MATHS 4 ALL

MATHEMATICS TODAY | APRIL ‘17 7

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Vol. XXXV No. 4 April 2017Corporate Office:Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-6601200e-mail : [email protected] website : www.mtg.inRegd. Office:406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029.Managing Editor : Mahabir SinghEditor : Anil Ahlawat

CONTENTS

1031

76

71

73

80 82

8

88

41

8 Maths Musing Problem Set - 172

10 Practice Paper - JEE Advanced

21 Logical Reasoning

24 Practice Paper 2017


41 Full Length Practice Paper - BITSAT


71 Math Archives

73 Olympiad Corner

76 Quantitative Aptitude

84 Challenging Problems

88 You Ask We Answer

89 Maths Musing Solutions

80 MPP

82 MPPClas

s XI

/XII

Com

peti

tion

Edg

e

MATHEMATICS TODAY | APRIL ‘178

JEE MAIN1. �e lines x = ay + b, z = cy + d and x = a′y + b′,

z = c′y + d′ are perpendicular. �en

(a) aa

cc′ ′

+ = −1 (b) aa

cc′

+′

= 1(c) aa′ + cc′ = –1 (d) aa′ + cc′ = 1

2. In an ellipse the distance between the foci is 8 and the distance between the directrices is 10. �e angle a that the latusrectum subtends at the centre is(a) cos−1 17

19 (b) cos−1 19

21(c) cos−1 17

23 (d) tan−1 17

193. If 2a + 3b + 6c = 0, a, b, c ∈ R, then the equation

ax2 + bx + c = 0 has a root in(a) (0, 1) (b) (2, 3)(c) (4, 5) (d) none of these

4. If 3 distinct numbers are chosen randomly from the �rst 100 natural numbers, then the probability that all three of them are divisible by both 2 and 3, is(a) 4

55 (b) 4

35 (c) 4

33 (d)

41155

5. If x dydx

y y y x3 3 2 2 2= + − and y = 1 when x = 1,

then y = 2 when x =

(a) 25

(b) 45

(c) 65

(d) 85

JEE ADVANCED

6. Let a > 0 the equation a ax x⋅ + ⋅ − =−2 2 2 02 2sin sin

has a root if a ∈

(a) (1, 2) (b) 12

1,

(c)

35

1,

(d) 45

1,

COMPREHENSIONLet ABC be a triangle with inradius r and circumradius R. Its incircle touches the sides BC, CA, AB at A1, B1, C1 respectively. �e incircle of triangle A1B1C1 touches its sides at A2, B2, C2 respectively and so on.7. In triangle A4B4C4, the value of A4 is

(a) 3

8π + A

(b) 58

π − A (c) 516

π − A (d) 516

π + A

Set 172

8. B CBC1 1 =

(a) sin B2

sin C2

(b) cos B2

cos C2

(c) cos B C−

2

– sin A2

(d) sin A2

+ cos B C−

2

INTEGER MATCH9. �e sum of three numbers in G.P. is 42. If each of

the extremes be multiplied by 4 and the mean by 5, then the products are in A.P. �e least of the original numbers is

MATRIX MATCH10. Match the following.

List-I List-II

P. I f y = cos sin13

1−

x , t h e n

(1 – x2) y2 – xy1 + ly = 0 where l =1. 7

4

Q. If the function

y = ax bx x

+− −( )( )4 1

has an extreme

value at (2, 1), then its maximum value is

2. 52

R. �e curve y = ax3 + bx2 + cx + 5 touches the x-axis at (–2, 0) and cuts the y-axis where its gradient is 3. �en a + b + c =

3.19

S. �e product of the abscissae of the points on the curve y = x3 – 7x2 + 6x + 5 at which tangents pass through the origin is

4. − 19

P Q R S(a) 3 3 1 2(b) 1 2 3 4(c) 4 3 2 1(d) 3 3 2 1

Maths Musing was started in January 2003 issue of Mathematics Today. The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material.

During the last 10 years there have been several changes in JEE pattern. To suit these changes Maths Musing also adopted the new pattern by changing the style of problems. Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our readers. It is heartening that we receive solutions of Maths Musing problems from all over India.Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them. We do hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced.

See Solution Set of Maths Musing 171 on page no. 89


SECTION-1

SINGLE CORRECT ANSWER TYPE

1. A committee of 6 is chosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. �e number of ways this can be done, if two particular women refuse to serve on the same committee, is(a) 8000 (b) 7800 (c) 7600 (d) 7200

2. If lnsin x cos x + lncos x sin x = 2, then x =

(a) n4

(b) ( )n 14

(c) ( )2 14

n (d) ( )4 14

n

3. �e slope of the straight line which is both tangent and normal to the curve 4x3 = 27y2 is

(a) ± 1 (b) 12

(c) 12

(d) 2

4. If the sum of the roots of the equation ax2 + bx + c = 0 is equal to the sum of the reciprocals

of their squares, then ac

ba

cb

, , are in

(a) A.P. (b) G.P. (c) H.P. (d) none of these

5. A person goes to o�ce either by car, scooter, bus

or train the probabilities of which being 17

37

27

17

, , ,

respectively. �e probability that he reaches o�ce late, if he takes car, scooter, bus or train is 29

19

49

19

, , , respectively. If he reached o�ce in time,

the probability that he travelled by car is

(a) 15

(b) 19

(c) 2

11 (d)

17

SECTION-2

MORE THAN ONE CORRECT ANSWER TYPE

6. Let f(x) = ln |x| and g(x) = sin x. If A is the range of f(g(x)) and B is the range of g(f(x)), then(a) A B = (–, 1] (b) A B = (–, )(c) A B = [–1, 0] (d) A B = [0, 1]

7. Let y = f (x) be a curve in the �rst quadrant such that the triangle formed by the co-ordinate axes and the tangent at any point on the curve has area 2. If y(1) = 1, then y(2) =

(a) 0 (b) 1 (c) 2 (d) 12

8. If

1 4 4

1 4 4

1 4 4

0

2 2

2 2

2 2

sin cos sin

sin cos sin

sin cos sin

then =

(a) 724 (b) 5

24 (c)

1124

(d) 24

9. Let P(x1, y1) and Q(x2, y2), y1 < 0, y2 < 0 be the ends of the latusrectum of the ellipse x2 + 4y2 = 4. �e equations of parabolas with latusrectum PQ are(a) x y2 2 3 3 3

(b) x y2 2 3 3 3

(c) x y2 2 3 3 3

(d) x y2 2 3 3 3 10. �e equation of the plane containing the line

2x – y + z – 3 = 0, 3x + y + z = 5 and at a distance of 16

from the point (2, 1, –1) is

(a) x + y + z – 3 = 0 (b) 2x – y – z – 3 = 0(c) 2x – y + z – 3 = 0 (d) 62x + 29y + 19z – 105 = 0

Exam on

21st May

PAPER-I


11. Tangents are drawn from a point P on the circle C : x2 + y2 = a2 to the circle C1 : x2 + y2 = b2. If these tangents cut the circle C at Q and R and if QR is a tangent to the circle C1, then the area of triangle PQR is

(a) 3 3 2b (b) 3 3

42a

(c) 3 3

2ab (d) 2 3 ab

12. �e internal bisector of A of triangle ABC meets side BC at D. A line drawn through D perpendicular to AD meets the side AC at E and the side AB at F. If a, b, c are the sides of triangle ABC, then(a) AE is H.M. of b and c

(b) AD bcb c

A

22

cos (c) EF bcb c

A

42

sin

(d) triangle AEF is isosceles.

13. Let a, b, c, d, e be �ve numbers such that a, b, c are in A.P., b, c, d are in G.P. and c, d, e are in H.P. If a = 2 and e = 18, then b = (a) 2 (b) – 2 (c) 4 (d) – 4

SECTION-3

INTEGER ANSWER TYPE

14. If n ∈ N, then the highest integer m such that 2m divides 32n + 2 – 8n – 9 is

15. If lim ( ) , then lim ( )/

x

x

xx f x

xe f x

x

0

13

0 21

16. Let px4 + qx3 + rx2 + sx + t = x x x x

x x xx x x

2 3 1 31 2 43 4 3

be an identity, where p, q, r, s, t are constants. �en t =

17. Let n ∈ N, n 5. If In = e x dxx n( ) 10

1 = 16 – 6e,

then n =

18. If is complex cube root of unity and a, b, c are

such that 1 1 1 2 2

a b c

and 1 1 1 22 2 2a b c

,

then 11

11

11a b c

SECTION-1

SINGLE CORRECT ANSWER TYPE

1. A plane moves such that the volume of the tetrahedron formed with coordinate planes is a constant k. �e locus of the foot of perpendicular from the origin O on it is (x2 + y2 + z2)3 =(a) 12 kxyz (b) 24 kxyz (c) 48 kxyz (d) 6 kxyz

2. An open box with a square base is to be made from 400 m2 of lumber. When its volume is maximum, the ratio of an edge of the base to the height is

(a) 1 (b) 12

(c) 2 (d) 2

3. S rnr

n

cos2

1

1

(a) n2

(b) n 1

2 (c)

n 22

(d) n 1

2

4. Let S be the sum, P be the product and R be the sum of reciprocals of n terms of a G.P. �en(a) R = S P1/n (b) R = S P2/n (c) S = R P1/n (d) S = R P2/n

PAPER -II

5. Let f (x) be a function such that f ′(x) = f (x), f (0) = 1 and g(x) be a function such that f (x) + g(x) = x2.

�en f x g x dx( ) ( )01

(a) e e 2

252

(b) e e 2

232

(c) e e 2

232

(d) e e 2

252

6. If the system of equations x + ky + 3z = 0, 3x + ky – 2z = 0, 2x + 3y – 4z = 0 has non trivial solution,

then xyz2

(a) 56

(b) 56

(c) 65

(d) 65

SECTION-2

MORE THAN ONE CORRECT ANSWER TYPE

7. If x + |y| = 2y, then y as a function of x is (a) de�ned for all x (b) continuous at x = 0(c) di�erentiable for all x

(d) dydx

x 13

0for


8. Let

a b c, , be unit vectors, equally inclined to each

other at angle , where 3 2 . If

a b c, , are the

position vectors of the vertices of a triangle and g is

the position vector of its centroid, then

(a) | |g 13

(b) | |g 1

(c) | |g 23

(d) | |g 1

9. Let z1 and z2 be two distinct complex numbers and let z = (1 – t) z1 + tz2, for some real number t with 0 < t < 1, then(a) |z – z1| + |z – z2| = |z1 – z2| (b) arg (z – z1) = arg |z – z2| (c) arg (z – z1) = arg (z2 – z1)

(d) z z z zz z z z

1 1

2 1 2 10

10. If a hyperbola passes through a focus of the ellipse x y2 2

25 161 and its transverse and conjugate axes

coincide with major and minor axes of the ellipse, and the product of their eccentricities is 1, then

(a) the hyperbola is x y2 2

9 161

(b) the hyperbola is x y2 2

9 251

(c) a focus of hyperbola is (5, 0)

(d) a focus of hyperbola is 5 3 0, 11. Let S n

n kn kn

k

n

2 2

1

and T nn kn k

nk

n

2 20

1

. �en

(a) Sn 3 3

(b) Sn 3 3

(c) Tn 3 3

(d) Tn 3 3

12. �e area of the triangle formed by the tangent to the curve y = x2 + bx – b at the point (1, 1) and the coordinate axes is 2. �en b = (a) 3 (b) –3 (c) 1 2 2 (d) 1 2 2

13. If the equations x + y = 1, (c + 2) x + (c + 4) y = 6, (c + 2)2 x + (c + 4)2 y = 36 are consistent, then c =(a) 1 (b) 2 (c) 3 (d) 4

14. If the solution of yd y

dx

dydx

x2

2

2

, y(0) = y(1) =1

is given by y2 = f(x) then (a) f(x) is monotonically increasing x ∈ (1, ) (b) f(x) = 0 has only one root(c) f(x) is neither even nor odd (d) f(x) has 3 real roots

SECTION-3

COMPREHENSION TYPE

Paragraph for Q. No. 15 and 16

For x ∈

04

, ,

let S x C xnr

nr

nr

nr

sin(sin ), cos(cos )1

21 3 2

1

21 3 1

and T xnr

r

n

tan(tan ),1 3

1

2

where n ∈ N and n 315. �e correct order of Sn, Cn and Tn is given by

(a) Sn > Tn > Cn (b) Sn < Cn < Tn

(c) Sn < Tn < Cn (d) Sn > Cn > Tn

16. �e value of ‘x’ for which Sn = Cn + Tn, is

(a) sin 5

(b) 2

5sin

(c) 210

sin

(d) sin

10

Paragraph for Q. No. 17 and 18The normal at any point (x1, y1) of curve is line perpendicular to tangent at (x1, y1). In case of parabola y2 = 4ax the equation of normal is y = mx – 2am – am2 (m is slope of normal). In case of rectangular hyperbola xy = c2 the equation of normal

at ct ct

,

is xt3 – yt – ct4 + c = 0 and the shortest

distance between any two curve always exist along the common normal.

17. If normal at (5, 3) of hyperbola xy – y – 2x – 2 = 0

intersects the curve again at (a, – 29), then a is

(a) 10 (b) 20 (c) 30 (d) 40


18. If the shortest distance between 2y2 – 2x + 1 = 0 and 2x2 – 2y + 1 = 0 is d then the number of solutions of sin ( [ , ])a a ∈ 2 2 2d is (a) 3 (b) 4 (c) 5 (d) none of these

SOLUTIONS

PAPER - I

1. (b) : Let the men be M1, M2, ..., M10 and women be W1, W2, ..., W7. Let W1 and W2 do not want to be on the same committee.�e six member committee can contain 4 men and 2 women or 3 men and 3 women.�e number of ways of forming 4M, 2W committee is

104

52

251

4200

. ...(1)

52

is the number of ways without W1 and W2

51

is the number of ways with W1, and without W2,

or with W2 and without W1. �e number of ways of forming 3M, 3W committee is

103

53

252

3600

. ...(2)

53

is the number of ways without W1 and W2. 52

is

the number of ways with W1 or W2 but not both. �e desired number is 4200 + 3600 = 7800.

2. (d) : ln cosln sin

ln sinln cos

xx

xx

2

tt

t 1 2 1

ln sinln cos

ln sin ln cosxx

x x 1 0

ln tan x = 0 tan tanx 14

x n n 4

4 14

( ) .

3. (d) : x = 3t2, y = 2t3, dydx

t

�e tangent at t, y – 2t3 = t(x – 3t2), tx – y = t3…(1)

�e normal at t1, t1y + x = 2 314

12t t ...(2)

(1), (2) are identical, t

tt

t t11

2 31

3

14

12

–t3 = 2t13 + 3t1, t1 =

1t

.

Eliminating t1, we get t6 = 2 + 3t2 t2 = 2, t 2 .

�e lines are y x 2 2( ) .

4. (c) : a a a a

ba

ca

, , 1 12 2

a a a

a aa

2 2

2

2

22

( )( )

( ) (a + ) (a)2 = (a + )2 – 2a

b

aca

ba

ca

b caa

.2

2

2

2

2

22 2

– bc2 = ab2 – 2ca2 ab2 + bc2 = 2ca2 ab2, ca2, bc2 are in A.P.

Dividing by abc, we get bc

ab

ca

, , are in A.P. ca

ab

bc

, ,

are in A.P., ac

ba

cb

, , are in H.P.

5. (d) : Let C, S, B, T be the events of the person going by car, scooter, bus, train respectively.

P C P S P B P T( ) , ( ) , ( ) , ( ) . 17

37

27

17

Let L be the event that the person reaching the o�ce in time

P L C P L S P L B P L T| | | | 79

89

59

89

, , ,

P C LP C P L C

P L|

|( ) ( )( )

.

. . . .

17

79

17

79

37

89

27

59

17

89

7

4917

6. (a, c) : �e range of |sin x| = [0, 1]. A = range of ln |sin x| = (– , 0]. �e range of ln |x| = (– , ). B = range of sin ln |x| = [–1, 1] A B = (– , 0] [–1, 1] = (– , 1], A B = (– , 0) [–1, 1] = [–1, 0].7. (a, d) : �e tangent at P(x, y) is Y – y –(X – x)y1 = 0

It meets x-axis at A x yy

1

0, and y-axis at B(0, y – xy1).


OAB = 2 OA OB = 4

x yp

y xp p dydx

y xp p

( ) , , ( )4 42

y xp p 2 y xp p 2

�e general solution is y cx c 2 ...(1) x = 1, y = 1 c = – 1, y = 2 – x, y(2) = 0

Di�erentiating (1) w.r.t c, 0 1

xc

...(2)

Eliminating c from (1), using (2)

y

x x xy 1 2 1 2 1

2, ( )

8. (a, c) : Subtracting R3 from R1 and R2,

1 0 10 1 1

1 4 4

02 2

sin cos sin

sin2 + cos2+ 1 + 4 sin 4 = 0

sin ,4 1

2724

1124

9. (b, c) : x y a b2 2

4 11 2 1 , ,

�e ends of latusrectum PQ are

a b b

a2 2

2,

P Q

3 12

3 12

, , ,

�e midpoint of PQ is M 0 12

,

which is the focus of the parabola.

�e vertex of the parabola is V 0 12

32

, .

�e parabolas are x y2 2 3 12

32

i e x y x y. ., and .2 22 3 3 3 2 3 3 3

10. (c, d) : �e plane is 2x – y + z – 3 + l (3x + y + z – 5) = 0 ...(i)

or (2 + 3l)x + (l – 1)y + (l + 1)z – (5l + 3) = 0

Its distance from (2, 1, –1) is 16

| |

( ) ( ) ( )

4 6 1 1 5 3

2 3 1 1

162 2 2

l l l l

l l l

6(l – 1)2 = 11l2 + 12l + 6 (5l + 24)l = 0l = 0, (i) 2x – y + z – 3 = 0

l 245

, (i)

5(2x – y + z – 3) – 24 (3x + y + z – 5) = 0 62x + 29y + 19z – 105 = 0.

11. (a, b, c) : All the sides of PQR touch C1. C1 is incircle and C is the circumcircle with same centre.�e triangle is equilateral.R = 2r a = 2b

PQR PR 34

2

O

P

Q R

C

b C1

But 3 32

b PR

PQR b b 34

4 3 3 32 2

32

3 3 34

2ab a since a = 2b.

12. (a, b, c, d) : ABC = ABD + ADC12

12 2

12 2

bc A c AD A b AD Asin sin sin

bc A b c AD 22

cos

AD bcb c

A22

cos ...(1)

AE A AD AE bcb c

cos2

2

AE is H.M. of b and c ...(2)

EF ED AD A bcb c

A

2 22

42

tan sin , using (1)


AD EF . AD is angle bisector AEF is isosceles, from (2) 13. (b, c) : a, b, c are in A.P. a + c = 2b ...(1) b, c, d are in G.P. bd = c2 ...(2)

c, d, e are in H.P.

d cec e2

...(3)

Since a = 2, e = 18, b c dc

c

2

236

18, .

Now (2)

18 2

18362 2c c

cc c( )

c b6 2 62

2 62

4 2, , .

14. (6) : 32n + 2 – 8n – 9 = 9n + 1 – 8(n + 1) – 1 = (8 + 1)n + 1 – 8(n + 1) – 1

8 8 81

11

11 2n n

nnn ... , which is divisible

by 82 = 26.

15. (2) : e x f xx

xx3

0

exp lim( )

3 1 2

0 2 0 2

lim ( ) lim ( )x x

f xx

f xx

16. (0) : x = 0 in the identity gives

t

0 1 31 0 43 4 0

12 12 0.

17. (3) : I x e dxnn x ( )1

0

1

( ) ( )x e n x e dxn x n x1 1

0

1 1

0

1

In = (– 1)n + 1 – nIn –1 I e dx ex0

0

11

I1 = 1 – I0 = 1 – (e – 1) = 2 – eI2 = – 1 – 2I1 = –1 – 2(2 – e) = 2e – 5I3 = 1 – 3I2 = 1 – 3(2e – 5) = 16 – 6e n = 3.

18. (2) : , 2 are the roots of the equation 1 1 1 2

a x b x c x x

...(i)

x (b + x) (c + x) = 2 (a + x) (b + x) (c + x)x3 – (ab + bc + ca) x – 2abc = 0coe�. x2 = 0 sum of the roots = 0Also, 1 + + 2 = 0

�e third root is 1

(i)

11

11

11

21

2a b c

.

PAPER-II

1. (d) : Let the plane be xa

yb

zc

1

Volume k abc abc k6

6 ....(i)

�e foot of perpendicular from O on it is x

ay

bz

ca b c

1 1 11

1 1 12 2 2

/ / /, say

a

x2 + y2 + z2 = a ax = by = cz = x2 + y2 + z2 ...(ii)

From (i), (ii)

62 2 2 3

kx y z

xyz( )

or (x2 + y2 + z2)3 = 6 kxyz.2. (c) : Let x be the length of an edge of the base and y be the height. Given, x2 + 4xy = 400 ...(i)

Volume, ( )V x y V x x dVdx

2 24

400 0

14

4004

2 02( ) ( )x x x

x x2 2

4100

20

x y xy

203

103

2,

3. (c) : S rnr

n

1

21 2

1

1cos

1

21 2

1

1n r

nr

ncos

T r

nr

n

cos 2

1

1

2 2 2

1

1sin sin cos

nT

nrnr

n

sin( ) sin2 1 2 1

1

1r

nr

nr

n


sin( ) sin sin2 1 2n

n n n

T S n n1 12

1 1 22

, ( ) .

4. (d) : Let a, ar, ar2,…… be the G.P. S a rr

n

( )11

�e reciprocals are in G. P. i.e. 1 1 1

2a ar ar, , , …..

Ra r

ra

rr r

n n

n

1 1 1

1 11 1

11

1

SR

a rn2 1 ....(1)

P = a · ar ar2 … arn – 1 = anr(1 + 2 + ...+ (n – 1)) = an r(n – 1)n/2

P2 = a2n rn(n – 1) = (a2rn – 1)n

SR

S R Pn

n2/ (by (1))

5. (c) : f ′(x) = f (x) f (x) = Aex f (0) = 1 f (x) = ex, g(x) = x2 – ex

I e x e dx x e e dxx x x x 0

1 2 2 201

( ) ( )

( )x x e ex

x2

2

0

1

2 22

e e e e2

212 2

32

2 2.

6. (b) : 1 33 22 3 4

0 332

kk k

By cross multiplication rule using the last 2 equations, x

ky

k6 4 83

9 2

x y z xyz15 2 6

3036

562 .

