

23

Answer (B)


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Hint : 23

1

1

cot 1 ( 1)n

k k

=23

1 1

1

tan ( 1) tan ( )n

k k

= tan1(24) tan1(1)

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

=1 24 1cot tan

1 24 1

=

1 23cot tan25

=25

23

45. A curve passes through the point

1,

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

sec , 0

y yx

x x.

Then the equation of the curve is

(A)

1sin log

2

yx

x(B)

cosec log 2

yx

x

(C)

2sec log 2

yx

x(D)

2 1cos log

2

yx

x

Answer (A)

Hint : secdy y y

dx x x

Put y= vx

secdv

v x v vdx

cos dx

v dv x

sin v= ln x+ c

It passes through

1,

6

1

2c

1sin ln2

y xx


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46. Let

1

: ,12

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

thatf(x) < 2f(x) and

11

2f . Then the value of

1

1/2

( )f x dxlies in the interval

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

(C)

1, 1

2

ee (D)

10,

2

e

Answer (D)

Hint : 2dy

ydx

2 22x xdy

e ye

dx 2( ) 0x

dye

dx

ye2xis decreasing function

As, 11

2x

e1> ye2x> y(1)e2

e2x1> y> y(1)e2x2

1 1 12 1 2 2

1/2 1/2 1/2

(1) 0x xe dx ydx y e

1

1/2

10

2

eydx

47. Let

3 2PR i j k and

3 4SQ i j k determine diagonals of a parallelogramPQRSand

2 3PT i j k

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

, andPT PQ PSis

(A) 5 (B) 20

(C) 10 (D) 30

Answer (C)

Hint :

1 3 2PR a b i j k d

S b( )

P

R

Q a( )

2 3 4SQ a b i j k d

2 3a i j k

2b i j k


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Volume of the required parallelopiped =

2 1 3

1 2 1

1 2 3

= |2(6 2) + 1(3 1) 3(2 2)|

= 10 cubic units

OR

Area of parallelogram =

1 2

1| |

2d d

Volume of parallelepiped =

3 1 21

1 3 42

1 2 3

= 10 cubic units.

48. Perpendiculars are drawn from points on the line

2 1

2 1 3

x y z

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

perpendiculars lie on the line

(A)

1 2

5 8 13

x y z(B)

1 2

2 3 5

x y z

(C)

1 2

4 3 7

x y z(D)

1 2

2 7 5

x y z

Answer (D)

Hint :

2 1, 3

2 1 3

x y zx 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

3

2

So, point of intersection of plane is

5 91, ,

2 2C.

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

2 1

1 1 1k .

Any general point on lineABis (k 2, k 1, k).

This will satisfy equation so, (k 2) + (k 1) + k= 3.

k= 2

Therefore, (, , ) (0, 1, 2).

So, equation of line passing throughBCis

1 2

7 51

2 2

x y z

1 2

2 7 5

x y z


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49. Four persons independently solve a certain problem correctly with probabilities1 3 1 1

, , ,2 4 4 8

. Then the

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

(A)

235

256 (B)

21

256

(C)3

256(D)

253

256

Answer (A)

Hint : 1 3 1 1

( ) , ( ) , ( ) , ( )2 4 4 8

P A P B P C P D

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

= 1 ( )P A B C D

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

= 1 1 3 7

12 4 4 8

= 21

1256

=235

256

50. Let complex numbers and

1

lie on circles 2 2 20 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)1

2(B)

1

2

(C)1

7(D)

1

3

Answer (C)

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

( , )x y0 0

r

2

1

lies on 2 2 20 0( ) ( ) 4x x y y r ... (2)

Let be 1+ i

1

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)


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

( )12 4

x yz r

2 2 22

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

From (3) (4), 2 2 2 22

01 (1 ) 1 4z r

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

Now, 22 02 1r z 2

2

0

21

r

z

Substituting in (5),

2 2

1 2(1 4 )

2 21 2 8

2

7 1

1

7




51. A line lpassing through the origin is perpendicular to the lines

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

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

Then, the coordinate(s) of the point(s) on l2at a distance of 17 from the point of intersection of land l1is

(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)

Hint : x y z

a b c(Equation of line l)

Equation of a line l1, 3 1 4

1 2 2

x y zt


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Equation of a line l2,

3 3 2

2 2 1

x y zs

Direction ratio of a line lis given by,

1 2 2

2 2 1

i j k

= 2 3 2i j k

Equation of a line lis2 3 2

x y z

Point of intersection of land 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

10

2,9

s

i.e., intersection points are (1, 1, 0) and

7 7 8, ,

9 9 9

52. Letf(x) = xsin x, x> 0. Then for all natural numbers n,f(x) vanishes at

(A) A unique point in the interval

1

, 2n n

(B) A unique point in the interval

1, 1

2n 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+ xcos x= 0 tan x= x


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54. For 3 3 matrices Mand N, which of the following statement(s) is (are) NOTcorrect?

(A) NTMNis symmetric or skew symmetric, according as Mis symmetric or skew symmetric

(B) MN NMis skew symmetric for all symmetric matrices Mand N

(C) MNis symmetric for all symmetric matrices Mand N

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

Hint :

(NTMN)T= NTMT(NT)T

NTMTN

(A) If Mis skew symmetric, (NTMN)T= NTMN,

Hence skew symmetric.

If Mis 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 Mand N, option (C) is incorrect.

(D) adj(MM) = adj(N) adjM,Option (D) is incorrect.

55. A rectangular sheet of fixed perimeter with sides having their lengths in the ratio 8 : 15 is converted into

an open rectangular box by folding after 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

rectangular 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

(6 5)( 3) 0

For = 3 lengths of sides are 45, 24


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This section contains 5 questions. The answer to each question is a single digit integer,ranging from 0 to 9

(both inclusive).

56. Consider the set of eight vectors V= : , , { 1,1}ai bj ck a b c

. Three non-coplanar vectors can be chosenfrom Vin 2pways. Thenpis

Answer (5)

Hint : Total 8 vectors are shown in the figure.

Total number of vectors = 8C3= 56

Number of coplaners = 2 (6 2) = 24

56 24 = 32 25

57. Of the three independent events E1, E

2and E

3, the probability that only E

1occurs is , only E

2occurs is

and only E3occurs is . Let the probability pthat none of events E

1, E

2or E

3occurs 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 ofProbability 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 E

P E

1

1

( )

( ) 2

P E

P E

1

1

1 ( )

( ) 2

P E

P E

1

1 1( ) 2P E


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59. A pack contains ncards 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 removed

cards is k, then k 20 =

Answer (5)

Hint : The smallest value of nfor which

( 1)1224

2

n n

n(n+ 1) > 2448

n> 49

For n= 50

( 1)

2

n n 1275

So, k+ (k+ 1) = 1275 1224 = 51

k= 25

k 20 = 5

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

14 3

x yat the pointsPand 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

88

5=

Answer (9)

Hint :

2 2

+ = 14 3

x yS

LetPand Qbe (h, ) and (h, )

So, Ris

4, 0

h

Now, =

1 4 2

2h

h

R

Q(h, )

h,( 0)

P h,( )

=

2 43 1

4

hh

h

= 2

3/ 2

4 3

2

h

h

So, ddh

< 0


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i.e. is decreasing

i.e. 1=

1

2=

15 45

8

and 2= (1) =

9

2

Now 15

28

8 9

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