7. (a, b, d) : For y 0, x + y = 2y y = x

For y < 0 , x – y = 2y y = x3 f(x) is de�ned for all x.

f(x) is continuous for all x. f(x) is di�erentiable for all

x except x = 0.

f x x f x x′ ′ ( ) , , ( ) ,1 0 1

30

f x x f x x′ ′ ( ) , , ( ) ,1 0 1

30

8. (a, c) :

a b b c c a cos

ga b c

g a b c3

13

2, | | | |

13

1 1 1 6 13

1 2cos cos ....(i)

3 2

0 12

cos ,

0 2 cos 1, 1 1 + 2 cos 2

∈

| | , by (i).g 1

323

9. (a, c, d) : C divides AB in the ratio t : 1 – t

A C B

zt 1 – tz1

z2

|z – z1| = t |z1 – z2| |z – z2| = (1 – t) |z1– z2|

Adding, |z – z1| + |z – z2| = |z1 – z2|AC and AB lie along the same line. arg (z – z1) = arg (z2 – z1) = , say

z – z1 = t(z2 – z1) = t|z2 – z1| ei = AB t ei

z z z zz z z z

AB t e ABt e

AB e AB e

i i

i i

1 1

2 1 2 1

A B t

e e

e e

i i

i i2 0

.

10. (a, c) : �e foci of the ellipse are

a b2 2 0 3 0, ( , )

�e eccentricity of the ellipse is

1 1 1625

35

2 b

a

�e eccentricity of hyperbola is 53

Let the hyperbola be xp

yq

2

2

2

2 1

It passes through (± 3, 0) p2 = 9


q p e2 2 2 1 9 259

1 16

( )

�e hyperbola is x y2 2

9 161

Foci are

p q2 2 0 5 0, ( , ).

11. (a, d) : S Sn k

nkn

n n n n k

n

lim lim 1 1

12

1

dxx x

dx

x1 1

234

201

201

2

32 1

323 3 6 3 3

1

0

1

tan x

f (x) decreases in (0, 1)

1 1

0

1

01

1nf r

nf x dx

nf r

nr

n

r

n( )

T dxx x

n1 3 320

1 .

12. (b, c, d) : �e tangent is y – 1 = (2 + b) (x – 1)

Its intercepts on the x and y-axis are 12

bb

and –(1 + b)12

12

1 2

bb

b b2 + 2b + 1 = ± 4 (2 + b)

(b + 3)2 = 0, b2 – 2b + 1 = 8 b 3 1 2 2,13. (b, d) : x + y = 1 ...(1)(c + 2) x + (c + 4) y = 6 ...(2)(c + 2)2 x + (c + 4)2 y = 36 ...(3)(2) – (c + 2) (1) (2), (3) – (c + 2)2 (1) (3) givesx + y = 1 ...(1)2y = 4 – c ...(2)4(c + 3) y = (c + 8) (4 – c) ...(3)(3) – 2 (c + 3) (2) (3) gives0 = [(c + 8) – 2 (c + 3)] (4 – c) (4 – c) (2 – c) = 0 c = 2, 4.

14. (a, b, c) : Given ddx

ydydx

x ydydx

x c

2

2

d

dxy x c

2 2

2 2

y x x y y23

30 1 1 1a given ( ) , ( )

a1 13

,

′ y f x x x f x x x23 23

33 1

30 1( ) , ( ) ( ) for

f(x) is monotonically increasing ∈ x ( , )1

f f13

13

0

f(x) = 0 has only one real root.15. (d) : Clearly Sn > Cn > Tn as ‘x’ is a proper fraction. x > x2 > x3 and so on.16. (c) : We have Sn = Cn + Tn

x xx

n(( ) )( )

3 2

31

1

x x

xx x

x

n n2

3 2

33

3 2

31

11

1(( ) )

( )(( ) )

( )

But x 1, as x ∈

04

, , so, we get x = x2 + x3

x2 + x – 1 = 0 (x 0)

∈

x 5 12

210

04

sin ,

17. (b) : Equation of hyperbola (x – 1)(y – 2) = 4 XY = 4

If normal at ct ct

,

intersect curve again at ct c

t′

′

,

then ′ tt13

2t = 4 t = 2

( , ) ,′ ′

X Y 14

16

( , ) , ,a a 29 3

414 3

415

so a 15 4

320 .

18. (a) : 2 1 12

1 12

ydydx

dydx y

y

So, d 116

116

12 2

So |sina| = 1

So number of solutions is 3.

y

xO


Direction (1 - 2) : If 'P @ Q' means 'P is not greater than Q', 'P % Q' means 'P is not smaller than Q', 'P Q' means 'P is neither smaller than nor equal to Q', 'P Q' means 'P is neither greater than nor equal to Q and 'P $ Q' means 'P is neither greater than nor smaller than Q'.In the following questions four statements showing relationship have been given, which are followed by four conclusions I, II, III and IV. Assuming that the given statements are true, find out which conclusion(s) is/are definitely true.

1. Statements : B % K, K $ T, T F, H F Conclusions : I. B $ T II. T B III. H K IV. F B (a) Only either I or II is true (b) Only III is true (c) Only III and IV are true (d) Only either I or II and III and IV are true2. Statements : W B, B @ F, F R, R $ M Conclusions : I. W F II. M B III. R B IV. M W (a) Only I and IV are true (b) Only II and III are true (c) Only I and III are true (d) Only II and IV are trueDirection (3 - 4) : B, M, K, H, T, R, D, W and A are sitting around a circle facing at the centre. R is third to the right of B. H is second to the right of A who is second to the right of R. K is third to the right of T, who is not an immediate neighbour of H. D is second to the left of T. M is fourth to the right of W. Now, answer the questions.

3. Who is third to the le� of M? (a) R (b) W (c) K (d) T4. In which of the following combinations is the third

person sitting in between the �rst and the second persons?

(a) WTR (b) BDT (c) MHD (d) WKR5. In the given question three statements followed by

three conclusions numbered I, II and III. You have to take the given statements to be true even if they seem to be at variance with commonly known of the given conclusions logically follows from the given statement, disregarding commonly known facts.

Statements : Some bags are plates. Some plates are chairs. All chairs are tables.

Conclusions : I. Some tables are plates. II. Some chairs are bags. III. No chair is bag. (a) Only I follows (b) Only either II or III follows (c) Only I and either II or III follow (d) None of these6. In the given question statement followed by three

assumptions numbered I, II and III. An assumption is something supposed or taken for granted. You have to consider the statement and the assumptions and decide which of the assumptions is implicit in the statement. �en decide which of the following options hold.

Statement : "Our school provides all facilities like school bus service, computer training, sports facilities. It also gives opportunity to participate in various extra-curricular activities apart from studies." An advertisement by a public school.


Assumptions : I. Nowadays extra-curricular activities assume more importance than studies.

II. Many parents would like to send their children to the school as it provides all the facilities.

III. Overall care of the child has become the need of the time as many women are working.

(a) Only I is implicit (b) Only II is implicit (c) Only I and II are implicit (d) All I, II and III are implicit

7. Of the �ve villages P, Q, R, S and T situated close to each other, P is to west of Q, R is to the south of P, T is to the north of Q, and S is to the east of T. �en R is in which direction with respect to S?

(a) North-West (b) South-East (c) South-West (d) Data Inadequate

8. A group of given digits is followed by four combinations of letters and symbols.Digits are to be coded as per the scheme and conditions given below. You have to �nd out which of the four combinations correctly represents the group of digits. The serial number of that combination is you answer.Digit : 5 1 4 8 9 3 6 2 7 0Letter/Symbol : Q T % # E F $ L W @Conditions:

(i) If the �rst digit is odd and the last digit is even, their codes are to be swapped.

(ii) If the �rst as well as the last digit is even, both are to be coded by the code for �rst digit.

(iii) If the �rst digit is even and the last digit is odd, both are to be coded by the code for odd digit.

384695 (a) F#%$EF (b) F#%$EQ (c) Q#%$EQ (d) None of these9. If it is possible to make a meaningful word from

the second, fourth, seventh and tenth letters of the word UNDENOMINATIONAL, using each letter only once, then the third letter of the word would be your answer. If more than one such word can be formed your answer would be 'X' whereas if no such word can be formed, your answer would be 'Z'.

(a) X (b) Z (c) M (d) N

10. Pointing to a man in the photograph a woman says, "He is the father of my daughter-in-law's brother-in-law". How is the man relate to the woman?

(a) husband (b) brother (c) brother-in-law (d) father11. A word and number arrangement machine

when given an input line of words and numbers rearranges them following a particular rule in each step. �e following is an illustration of input and steps of rearrangement.Input : sky forward 17 over 95 23 come 40Step I: come sky forward 17 over 95 23 40Step II: come 95 sky forward 17 over 23 40Step III: come 95 forward sky 17 over 23 40Step IV: come 95 forward 40 sky 17 over 23Step V: come 95 forward 40 over sky 17 23Step VI: come 95 forward 40 over 23 skyStep VI is the last step of the rearrangement of the above input.As per the rules followed in the above steps. Which of the following will be step II for the given input?

(a) against 85 hire machine for 19 21 46 (b) against 85 machine 19 hire for 21 46 (c) against 85 machine hire for 19 21 46 (d) Cannot be determined12. In the given question, two rows of numbers are

given. �e resultant number in each row is to be worked out separately based on the following rules and the question below the rows of numbers is to be answered. �e operations of numbers progress from le� to right. Rules:

(i) If an odd number is followed by another composite odd number, they are to be multiplied.

(ii) If an even number number is followed by an odd number, they are to be added.

(iii) If an even number is followed by a number which is a perfect square, the even number to be subtracted from the perfect square.

(iv) If an odd number is followed by an even number, the second one is to be subtracted from the �rst one.

(v) If an odd number is followed by a prime odd number, the �rst number is to be divided by the second number.

27 18 3 21 x 9


If x is the resultant of the �rst row, what will be the resultant of the second row?

(a) 63 (b) 36 (c) 3 (d) 16

13. Find the missing number, if the given matrix follows a certain rule row-wise or column-wise.

(a) 3 (b) 4 (c) 5 (d) 6

7 4 58 7 63 3 ?

29 19 31

14. Out of the car manufacturing companies A, B, C, D and E, the production of company B is more than that of company A but not more than that of company E. Production of company C is more than the production of company B but not as much as the production of company D. Considering the information to be true, which of the following is de�nitely true?

(a) Production of company D is highest of all the �ve companies.

(b) Company C produces more number of cars than Company E.

(c) �e numbers of cars manufactured by companies E and C are equal

(d) Company A produces the lowest number of cars.

15. Four di�erent positions of a dice have been shown below. What is the number opposite 1?

3 262 6 44 4 26 5 1

(a) 2 (b) 3 (c) 5 (d) 6

ANSWER KEY

1. (d) 2. (b) 3. (a) 4. (d) 5. (c) 6. (b) 7. (c) 8. (b) 9. (a) 10. (a) 11. (c) 12. (a) 13. (c) 14. (d) 15. (d)


1. If (1 + tan 1°) (1 + tan 2°) (1 + tan 3°)…(1 + tan 45°) = 2n, then n is equal to (a) 21 (b) 24 (c) 23 (d) 22

2. �e value of ‘a’ for which the equation 4cosec2((a + x)) + a2 – 4a = 0 has a real solution is(a) a = 1 (b) a = 2(c) a = 3 (d) none of these

3. If x ∈

32

2

, , then value of the expression

sin–1(cos (cos–1(cos x) + sin–1(sin x))), is equal to (a) –/2 (b) /2 (c) 0 (d) none of these

4. ABC is a triangle whose medians AD and BE are perpendicular to each other. If AD = p and BE = q, then area of ABC is

(a) 23

pq (b) 32

pq

(c) 43

pq (d) 34

pq

5. �e value of ‘a’ so that the sum of the squares of the roots of the equation x2 – (a – 2)x – a + 1 = 0 assume the least value, is (a) 2 (b) 0 (c) 3 (d) 1

6. �e number of numbers lying between 100 and 500 that are divisible by 7 but not by 21 is (a) 19 (b) 38 (c) 57 (d) 76

7. If a, b, c are in H.P., then straight line

xa

yb c

1

0

always passes through a �xed point P with coordinates(a) (–1, – 2) (b) (–1, 2)(c) (1, – 2) (d) none of these

8. Integral part of ( )5 5 11 2 1 n is(a) even (b) odd (c) cannot say anything (d) neither even nor odd

9. Number of rectangles in �gure shown which are not squares is

(a) 159 (b) 136(c) 161 (d) none of these

10. �ree equal circles each of radius r touch one another. �e radius of the circle touching all the three given circles internally, is

(a) ( )2 3 r (b) ( )2 3

3

r

(c) ( )2 33

r (d) ( )2 3 r

11. �e abscissae and ordinates of the end points A and B of the focal chord of the parabola y2 = 4x are respectively the roots of x2 – 3x + a = 0 and y2 + 6y + b = 0. �e equation of the circle with AB as diameter is (a) x2 + y2 – 3x + 6y + 3 = 0(b) x2 + y2 – 3x + 6y – 3 = 0 (c) x2 + y2 + 3x + 6y – 3 = 0 (d) x2 + y2 – 3x – 6y – 3 = 0

12. A tangent having slope –4/3 to the ellipse x y2 2

18 321

intersects the major and minor axes at A and B. If O is the origin, then the area of OAB is (a) 48 sq. units (b) 9 sq. units (c) 24 sq. units (d) 16 sq. units

13. Let f xax bcx d

( )

, then f [f(x)] = x provided that

(a) d = –a (b) d = a (c) a = b = 1 (d) a = b = c = d = 1

14. If f(x) is di�erentiable and strictly increasing

function, then the value of lim( ) ( )( ) ( )x

x xx

0

2

0f ff f

is

PRACTICE PAPER 2017Useful for All National/State Level Entrance Examinations

Sanjay Singh Mathematics Classes, Chandigarh, Ph : 9888228231, 9216338231


(a) 1 (b) 0 (c) –1 (d) 2

15. Let f : R R is a real valued function x, y ∈ R such that | f(x) – f(y)| |x – y|2 . �e function h x f x dx( ) ( ) is (a) everywhere continuous (b) discontinuous at x = 0 only(c) discontinuous at all integral points(d) h(0) = 0

16. If f x x x x x( ) , 3 4 1 8 6 1 then f ′(x) at x = 1.5 is(a) 0 (b) 2 (c) 3 (d) –4

17. �e eccentricity of the ellipse x y2 2

4 31 is changed

at the rate of 0.1 m/sec. �e time at which it will

co-incident with the auxiliary circle is(a) 2 seconds (b) 3 seconds(c) 6 seconds (d) 5 seconds

18. �e greatest of the numbers 1, 21/2, 31/3, 41/4, 51/5, 61/6 and 71/7 is (a) 21/2 (b) 31/3 (c) 71/7 (d) all are equal

19. If is the angle (semi-vertical) of a cone of maximum volume and given slant height, then tan is given by(a) 2 (b) 1 (c) 2 (d) 3

20. If logsin ,/

x dx k0

2

then log( cos )10

x dx

is given

by(a) log2 + 4k (b) log2 + 2k(c) log2 + k (d) none of these

21. �e area of the �gure bounded by two branches of the curve (y – x)2 = x3 and the straight line x = 1 is (a) 1/3 sq. units (b) 4/5 sq. units(c) 5/4 sq. units (d) 3 sq. units

22. �e di�erential equation ydydx

x k k R ∈( )represents (a) family of circles centered at y-axis(b) family of circles centered at x-axis(c) family of rectangular hyperbolas(d) family of parabolas whose axis is x-axis

23. If |z2 – 1| = |z|2 + 1, then z lies on a

(a) circle (b) parabola(c) ellipse (d) none of these

24. �e probability that a particular day in the month of July is a rainy day is 3/4. Two persons whose

credibility are 45

23

and respectively, claim that

15th July was a rainy day. �e probability that it was really a rainy day is ______.

(a) 34

(b) 2425

(c) 89

(d) none of these

25. In a quadrilateral ABCD,

let

cos sin cos( )cos sin cos( )cos sin cos( )

A A A DB B B DC C C D

, then is

(a) independent of A and B only(b) independent of B and C only(c) independent of A, B and C only(d) independent of A, B, C and D all

26. If A

0 50 0

and f(x) = 1 + x + x2 + ..... + x16, then

f(A) is equal to

(a) 0 (b)

1 50 1

(c)

1 50 0

(d)

0 51 1

27. A vector of magnitude 3, bisecting the angle between the lines in direction of the vectors a i j k 2^ ^ ^ and

b i j k 2 and making an obtuse angle with

b is

(a) 36

i j^ ^ (b) i j k^ ^ ^ 3 214

(c) 3 3 2

14( )^ ^ ^i j k

(d) 3

10i j^ ^

28. Equation of a plane through the line of intersection of planes 2x + 3y – 4z = 1 and 3x – y + z + 2 = 0 and parallel to 12x – y = 0 is (2x + 3y – 4z –1) + l(3x – y + z + 2) = 0 then l is (a) –4 (b) 1/4 (c) 4 (d) –1/4

29. Negation of the proposition “If we control population growth, we prosper”.


(a) If we do not control population growth, we prosper.

(b) It we control population growth, we do not prosper.

(c) We control population but we do not prosper.(d) We do not control population, but we prosper.

30. �e variance of the �rst n natural numbers is

(a) n2 112 (b) n2 1

6

(c) n2 16 (d) n2 1

12

SOLUTIONS

1. (c) : (1 + tank°)(1 + tan(45 – k)°) = 2 n is equal to 23.2. (b) : We have, 4 cosec2((a + x)) + a2 – 4a = 0

cosec224

4( ( )) a x

a a

4

41 4 4

22a a

a a

a2 – 4a + 4 0 (a – 2)2 0 a = 2

3. (b) : For x ∈

32

2

, ,

cos–1(cos x) = cos–1(cos(2 – (2 – x))) = cos–1 (cos(2 – x)) = 2 – xand sin–1(sin x) = sin–1(sin(2 – (2 – x))) = x – 2 cos–1(cos x) + sin–1(sin x) = (2 – x) + (x – 2) = 0. sin–1(cos(cos–1(cosx) + sin–1(sinx))) = sin–1 (cos (0)) = sin–1(1) = /24. (a) : Since M is centroid, AM : MD = 2 : 1. Also BM : ME = 2 : 1

AM p MD p BM q ME q 23

13

23

13

, , and

Area of triangle (ABM) A

E

CDBM

= 12 AM BM

12

23

23

29

p q pq

Area of ABC = 3area(ABM) = 23

pq

5. (d) : x2 – (a – 2)x – a + 1 = 0 Let a, be the roots. a + = (a – 2), a= 1 – aa2 + 2 = (a + )2 – 2a = (a – 2)2 – 2(1 – a) = a2 – 4a + 4 – 2 + 2a = a2 – 2a + 2 = (a – 1)2 + 1It is minimum when a = 1

6. (b) : �e numbers which are divisible by 7 and lying between 100 and 500 are 105, 112, 119,...., 497.

Total numbers

497 1057

1 57

�e numbers which are divisible by 21 are 105, 126, 147, …., 483

Total numbers

483 105

211 19

So, required number = 57 – 19 = 38

7. (c) : We have, 21 1 1 b a c

1 2 1

01

0a b c

xa

yb c

x = 1, y = – 28. (a) : ( ) & ( )5 5 11 5 5 112 1 2 1 ′ n nI f flet

Also I + f –f ′ = ( ) ( )5 5 11 5 5 11 22 1 2 1 n n k i.e., Even integer Hence f –f ′ = 2k – I is an integerBut –1 < f – f ′ < 1, therefore f – f ′ = 0 I = 2k = even integer9. (d) 10. (b) : DEF is equilateral with s ide 2r. If radius of circumcircle of DEF is R1, then

Area of DEF = 34

2 32 2( )r r

32 2 2

4 42

1r

r r rR

abcR

Rr

12

3 Radius of the circle touching all the three given circles

= r + R1

rr r23

2 33

( )

11. (b) : t1t2 = –1 as AB is focal chord x2 – 3x + a = 0 ; x1 + x2 = 3 and x1x2 = a y2 + 6y + b = 0 ; y1 + y2 = – 6 and y1y2 = b

x x

tt a1 2

12 1

211

y x2 = 4

A t t( , 2 )12

1

(1, 0)

B t t( , 2 )22

2 y y t

tb1 2 1

12

24

a = 1, b = – 4Equation of circle AB is diameter x2 + y2 – 3x + 6y – 3 = 0


12. (c) : Any point on the ellipse x y2

2

2

23 2 4 21

( ) ( )

can be taken as ( cos , sin )3 2 4 2 and the slope of

the tangent b xa y

2

2 32 3 218 4 2

43

( cos )( sin )

cot

...(1)

Given slope of the tangent 43

...(2)

From (1) and (2), we have cot = 1 = /4

Hence, the equation of the tangent is x y. .1

23 2

12

4 21

i ex y

. .6 8

1

Hence, A = (6, 0), B = (0, 8)

Area of OAB = 12

6 8 24 sq. units

13. (a) : fof xa ax b

cx db

c ax bcx d

dx( )

(ac + dc)x2 + (bc + d2 – bc – a2)x – ab – bd = 0 ac + dc = 0, a2 – d2 = 0, ab + bd = 0 So a = –d

14. (c) : lim ( ) ( )( ) ( )x

x xx

0

2

000

f ff f

form

Apply L'Hospital rule, we get

lim

( )( ) ( )( )x

x x xx

′ ′′0

2 2f ff

f(x) is strictly increasing function. So, f ′(x) > 0

′ ′′

′′

f ff

ff

( )( ) ( )( )

( )( )

0 2 0 00

00

1

15. (a) : Since, |f(x) – f(y)| |x – y|2, x y

f x f y

x yx y

( ) ( )| |

Taking lim as y x, we get

lim

( ) ( )lim | |

y x y x

f x f yx y

x y

lim( ) ( )

lim ( )y x y x

f x f yx y

x y

| f ′(x)| 0 | f ′(x)| = 0 (Q | f ′(x) 0|) f ′(x) = 0 f (x) = c (constant)

h x f x dx cdx cx d( ) ( )where d is constant of integration. h(x) of a linear function of x which is continuous for all x ∈ R.

16. (b) : f x x x x x( ) ( ) ( ) 1 4 4 1 1 9 6 1

x x1 2 1 3 f x x x x( ) ( ) ( ) 2 1 3 1 5 2 1

′

f xx

( )2

2 1

′

f ( . )1 5

132

12

17. (d) : dedt

0 1. m/sec

et = – 0.1 t + e0 We have, 3 = 4(1 – e0

2)

34

1120

20 e e

e t et t 0 1 12

0. , given

0 112

5. t tor seconds

18. (b) : Assuming the function f(x) = x1/x {x > 0}

′

f x xd

dx xxx( ) log/1 1

′

f x xx

x xx( )

log/12 2

1

′ f xxx

xx

( ) ( log )/1

2 1

Clearly when x > e, f ′(x) will be negative f(x) is decreasing.When 0 < x < e, f ′(x) will be positive f(x) is increasing.It means 31/3 > 41/4 > 51/5 > 61/6 > 71/7 Now compare (1) and (2)1/2 and (3)1/3 Apply same power on given three numbers. (1)6 (2)1/2 × 6 (3)1/3 × 6

1 23 32

1 < 8 < 9 (1) < (2)1/2 < (3)1/3 So maximum number is (3)1/3.19. (c) : Let OB = l, OA = l cos and AB = l sin (0 /2). �en

V AB OA

3

2( ) ( )

3

3 2l sin cos


dVd

l

33 13 2sin ( cos )

l

A B

O

For maximum,

dVd

= 0 = 0 or cos 13

.

Also V(0) = 0, V (/2) = 0

and V lcos

1

313

29 3

Hence V is maximum when cos . 13

i e. . tan 2

20. (d) : I x dx log( cos )10

log cos2

22

0

xdx

log log cos2

20

2

0dx

xdx

log log cos2 2

20

xdx

21. (b) : Given curve is (y – x)2 = x3

y x x x

y x x x

Required area

= [( ( )]x x x x x x dx 0

1

2 2

25

45

3 2

0

1x dx/ . sq. units

22. (b) : We have, ydydx

k x ( ) y dy = (k – x)dx

Integrate both sides, we get

y dy k x dx ( ) y k xc

2 2

2 2

( )

( ) ,x k y c

2 2

2 2 represents family of circles whose

centre is at (k, 0).23. (d) : On putting z = x + iy, the equation is same as | x2 – y2 + 2ixy – 1| = x2 + y2 + 1 (x2 – y2 – 1)2 + 4x2y2 = (x2 + y2 + 1)2 x = 0 z lies on imaginary axis, so (a), (b), (c) are ruled out. 24. (b) : A : Event that �rst man speaks truthB : Event that second man speaks truthR : Day is rainy

P RP A B P R

P A B P R P A B P R( )

( ) ( )( ) ( ) ( ) ( )

′ ′ ′

45

23

34

45

23

34

15

13

14

2425

. .

. . . .

25. (d) : C3 C3 – C1 cosD + C2 sinD = 0So = 0, hence is independent of A, B, C, D all.26. (b) : f(A) = I + A + A2 + .... + A16

A A

0 50 0

0 50 0

0 50 0

0 00 0

2

A A A3 2 0 0

0 0

Similarly A4 = A5 = ....... = A16 = 0 00 0

f A( ) ....

1 00 1

0 50 0

0 00 0

0 00 0

1 50 1

27. (c) : A vector bisecting the angle between

a b and

is

aa

bb| | | |

.

In this case, 2

62

6i j k i j k^ ^ ^ ^ ^ ^

i ei j i j k

. .^ ^ ^ ^ ^3

63 2

6

or

A vector of magnitude 3 along these vectors is

3 310

3 3 214

( ) ( )^ ^ ^ ^ î j i j k or

Now, 314

3 2 2( ) ( )^ ^ ^ ^ ^ î j k i j k is negative and hence

314

3 2( )^ ^ î j k makes an obtuse angle with

b .

28. (c) : (2 + 3l)x + (3 – l)y + (–4 + l)z = 1 – 2lIt is parallel to 12x – y = 0 if

2 3

123

14

04

l l l

l

29. (c) : p : we control population, q : we prosper. we have p qIts negation is ~ (p q) i.e. p ~ qi.e. we control population but we do not prosper

30. (a) : Variance =( ) ;S.D. 2 221

n

xx

nx

xn

n n nn

n nn

n( ) ( ) ( )1 2 16

12

112

2 2


ONLY ONE OPTION CORRECT TYPE

1. �e number of ways in which the squares of a 8 × 8 chess board can be painted red or blue so that each 2 × 2 square has two red and two blue squares is (a) 29 (b) 29 – 1(c) 29 – 2 (d) none of these

2. Box A contains black balls and box B contains white balls take a certain number of balls from A and place them in B, then take same number of balls from B and place them in A. �e probability that number of white balls in A is equal to number of black balls in B is equal to (a) 1/2 (b) 1/3(c) 2/3 (d) none of these

3. �e number of planes that are equidistant from four non-coplanar points is (a) 3 (b) 4 (c) 7 (d) 9

4. 5 di�erent games are to be distributed among 4 children randomly. �e probability that each child get atleast one game is

(a) 14

(b) 1564

(c) 2164 (d) none of these

5. Let a, b, c be distinct non-negative numbers and the vectors a i a j c k^ ^ ^ , i k^ ^ , c i c j b k^ ^ ^ lie in a plane, then the quadratic equation ax2 + 2cx + b = 0 has(a) real and equal roots (b) real and unequal roots (c) both roots real and positive (d) none of these

6. If three equations are consistent (a + 1)3x + (a + 2)3y = (a + 3)3; (a + 1)x + (a + 2)y = a + 3, x + y = 1, then a = (a) 1 (b) 2 (c) –2 (d) 3

7. If three one by one squares are selected at random from the chessboard, then the probability that they form the letter ‘L’ is

(a)

19664

3C (b)

4964

3C (c)

3664

3C (d)

9864

3C

8. lim( )

/

x

x tx t

dt

2

2 20

2

1

2

is equal to

(a) 1/4 (b) 1/2(c) 1 (d) none of these

9. If a, b, c > 0, then the minimum value of

(a + b + c) 1 1 1a b c

must be

(a) 6 (b) 8 (c) 3 (d) none of these

10. �e value of ( )

( )cot

x

x x xx

dx2

4 2 1

1

1 1

will be

(a)

ln cot

1 1x

xc

(b)

ln cot 1 1x

xc

(c)

ln cot

1 1x

xc

(d) none of these

*ALOK KUMAR, B.Tech, IIT Kanpur

* Alok Kumar is a winner of INDIAN NAtIoNAl MAtheMAtIcs olyMpIAD (INMo-91). he trains IIt and olympiad aspirants.

MATHEMATICS TODAY | APRIL ‘1732 MATHEMATICS TODAY | APRIL ‘1732

11. �e highest power of 3 contained in (70)! must be equal to (a) 16 (b) 24 (c) 32 (d) 48

12. If A is a nilpotent matrix of index 2, then for any positive integer n, A(I + A)n is equal to (a) A–1 (b) A (c) An (d) In

13. If r r ra b c, , are unit vectors such that r r r ra b a c 0

and the angle between r rb c and is /3. �en the value of | | r r r ra b a c is (a) 1/2 (b) 1(c) 2 (d) none of these

14. �e curve xy = c(c > 0) and the circle x2 + y2 = 1 touch at two points, then distance between the points of contact is (a) 1 (b) 2(c) 2 2 (d) none of these

15. A variable circle is drawn to pass through (1, 0) and to touch the line x + y = 0. Let S = 0 represent the locus of the centre of the circle. �en S = 0 represents (a) pair of parallel lines (b) circle (c) parabola (d) none of these

16. If a, and are the roots of the equation x2(px + q) = r(x + 1). �en the value of determinant

1 1 11 1 11 1 1

a

is

(a) a (b) 1 1 1 1 a


17. If I x nx dxm nm

, cos cos , then the value of

(m + n)Im, n – mIm – 1, n – 1 (m, n ∈ N) is equal to

(a) cos cosm x nx

nc (b) cosmx sinnx + c

(c) cos cosm x nxn

c

1

(d) –cosmx cosnx + c

18. If I nx xdx cos logcos/

20

2. �en I =

(a) cot cos/

x nxdx20

2

(b) cot sin/

x nxdx20

2

(c) tan cos/

x nxdx20

2

(d) 1

22

0

2

nx nxdxtan sin

/

19. If y e x x e {(cos cos ... ) log }2 4 2 satis�es the equation x2 – 9x + 8 = 0, then �nd the value of

sin

sin cos, .x

x xx

0

2

(a)

3 12 (b)

13 1

(c) 1

3 1 (d) none of these

20. If p is a prime number (p 2), then the di�erence [( ) ]2 5 2 1 p p is always divisible by (where [.] denotes the greatest integer function)(a) p + 1 (b) p(c) 2p (d) 5p

21. Let a, b, c ∈ R such that no two of them are

equal and satisfy 2

22

0a b cb c ac a b

, then equation

24ax2 + 4bx + c = 0 has (a) atleast one root in [0, 1/2] (b) atleast one root in [–1/2, 0)(c) atleast one root in [1, 2](d) atleast two roots in [0, 2]

22. I d

( cos )( cos )

/

/11

2 7

9 7

is equal to

(a) tan 2 c (b) 7

11 2

11 7tan

/

c

(c) 72

7tan

c (d) none of these

23. Number of solutions of the equation 1 + e|x| = |x| is/are (a) 0 (b) 1 (c) 2 (d) 4

24. �e value of limn n

nr

rt

t

t

r

r

nC C

15

30

1

1 is equal

to (a) 4 (b) 3(c) 1 (d) none of these

MATHEMATICS TODAY | APRIL ‘17 33MATHEMATICS TODAY | APRIL ‘17 33

ONE OR MORE THAN ONE OPTION CORRECT TYPE

25. �e values of afor which the integral

x a dx 0

21 is satis�ed are

(a) [2, ) (b) (–, 0](c) (0, 2) (d) none of these

26. �e number of ways of choosing triplet (x, y, z) such that z max {x, y} and x, y, z ∈(1, 2, ......n, n +1) is

(a) n + 1C3 + n + 2C3 (b) 16

1 2 1n n n( )( )

(c) 12 + 22 + ... + n2 (d) 2(n + 2C3) – n + 1C2

27. If inside a big circle exactly 24 small circles, each of radius 2, can be drawn in such a way that each small circle touches the big circle and also touches both its adjacent small circles, then radius of the big circle is

(a) 2 14

cosec (b) 1

24

24

tan

cos

(c) 2 112

cosec (d)

248 48

24

2sin cos

sin

28. Consider f (x) = (cos + xsin)n – cosn – xsinn, (n ∈ N) then f(x) is always divisible by (a) x + i (b) x – i (c) x2 + 1 (d) none of these

29. If a, b, c are the sides of a triangle, then a

c a bb

a b cc

b c a

can take value(s)

(a) 1 (b) 2 (c) 3 (d) 4

30. �e triangle formed by the normal to the curve f (x) = x2 – ax + 2a at the point (2, 4) and the coordinate axes lies in second quadrant. If its area is 2 sq. units, then a can be (a) 2 (b) 17/4 (c) 5 (d) 19/4

31. If ( )x

x x x

x x x

x x x

2

2

2

4 3 2 4 13

2 5 9 4 5 26

8 6 1 16 6 104

= ax3 + bx2 + cx + d, then

(a) a = 3 (b) b = 0 (c) c = 0 (d) none of these

32. If f x tt t

dtx

( ) ln( )

11

, then

(a) fx

tt t

dtx1

11

ln

( )

(b) fx

tt t

dtx1

121

ln( )

(c) f x fx

( )

1 0

(d) f x fx

x( ) (ln )

1 1

22

33. A, B, C are three events for which P(A) = 0.4, P(B) = 0.6, P(C) = 0.5, P(A B) = 0.75,

P(A C) = 0.35 and P(A B C) = 0.2. If P(A B C) 0.75, then P(B C) can take

values (a) 0.1 (b) 0.2 (c) 0.3 (d) 0.5

COMPREHENSION TYPE

Paragraph for Q. No. 34 to 36

Let n be a positive integer such that I x a x dxnn 2 2 .

34. �e value of I1 is

(a) 23

2 2 1 2( ) /a x C

(b) 13

2 2 3 2( ) /a x C

(c) 23

2 2 3 2( ) /a x C

(d) 13

2 2 3 2( ) /a x C

35. �e value of the expression

x a x dx

x a x dx

a

a

4 2 2

0

2 2 2

0

is equal

to

(a) a2

6 (b) 3

2

2a

(c) 34

2a (d) a2

2


36. If I x a xn

kIn

n

n

1 2 2 3 2

22( ) /

, then the values k is

(a) nn

12

(b) nn

21

(c) nn

a

12

2 (d) nn

a

21

2

MATRIX MATCH TYPE

37. Match the following .

List-I List-IIA. Normal of parabola y2 = 4x at P and

Q meets at R (x2, 0) and tangents at P and Q meets at T(x1, 0), then if x2 = 3, then the area of quadrilateral PTQR is

(i) 3

B. The length of latus rectum plus tangent PT will be

(ii) 6

C. The quadrilateral PTQR can be inscribed in a circle, then the value

of

circumference4

will be

(iii) 1

D. The number of normals that can be drawn to the parabola from R(x2, 0)

(iv) 8

38. Match the following.List-I List-II

A. Area bounded by the parabola y2 = 4x and the line y = x is

(i) 1/2

B. (sin cos )5 5

0x x dx

is equal to (ii) –11

C. If S = 0 be the equation of the hyperbola x2 + 4xy + 3y2 – 4x + 2y + 1 = 0 then the value of k for which S + k = 0 represents its asymptotes is

(iii) 8/3

D. The value of a for which a i j k^ ^ ^ 2 ,

lies in the plane containing three points i j k i j k

^ ^ ^ ^ ^ ^, 2 2 and

3 i k^ ^ is

(iv) 1615

(v) –22

INTEGER ANSWER TYPE

39. �e number of real solutions of the equation esin x – e– sin x – 4 = 0 is _____.40. �e �gures 4, 5, 6, 7, 8 are written in every possible

order. �e sum of the number of numbers greater than 56000 is _____.

SOLUTIONS

1. (c) : Number of ways when the squares are alternating colour in �rst column is 28. Number of ways when the squares in the �rst column are not alternating colour = 28 – 2 Required number of ways = 29 – 22. (d)3. (c) : Let the points be A, B, C and D.�e number of planes which have three points on one side and the fourth point on the other side = 4. �e number of planes which have two points on each side of the plane = 3 Hence, number of planes = 7. 4. (b) : Total ways of distribution = 45

Total ways of distribution so that each child get atleast one game 4 3 2 2405 4

15 4

25 4

3C C C

Required probability 2404

15645

5. (a) :

a a c

c c b1 0 1 0

c2 – ab = 0 and from ax2 + 2cx + b = 0, we haveD = 4c2 – 4ab = 4(c2 – ab) = 0So, roots are real and equal. 6. (c) : Since the equations are consistant. = 0

( ) ( ) ( )( ) ( ) ( )a a aa a a

1 2 31 2 3

1 1 10

3 3 3

Put u = a + 1, v = a + 2 and w = a + 3 u – v = –1, v – w = – 1, w – u = 2 Also, u + v + w = 3a + 6

u v wu v w u v v w w u u v w

3 3 3

1 1 10 0( )( )( )( )

(–1)(–1)(2)(3a + 6) = 0 a = –2


7. (a) : Total number of outcomes = 64C3.Let ‘E’ be the event of selecting 3 squares which form the letter ‘L’.No. of ways selecting squares consisting of 4 unit squares = 7C1 × 7C1 = 49Each square with 4 unit squares forms 4L-shapes consisting of 3 unit squares.

n E P EC

( ) , ( )7 7 4 196 19664

3

8. (b) : lim( )

/

x

x tx t

dt

2

2 20

2

1

2

lim ( ) lim

/

x

x

x

tt

dt

x

x

x

2

20

2

2

2 2

2

1 2

1

2

form

222

22

2

x

d x

lim lim

x x

xx

x

xx

x

4

4 4

44

2 2 812

9. (d)

10. (b) : Let I x

x x xx

dx

( )

( )cot

2

4 2 1

1

1 1

I x

xx

xx

dx1 1

1 2 1 12

22

1cot

Put xx

tx

dx dt

1 1 12

Let cot–1t = u

I dtt t t

dt du( )cot ( )2 1 21

11

I du

u

ln | | ln cotu c x

xc1 1

11. (c) : 703

709

7027

23 7 2 32

12. (b) : A2 = 0 · A3 = A4 = .... = An = 0then A(I ± A)n = A(I ± nA) = A ± nA2 = A

13. (b) : r r r ra b a b 0 ,

r r r ra c a c 0

r r ra b c

r r r r r r ra b c a b a c( ) | |

| | | | | | r r r r ra b c b c

Now, r r r r r rb c b c b c 2 2 2 2

32 2 1

21| | | | | | | | cos

r rb c 1

14. (b) : �e curve xy = c and the circle x2 + y2 = 1 touches each other, so

x cx

x x c22

24 2 21 0 0 will have equal roots

so, (–1)2 – 4c2 = 0

4 1 14

12

2 2c c c

Roots of the equation x x4 2 14

0 are

x y 1

212

Hence, distance between the points of contact = 2 units. 15. (c) : Let P(h, k) be the centre of the circle. Since the circle passes through (1, 0).

Its radius ( )h k1 2 2 It also touches the line x + y = 0,

so h k h k 2

1 2 2( )

Hence locus of the centre P(h, k) is S x2 + y2 – 2xy – 4x + 2 = 0Here, h2 = ab and 0. So it represent a parabola. 16. (c) : Given equation, px3 + qx2 – rx – r = 0 having roots a, , . So, we have

a a a a q

prp

rp

; ;

a

a a a a a

a1 1 1 0


17. (b) : I x nxdxm nm

, cos cos

cos sin cos sin sinm

mx nxn

mn

x x nxdx1

cos sin cos (cos( )

mmx nx

nmn

x n x1 1

– cosnx cosx)dx

I x nxn

mn

x n xdx mn

Im n

mm

m n, ,cos sin cos cos( ) 1 1

m n

nI x nx

nmn

I cm n

m

m n, ,cos sin

1 1 1

( ) cos sin, ,m n I mI x nx cm n m nm

1 1

18. (d) : We have, I nx xdx cos log cos/

20

2

On integrating by parts, we get,

log cos . sin ( tan ) sin/ /

x nxn

x nxn

dx22

220

2

0

2

Since, lim log cos . sin/x

x xx

2

22

0

I x nxn

dx(tan ) sin/ 220

2

19. (c) : Since 02

1 x x and cos ,

y e ex

x x xee

coscos

logcot log cot

2

2 2 21

22 2

Now y satis�es x2 – 9x + 8 = 0 y2 – 9y + 8 = 0 (y – 1)(y – 8) = 0 y = 1, y = 8

If y = 1 then, 2 1 22 0cot x

cot2x = 0 x 2

( )not possible 02

x

If y = 8 then 2 8 2 32 3cot cotx x

x x x

6 6 not possible 0

2 and cot

is positive

x 6

Now sinsin cos

xx x

12

12

32

13 1

20. (b) : We have,

( ) ( ) [2 5 2 5 2 2 2 522 p p p p pC

p p pp

pC C44 2

11 22 5 2 5... ]( )/

...(1)

From (1), ( ) ( )2 5 2 5 p p is an integer

and 1 2 5 ( )p 0 (as p is odd), so

[( ) ] ( ) ( )2 5 2 5 2 5 p p p

2 2 5 2 512

11

2 1 2p p p pp

pC C... ( )/

[( ) ] [2 5 2 2 2 5 2 512

24

4 2p p p p p pC C

... ]( )/p

ppC 1

1 22 5

Now, p pC p p C p p p p2 4

11 2

1 2 31 2 3 4

( ) , ( )( )( ) ,

...., ppC p 1

are divisible by the prime p.

�us R.H.S. is divisible by p. 21. (a) : We have,2a(bc – 4a2) + b(2ac – b2) + c(2ab – c2) = 0 6abc – 8a3 – b3 – c3 = 0 (2a + b + c)[(2a – b)2 + (b – c)2 + (c – 2a)2] = 0 2a + b + c = 0 [as b c]Let f (x) = 8ax3 + 2bx2 + cx

f f a b c a b c( ) ,0 0 1

2 2 22

20

So, f(x) satis�es the Rolle’s �eorem condition so, f ′(x) = 0 has atleast one root in [0, 1/2].

22. (b) : I d

( cos )( cos )

sin

cos

/

/

/

/11

12

2

2

2 7

9 7

4 7

18

77 d

Put 2

2 t d dt

I tt

dt t tdt(sin )(cos )

(tan ) sec/

//

4 7

18 74 7 2

Put, tant = u sec2tdt = du

I u du c4 711 77

11 2/

/tan


23. (a) : For x > 0, f (x) = 1 + ex – x, f ′(x) = ex – 1 = 0 for x = 0 and for x > 0, f ′(x) > 0 f (x) is increasing. f (x) > f (0) f (x) > 2. Hence no solution. Also, curve is symmetrical about y-axis, hence, no solution for x < 0 also. Hence, no solution.

24. (c) : lim .n n

nr

rt

t

t

r

r

nC C

15

30

1

1

lim . ( )n n

nr

r

nr rC1

54 3

1

lim lim ( )

n nn

rr n

rr

r

n

r

n

n nn nC C1

54 3 1

55 4 1

11

25. (a, b, c) : For a 0 given equation becomes

( )x a dx a a 1 1

20

0

2

For 0 < a < 2

| | ( ) ( )x a dx a x dx x a dx

a

a 1 1

0

2

0

2

a a a a a2 2

22

2 22

1 2 1 0

(a – 1)2 0.

For a 2, x a dx 10

2

( )a x dx a a a1 2 2 1 32

20

2

26. (b, c, d) : For a given z, say z = k, we have x, y k, hence there are k2 ordered pairs (x, y).So the total number of numbers

k n n n

k

n2

1

1 2 16

( )( )

27. (a, d) : sin 24

22

2 124

R

R cosec(a) is true.

sin cos

sin

a a

a2 2 1

2

A

cosec (d) is true.

28. (a, b, c) : (x2 + 1) = (x + i)(x – i) f (i) = f (–i) = (cos + isin)n – (cosn + isinn) = 0 (By Demoiver theorem)

29. (c, d) : c + a – b, b + c – a, a + b – c are all positive.

a

c a bb

a b cc

b c a3

abcc a b a b c b c a( )( )( )

/1 3

...(1)

Also, a2 a2 – (b – c)2 a2 (a + b – c)(a – b + c)Similarly, b2 (b + c – a)(b – c + a), c2 (c + a – b)(c – a + b) a2b2c2 (a + b – c)2(b + c – a)2(c + a – b)2

�us abc (a + b – c)(b + c – a)(c + a – b)

abcc a b a b c b c a( )( )( )

1

So, from (1) ac a b

ba b c

cb c a

3

30. (b, c) : We have, f(x) = x2 – ax + 2a f ′(x) = 2x – a At (2, 4), f ′(x) = 4 – a Equation of normal at (2, 4) is

( )

( )( )y

ax

4 1

42

Let point of intersection with x and y axis be A and B respectively, then A a B a

a

( , ), ,4 18 0 0 4 184

. Hence a 92

Area of triangle 12

( ) ( )( )

4 18 4 184

2a aa

( )( )4 17 5 0 5 174

a a a or

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31. (b, c) :

′

xx xx xx x

x x

x x2 4 2 4 134 5 4 5 26

16 6 16 6 104

4 3 2 13

2 5 9 4 2

2

2 66

8 6 1 16 104

02x x

(x) = constant a = 0, b = 0, c = 0.

32. (d) : f x tt t

dtx

( ) ln( )

11

fx

u uu

uu

du tt

dtx x1

11

121

1

1

ln ln

( )

/

Now f x fx

tt t

dt tt

dtxx

( ) ln( )

ln

1

11 1

11

(ln ) (ln )t xx2

1

2

2 2

33. (a, b, c) : P A B C P A P B P C( ) ( ) ( ) ( )

P A B P B C P C A P A B C( ) ( ) ( ) ( )

0.4 + 0.6 + 0.5 + 0.75 – (0.4 + 0.6) – P(B C) – 0.35 + 0.2

[since P(A B) = P(A) + P(B) – P(A B)] = 1.1 – P(B C) But, 0.75 P(A B C) 1 0.75 1.1 – P(B C) 1 0.1 P(B C) 0.35

34. (d) : I x a x dx12 2 ,

Put a2 – x2 = t xdx dt12

I tdt t a x C13 2 2 2 3 21

213

13

/ /( )

35. (d) : x a x dxa

4 2 2

0

x a x x a x dxa a2 2 2 3 2

0

2 2 2 3 2

03( ) ( )

//

( )a x x a x dx

a2 2 2 2 2

0

a x a x dx x a x dxa a

2 2 2 2

0

4 2 2

0

a x a x dxa2

2 2 2

02

x a x dx

x a x dx

a

a

a

4 2 2

0

2 2 2

0

2

2

36. (c) : I x a x dx x x a x dxnn n 2 2 1 2 2( )

x a x n x a x a x dxn n1 2 2 3 2 2 2 2 2 21

31

3( ) ( )/

13

13

13

1 2 2 3 2 22x a x n a I n In

n n( ) /

1 1

313

13

1 2 2 3 22

2n I x a x a n In

nn( ) ( )/

37. (A) - (iv) ; (B) - (ii) ; (C) - (iii) ; (D) - (i)

(A) y2 = 4x ...(1) Clearly, if P(at2, 2at) then by symmetry, Q(at2, –2at) Equation of tangent is ty = x + at2 For t, y = 0, x1 = –at2 and equation of normal is y = –tx + 2at + at3 For R, y = 0, x2 = (2a + at2) ...(2) Here, a = 1 x1 = –t2 and x2 = (2 + t2) 3 = 2 + t2 t2 = 1 t = ±1 Take t = 1, then x1 = –1. PM = 2at = 2 × 1 × 1 = 2 RT = x1 + x2 = (1 + 3) = 4

Area of quadrilateral PTQR = 2

12

4 2

= 8 sq. units(B) Length of latus rectum + tangent PT

4 4 1 0 4 1 4 2 612a S ( ) units

(C) Clearly, RT will be the diameter of circle

Circumference diameter( )

RT 44

1

(D) Since in the part (1), we have found x2

= (2a + at2) > 2a, (if t 0) For three real normals, x2 > 2a = 2 × 1 = 2i.e. x2 = 2 of straight lines otherwise a rectangular hyperbola.


38. (A) - (iii) ; (B) - (iv) ; (C) - (v) ; (D) - (i)

(A) Required area 2 12

4 40

4xdx

4

38 8

33 2

0

4x /

(B) (sin cos ) sin/

5 5

0

5

0

22x x dx xdx

2 4

523

1615

(C) For equation S + k = 0 to represent pair of lines

1 2 22 3 12 1 1

0

k

3 1 1 2 2 2 2 2 2 6 0 22( ) ( ) ( )k k k

(D) Let p.v. of given points be A i j k( )^ ^ ^ , B i j k( )^ ^ ^2 2

and C i k( )^ ^3 so that two vectors in the plane may be

AB i j

^ ^ and AC i j k

2 2^ ^ ^

�us, a

a a2 1

1 1 02 1 2

0 2 2 2 1 2 0 12

( ) ( )

39. (0) : Giv en, sin sine ex x 4 0L et esinx = y [ y > 0x ∈ R]then, y – 1/y – 4 = 0 y2 – 4y – 1 = 0

y

4 16 4

22 5

y can never be negative, 2 5 can not be accepted.Now, 2 5 e and maximum value of esin x = eHence, e xsin 2 5 i.e. there is no solution. 40. (9) : Case I : When we use 6, 7 or 8 at ten thousand place then number of numbers = 3 × 4P4 = 72Case II: When we use 5 at ten thousand place and 6, 7 or 8 at thousand place then number of numbers is = 1 × 3 × 3P3 = 18Hence, the required numbers of numbers is 72 + 18 = 90.


SECTION-I (PHYSICS)

1. Two masses of 5 kg and 3 kg are

3 kg

5 kgT1

T2

suspended with the help of a massless inextensible strings as shown in �gure. �e whole system is going upwards with an acceleration of 2 m s–2. �e tensions T1 and T2 are (Take g = 10 m s–2)(a) 96 N, 36 N (b) 36 N, 96 N(c) 96 N, 96 N (d) 36 N, 36 N

2. �e dimensions of ab

in the equation, P a tbx

2,

where P is pressure, x is distance and t is time are(a) [M2LT–3] (b) [ML0T–2](c) [ML3T–1] (d) [MLT–3]

3. A particle is projected from ground at some angle with the horizontal. Let P be the point at maximum height H. At what height above the point P the particle should be aimed to have range equal to maximum height? (a) H (b) 2H (c) H/2 (d) 3H

4. �e dimensions of [Angular momentum][Magnetic moment]

are

(a) [M3LT–2A2] (b) [MA–1T–1](c) [ML2A–2T] (d) [M2L–3AT2]

5. A wire of length l and mass m is bent in the form

of a rectangle ABCD with ABBC

2 . �e moment of

inertia of this wire frame about the side BC is

(a) 11

2522ml

(b) 8

2032ml

(c) 5

1362ml (d) 7

1622ml

6. A body is rolling down an inclined plane. If kinetic energy of rotation is 40% of kinetic energy in translatory state, then the body is a

(a) ring (b) cylinder(c) hollow ball (d) solid ball

7. For two vectors

A B A B A Band ,| | | | is always true when(a) | | | |

A B 0 (b)

A B(c) | | | |

A B 0 and

A Band are parallel(d) | | | |

A B= 0 and

A Band are antiparallel8. An ideal massless spring S can be compressed

2 m by a force of 200 N. �is spring is placed at the bottom of the frictionless inclined plane which makes an angle = 30° with the horizontal. A 20 kg mass is released from rest at the top of the inclined plane and is brought to rest momentarily a�er compressing the spring 4 m. �rough what distance does the mass slide before coming to rest?(a) 2.2 m (b) 4 m (c) 8.17 m (d) 1.9 m

9. Six moles of O2 gas is heated from 20°C to 35°C at constant volume. If speci�c heat capacity at constant pressure is 8 cal mol–1 K–1 and R = 8.31 J mol–1 K–1. What is the change in internal energy of the gas?(a) 180 cal (b) 300 cal (c) 360 cal (d) 540 cal

10. A small mass attached to a string rotates on a frictionless table top as shown. If the tension in the string is increased by pulling the string causing the radius of the circular motion to decrease by a factor of 2, the kinetic energy of the mass will(a) decrease by a factor of 2 (b) remain constant(c) increase by a factor of 2(d) increase by a factor of 4

11. A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. �e mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. �e minimum


value of u so that the particle does not return back to the earth, is

(a) 2

2GMR

(b) 2GM

R

(c) 2

2gM

R (d) 2 2gR

12. A solid sphere of mass 10 kg is placed over two smooth inclined planes as shown in �gure. Normal reaction at 1 and 2 will be (Take g = 10 m s–2)

60° 30°2

1

(a) 50 3 50N N, (b) 50 N, 50 N(c) 50 50 3N N, (d) 60 N, 40 N

13. A particle of mass M is situated at the centre of a spherical shell of same mass and radius a. �e magnitude of the gravitational potential at a point situated at a/2 distance from the centre, will be

(a) GM

a (b) 2GM

a (c)

3GMa

(d) 4GM

a14. A mass of diatomic gas ( = 1.4) at a pressure of 2

atmospheres is compressed adiabatically so that its temperature rises from 27°C to 927°C. �e pressure of the gas in the �nal state is(a) 8 atm (b) 28 atm(c) 68.7 atm (d) 256 atm

15. If 4 moles of an ideal monatomic gas at temperature 400 K is mixed with 2 moles of another ideal monatomic gas at temperature 700 K, the temperature of the mixture is(a) 550°C (b) 500°C (c) 550 K (d) 500 K

16. In the case of a hollow metallic sphere, without any charge inside the sphere, electric potential (V) changes with respect to distance (r) from the centre as

(a) V

(b) V

(c) V

(d) V

17. In the circuit shown, the R

R R

RR

15 V

C = 3 F

cell is ideal, with emf = 15 V. Each resistance R is of 3 . �e potential di�erence across the capacitor of capacitance 3 F is(a) Zero (b) 9 V (c) 12 V (d) 15 V

18. An object of mass 0.2 kg executes simple harmonic oscillations along the x-axis with a frequency 25

Hz. At the position x = 0.04 m, the object has

kinetic energy 0.5 J and potential energy 0.4 J. �e amplitude of oscillation is

(Potential energy is zero at mean position) (a) 6 cm (b) 4 cm (c) 8 cm (d) 2 cm

19. As the switch S is closed in 10 V 4 5 V2

2

S

the circuit shown in �gure, current passed through it is(a) Zero(b) 1 A(c) 2 A(d) 1.6 A

20. In a sonometer wire, the tension is maintained by suspending a 50.7 kg mass from the free end of the wire. �e suspended mass has a volume of 0.0075 m3. �e fundamental frequency of the wire is 260 Hz. If the suspended mass is completely submerged in water, the fundamental frequency will become(a) 200 Hz (b) 220 Hz(c) 230 Hz (d) 240 Hz

21. A body of mass 0.01 kg executes F(N)

800.20

–0.2–80

x(m)

simple harmonic motion (SHM) about x = 0 under the in�uence of a force as shown in the adjacent �gure. �e period of the SHM is(a) 1.05 s (b) 0.52 s (c) 0.25 s (d) 0.03 s

22. A uniform magnetic �eld of

B

Iy

x

z 1000 G is established along

the positive z-direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop as shown in the �gure? (a) Zero (b) 1.8 × 10–2 N m(c) 1.8 × 10–3 N m (d) 1.8 × 10–4 N m


23. A transverse wave in a medium is described by the equation y = A sin 2(t – kx). �e magnitude of the maximum velocity of particles in the medium is equal to that of the wave velocity, if the value of A is

(a) l2

(b) l4

(c) l

(d) 2l

24. A small square loop of wire of side l is placed inside a large square loop of wire of side L (>> l). �e loops are coplanar and their centres coincide. What is the mutual inductance of the system?

(a) 2 2 02

lL

(b) 8 2 02

lL

(c) 2 22

02

lL

(d) None of these

25. An unpolarized light beam is incident on a surface at an angle of incidence equal to Brewster’s angle. �en,(a) the re�ected and the refracted beams are both

partially polarized(b) the re�ected beam is partially polarized and

the refracted beam is completely polarized and are at right angles to each other

(c) the re�ected beam is completely polarized and the refracted beam is partially polarized and are at right angles to each other

(d) both the re�ected and the refracted beams are completely polarized and are at right angles to each other.

26. �e range of voltmeter of resistance 300 is 5 V. �e resistance to be connected to convert it into an ammeter of range 5 A is (a) 1 in series (b) 1 in parallel(c) 0.1 in series (d) 0.1 in parallel

27. A square frame of side 1 m carries a current I, produces a magnetic �eld B at its centre. �e same current is passed through a circular coil having the same perimeter as the square. �e magnetic �eld at the centre of the circular coil is B′. �e ratio B/B′ is

(a) 82

(b) 8 22

(c) 162

(d) 162 2

28. A conducting circular loop is placed in a uniform magnetic �eld, B = 0.025 T with its plane perpendicular to the loop. �e radius of the loop is made to shrink at a constant rate of 1 mm s–1. �e induced emf when the radius is 2 cm, is

(a) 2p mV (b) p mV (c) 2

V (d) 2 mV

29. In an ac circuit, the current is

i t

5 1002

sin ampere and the potential

di�erence is V = 200sin(100t) volt. �en the power consumed is(a) 200 W (b) 500 W (c) 1000 W (d) Zero

30. �e resistance of the wire in the platinum resistance thermometer at ice point is 5 and at steam point is 5.25 . When the thermometer is inserted in an unknown hot bath its resistance is found to be 5.5 . �e temperature of the hot bath is(a) 100°C (b) 200°C (c) 300°C (d) 350°C

31. A series LCR circuit is connected to an ac source of variable frequency. When the frequency is increased continuously, starting from a small value, the power factor(a) goes on increasing continuously(b) goes on decreasing continuously(c) becomes maximum at a particular frequency(d) remains constant

32. A coil has 1000 turns and 500 cm2 as its area. �e plane of the coil is placed perpendicular to a uniform magnetic �eld of 2 × 10–5 T. �e coil is rotated through 180° in 0.2 seconds. �e average emf induced in the coil, in mV is(a) 5 (b) 10 (c) 15 (d) 20

33. Two parallel light rays are incident at one surface of a

prism as shown in the �gure. �e prism is made of glass of refractive index 1.5. �e angle between the rays as they emerge is nearly(a) 19° (b) 37° (c) 45° (d) 49°

34. What is the conductivity of a semiconductor if electron density = 5 1012 cm–3 and hole density = 8 1013 cm–3?

(e = 2.3 V–1 s–1 m2 and h = 0.01 V–1 s–1 m2)(a) 5.634 –1 m–1 (b) 1.968 –1 m–1

(c) 3.421 –1 m–1 (d) 8.964 –1 m–1

35. If l1 and l2 are the wavelengths of the �rst members of the Lyman and Paschen series respectively, then l1 : l2 is(a) 1 : 3 (b) 1 : 30 (c) 7 : 50 (d) 7 : 108

36. �e half life of radioactive radon is 3.8 days. �e

time at the end of which 120

t hof the radon sample

will remain undecayed is


(a) 3.8 days (b) 16.5 days(c) 33 days (d) 76 days

37. A hydrogen atom and a Li2+ ion are both in the second excited state. If lH and lLi are their respective electronic angular momenta, and EH and ELi their respective energies, then(a) l l E EH Li H Li and | | | |(b) l l E EH Li H Li and | | | |

(c) l l E EH Li H Li and | | | |(d) l l E EH Li H Li and | | | |

38. �e ratio of de Broglie wavelength of a proton and an a particle accelerated through the same potential di�erence is(a) 3 2 (b) 2 2 (c) 2 3 (d) 2 5

39. �e temperature of the sink of a Carnot engine is 27°C and its e�ciency is 25%. �e temperature of the source is(a) 227°C (b) 27°C (c) 327°C (d) 127°C

40. �e re�ecting surfaces of two M2

M1

mirrors M1 and M2 are at an angle (angle between 0° and 90°) as shown in the �gure. A ray of light is

incident on M1. �e emerging ray intersects the incident ray at an angle . �en,(a) = (b) = 180° – (c) = 90° – (d) = 180° – 2

SECTION-II (CHEMISTRY)

41. A nuclide of an alkaline earth metal undergoes radioactive decay by emission of a-particle in succession to give the stable nucleus. �e group of the periodic table to which the resulting daughter element would belong is(a) Group 4 (b) Group 6(c) Group 16 (d) Group 14

42. �ree cyclic structures of monosaccharides are given below. Which of these are anomers?

H OHH OH

HH OHH

CH OH2

O

(I)

HH OH

HH OHH

CH OH2

O

(II)

HHOHH

H

CH OH2

O

(III)

H

(a) I and II (b) II and III(c) I and III(d) III is anomer of I and II.

43. �e electronegativity of the following elements increases in the order(a) C < N < Si < P (b) N < Si < C < P(c) Si < P < C < N (d) P < Si < N < C

44. �e atomic masses of He and Ne are 4 and 20 amu respectively. �e value of the de Broglie wavelength of He gas at –73°C is ‘M’ times that of the de Broglie wavelength of Ne at 727°C. ‘M’ is(a) 2 (b) 3 (c) 4 (d) 5

45. A �xed mass of a gas is subjected to transformations of state from K to L to M to N and back to K as shown. �e pair of isochoric processes among the transformations of state is(a) K to L and L to M K

N

L

MPV

(b) L to M and N to K(c) L to M and M to N (d) M to N and N to K

46. For the estimation of nitrogen, 1.4 g of an organic compound was digested by Kjeldahl’s method and the evolved ammonia was absorbed in 60 mL of M10

sulphuric acid. �e unreacted acid required

20 mL of M10

sodium hydroxide for complete

neutralisation. �e percentage of nitrogen in the compound is(a) 5% (b) 6% (c) 10% (d) 3%

47. Which one of the following sequences represents the correct increasing order of bond angles in the given molecules?(a) H2O < OF2 < OCl2 < ClO2(b) OCl2 < ClO2 < H2O < OF2(c) OF2 < H2O < OCl2 < ClO2(d) ClO2 < OF2 < OCl2 < H2O

48. Which one of the following is the correct statement?(a) B2H6·2NH3 is known as ‘inorganic benzene’.(b) Boric acid is a protonic acid.(c) Beryllium exhibits coordination number of

six.(d) Chlorides of both beryllium and aluminium

have bridged chloride structures in solid phase.

49. van der Waals’ equation for 0.2 mol of a gas is


(a) P aV

2

+ (V – 0.2b) = 0.2RT

(b) P aV

0 04 2.

(V – b) = 0.02RT

(c) P aV

0 22

. (V – 0.2b) = 0.2RT

(d) P aV

0 042

.(V – 0.2b) = 0.2RT

50. If l0 is the threshold wavelength for photoelectric emission, l the wavelength of light falling on the surface of a metal and m the mass of the electron, then the velocity of ejected electron is given by

(a)

20

1 2h

m( )

/l l

(b)

20

1 2hcm

( )/

l l

(c)

2 0

0

1 2hcm

l ll l

/

(d)

2 1 10

1 2h

m l l

/

51. �e �nal product ‘Q’ in the following reaction is

(a)

(b)

(c)

(d)

52. Which one of the following is a quaternary salt?

(a)

(b) (CH3)3N+

HCl–

(c)

(d) (CH3)4N+

Cl–

53. Silver is monovalent and has an atomic mass of 108. Copper is divalent and has an atomic mass of 63.6. �e same electric current is passed, for the same length of time through a silver coulometer and a copper coulometer. If 27.0 g of silver is deposited, then the corresponding amount of copper deposited is(a) 63.60 g (b) 31.80 g (c) 15.90 g (d) 7.95 g

54. A compound contains atoms X, Y and Z. �e oxidation number of X is +3, Y is +5 and Z is –2. �e possible formula of the compound is

(a) XYZ2 (b) Y2(XZ3)2(c) X3(YZ4)3 (d) X3(Y4Z)2

55. Which of the following values of stability constant K corresponds to the most unstable complex compound?(a) 1.6 × 107 (b) 4.5 × 1014

(c) 2.0 × 1027 (d) 5.0 × 1033

56. 0.22 g of an organic compound CxHyO which occupied 112 mL at STP, and on combustion gave 0.44 g of CO2. �e ratio of x to y in the compound is (a) 1 : 1 (b) 1 : 2 (c) 1 : 3 (d) 1 : 4

57. Which of the following is pseudo alum? (a) (NH4)2SO4.Fe2(SO4)3.24H2O(b) K2SO4.Al2(SO4)3.24H2O(c) MnSO4.Al2(SO4)3.24H2O(d) None of these.

58. Which of the following is not the characteristic property of alcohols?(a) Lower members have pleasant smell, higher

members are odourless, tasteless.(b) Lower members are insoluble in water, but it

regularly increases with increase in molecualar weight.

(c) Boiling point rises regularly with the increase in molecular weight.

(d) Alcohols are lighter than water.

59. For the reaction A + 2B C + D, the equilibrium constant is 1.0 × 108. Calculate the equilibrium concentration of A if 1.0 mole of A and 3.0 moles of B are placed in 1 L �ask and allowed to attain the equilibrium.(a) 1.0 × 10–2 mol L–1 (b) 2.1 × 10–4 mol L–1

(c) 5 × 10–5 mol L–1 (d) 1.0 × 10–8 mol L–1

60. �e correct order of basicities of the following compounds is

CH CH NH3 2 2CH C3 NH2

NH

I II

(CH ) NH3 2III IV

CH C NH3 2

O

(a) II > I > III > IV (b) I > III > II > IV(c) III > I > II > IV (d) I > II > III > IV


61. CH C CH CH3 2

CH3

CH3

B H , THF2 6

H O , OH2 2– ( )A

�e product (A) is

(a) CH C CH CH3 3

CH3

CH3 OH

(b) CH C CH CH3 3

OH

CH3 CH3

(c) CH C CH CH OH3 2 2

CH3

CH3

(d) CH C CH CH2 2 3

OH CH3

CH3

62. Which one of the following statements is not true?(a) For �rst order reaction, straight line graph

of log (a – x) versus t is obtained for which slope = – k/2.303.

(b) A plot of log k vs 1/T gives a straight line graph for which slope = – Ea/2.303R.

(c) For third order reaction, the product of t1/2 and initial concentration a is constant.

(d) Units of k for the first order reaction are independent of concentration units.

63. If K1 and K2 are the ionization constants of H3N

+ CHRCOOH and H3N+ CHRCOO– respectively,

the pH of the solution at the isoelectric point is(a) pH = pK1 + pK2 (b) pH = (pK1 pK2)1/2

(c) pH = (pK1 + pK2)1/2 (d) pH = (pK1 + pK2)/264. �e rate of change of concentration of (A) for

reaction: A B is given by

d Adt

k A[ ]

[ ] /1 3

�e half-life period of the reaction will be

(a) 3 2 1

20

2 3 2 3 2

5 3[ ] [( ) ]

( )

/ /

/A

k

(b)

32

2 102 3 2 3[ ] [( ) ]/ /A

k

(c) 3 2 1

20

2 3 2 3

5 3[ ] [( ) ]

( )

/ /

/A

k

(d)

23

2 103 2 2 3[ ] [( ) ]/ /A

k

65. Which of the following compounds have two lone pairs of electrons?

NO2– , OF2, XeF2, ICl3

I II III IV(a) I and III (b) II and III(c) II and IV (d) I and IV

66. Given: G° = –nFE°cell and G° = – RT lnK �e value of n = 2 will be

given by the slope of which line in the given �gure?(a) OA(b) OB(c) OC (d) OD

67. An amide, C3H7NO upon acid hydrolysis gives an acid, C3H6O2 which on reaction with Br2/P gives a-bromo acid which when boiled with aq. OH– followed by acidi�cation gives lactic acid. What is the structural formula of amide?

(a) O

CH C NH CH3 3 (b) O

H C NH CH CH2 3

(c) O

CH CH C NH3 22

(d) O

H C NCH3

CH368. Certain volume of a gas exerts some pressure on

walls of a container at constant temperature. It has been found that by reducing the volume of the gas to half of its original value, the pressure becomes twice that of initial value at constant temperature. �is is because(a) the weight of the gas increases with pressure(b) velocity of gas molecules decreases(c) more number of gas molecules strike the

surface per second(d) gas molecules attract one another.

69. Choose the correct comparison of heat of hydrogenation for the following alkenes:

I II III

IV V

(a) II < IV < III < V < I(b) III < IV < I < V < II(c) V < IV < III < I < II(d) IV < V < I < III < II


70. �e data for the reaction: A + B C, is given below:Ex. [A]0 [B]0 Initial rate1. 0.012 0.035 0.102. 0.024 0.070 0.803. 0.024 0.035 0.104. 0.012 0.070 0.80

�e rate law which corresponds to the above data is(a) rate = k[B]3 (b) rate = k[B]4

(c) rate = k[A][B]3 (d) rate = k[A]2[B]2

71. Which one of the following octahedral complexes will not show geometrical isomerism?

(A and B are monodentate ligands.)(a) [MA2B2] (b) [MA3B3](c) [MA4B2] (d) [MA5B]

72. �e correct IUPAC name of the following compound is

(a) 3,7,7-trimethylhepta-2,6-dien-1-al(b) 2,6-dimethylocta-3,7-dien-1-al(c) 2,6,6-trimethylhepta-3,7-dien-1-al(d) 3,7-dimethylocta-2,6-dien-1-al.

73. Catenation, i.e., linking of similar atoms depends on size and electronic con�guration of atoms. �e tendency of catenation in group 14 elements follows the order(a) C > Si > Ge > Sn (b) C >> Si > Ge Sn(c) Si > C > Sn > Ge (d) Ge > Sn > Si > C

74. In a solid, oxide ions are arranged in ccp. One-sixth of the tetrahedral voids are occupied by cations X while one-third of the octahedral voids are occupied by the cations Y. �e formula of the compound is(a) X2YO3 (b) XYO3 (c) XY2O3 (d) X2Y2O3

75. In an atom, an electron is moving with a speed of 600 m s–1 with an accuracy of 0.005%. Certainty with which the position of the electron can be located is (h = 6.6 × 10–34 kg m2 s–1, mass of electron, me = 9.1 × 10–31 kg)(a) 1.52 × 10–4 m (b) 5.10 × 10–3 m(c) 1.92 × 10–3 m (d) 3.84 × 10–3 m

76. Which of the following cannot be prepared by Williamson’s synthesis?

(a) Methoxybenzene (b) Benzyl p-nitrophenyl ether(c) Methyl tert.-butyl ether(d) Di-tert.-butyl ether

77. Which of the following reactions is an example of SN2 reaction?(a) CH3Br + OH– CH3OH + Br–

(b)

(c) CH3CH2OH CH2 CH2(d) (CH3)3C—Br + OH– (CH3)3C—OH + Br–

78. �e solubility of metal halides depends on their nature, lattice enthalpy and hydration enthalpy of the individual ions. Amongst �uorides of alkali metals, the lowest solubility of LiF in water is due to(a) ionic nature of lithium �uoride(b) high lattice enthalpy(c) high hydration enthalpy for lithium ion(d) low ionisation enthalpy of lithium atom.

79. Which of the following statements is incorrect?(a) �e covalent bonds have some partial ionic

character.(b) �e ionic bonds have some partial covalent

character.(c) �e larger the size of the cation and the smaller

the size of the anion, the greater is the covalent character of the ionic bond.

(d) �e greater the charge on the cation, the greater is the covalent character of the ionic bond.

80. Which of the following is a constituent of nylon?(a) Adipic acid (b) Styrene(c) Te�on (d) None of these

SECTION-III (ENGLISH AND LOGICAL REASONING)

PASSAGE

Direction (Questions 81 to 84): Read the passage and answer the following questions.Democratic societies from the earliest times have expected their governments to protect the weak against the strong. No ‘era of good feeling’ can justify discharging the police force or giving up the idea of public control over concentrated private wealth. On the other hand, it is obvious that a spirit of self-denial and moderation on the part of those who hold economic power will greatly so�en the demand for absolute equality. Men are more interested in freedom and security than in an


equal distribution of wealth. �e extent to which the government must interfere with business, therefore, is not exactly measured by the extent to which economic power is concentrated into a few hands. �e required degree of government interference depends mainly on whether economic powers are oppressively used, and on the necessity of keeping economic factors in a tolerable state of balance.But with the necessity of meeting all these dangers and threats to liberty, the powers of the government are unavoidably increased, whichever political party may be in o�ce. �e growth of government is a necessary result of the growth of technology and of the problems that go with the use of machines and science. Since the government in our nation, must take on more powers to meet its problems, there is no way to preserve freedom except by making democracy more powerful.81. �e advent of science and technology has increased

the _____.(a) powers of the governments(b) chances of economic inequality(c) freedom of people(d) tyranny of the political parties

82. A spirit of moderation on the economically sound people would make the less privileged _____.(a) unhappy with their lot(b) clamour less for absolute equality(c) unhappy with the rich people(d) more interested in freedom and security

83. �e growth of government is necessitated to _____.(a) monitor science and technology(b) deploy the police force wisely(c) make the rich and the poor happy(d) curb the accumulation of wealth in a few hands

84. ‘Era of good feeling’ in sentence 2 refers to _____.(a) time without government(b) time of police atrocities(c) time of prosperity(d) time of adversity

Direction (Questions 85 to 86): Choose the correct antonym for each of the following words.

85. Autonomy(a) Subordination (b) Slavery(c) Submissiveness (d) Dependence

86. Laconic(a) Proli�c (b) Bucolic(c) Prolix (d) Pro�igate

Direction (Questions 87 to 89): In each of the following questions. Find out which part of the sentence has an error. If there is no mistake, the answer is ‘No error’.87. (a) Having deprived from their homes

(b) in the recent earthquake(c) they had no other option but(d) to take shelter in a school.

88. (a) �e Ahujas (b) are living in this colony(c) for the last eight years.(d) No error.

89. (a) She sang (b) very well(c) isn’t it? (d) No error.

Direction (Questions 90 to 93): Rearrange the given �ve sentences A, B, C, D, E in the proper sequence so as to form a meaningful paragraph and then answer the given questions.A. Many consider it wrong to blight youngsters by

recruiting them into armed forces at a young age.B. It is very di�cult to have an agreement on an

issue when emotions run high.C. �e debate has again come up whether this is right

or wrong.D. In many countries military service is compulsory

for all.E. Some of these detractors of compulsory dra� are

even very angry.90. Which sentence should come fourth in the

paragraph?(a) A (b) B (c) C (d) D

91. Which sentence should come �rst in the paragraph?(a) A (b) B (c) C (d) D

92. Which sentence should come last in the paragraph?(a) A (b) B (c) C (d) D

93. Which sentence should come third in the paragraph?(a) A (b) B (c) C (d) E

Direction (Questions 94 to 95): In each of the following questions, a word has been written in four di�erent ways out of which only one is correctly spelt. Choose the correctly spelt word.94. (a) Sepalchral (b) Sepulchrle

(c) Sepulchral (d) Sepalchrle95. (a) Overleped (b) Overelaped

(c) Overlapped (d) Overlaped


96. How many such pairs of letters are there in the word RECRUIT each of which has as many letters between them in the word as they have in the English alphabet series?(a) None (b) One (c) Two (d) �ree

97. Which of the following forms the mirror image of given word, if the mirror is placed vertically to the le�?

1 B F 9 8 l 6 i F(a)

1 B F 9 8 l 6 i F (b) 1 B F 9 8 l 6 i F

(c) F B F 9 8 l 6 i 1 (d)

F B F 9 8 l 6 i 1

98. In a certain code language,I. 'she likes apples' is written as 'pic sip dip'.II. 'parrot likes apples lots' is written as 'dip pic

tif nif '.III. 'she likes parrot' is written as 'tif sip dip'.

How is 'parrot' written in that code language?(a) pic (b) dip (c) tif (d) nif

99. Find the �gure from the options which will continue the series established by problem �gures.

Problem Figures

(a) (b)

(c) (d)

100. X's mother is the mother-in-law of the father of Z. Z is the brother of Y while X is the father of M. How is X related to Z?(a) Paternal uncle (b) Maternal uncle(c) Cousin (d) Grandfather

101. Read the following information carefully and answer the question given below it :1. �ere is a group of six persons A, B, C, D, E and

F in a family. �ey are Psychologist, Manager, Lawyer, Jeweller, Doctor and Engineer.

2. �e doctor is the grandfather of F who is a Psychologist.

3. �e Manager D is married to A.4. C, the Jeweller, is married to the Lawyer.5. B is the mother of F and E.6. �ere are two married couples in the family.

What is the profession of E?

(a) Doctor (b) Engineer(c) Manager (d) Psychologist

102. Study the information given below carefully and answer the question that follows.

On a playing ground, Mohit, Kartik, Nitin, Atul and Pratik are standing as described below facing the North.I. Kartik is 40 metres to the right of Atul.II. Mohit is 70 metres to the south of Kartik.III. Nitin is 35 metres to the west of Atul.IV. Pratik is 100 metres to the north of Mohit.

Who is to the north-east of the person who is to the le� of Kartik?(a) Mohit (b) Nitin (c) Pratik (d) Atul

103. Select a �gure from the options which will continue the same series as established by the Problem Figures.

?

Problem Figures

(a) (b)

(c) (d)

104. How many straight lines are required to draw the given �gure?(a) 8(b) 9(c) 10(d) 11

105. Select a �gure from the options which forms the water image of the given combination of letters/numbers.

8C185e2F9(a) (b) (c) (d)

SECTION-IV (MATHEMATICS)

106. �ree numbers, the third of which being 12, form decreasing G.P. If the last term was 9 instead of 12, the three numbers would have formed an A.P. �e common ratio of the G.P. is(a) 1/3 (b) 2/3 (c) 3/4 (d) 4/5

107. In a plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the


point B. Besides, no three lines pass through one point, no lines passes through both points A and B, and no two are parallel, then the number of intersection points the lines have is equal to(a) 535 (b) 601 (c) 728 (d) 963

108. �e value of the determinant

11

1

3 4

3 2 3

4 2 3

e ee ee e

i i

i i

i i

/ /

/ /

/ /,

where i = 1, is

(a) 2 2 (b) ( )2 2(c) 2 3 (d) 2 3

109. A ri�e man �ring at a distant target and has only 10% chance of hitting it. �e minimum number of rounds he must �re in order to have 50% chance of hitting it atleast once is(a) 6 (b) 7 (c) 8 (d) 9

110. limcos cos cos

sinx

x x x

x

0 2

1 2 3

2 is equal to

(a) 7/2 (b) 7/3 (c) 7/4 (d) 7/5111. If the tangent at the point P on the circle x2 + y2 +

6x + 6y = 2 meets the straight line 5x – 2y + 6 = 0 at a point Q on the y-axis, then the length of PQ is(a) 4 (b) 2 5 (c) 5 (d) 3 5

112. �e points A, B and C represent the complex numbers z1, z2, (1 – i)z1 + iz2 (where i = 1 ) respectively on the complex plane. �e triangle ABC is(a) isosceles but not right angled(b) right angled but not isosceles(c) isosceles and right angled(d) none of these

113. �e area of the �gure bounded by the curves y = x – 1 and y = 3 – x is(a) 2 sq. units (b) 3 sq. units(c) 4 sq. units (d) 1 sq. unit

114. �e direction cosines of the line drawn from P(– 5, 3, 1) to Q(1, 5, – 2) is(a) (6, 2, – 3) (b) (2, – 4, 1)

(c) (– 4, 8, –1) (d) 67

27

37

, ,

115. 18

1 38

1 58

1 78

cos cos cos cos

is equal to

(a) 1/2 (b) cos /8

(c) 1/8 (d) 1 22 2

116. �e two vectors { , }

a i j k b i j k 2 3 4 6l are parallel if l is equal to(a) 2 (b) – 3 (c) 3 (d) – 2

117. Let f(x) =

a x x

x xx

b xx xx

x

| |,

,[ ] ,

2

22

22

2

22

([] denotes the greatest integer function)If f(x) is continuous at x = 2, then(a) a = 1, b = 2 (b) a = 1, b = 1(c) a = 0, b = 1 (d) a = 2, b = 1

118. Let a, b, c ∈ R and a 0. If a is a root of a2x2 + bx + c = 0, is a root of a2x2 – bx – c = 0 and 0 < a < , then the equation a2x2 + 2bx + 2c = 0 has a root that always satis�es(a) = a (b) = (c) = (a + )/2 (d) a < <

119. �e term independent of x in the expansion of

xx33

2 2

10

is

(a) 5/12 (b) 1(c) 1/3 (d) None of these

120. If a < 0, the function f(x) = eax + e–ax is a monotonically decreasing function for values of x given by (a) x > 0 (b) x < 0(c) x > 1 (d) x < 1

121. �e number of solutions of the equation tan x + sec x = 2cos x lying in the interval [0, 2]

is(a) 0 (b) 1 (c) 2 (d) 3

122. Let R be a relation on the set of all lines in a plane de�ned by (l1, l2) ∈ R such that l1 l2 then R is(a) re�exive only (b) symmetric only(c) transitive only (d) equivalence

123. Let P(n) = 5n – 2n, P(n) is divisible by 3l, where l and n both are odd positive integers then the least value of n and l will be(a) 13 (b) 11 (c) 1 (d) 5


124. �e rank of 4 0 00 3 00 0 5

is equal to

(a) 4 (b) 3 (c) 5 (d) 1125. A single letter is selected at random from the word

'PROBABILITY'. �e probability that it is a vowel, is(a) 2/11 (b) 3/11(c) 4/11 (d) none of these

126. Let A {1, 2, 3, 4}, B {a, b, c}, then number of functions from A B, which are not onto is(a) 8 (b) 24 (c) 45 (d) 6

127. �e minimum value of the function de�ned by f(x) = maximum {x, x + 1, 2 – x} is

(a) 0 (b) 12

(c) 1 (d) 32

128. For parabola x2 + y2 + 2xy – 6x – 2y + 3 = 0, the focus is(a) (1, – 1) (b) (– 1, 1)(c) (3, 1) (d) none of these

129. If A = {x : x = 4n + 1, 2 n 6}, then number of subsets of A are(a) 22 (b) 23 (c) 25 (d) 26

130. �e mean deviation and S.D. about actual mean of the series a, a + d, a + 2d, ..., a + 2nd are respectively

(a) n n d

nn n d( ) , ( )

1

2 11

3

(b) n n n nn

d( ) , ( ) 13

12

(c) n n d

nn n d( )

( ), ( )

1

2 11

3

(d) n n d

nn n d( ) , ( )

1

2 11

3131. If the arithmetic progression whose common

di�erence is non zero, the sum of �rst 3n terms is equal to the sum of the next n terms. �en the ratio of the sum of the �rst 2n terms to the next 2n terms is(a) 1 : 5 (b) 2 : 3(c) 3 : 4 (d) none of these

132. lim ( !) /

n

nnn

1 equals

(a) e (b) e–1


133. A man of height 2 m walks directly away from a lamp of height 5 m, on a level road at 3 m/s. �e rate at which the length of his shadow is increasing is(a) 1 m/s (b) 2 m/s (c) 3 m/s (d) 4 m/s

134. If the foci of the ellipse x y

b

2 2

225 = 1 and the

hyperbola x y2 2

144 811

25 coincide, then the value

of b2 is(a) 3 (b) 16 (c) 9 (d) 12

135. If sin–1 x + sin–1 y = 23 , then cos–1 x + cos–1 y is

equal to

(a) 23 (b)

3 (c)

6 (d)

136. If the sum of the binomial coe�cients in the

expansion of xx

n

is 64, then the term

independent of x is equal to(a) 10 (b) 20 (c) 40 (d) 60

137. Solution of the di�erential equation

sin ydydx

= cos y(1 – x cos y) is

(a) sec y = x – 1 – cex (b) sec y = x + 1 + cex

(c) sec y = x + ex + c (d) none of these138. �e function f(x) = |x2 – 3x + 2| + cos|x| is not

di�erentiable at x is equal to(a) –1 (b) 0 (c) 1 (d) 2

139. From 6 di�erent novels and 3 di�erent dictionaries, 4 novels and 1 dictionary are to be selected and arranged in a row on a shelf so that the dictionary is always in the middle. �en the number of such arrangements is (a) at least 500 but less than 750(b) at least 750 but less than 1000(c) at least 1000 (d) less than 500

140. A normal to y2 = 4ax at t touches x2 – y2 = a2, then (t2 + 1)3 is (a) < 0 (b) > 0(c) 0 (d) nothing can be said

141. If f : X Y de�ned by f(x) = 3 sin x + cos x + 4 is one-one and onto, then Y is(a) [1, 4] (b) [2, 5] (c) [1, 5] (d) [2, 6]

142. If

a b c, , be unit vectors such that

a b c , is inclined at the same angle to both

a b c pa qb r a band and ( ), then which of the following is true?


(a) p = q (b) |p| 1(c) |q| 1 (d) All of these

143. Area of the region bounded by the curves, y = ex, y = e–x and the straight line x = 1 is given by(a) (e – e–1 + 2) sq. units(b) (e – e–1 – 2) sq. units(c) (e + e–1 – 2) sq. units(d) none of these

144. If the normal at the end of latus rectum of the

ellipse xa

y

b

2

2

2

2 = 1 passes through (0, –b), then

e4 + e2 (where e is eccentricity) equals

(a) 1 (b) 2

(c) 5 12 (d) 5 1

2

145. A variable plane passes through the �xed point (a, b, c) and meets the axes at A, B, C. �e locus of the point of intersection of the planes through A, B, C and parallel to the coordinate planes is

(a) ax

by

cz

2 (b) ax

by

cz

1

(c) ax

by

cz

2 (d) ax

by

cz

1

146. If a ,

b and c are unit coplanar vectors, then the scalar triple product

[ ]2 2 2

a b b c c a is equal to(a) 0 (b) 1 (c) 3 (d) 3

147. �e total number of numbers that can be formed by using all the digits 1, 2, 3, 4, 3, 2, 1, so that the odd digits always occupy the odd places, is(a) 3 (b) 6 (c) 9 (d) 18

148. Number of real roots of the equation

x x x ( )1 = 1 is(a) 0 (b) 1 (c) 2 (d) 3

149. xex

dxx

( )1 2 is equal to

(a) ex

cx

1 (b) ex(x + 1 ) + c

(c)

ex

cx

( )1 2 (d) ex

cx

1 2

150. If y = x y x y ... , then dydx

is equal

to

(a) 1

2 1y (b) y x

y xy

2

32 2 1

(c) 2y – 1 (d) none of these


SOLUTIONS

1. (a) 2. (b)

3. (a) :

M

/2

From �gure, AC = R/2, PC = H, h = MP = ?

Given, R = H or ug

ug

2 2 222

sin sin

or tan = 4

In AMC, tan MCAC

MP PCAC

42

h HR /

4 2( )h HH

h H

4. (b) : [Angular momentum][Magnetic moment]

2mq

2mI t

MAT

= [MA–1 T–1]

5. (d) : ABBC

AB DC l 23

and BC AD l 6

Similarly, m m mAB DC

3

and m m mBC AD

6Moment of inertia about the desired axis isI = IAB + IAD + IDC + IBC

13 3 3 6 3

13 3 3

2 2 2m l m l m l 0 7162

2ml

6. (d) : Rotational kinetic energy is

K I MK vR

I MK v RR

12

12

2 22

2 ( ) and

12

22

2Mv K

RTranslational kinetic energy, K MvT 1

22

As per question, KR = 40% KT

12

40 12

22

22Mv K

RMv% K

R

2

225

For solid sphere, KR

2

225

Hence, the body is a solid ball.

7. (b)8. (c) : As the spring is compressed by 2 m with the

application of a force of 200 N, hence its spring constant k is given by

= 30°

20 kg

hk F 2

2002

100 1Nm

N m

Suppose l be the distance along the inclined plane. Applying the conservation of energy,

12 1

2kx mgh mgl sin

or 12

100 4 20 9 8 12

2 . l

l 80098

8 17. m

9. (d) 10. (d)11. (b) : According to law of conservation of

mechanical energy12

02mu GMmR

u GMR

2 2

u GMR

gR g GMR

2 2 2

12. (a)13. (c) : Here, mass of a particle = M

Mass of a spherical shell = M Radius of a spherical shell = a Gravitational potential at point P due to Q particle at O,

V GM

a1 2

/

Gravitational potential at point P due to spherical

shell, V GMa2

Hence, total gravitational potential at point P is

V = V1 + V2

GMa

GMa/ 2

3GMa

| |V GMa

3

14. (d)15. (d) : Here, n1 = 4, T1 = 400 K, n2 = 2, T2 = 700 K

�e temperature of the mixture is

Tmixture = n T n T

n n1 1 2 2

1 2

1600 14006

500K K

K


16. (b) 17. (c) : When capacitor is fully charged, it draws

no current. Hence potential difference across capacitor = Potential di�erence across C and F.E�ective resistance of the network between A and D is

R13 3 33 3 3

2

( )( )

Total resistance of the circuit= 2 + 3 = 5

I1I1

I2 II I

B

A

EC

F

GH

3

3 3

15 V

3 3 D

3 F

Current, I 155

V

= 3 A

I13 33 6

1( )( )A

A

and

AAI2

3 63 6

2

( )( )

Potential di�erence across A and D = 3 2A = 6 VPotential di�erence across D and F, VD – VF = 3 3 A = 9 VPotential di�erence across C and D, VC – VD = 3 1 A= 3 VPotential di�erence across capacitor = VC – VF = (VC – VD) + (VD – VF) = 3 V + 9 V = 12 V

18. (a)19. (c) : �e currents through various arms will be

as shown in �gure.

S

10 V4

5 V2

2 31 2

Let V be the potential at C.Applying Kirchho� ′s �rst law at C, we get I1 + I2 = I3

10

45

20

2 V V V

V = 4 V

I342

2V

A

20. (d) : 260 12

1 l

T

T1 = 50.7g = 507 N

When the mass is submerged, upthrust = (0.0075 m3)(103 kg m–3) (10 m s–2) = 75 NNew tension, T2 = (507 – 75) N = 432 NNew fundamental frequency,

1

22

lT

260

432507

1213

2

1

TT

or = 240 Hz

21. (d) : From the given graph, k = Slope of F-x graph

( )( . )

80 00 2 0

400 1Nm

N m

Time period of SHM is

T mk

2

20 01

4000 03

1

..

kg

N ms

22. (a)23. (b) : �e given equation of the transverse wave

is y = A sin2(t – kx)

Velocity of the particle =dydt

= 2Acos2(t – kx)Maximum velocity = 2A

Velocity of the wave = Coefficient ofCoefficient of

tx k k 2

2

As per question,

2 2 12 4

Ak

Ak

A l

l

or

24. (a) 25. (c)26. (b) : A voltmeter is a galvanometer having a high

resistance connected in series with it. �e current through the galvanometer is

Ig 5

3001

60V

A

An ammeter is a galvanometer having a low resistance connected in parallel with it. �e shunt resistance S is determined from

I

I ISG

g

g

Given G = 300 , I = 5 A, we get1 60

5 1 60 3001/

( / ) S S

27. (d)28. (b) : Magnetic �ux linked with the loop is

= BAcos = B(r2)cos0° = Br2

�e magnitude of the induced emf is

d

dtddt

B r B r drdt

2 2

= 0.025 × × 2 × 2 × 10–2 × 1 × 10–3 = × 10–6 V = V


29. (d) : Here, i t

5 1002

sin A

V = 200sin (100t) V

Phase di�erence between V and i is, 2

Power consumed, P = Vrms irms cos = Vrms irms cos90° = zero

30. (b) 31. (c) 32. (b)33. (b) :

48.59°

48.59°

30° 18.59°18.59°18.59°

30°

30°

30°

18.59°

A

B

C N

N

At the �rst surface AC, the ray will pass undeviated.Applying the Snell’s law at second surface AB, we get

sin30° = 1sinr 1 5 12

. sin r

sinr = 0.75 r = 48.59 �e angle between the incident horizontal ray and the emergent ray is 18.59°. �e angle between the two emergent rays is nearly 37° i.e. 2 × 18.59°.

34. (b) : = e[nee + nhh]= 1.6 10–19[5 1018 2.3 + 8 1019 0.01]= 1.968 –1 m–1

35. (d)

36. (b) : As N = N0e–lt or NN

e t

0 l

Taking natural logarithm on both sides, we get

l lt N

Nt

NN

ln ln0

0or or t

NN

1 0l

ln

Here, l ln ..

. ,/

2 0 6933 8

0 182 1201 2 0T

NN

t 10 182

20 16 5.

ln( ) . days

37. (a) : In second excited state, n = 3,

So, l l hH Li

32

while E Z2 and ZH = 1, ZLi = 3So, E E E ELi H H Lior 9

38. (b) : de Broglie wavelength associated with charged particle is given by,

or l

l

laa a h

mqVm q Vm q V

p

p p222

,

As andm

mqq

p

pa

a 14

21

l

la

p 4 2 2 2

39. (d) : E�ciency of a Carnot engine, 1 2

1

TT

Here, = 25% , T2 = 27°C = 300 K

25100

1 300

1T T1 = 400 K = 127°C

40. (d) 41. (d) : Alkaline earth metals are Group 2 elements.

Emission of a-particle (24He) will reduce its atomic

number by 2 units and thus, displaces the daughter nuclei two positions le� in the periodic table, thus to group 18, 16, 14, 12, 10, 8, 6 or 4, etc. As the last stable daughter nuclei formed could be either Pb (Group 14) or Bi (Group 15) therefore, the daughter nuclei would belong to Group 14.

42. (a) : Structures I and II di�er only in the position of OH at C–1 and hence are anomers.

43. (c)

44. (d) : l = h

m KE2 ( )

ll

He

Ne

Ne Ne

He He

m KEm KE

( )( )

…(i)

mm

Ne

He 20

4 = 5 …(ii)

KE T

( )( )KEKE

Ne

He

727 27373 273

1000200

= 5 …(iii)

Put values from eqns. (ii) and (iii) in eqn. (i)

ll

He

Ne 5 5 = 5

45. (b) : For transformations L M and N K, volume is constant.

46. (c) : Mass of an organic compound = 1.4 g

% of NMeq. of acid consumed

Mass of compound taken

1 4.

Meq. of acid consumed


60 1

102 20 1

101 = 10

[Basicity of acid = 2]

% of N 1 4 10

1 4.

. = 10%

47. (c) 48. (d)49. (d) : For ‘n’ moles of gas, van der Waals’ equation is :

P anV

2

2(V – nb) = nRT

For 0.2 mol, P aV

0 042

.(V – 0.2b) = 0.2RT

50. (c) :

12

20

0mv h h c c

( )

l l

hc hc1 1

0

0

0l ll ll l

or v hcm

2 0

0

2

l ll l

; or v hcm

2 0

0

1 2l ll l

/

51. (d) :

52. (d) : N is attached to four carbon atoms.53. (d) : When the same quantity of electricity is passed

through the solutions of di�erent electrolytes, the weights of the elements deposited are directly proportional to their chemical equivalents.

i.e., Wt. of Cu depositedWt. of Ag deposited

Eq. wt. of CuEq. wt. of Ag

⇒ WCu

2763 6 2108 1

. //

WCu g

63 6 272 108

7 95.

.

54. (c) : Sum of oxidation numbers of all the atoms in X3(YZ4)3 is zero i.e.3 × (+3) + 3 [1 × (+5) + 4 × (–2)] = 9 – 9 = 0In all other compounds, the sum of oxidation numbers is a �nite number.

55. (a)

56. (b) : Mol. wt. of compound 0 22 22400112

44. g

% of C 1244

0 44 1000 22

54 54..

.

Amount of C in compound 44 54 54100

.= 24 g

Molecular formula is C2HyO.Now, corresponding to mol. wt. 44, molecular formula will be C2H4O.Hence x : y is 1 : 2.

57. (c)58. (b) : Lower members, only, are soluble in water.59. (d) : A + 2B C + D;

x = mol of A remaining at equilibriumMoles of A reacted = 1 – x;Moles of B reacted = 2(1 – x)Moles of B remaining = 3 – 2(1 – x) = 1 + 2x 1 (K is very large, x << 1)Moles of C = Moles of D = 1 – x 1

Hence, K = 1.0 × 108

[ ][ ][ ][ ] ( )

C DA B x x2 2

1 11

1

x = 1.0 × 10–8 mol L–1

60. (b) :

The conjugate acid obtained by addition of a proton to (I) is stabilized by two equivalent resonance structures and hence compound (I) is the most basic. Further 2° amines are more basic than 1° amines while amides are least basic due to delocalization of lone pair of electrons of N over the CO group. �us the order isI > III > II > IV.

61. (c) : It is a hydroboration-oxidation reaction. It is the addition of H2O according to anti-Markownikoff ’s rule. Hence, terminal carbon gets the – OH group.

62. (c)

63. (d) :


K1 [H NCH COO ][H ]

[H NCH COOH]3

+ +

3+

R

R

K2

[ ]

[ ]

H NCH COO [H ]

H NCH COO2

+

3+

R

R

�us, K K1 2 [H NCH COO ][H ]

[H NCH COOH]2

+ 2

3+

R

R

At the isoelectric point,[H2NCHRCOO–] = [H3N+CHRCOOH] K1K2 = [H+]2; 2log[H+] = logK1 + logK2 –2log [H+] = –logK1 – logK2 2pH = pK1 + pK2or pH = (pK1 + pK2)/2

64. (c) :

d Adt

k Ad A

Ak dt

[ ][ ]

[ ]

[ ]/

/1 3

1 3

32

2 3[ ] /A kt C …(i)

At t = 0; [A] = [A0]

�en, 32 0

2 3[ ] /A C

Putting the value of C in eq. (i), we get

32

32

2 30

2 3[ ] [ ]/ /A kt A

kt A A32

320

2 3 2 3[ ] [ ]/ /

At t = t1/2; [ ][ ]

AA

02

kt AA

1 2 02 3 0

2 332

32 2/

//

[ ]

kt AA

1 2 02 3 0

2 3

2 332 2

//

/

/[ ]

[ ]

( )

t

kA1 2 0

2 32 3

32

1 12

//

/[ ]

( )

t

Ak1 2

02 3 2 3

2 33

22 1

2/

/ /

/[ ]

t

Ak1 20

2 3 2 3

5 3

3 2 12

/

/ /

/

[ ]

65. (c) 66. (b)

67. (c) :

68. (c) : As the volume is decreased, pressure increases

because more number of molecules strike the surface per second.

69. (c) : Greater is the stability of alkene, lower the heat of hydrogenation.Out of cis and trans isomers, trans isomer is more stable than cis isomer in which two alkyl groups lie on the same side of the double bond and hence cause steric hindrance, therefore, heat of hydrogenation of trans isomer is less than that of cis isomer.

70. (a)71. (d) : Octahedral complexes of type [MA5B] does

not show geometrical isomerism.

72. (d) :

3,7-Dimethylocta-2,6-dien-1-al

73. (b) : As we move down the group, the atomic size increases and electronegativity decreases, and, thereby, tendency to show catenation decreases. �e order of catenation is C >> Si > Ge Sn. Lead does not show catenation.

74. (b) : Suppose number of O2– ions = n. Then number of octahedral voids = n and number of tetrahedral voids = 2nNo. of cations X present in tetrahedral voids

16

23

n n

No. of cations Y present in octahedral voids

13 3

n n


Ratio X : Y : O2– n n n

3 31 1 3: : : :

Hence, formula is XYO3.

75. (c) : v 0 005100

600 3 101 2 1. m s m s

x m v h 4

;

x hm v4

x

6 6 10

4 3 14 9 1 10 3 10

34 2 1

31 2 1.

. .

kg m s

kg m s = 1.92 × 10–3 m

76. (d) : Di-tert.-butyl ether cannot be prepared by Williamson’s synthesis, since tert.-alkyl halides prefer to undergo elimination rather than substitution, i.e.,

77. (a) : 1° alky halides (i.e., CH3Br) undergo SN2 reaction.

78. (b) : �e low solubility of LiF in water is due to its high lattice enthalpy. Smaller Li+ ion is stabilised by smaller F– ion.

79. (c) : �e smaller the size of the cation and the larger the size of the anion, the greater is the covalent character of an ionic bond.

80. (a) 81. (b) 82. (b) 83. (c) 84. (c)85. (d) 86. (c) 87. (a) 88. (b) 89. (c)90. (c) 91. (d) 92. (b) 93. (d) 94 (c)95. (c) 96. (b) 97. (b) 98. (c) 99. (a)100. (b) 101. (b) 102. (c) 103. (a) 104. (b)105. (c)106. (b) : Numbers a, b, 12 are in G.P.

b2 = 12a ...(i)and a, b, 9 are in A.P. 2b = a + 9 or a = 2b – 9 ...(ii)From Eqs. (i) and (ii), we get b2 = 12(2b – 9) b2 – 24b + 108 = 0 (b – 18)(b – 6) = 0 b = 6, 18From Eq. (ii), a = 3, 27 respectively.

Common ratio = ba 6

3 and 18

27= 2 and 2

3

Common ratio = 23

(for decreasing G.P. common ratio 2)

107. (a) : A

13 pass through A 11 pass through Number of intersection points = 37C2 – 13C2 – 11C2 + 2 = 535

( Two points A and B)108. (b)

109. (b) : �e probability of hitting in one shot

= 10100

110

If he fires n shots, the probability of hitting atleast once

= 1 – 1 110

n

= 1 – 9

10

n

= 50% = 50100

12

9

1012

n

n{log 9 – log 10} = log 1 – log 2 n{2 log 3 – 1} = 0 – log 2

n = log

log.

.2

1 2 30 3010300

1 2 0 4771213

= 6.6

Hence, n = 7110. (c)111. (c) : Q lies on y-axis,

Put x = 0

in 5x – 2y + 6 = 0

y = 3

Q (0, 3)

PQ = 0 3 0 6 3 2 252 2 = 5112. (c) : Since, A z1, B z2, C (1 – i)z1 + iz2

AB = |z1 – z2|, BC = |(1 – i)z1 + iz2 – z2| = |1 – i| |z1 – z2| = 2 |z1 – z2|and CA = |(1 – i)z1 + iz2 – z1| = |i| |–z1 + z2| = |z1 – z2| AB = CA and (AB)2 + (CA)2 = (BC)2.

113. (c) : Since, y = |x – 1| = x x

x x

1 11 1

,,

...(i)


and y = 3 – |x| = 3 03 0

x xx x

,,

...(ii)

Solving eqs. (i) and (ii), we get x = 2 and x = – 1

Required area = ( | | | |)3 11

2

x x dx

= ( ) ( )2 2 2 4 21

001

12

x dx dx x dx

= 4 sq. units.114. (d) : D.R.’s of PQ are 1 – (–5), 5 – 3, – 2 – 1

i.e., 6, 2, –3

and 6 2 32 2 2 ( ) = 7 D.C.’s are 67

27

37

, ,

115. (c) : (1 + cos /8) (1 + cos 3/8) (1 + cos 5/8) (1 + cos 7/8)= 2cos2(/16) 2cos2(3/16) 2cos2(5/16) 2cos2(7/16)= 16[cos (/16) cos (3/16) cos (5/16) cos (7/16)]2

= [2 cos (7/16) cos (/16)]2 [2cos (5/16) cos (3/16)]2

= [cos (/2) + cos (3/8)]2 [cos (/2) + cos (/8)]2

= cos2(3/8) cos2(/8)

= 14

(cos /2 + cos /4)2 = 18

116. (d) : For parallel,

a b = 0

i j k

2 1 34 6l

= 0

i j k ( ) ( ) ( )6 3 0 2 4 0 l l 6 + 3l = 0 l = –2

117. (b) : We have,

f(x) =

a x x

x xx

b xx xx

x

| |,

,[ ] ,

2

22

22

2

22

f(x) =

a x xx x

x

b xx xx

x

( ) ,

,[ ] ,

2

22

22

2

22

f(x) =

a xb x

x xx

x

,,

[ ] ,

22

22

Now,L.H.L. = lim ( ) lim ( )

x hf x f h

2 02 = lim

ha a

0

R.H.L. = lim ( ) lim ( )x h

f x f h

2 0

2

= lim [ ]h

h hh

0

2 22 2

= limh

hh

0

2 22 2

= 1

Value of the function = f(2) = b. Since f(x) is continuous at x = 2 L.H.L. = R.H.L. = Value of f(x) a = b = 1

118. (d) : Let f(x) = a2x2 + 2bx + 2cFrom the question, a2a2 + ba + c = 0 and a22 – b – c = 0Now, f(a) = a2a2 + 2ba + 2c = ba + c = –a2a2

f() = a22 + 2b + 2c = 3(b + c) = 3a22

But 0 < a < a, are real f(a) < 0, f() > 0Hence, a < < .

119. (d) : In the given expansion, (r + 1)th term = Tr + 1

= 10Crx

x

r r

33

2

10

2

= 10Crx

x

r r

33

2

52

22

= 10Cr

xr r

r r r

52

52 2 23 2

For independent of x,

Put 5 – r2

– r = 0

5 = 32r ⇒ r = 10

3, impossible

r whole number120. (b) 121. (c)122. (d) : Let each line ∈ set of the lines (L)

(i) As (, ) ∈ R ∈ L R is re�exive.(ii) Let 1, 2 ∈ L such that (1, 2) ∈ R then 1 2 2 1 (2, 1) ∈ R R is symmetric.(iii) Let 1, 2, 3 ∈ L such that (1, 2) ∈ Rand (2, 3) ∈ R 1 2 3 (1, 3) ∈ R R is transitive.Since a relation R, which is re�exive, symmetric and transitive is known as equivalence relation. Given relation is an equivalence relation.


123. (c) : P(n) = 5n – 2n

Let n = 1 P(1) = 3l = 3 l = 1Similarly n = 5 P(5) = 55 – 25 = 3125 – 32 = 3093 = 3 × 1031In this case, l= 1031Similarly, we can check the result for other cases and �nd that the least value of land n is 1.

124. (b) : Rank of diagonal matrix = Order of matrix = 3.

125. (b) : �ere are 11 letters in the word 'PROBABILITY' out of which 1 can be selected in 11C1 ways. Exhaustive number of cases = 11C1 = 11�ere are three vowels viz AIO.�erefore, favourable number of cases = 3C1 = 3

Hence, the required probability = 311

126. (c) : Total number of functions from A B = 34 = 81Number of onto mappings = Coe�cient of x4 in 4!(ex – 1)3.= Coe�cient of x4 in 4!(e3x – 3e2x + 3ex – 1)

= 4!34

3 24

3 14

04 4 4

! ! !

= 81 – 48 + 3 = 81 – 45 = 36 Number of functions from A B, which are not onto is 81 – 36 = 45

127. (d) : f(x) = maximum {x, x + 1, 2 – x}

Minimum value of function = 3/2128. (d)129. (c) : A = {x : x = 4n + 1 ∀ 2 n 6}

or A = {9, 13, 17, 21, 25} Number of elements of A = 5 Number of subsets of A = 2n = 25

130. (c) : Since, x = Mean of the series

= a a d a ndn

( ) ..... ( )22 1

= a + nd

xi d = xi – x d2 = Da nd n2d2

a + d (n – 1)d (n – 1)2d2

a + (n – 2)d 2d 4d2

a + (n – 1)d d d2

a + nd 0 0a + (n + 1)d d d2

a + (n + 2)d 2d 4d2

a + 2nd nd n2d2

d =

2 12

dn n

d2 = 2d2

[12 + 22 + ... + n2]

We have d = n(n + 1) d and

d2 = 2 1 2 1

6

2d n n n( )( )( )

Now, M.D. = dN

and 2 = d

N

2

M.D. = n n dn

( )1

2 1 and 2 =

2 1 2 16 2 1

2d n n nn

( )( )

( )

= n n d( )1

3

2

S.D. = 2 13

n n d( )

131. (a) : Let Sn = Pn2 + Qn = Sum of �rst n termsAccording to question,Sum of �rst 3n terms = Sum of the next n terms S3n = S4n – S3n or 2S3n = S4nor 2[P(3n)2 + Q(3n)] = P(4n)2 + Q(4n) 2Pn2 + 2Qn = 0 or Q = –nP ...(i)�en,

Sum of first termsSum of next terms

22

2

4 2

nn

SS S

n

n n

= P n Q nP n Q n P n Q n

( ) ( )[ ( ) ( )] [ ( ) ( )]

2 24 4 2 2

2

2 2

= 26 5

15

nP QPn Q

nPnP

[From eq. (i)]

132. (b) : Let P = lim ( !) /

n

nnn

1 = lim ! /

n n

nnn

1


= lim ...........

/

n

nnn n n n

1 2 3 4 1

= lim ...../

n

n

n n nnn

1 2 3 1

ln P =

lim ln ln ln ... lnn n n n n

nn

1 1 2 3

= lim lnn r

n

nrn

1

1 = ln x dx

01

= [ln x x – x]01 = 0 – 1 = – 1 P = e–1

133. (b) 134. (b) : If eccentricities of ellipse and hyperbola are

e and e1. Foci are (ae, 0) and (a1e1, 0).

Here, ae = a1e1 a2e2 = a12e1

2

a b

aa

b

a2

2

2 12 1

2

12

1 1

a2 – b2 = a12 + b1

2 25 – b2 = 14425

8125

= 9

b2 = 16

135. (b) : cos–1x + cos–1y = 2

– sin–1x + 2

– sin–1y

= – (sin–1x + sin–1y) = – 23 =

3 (given)

136. (b) : 1 11

n

= 64 2n = 26 n = 6

General term Tr + 1 = nCr(x)n – r 1x

r

= 6Cr x6 – 2r

For independent of x, 6 – 2r = 0 r = 3 T3 + 1 = 6C3x0 = 6C3 = 20

137. (b) : sin y dydx

= cosy(1 – x cos y)

tan y dydx

= 1 – x cos y

sec y tan y dydx

– sec y = – x ...(i)

Put sec y = v sec y tan y dydx

= dvdx

�en, from eq. (i), dvdx

– v = – x

IF = e dx 1 = e–x

�en, the solution is

v . (e–x) = ( ) x e dxx

v e–x = (–x)(–e–x) + e–x + cor v = x + 1 + cex or sec y = x + 1 + cex

138. (c)139. (c) : Out of 6 novels, 4 novels can be selected in

6C 4 ways.Also out of 3 dictionaries, 1 dictionary can be selected in 3 C 1 ways.Since the dictionary is �xed in the middle, we only have to arrange 4 novels which can be done in 4! ways.Then the number of ways = 6 C 4 · 3 C 1 · 4!

6 52

3 24 1080

140. (a) : Normal at t, i.e. (at2, 2at) is tx + y = 2 at + at3 o r , y = – tx + ( 2 at + at 3 )

T h i s w i l l t o u c h x2 – y2 = a 2 o r xa

ya

2

2

2

2 1 if

( 2 at + at 3 ) 2 = a2 ( – t) 2 – a2 [ U s i n g c 2 = a2 m2 – b2 ] 4 t2 + t6 + 4 t 4 = t2 – 1 t6 + 3 t4 + 3 t 2 + 1 = – t4 ( t2 + 1) 3 = – t 4 < 0[ H er e, i f t = 0, t h en n o r m al t o y 2 = 4 ax w i l l b e x - ax i s at ( 0, 0) an d x - ax i s c an ’ t t o u c h r ec t an g u l ar h y p er b o l a x2 – y2 = a 2 t 0]

141. (d) : Rewrite f(x) = 2 sin(x + /6) + 4

or f(x) = 2cos x

3

+ 4

Y = [2, 6] ( min and max values of sin and cos are –1 and +1)

142. (d) :

a b c 1Let be the angle between

a c b c& &and

cos.

| || |.

a ca c

a c ...(i)

Similarly, cos .

b c ...(ii)

a b a b . 0

c pa qb r a b ( )

a c pa a q a b r a a b. . . .( ) = p + 0 + 0 = p cos = p [from (i)] |p| = |cos | 1 Similarly, cos = q |q| 1Also, p = q


143. (c) :

xx′

y′

y

Required area = ( )e e dxx x 01

= [ex + e–x]0–1 = (e + e–1) – (e0 + e–0)

= (e + e–1 – 2) sq. units.144. (a) : Normal at the extremity of latus rectum in

the �rst quadrant (ae, b2/a) isx aeae a

y b a

b ab /

/

/2

2

2 2

As it passes through (0, –b)

aeae a

b b aa/

//2

2

1

–a2 = –ab – b2 a2 – b2 = ab a2e2 = ab or e2 = b/a

e4 = ba

2

2 = 1 = e2 e4 + e2 = 1

145. (b) : Let plane is xx

yy

zz1 1 1

= 1

which passes through (a, b, c). a

xby

cz1 1 1

1

Locus of (x1, y1, z1) is ax

by

cz

1

146. (a) : Given, [ ]

a b c = 0Also, | |a = 1, | |

b = 1 and | |c = 1 [ ]2 2 2

a b b c c a = ( ) {( ) ( )}2 2 2

a b b c c a = ( ) { }2 4 2 2

a b b c b a c c c a = ( ) { ( ) ( ) ( )}2 4 2 0

a b b c a b c a = 8 4 2

a b c a a b a c a ( ) ( ) ( ) 4 2

b b c b a b b c a( ) ( ) ( ) = 8 0 0 0 0[ ] [ ]

a b c a b c = 7[ ]

a b c = 0 ( [ ] )

a b c 0

147. (d) : Required numbers = 42 2

32

18!! !

!!

148. (b) : Given x x x ( )1 1 ...(i)

x x x ( )1 1 ...(ii)Squaring (ii) both sides, we get

x – 1 x = 1 + x – 2 x

( )1 x = 1 – 2 x ...(iii)Squaring (iii) both sides, we get 1 – x = 1 + 4x – 4 x 4 x = 5x

25x2 – 16x = 0 x = 0, 1625

x = 1625

( x = 0 does not satisfy eq. (i))

149. (a) : Let I = xe

xdx

x

( )1 2

= e xx

dxx ( )( )1 11 2

= e

x xdxx 1

11

1 2

( )

= e

xc

x

( )1

( {( ( ) ( )} ( ) ) e f x f x dx e f x cx x ′

150. (b) : We have, y = x y y

(y2 – x) = 2y (y2 – x)2 = 2y

Di�erentiating both sides w.r.t. x, we get

2(y2 – x) 2 1y dydx

= 2 dydx

dydx

y x

y xy

( )2

32 2 1

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WEST BENGAL at


SECTION-1SINGLE CORRECT ANSWER TYPE

1. Suppose tana pq

where p, q are integers and

q 0 and tan3a = tan2 such that tan is rational, then p2 + q2 is a(a) perfect square (b) perfect cube(c) perfect 5th power (d) perfect 6th power of an integer

2. nk

kk

kk

nC x

xtan

( tan / )21 2 1

1 2

2

20

(a) sec sec22

n nx x

(b) cosec2 22

n nx x

cosec

(c) tan tan22

n nx x

(d) cot cot22

n nx x

3. �e number of positive roots of the equation

xx 12

is/are

(a) 0 (b) 1 (c) 2 (d) 4

4. Consider the equation x5 + 5lx4 – x3 + (la – 4)x2 – (8l + 3)x + la – 2 = 0. �e value of a such that the given equation has exactly 2 roots independent of lis (a) –2 (b) –3 (c) 2 (d) 3

5. Let S be the set of all points (x, y) in the plane such that x, y ∈[0, /2]. �e area of the subset of S for which sin2x + sin2y – sinxsiny 3/4 is

(a) 2

3 (b)

2

4 (c) 2

5 (d)

2

6

6. Let ABCD be a tetrahedron. Let ‘a’ be the length of edge AB and let be the area of projection of the tetrahedron on a plane, perpendicular to AB then volume of the tetrahedron is

(a) a2

(b) a3

(c) a4

(d) a5

7. Let z1, z2, z3 be distinct complex numbers such that |z1| = |z2| = |z3| = r, z2 z3, a ∈ R, then minimum value of |az2 + (1 – a)z3 – z1| is

(a) | || |z z z zr

1 2 1 3 (b) | || |z z z zr

1 2 1 32

(c) | || |z z z zr

1 2 2 3 (d) | || |z z z zr

1 2 2 32

8. De�ne a recursive sequence aa

nnn

1

201 ,

and a0 ∈ (–1, 1). Let An = 4n(1 – an) then limn nA

(a) (cos–1a0)2 (b) (cos )10

2

2a

(c) (cos )10

2

3a (d) (cos )1

02

4a

9. Given a point (a, b) with 0 < b < a, the minimum perimeter of a triangle with one vertex at (a, b), one on the x-axis and one on the line y = x is

(a) a b2 2 (b) 2 2 2( )a b

(c) a b2 2

2 (d) ab

10. 12 points are arranged on a semi-circle as shown. 5 on the diameter and 7 on the

arc. If every pair of these points is joined by a straight line segment, then no-three of these line segments intersect at a common point inside the semi-circle.

How many points are there inside the semi-circle where two of these line segments intersect?

By : Tapas Kr. Yogi, Mob : 9533632105.


(a) 240 (b) 360 (c) 420 (d) 560

11. �e number of integral points (x, y) on the hyperbola x2 – y2 = (2000)2 are (a) 49 (b) 98 (c) 48 (d) 96

SECTION-2MULTIPLE CORRECT ANSWER TYPE

12. For positive integers n and k, we have

( ) ( ) ( )( ) ( )

nn

nk

nn

nk

nC C C

C Cn

16 2 6 3

13

2 2 32

2

3then

(a) n = 6 (b) n = 10(c) k = 2 (d) k = 4

13. Consider the equation

x x x x k 2 1 2 1

(a) If k x ∈

2 12

1, ,then

(b) If k = 1 then x = 12

(c) If k x 2 32

, then

(d) If k = 2 then x = 3/2

14. If 22

1 1sin cos , x x then x is equal to

(a) 12

(b) 12

(c) 32

(d) 32

15. Consider the limit, L xx

x

lim (cos )|sin |

0

a as a

ranges from (0, )(a) L = 1 if 0 < a < 2 (b) L = e–1/2 if a = 2(c) L = 0 if a> 2 (d) None of these

SECTION-3COMPREHENSION TYPE

Passage-1Let A be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.

16. �e number of matrices in A is(a) 2 (b) 6 (c) 9 (d) None of these

17. �e number of matrices A in A, for which the

system of equations Axyz

100

has a unique solution, is

(a) less than 4(b) at least 4 but less than 7(c) at least 7 but less than 10(d) at least 10

Passage-2�e curve y = f (x) passes through the point (0, 1) and

the curve y = g(x) = f t dtx

( ) passes through the point

0 12

,

. �e tangents drawn to the curves at the point

with equal abscissae intersect on the x-axis. �en

18. y′(0) =

(a) 0 (b) 12

(c) 2 (d) 1

19. g(0) =

(a) 0 (b) 12

(c) 2 (d) 1

SECTION-4MATRIX-MATCH TYPE

20. Consider the ellipse x2 + 2y2 = 2. Let L be the end of the latus rectum in the �rst quadrant. �e tangent at L to the ellipse meets the right side directrix at T. �e normal at L meets the major axis at N and the le� side directrix at M.(i) �e area of LNT (p) 1/5

(ii) �e circumradius of LNT (q) 12

(iii) �e ratio LN : LT (r) 3/4

(iv) �e ratio LN : NM (s) 34 2

(i) (ii) (iii) (iv)(a) s r p q(b) s p r q(c) p s r q(d) s r q p

SOLUTIONS

1. (a) : According to question, tana is rational. Let = – a. So, a = 2. Now, tan is rational if tan is rational.

Let tan = r then tan2 21 2

pq

rr

i.e. pr2 + 2qr – p = 0.�is equation has rational solutions i� D = 4(p2 + q2) is a perfect square.


2. (a) : �e given sum can be rewritten as

nk

k

k

nn

k

k

kC x C x

x

tan tan ( / )

tan ( / )2

0

2

202

2 21 2

nn

1

21 12tan cos

cosx x

x

n n

sec sec ( )2

2n nx x

3. (c) : xx 12

can be written as x xlog log

12

12

i e f x f. . ( )

12

where f (x) = xlogx, f ′(x) = 1 + logxSo, f (x) is decreasing in (0, 1/e) and increasing in (1/e, ).

So, f x f x( )

1

212

14

or

Hence two solutions.4. (b) : We rewrite the given equation as x5 – x3 – 4x2 – 3x – 2 + l(5x4 + ax2 – 8x + a) = 0 A root of this equation is independent of l i� it

is a common root of the equation x5 – x3 – 4x2 – 3x – 2 = 0 and 5x4 + ax2 – 8x + a = 0

First equation has roots x = 2, and 2. So, for a = –3, there are two roots independent

of l, x1 = and x2 = 2.5. (d) : Consider the corresponding equation,

sin sin cosx y y

234

02

2

i.e. sinx = sin(y + 60°) sinx = sin(y – 60°)i.e. x = y + 60° or 120° – y x = y – 60° or 240° – ySo, the area of subset of S is bounded by x = y + 60°, x = 120° – y and x = y – 60°.

So, the required area is 2

6.

6. (b) : Let A(0, 0, 0), B(0, 0, a), C(p, b, c), D(a,, ).Let projection of C on XY plane is C ′ (p, b, 0) and projection of D on XY plane is D′ (a, , 0).

Area of DAD C AD ACi j k

p b′ ′ ′ ′ 1

212

00

^ ^ ^

a

|ab –p| = 2 ... (1)So, volume of tetrahedron ABCD,

V AB AC AD 1

6[ ]

Va

p b c a 16

0 0

62 1

a ( . ( ))from eqn

Hence, V a 3

.

7. (b) : Let A1(z1), A2(z2) and A3(z3) and A(z) be any point on the line A2A3 such that z = az2 + (1 – a)z3.Let P be the foot of altitude from A1 on line A2A3.So, AA1 A1Pi.e. min. |z – z1| = min. |az2 + (1 – a)z3 – z1| = A1P = h

So, A A A h z z1 2 3 2 312

| | 1

2 1 2 1 3 1( )( )(sin )A A A A A

12 21 2 1 3

2 3| || || |

z z z zz z

r

So, hz z z z

r

| || |1 2 1 32

8. (b) : Let a0 = cos, ∈ (0, ) then a1 = cos(/2)Similarly, a2 = cos(/4), ..., an = cos(/2n)So, An = 4n[1 – cos(/2n)]

2 2

1 22

2cos( / )sin( / )

/n

n

n

So, as n , A a 2 2

10

22

12

12

( ) (cos )

9. (b) : Let A(a, b), B(x, 0) and let E(b, a) be the reflection of A(a, b) in the line y = x, D(a, –b) be the re�ection of A in x-axis. Let C be the point on y = x line then AB = DB and AC = CE.So, perimeter of ABC = AB + BC + CA = DB + BC + CE DE

( ) ( ) ( )a b a b a b2 2 2 22

10. (c) : 5 points on diameter line and 7 points on arc.Two line segments can intersect at an interior point in following ways.(i) Line segments have all 4 end points on the arc = 7C4 = 35

ANSWER KEYMPP CLASS XI1. (a) 2. (b) 3. (c) 4. (a) 5. (d)6 . (b) 7 . (a, b, c) 8 . (a, c, d) 9 . (a, b, d) 1 0 . (a, d)1 1 . (a, b, c, d) 12. (a, b) 13. (a, b, c) 14. (a) 15. (b) 1 6 . (b) 1 7 . (1) 1 8 . (2) 1 9 . (5) 2 0 . (5)


(ii) Each line segment has one end point on diameter and one end point on arc = 7C2 × 5C2 = 210.(iii) One line segment has both end points on the arc and the other has 1 end point on diameter and one on the arc = 7C3 × 5C1 = 175Total = 35 + 210 + 175 = 420.11. (b) : (x + y)(x – y) = 28 · 56

So, ( x + y) and (x – y) both are even.Let us �rst assign a factor of 2 to both (x + y) and (x – y). We have 26 × 56 le�. Since there are (6 + 1)·(6 + 1) = 49 factors of 26 × 56 and since both x and y can be negative. So, total integral points = 2 × 49 = 9812. (a, c) : Use A.M. G.M. on (nCn – 1)6 · (n – 2Ck)6 and (n + 3C2)3 with equality occuring when the numbers are equal.

13. (a, d) : Let

2 1 12

12

x t k t t then

14. (d) : 2 sin–1 x – cos–1 x = /2 ...(i)Also, sin–1 x + cos–1 x = / 2 ...(ii)Adding (i) and (ii), 3 sin–1 x = sin–1 x = / 3 x = sin(/ 3) x 3 2/15. (a, b, c) : �e given limit is

ex xxlim

|sin |log(cos )

0

1a

= ex

xx

xx

xx

lim|sin |

cos log(cos )cos

0

2

21

1a

= 1 for 0 < a < 2 = e–1/2 for a = 2 = 0 for a > 2

16. (d) : 1

11

h gh fg f

where only one of f, g, h is 1, (3 ways)

1

00

h gh fg f

where only one of f, g, h is 0. (3 ways)

0

10

h gh fg f

(3 ways), 0

01

h gh fg f

(3 ways)

Required number of matrices = 12 ways.

17. (b) : Da h gh b fg f c

= abc + 2fgh – af 2 – bg2 – ch2

a = b = 0, c = 1 ⇒ D 0 if g = 0 or f = 0, 2 matrices a = c = 0, b = 1 ⇒ D 0 if f = 0 or h = 0, 2 matrices b = c = 0, a = 1 ⇒ D 0 if g = 0 or h = 0, 2 matrices a = b = c = 1 ⇒ D = 0 if f or g or h is 1. �e number of matrices is 6.

18. (c) 19. (b)

20. (d) : (i) → (s), (ii) → (r), (iii) → (q), (iv) → (p)


1. If |z – 1| + |z + 3| 8, then the range of values of |z – 4| is (z is a complex number in the argand plane)(a) (0, 8) (b) [0, 8] (c) [1, 9] (d) [5, 9]

2. �e value of ( )cos ( )n k kn

nk

n

1

2 3 is

(a) n2

(b) 0 (c) n2

(d) –n

3. Let g x f xx

( ) ( )1

where f (x) is di�erentiable on

[0, 5] such that f (0) = 4, f (5) = –1. �ere exists c ∈ such that g′(c) is

(a) 16

(b) 16

(c) 56

(d) –1

4. If 4x4 + 9y4 = 64, then the maximum value of x2 + y2 is (where x and y are real)

(a) 43

(b) 43

13 (c) 323

(d) 3213

5. A chord of the circle x2 + y2 – 4x – 6y = 0 passing through origin subtends an angle tan–1(7/4) at the point where the circle meets positive y-axis, then equation of the chord is (a) 2x + 3y = 0 (b) x + 2y = 0 (c) x – 2y = 0 (d) 2x – 3y = 06. �e number of points of extremum of the function f (x) = (x – 2)2/3 (2x + 1) is(a) 1 (b) 0 (c) 2 (d) 3

7. Let the vector a i j k be vertical. �e line of greatest slope on a plane with normal

b i j k 2 is along the vector ______.

(a) i j k 4 2 (b) i j k 4 3(c) 4i j k (d) none of these8. �e probability that the triangle formed by choosing any three vertices from the vertices of a cube is equilateral, is(a) 3/7 (b) 6/7 (c) 4/7 (d) 1/7

9. Let Aa aa a

Xxx

Yyy

11 12

21 22

1

2

1

2, , (X′, Y′

denote the transpose of X and Y).Statement-1 : If A is symmetric, then X′AY = Y′AX for each pair of X and Y.Statement-2 : If X′AY = Y′AX for each pair of X and Y, then A is symmetric.Which of the following options holds?(a) Only Statement-1 is true.(b) Only Statement-2 is true.(c) Both Statement-1 and Statement-2 are true. (d) Both Statement-1 and Statement-2 are false.10. Consider the following relations: R = {(x, y)/x, y are real numbers and x = wy for

some rational number w}

S mn

pq

, m, n, p, q are integers such that

nq 0 and qm = pn}Statement-1 : S is an equivalence relation but R is not an equivalence relation.Statement-2 : R and S both are symmetric.(a) Both Statement-1 and Statement-2 are true

and Statement-2 is the correct explanation of Statement-1.

Math Archives, as the title itself suggests, is a collection of various challenging problems related to the topics of IIT-JEE Syllabus. This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for IIT-JEE. In every issue of MT, challenging problems are offered with detailed solution. The readers’ comments and suggestions regarding the problems and solutions offered are always welcome.

BESTPROBLEMS10

B y : Prof. Shyam Bhushan, D i r e c t o r , N a r a ya n a I I T A ca d e m y , Ja m sh e d p u r . M o b . : 0 9 3 3 4 8 7 0 0 2 1

MATHEMATICS TODAY | APRIL‘1772

(b) Both Statement-1 and Statement-2 are true but Statement-2 is not the correct explanation of Statement-1.

(c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true.

SOLUTIONS

1. (c) : z lies on or inside an ellipse with foci (–3, 0) and (1, 0) and (4, 0) is a point outside the ellipse. Its minimum and maximum distances from any point on the ellipse are 1 and 9.

2. (a) : Let S nn

nn

( )cos ( )cos1 2 2 4

( )cos ...... cos ( )nn

nn

3 6 2 1

...(i)

Sn n

n nn

1 2 2 4 1 2 1cos cos ..... ( )cos ( )

...(ii)Adding (i) and (ii), we get

2 2 4 2 1S n

n nn

n

cos cos ..... cos ( )

2

1 2 2 1

2S n

nn

n

nn

n nsin( )

sincos

( )

3. (c) : ′

g c

f f

( )

( ) ( )

( )

56

01

5from L.M.V.T

16

4

525

6 556

4. (b) : Let x2 = 4sin, y2 83

cos

x y2 2 4 83

sin cos

Maximum value 16 649

4 133

5. (c) : Equation of tangent at origin is –2(x + 0) – 3(y + 0) = 0 2x + 3y = 0

(2, 3)

x

y

Otan = 7

4

23

1 23

74

m

m

3 23 2

74

12 8 21 14mm

m m( )

26 13 12

2 29 292

m m m mor

y x x y y x x y12

2 0 292

29 2 0or

y x x y y x x y12

2 0 292

29 2 0or

6. (c) : ′ f x x x x( ) ( ) ( ) ( )/ /2 2 2 1 23

22 3 1 3

6 2 2 2 1

3 210 10

3 21 3 1 3( ) ( )

( ) ( )/ /x x

xx

xx = 1 is a point of maximum and x = 2 is a point of minimum. Number of extremum points is 2.7. (d)8. (d) : n(S) = 8C3 = 56 Choosing a vertex 3 triangles can be formed from the diagonals of the faces not passing through the vertex. But each triangle is repeated 3 times.

n E P E( ) ( )8 3

38 8

5617

n E P E( ) ( )8 3

38 8

5617

9. (c) : A is symmetric A′ = ANow, X ′AY = ((X ′AY)′) ( X′AY is 1 × 1 matrix) = (Y ′A′ X) = (Y ′AX)

Let E E1 210

01

,

If X = E1 and Y = E2 then

E′1AE2 = E′2AE1 a12 = a21 A is symmetric.10. (c) : For (x, y) ∈ R x = wybut (y, x) ∈R y = xw x = y R is not symmetric.

ANSWER KEYMPP CLASS XII1. (c) 2. (b) 3. (a) 4. (a) 5. (c)6. (d) 7. (b, d) 8. (a, b) 9 . (a, b, c, d)10. (b, d) 1 1 . (a, b, c) 12. (a, d) 13. (a, b, c, d)14. (a) 15. (b) 1 6 . (a) 1 7 . (3) 1 8 . (5)1 9 . (5) 2 0 . (1)


1. Let 34

1 a . Prove that the equation

x3(x + 1) = (x + a)(2x + a) has four distinct real solutions and �nd these solutions in explicit form.

2. �e vertices of an equilateral triangle lie on three parallel lines which are 3 units and 1 unit apart as shown. Find the area of this triangle.

3 units

1 unit

3. Find all real numbers x such that

x x

x x

1 1 11 2 1 2/ /

4. A railway line is divided into 10 sections by the stations A, B, C, D, E, F, G, H, I, J and K. �e distance between A and K is 56 km. A trip along two successive sections never exceeds 12 km. A trip along three successive sections is at least 17 km. What is the distance between B and G?

A B C D E F G H I J K

5. Given is a triangle ABC, A = 90. D is the midpoint of BC, F is the midpoint of AB, E the midpoint of AF and G the midpoint of FB.

AD intersects CE, CF and CG respectively in P, Q

and R. Determine the ratio PQQR

.

A E F G B

P Q RD

C

SOLUTIONS

1. Look at the given equation as a quadratic equation in a:

a2 + 3xa + 2x2 – x3 – x4 = 0 �e discriminant of this equation is 9x2 – 8x2 + 4x3 + 4x4 = (x + 2x2)2

�us a x x x 3 22

2( )

�e �rst choice a = –x + x2 yields the quadratic equation x2 – x – a = 0, whose solutions are

x a ( )1 1 4

2 �e second choice a = –2x – x2 yields the quadratic

equation x2 + 2x + a = 0, whose solutions are 1 1 a . �e inequalities

1 1 1 1 1 1 4

21 1 4

2a a a a

show that the four solutions are distinct.

Indeed 1 1 1 1 42

a a

reduces to 2 1 3 1 4 a a

which is equivalent to 6 1 4 6 8 a a

or 3a < 4a2, which is evident.2. Using Pythagoras' �eorem on the triangle marked

PQR we have

4 1 92 2 2

22 x x x

i.e., 16 + x2 – 1 + x2 – 9

2 1 92 2 2x x x �is rearranges to give


x x x2 2 26 2 1 9 , which, on squaring both

sides, becomes x4 + 12x2 + 36 = 4(x4 – 10x2 + 9) = 4x4 – 40x2 + 36 �is simpli�es to 3x4 – 52x2 = 0, or x2 = 52/3. �e area

of an equilateral triangle of side length x is half base

times height, i.e., 1/2 × x × 3 2x / or x2 × 3 4/ . �e area of this triangle is thus 52 3 12 13 3 3/ /

Alternative method: Let the side lengths of the equilateral triangle be x

and the angle be as marked in the diagram. First we note that cos = 4/x while

sin . 1 16

2x Further,

sin( ) cos sin302

32

1

4xcos .

So cos = 2 3 sin or cos2 = 12 sin2 = 12 – 12 cos2 giving 13 cos2 = 12 or cos2 = 12/13. So

x2

216 16 13

124 13

3

cos( )( ) ( ) .

�e required area is

12

60 34

34

4 133

13 33

2 2x xsin ,

3. Since xx

1 01 2/

and 1 1 01 2

x

/

, then

0 1 1 11 2 1 2

xx x

x/ /

Note that x 0. Else, 1x

would not be de�ned, so x 0.

Squaring both sides gives,

x x

x xx

x x2 1 1 1 2 1 1 1

x x

xx

x x2

21 2 2 1 1 1

Multiplying both sides by x and rearranging, we get

x

x1

30 –60

4

x x x x x x3 2 3 22 2 1

( )

( )

x x x x x x

x x x x x x

3 2 3 2

3 2 2 3 2

1 2 1 1 0

1 1 0 1 1

x x x x x x

x x

3 2 2

2

1 1 1 0

1 0

(Since x ≠ 0)

�us x 1 52

. We must check to see if these are

indeed solutions.

Let a 1 52

1 52

, . Note that a+ = 1,

a = –1 and a> 0 > .

Since < 0, is not a solution.

Now, if x = a, then

a

a a

1 1 11 2 1 2/ /

= (a + )1/2 + (1 + )1/2 (Since a = –1) = 11/2 + (2)1/2 (Since a + = 1 and 2 = + 1) = 1 – (Since < 0) = a (Since a + = 1) So x = a is the unique solution to the equation.

4. A B C D E F G H I J K

Now AK 56 and AK AD DG GJ JK . We

know that AD DG GJ, , 17 . �us JK 5 to satisfy

AK 56 . We know HK 17 , and since JK 5 ,

HJ 12 . But, we also know HJ 12 . �us HJ 12 .

Since HK 17 and HJ 12 , JK 5 . The only

possibility is that JK 5 .

Symmetrically we �nd that AB 5 and BD 12 .

Now, DH AK AB BD HJ JK = 56 – 5 – 12 – 5 – 12 = 22

Now GJ 17 but HJ 12 . Hence GH 5 . Since

DG 17 a nd DH DG GH 22 , we obt a i n

DG 17 and GH 5 .

Now, BG BD DG 12 17 29


5. We know that two medians in a triangle divide each other in 2 : 1 ratio, or in other words the point of

intersection is 23

the way from the vertex.

A E F G B

P QR

D

C

Since CF and AD are both medians in ABC, then

AQQD

21

, where Q is the point of intersection.

Also, since D is the midpoint of the hypotenuse in the right triangle ABC, then it is the centre of the circumscribed circle with radius DA = DC = DB.

Drop a perpendicular from D onto sides AB and CA. �e feet of the perpendiculars will be F and J, respectively, where J is the midpoint of AC, since DF and DJ are altitudes in isosceles triangles ADB and ADC, respectively. Now consider CFB. �e segments CG and FD are medians and therefore

intersect at H say in the ratio 2 : 1 so, HDFD

13

.

From here it can be seen that ARC and DRH are similar, since their angles are the same. Also,

since we know that FD JA , and 2JA AC then

HD CA 16

and ARC is 6 times bigger than

DRH. Now we can see that ARRD

61

and since

AR RD AD , then RDAD

17

.

Similarly APE KPD, where medians DJ and

CE meet at K. We know that AE AB 14

, so then

JK JD 14

, since JD is parallel to AB. It now fol-

lows that AEKD

23

, and from the similarity of the

triangles APPD

23

. Also, since AP PD AD ,

then APAD

25

. Combining these results we have

AP AD AQ AD QD AD 25

23

13

, ,

and RD AD 17

.

�us PQ AQ AP AD AD AD 23

25

415

and QR QD RD AD AD AD 13

17

421

From these PQQR

75

.


1. �e maximum number of students among them 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is(a) 91 (b) 910 (c) 1001 (d) 1911

2. Each boy contributed rupees equal to the number of girls and each girl contributed rupees equal to the number of boys in a class of 60 students. If the total contribution thus collected is `1600, how many boys are there in the class? (a) 25 (b) 30 (c) 50 (d) Data inadequate

3. �e average temperature of the town in the �rst four days of a month was 58 degrees. �e average for the 2nd, 3rd, 4th and 5th days was 60 degrees. If the temperature of the 1st and 5th days were in the ratio 7 : 8, then what is the temperature on the 5th day?(a) 64 degrees (b) 62 degress (c) 56 degrees (d) None of these

4. �e sum of four numbers is 64. If you add 3 to the �rst number, 3 is subtracted from the second number, the third is multiplied by 3 and the fourth is divided by 3, then all the results are equal. What is the di�erence between the largest and the smallest of the original numbers?(a) 21 (b) 27(c) 32 (d) None of these

5. A spider climbed 62 12

% of the height of the pole

in one hour and in the next hour it covered 12 12

%

of the remaining height. If the height of the pole is

192 m, then distance climbed in second hour is(a) 3 m (b) 5 m (c) 7 m (d) 9 m

6. Samant bought a microwave oven and paid 10% less than the original price. He sold it with 30% pro�t on the price he had paid. What percentage of pro�t did Samant earn on the original price?(a) 17% (b) 20% (c) 27% (d) 32%

7. �e electricity bill of a certain establishment is partly �xed and partly varies as the number of units of electricity consumed. When in a certain month 540 units are consumed, the bill is ̀ 1800. In another month 620 units are consumed and the bill is ̀ 2040. In yet another month 500 units are consumed. �e bill for that month would be(a) `1560 (b) `1840 (c) `1680 (d) `1950

8. Which of the following statements is/are necessary to answer the given question.�ree friends, P, Q and R started a partnership business investing money in the ratio of 5 : 4 : 2 respectively for a period of 3 years. What is the amount received by P as his share in the total pro�t? Statement (I) : Total amount invested in the business in `22,000.Statement (II) : Pro�t earned at the end of 3 years

is 38

of the total investment.

Statement (III) : �e average amount of pro�t earned per year is `2750.(a) Statement (I) or Statement (II) or Statement (III)(b) Either Statement (III) only, or Statement (I) and

Statement (II) together(c) Any two of the three (d) All Statement (I), Statement (II) and Statement (III)


9. Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two di�erent computers to type an assignment of 110 pages? (a) 7 hours 30 minutes (b) 8 hours(c) 8 hours 15 minutes (d) 8 hours 25 minutes

10. A person borrowed `500 @ 3% per annum S.I. and

`600 @ 4 12

% per annum on the agreement that the

whole sum will be returned only when the total interest becomes `126. �e number of years, a�er which the borrowed sum is to be returned is(a) 2 (b) 3(c) 4 (d) 5

11. A park square in shape has a 3 metre wide road inside it running along its sides. �e area occupied by the road is 1764 square metres. What is the perimeter along the outer edge of the road ?(a) 576 metres (b) 600 metres(c) 640 metres (d) Data inadequate

12. Two metallic right circular cones having their heights 4.1 cm and 4.3 cm and the radii of their bases 2.1 cm each, have been melted together and recast into a sphere. Find the diameter of the sphere.(a) 7.1 cm (b) 4.2 cm(c) 3.1 cm (d) 6.4 cm

13. In a class, 30% of the students o�ered English, 20% o�ered Hindi and 10% o�ered both. If a student is selected at random, what is the probability that he has o�ered English or Hindi ?

(a) 25

(b) 34

(c) 35

(d) 310

Direction (14-16) : Study the table and answer the given questions.

Expenditures of a company (in Lakh rupees) per annum over the given years.

Items of expen-diture

Years

Salary Fuel and Transport

Bonus Interest on

Loans

Taxes

1998 288 98 3.00 23.4 831999 342 112 2.52 32.5 1082000 324 101 3.84 41.6 742001 336 133 3.68 36.4 882002 420 142 3.96 49.4 98

14. �e ratio between the total expenditure on Taxes for all the years and the total expenditure on Fuel and Transport for all the years respectively is approximately.(a) 4 : 7 (b) 10 : 13(c) 15 : 18 (d) 5 : 8

15. What is the average amount of interest per year which the Company had to pay during this period?(a) `32.43 lakhs (b) `33.72 lakhs(c) `34.18 lakhs (d) `36.66 lakhs

16. Total expenditure on all these items in 1998 was approximately what percent of the total expenditure in 2002 ?(a) 62% (b) 66% (c) 69% (d) 71%

17. A booster pump can be used for �lling as well as for emptying a tank. �e capacity of the tank is 2400 m3. �e emptying capacity of the tank is 10 m3 per minute higher than its �lling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to �ll it. What is the �lling capacity of the pump ?(a) 50 m3/min. (b) 60 m3/min.(c) 72 m3/min. (d) None of these

18. �e number of students in each section of a school is 24. A�er admitting new students, three new sections were started. Now, the total number of sections is 16 and there are 21 students in each section. �e number of new students admitted is(a) 14 (b) 24 (c) 48 (d) 114

19. 4 mat-weavers can weave 4 mats in 4 days. At the same rate, how many mats would be woven by 8 mat-weavers in 8 days ?(a) 4 (b) 8 (c) 12 (d) 16


20. In dividing a number by 585, a student employed the method of short division. He divided the number successively by 5, 9 and 13 (factors of 585) and got the remainders 4, 8, 12 respectively. If he had divided the number by 585, the remainder would have been(a) 24 (b) 144 (c) 292 (d) 584

SOLUTIONS

1. (a) : Required number of students = H.C.F. of 1001, and 910 = 91.

2. (d) : Let number of boys = x�en, number of girls = (60 – x) x(60 – x) + (60 – x) x = 1600 2x2 – 120x + 1600 = 0 x2 – 60x + 800 = 0 (x – 40) (x – 20) = 0 x = 40 or x = 20Hence, data is inadequate.

3. (a) : Sum of temperature on 1st, 2nd, 3rd and 4th days = (58 × 4) = 232 degrees ...(i)Sum of temperature on 2nd, 3rd, 4th and 5th days = (60 × 4) = 240 degrees ...(ii)Subtracting (i) from (ii), we getTemp. on 5th day – Temp. on 1st day = 8 degreesLet the temperature on 1st and 5th days be 7x and 8x degrees respectively.�en, 8x – 7x = 8 x = 8 Temperature on the 5th day = 8x = 64 degrees.

4. (c) : Let the four numbers be A, B, C and D. Let

A + 3 = B – 3 = 3C = D3

= x

�en, A = x – 3, B = x + 3, C = x3

and D = 3x

A + B + C + D = 64 (x – 3) + (x + 3) + x3

+ 3x = 64

16x = 192 x = 12�us, the numbers are 9, 15, 4 and 36 Required di�erence = (36 – 4) = 32.

5. (d) : Height climbed in second hour = 12 12

% of

100 62 12

% of 192 m

= 252

1100

752

1100

192 9

m m

6. (a) : Let original price = ` 100�en, C.P. = ` 90

S.P. = 130% of ` 90 = ` 130100

90

= ` 117

Required percentage = (117 – 100)% = 17%

7. (c) : Let the �xed amount be ` x and the cost of each unit be ` y. �en, 540y + x = 1800 ...(i) 620y + x = 2040 ...(ii)On subtracting (i) from (ii), we get 80y = 240 y = 3Putting y = 3 in (i), we get540 × 3 + x = 1800 x = 180 Fixed amount = ` 180, Cost per unit = ` 3Total cost for consuming 500 units = ` (180 + 500 × 3) = ` 1680.

8. (b) : Statement (I) and Statement (II) give, pro�t

a�er 3 years = ` 38

22000

= ` 8250

From Statement (III) also, pro�t a�er 3 years = ` (2750 × 3) = ` 8250

P’s share = ` 8250 511

= ` 3750

�us, (either Statement (III) is redundant) or (Statement (I) and Statement (II) are redundant). Correct answer is (b).

9. (c) : Number of pages typed by Ronald in 1 hour

32

6163

Number of pages typed by Elan in 1 hour 405

8

Number of pages typed by both in 1 hour

16

38 40

3

Time taken by both to type 110 pages

110 340

8 14

hours hours = 8 hours 15 minutes.

10. (b) : Let the time be x years

�en, 500 3100

600 9100 2

126

x x

15x + 27x = 126 x = 3

Required time = 3 years.11. (b) : Let the length of the outer edge be x m.

�en length of the inner edge = (x – 6) m x2 – (x – 6)2 = 1764 x2 – (x2 – 12x + 36) = 1764 12x = 1800 x = 150 Required perimeter = (4x) m = (4 × 150) m = 600 m.


12. (b) : Volume of sphere = Volume of 2 cones

13

2 1 4 1 13

2 1 4 32 2 ( . ) . ( . ) . cm3

13

2 1 8 42 3 ( . ) ( . )cm

Let the radius of the sphere be R.

43

13

2 1 43 3 R ( . ) R = 2.1 cm

Hence, diameter of the sphere = 4.2 cm.

13. (a) : P( )E 30100

310

, P( )H 20100

15

and P( )E H 10100

110

P (E or H) = P (E H) = P (E) + P (H) – P (E H)

= 3

1015

110

410

25

.

14. (b) : Required ratio

( )( )

83 108 74 88 9898 112 101 133 142

451

5861

1 310 13

.:

15. (d) : Average amount of interest paid by the company during the given period

`23 4 32 5 41 6 36 4 49 4

5. . . . .

lakhs

`183 3

5.

lakhs = ` 36.66 lakhs

16. (c) : Required percentage

288 98 3 00 23 4 83420 142 3 96 49 4 98

100. .. .

%

495 4713 36

100 69 45..

% . % 69% (approx.)

17. (a) : Let the �lling capacity of the pump be x m3/min.�en, emptying capacity of the pump = (x + 10) m3/min

According to question, 2400 240010

8x x

( )

x2 + 10x – 3000 = 0

(x – 50) (x + 60) = 0 x = 50 [neglecting the – ve value of x]So, �lling capacity of the pump = 50 m3/min.

18. (b) : Original number of sections = (16 – 3) = 13

Original number of students = (24 × 13) = 312Present number of students = (21 × 16) = 336 Number of new students admitted = (336 – 312) = 24

19. (d) : Let the required number of mats be x.More weavers, More mats (Direct Proportion)More days, More mats (Direct Proportion)WeaversDays

4 84 8

4::

:: :

x

4 × 4 × x = 8 × 8 × 4 x

8 8 44 4

16( )

So, the required number of mats = 16.20. (d) : z = 13 × 1 + 12 = 25,

y = 9 × z + 8 = 9 × 25 + 8 = 233,x = 5 × y + 4 = 5 × 233 + 4 = 1169Hence, on dividing 1169 by 585, remainder = 584.

5 x9 y – 4

13 – 8

1 – 12


Only One Option Correct Type

1. �e solution of the equation(3|x| – 3)2 = |x| + 7 which belongs to the domain of

x x( ) 3 are given by

(a) 1

9,

(b)

19

3,

(c) 16

, (d) 19

0,

2. In a G.P. if the sum of in�nite terms is 20 and the sum of their squares is 100, then the common ratio is

(a) 5 (b) 35 (c) 2

5 (d) 15

3. Let S be a non-empty subset of R. Consider the following statement:P : �ere is a rational number x S such that x > 0.Which of the following statements is the negation of the statement P ?(a) �ere is a rational number x S such that

x 0.(b) �ere is no rational number x S such that

x 0.(c) Every rational number x S satis�es x 0.(d) x S and x 0. x is not rational.

4. Six boys and six girls sit in a row randomly. �e probability that the six girls sit together or the boys and girls sit alternately is

(a) 3

308 (b) 1

100 (c) 2

205 (d) 4

407

5. �e general solution of the equation sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x is

(a) n 4 (b)

n 2 4

(c) n 8 (d) n

2 8

6. Let P(n) = 32n n ∈ N, when divided by 8, leaves the remainder (a) 2 (b) 1 (c) 4 (d) 7

One or More Than One Option(s) Correct Type

7. �e real value of for which the expression 1

1 2

iicoscos

( )where i 1 is a real number is

(a) 22

n n I ∈,

(b) 22

n n I ∈,

(c) 22

n n I ∈,

(d) 24

n n I ∈,

8. �e number of ways of arranging seven persons (having A, B, C and D among them) in a row so that A, B, C and D are always in order A-B-C-D (not necessarily together) is (a) 210 (b) 5040(c) 6 × 7C4 (d) 7P3

9. If n is a positive integer and ( ) ,3 3 5 2 1 n a where a is an integer and 0 < < 1, then(a) a is an even integer(b) (a + )2 is divisible by 22n+1

(c) the integer just below ( )3 3 5 2 1 n divisible by 3(d) a is divisible by 10

10. If the lines x – 2y – 6 = 0, 3x + y – 4 = 0 and lx + 4y + l2 = 0 are concurrent, then

This specially designed column enables students to self analyse

their extent of understanding of specified chapters. Give

yourself four marks for correct answer and deduct one mark for

wrong answer.

Self check table given at the end will help you to check your

readiness.

Total Marks : 80 Time Taken : 60 Min.

Class XI

MATHEMATICS TODAY | APRIL‘17 81

(a) l = 2 (b) l = –3(c) l = 4 (d) l = –4

11. �e equation sin x = [1 + sin x] + [1 – cos x] has (where [x] is the greatest integer less than or equal to x)

(a) no solution in

2 2

,

(b) no solution in 2

,

(c) no solution in , 32

(d) no solution for x ∈ R

12. In a ABC tan A and tan B satisfy the inequation

3 4 3 02x x , then(a) a2 + b2 + ab > c2 (b) a2 + b2 – ab < c2

(c) a2 + b2 > c2 (d) none of these

13. Let f x xx

( ) sinsin

.5

1 3 If D is the domain of f, then

D contains(a) (0, ) (b) (–2, –)(c) (2, 3) (d) (4, 6)

Comprehension Type

ABC is a triangle right angled at A, AB = 2AC. A = (1, 2), B = (–3, 1). ACD is an equilateral triangle. �e vertices of the two triangles are in anti-clockwise sense. BCEF is a square with vertices in clockwise sense.

14. ACF =

(a) 518 (b)

514 (c)

315 (d)

214

15. DE =

(a) 17 4 3 (b) 172

8 3

(c)

172

4 3

(d)

15 4 3

Matrix Match Type

16. A right angled triangle has sides 1 and x. �e hypotenuse is y and the angle opposite to the side x is .

x

y1

Column I Column II

P. lim

2

y x 1. 0

Q. lim

2

y x 2. 12

R. lim

2

2 2y x 3. 1

S. lim

2

3 3y x 4.

P Q R S(a) 4 3 2 1(b) 1 1 3 4(c) 1 2 3 4(d) 2 2 4 3

Integer Answer Type

17. sin212° + sin221° + sin239° + sin248° – sin29° –sin218° is

18. If a is a complex seventh root of unity, then the equation whose roots are a + a2 + a4 and a3 + a5 + a6 is x2 + x + c = 0, where c is

19. If the integer k is added to each of the numbers, 36, 300, 596, one obtains the squares of three consecutive terms of an A.P., then the last digit of k is

20. If log245 175 = a, log1715 875 = b, then the value of 1ab

a b is

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< 60% NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

No. of questions attempted ……

No. of questions correct ……

Marks scored in percentage ……


Only One Option Correct Type

1. Consider the following relations : R = {(x, y) : x, y are real and x = wy for some non-

zero rational number w}

S mn

pq

m n p q

, : , , , are integers, such that

n, q 0, mq = np} then (a) R is an equivalence relation, but S is not. (b) S is an equivalence relation, but R is not. (c) both R and S are equivalence relations. (d) neither R nor S is an equivalence relation. 2. �e points on the curve y3 + 3x2 = 12y where tangent

is vertical, are

(a)

43

2,

(b)

43

2,

(c) (0, 0) (d)

113

0,

3. For the set of linear equationslx – 3y + z = 0, x + ly + 3z = 1, 3x + y + 5z = 2

the value of l, for which the equations does not have a unique solution.

(a) 1 115

, (b) 1 115

, (c) 115

1, (d) 1 115

,

4. If from each of 3 boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn then the probability of drawing 2 white and 1 black ball is

(a) 1332

(b) 14

(c) 1

32 (d)

316

5. sin–1 (cos sin–1 x) + cos–1 (sin cos–1 x) =

(a) 0 (b) 4

(c) 2

(d) 34

6. If f is a continuous function and f t dt xx

( ) | | ,0 as as

f t dt xx

( ) | | ,0 as then the number of points in which the

line y = mx intersects the curve y f t dtx2

02 ( ) is

(a) 0 (b) 1 (c) atmost 1 (d) atleast 1One or More Than One Option(s) Correct Type

7.

Let thenx

xdx gof x c

1 2

31

23

/

( )( ) ,

(a) f x x( ) (b) f(x) = x3/2

(c) f(x) = x2/3 (d) g(x) = sin–1x8. A normal is drawn at a point P(x, y) of a curve.

It meets the x-axis at Q. If PQ is of constant length k. Such a curve passing through (0, k) is(a) a circle with centre (0, 0)(b) x2 + y2 = k2

(c) (1 – k)x2 + y2 = k2 (d) x2 + (1 + k2)y2 = k2

9. A student appears for tests I, II and III. �e student is successful if he passes either in tests I and II or tests I and III. �e probabilities of the student passing in tests I, II, III are p, q and 1/2 respectively. If the probability that the student is successful is 1/2, then(a) p = 1, q = 0 (b) p = 2/3, q = 1/2(c) p = 3/5, q = 2/3 (d) there are in�nitely many values of p and q

10. �e determinant

b c b cc d c d

b c c d a c

ll

l l l l3 3

is equal to zero, if

This specially designed column enables students to self analyse

their extent of understanding of specified chapters. Give

yourself four marks for correct answer and deduct one mark

for wrong answer.

Self check table given at the end will help you to check your

readiness.

Total Marks : 80 Time Taken : 60 Min.

Class XII


(a) b, c, d are in AP (b) b, c, d are in GP(c) b, c, d are in HP(d) l is a root of ax3 – bx2 + cx – d = 0

11. If P(2, 3, 1) is a point and L x – y – z – 2 = 0 is a plane then(a) origin and P lie on the same side of the plane.

(b) distance of P from the plane is 43

.

(c) foot of perpendicular is 103

53

13

, , .

(d) image of point P by the plane is 103

53

13

, , .

12. If (x) = f(x) + f(2a – x) and f(x) > 0, a > 0, 0 x 2a, then (a) (x) increases in (a, 2a) (b) (x) increases in (0, a) (c) (x) decreases in (a, 2a) (d) (x) decreases in (0, a)

13. Which of the following functions are periodic?(a) f(x) = sin x + |sinx|

(b) g x x xx x

( ) ( sin )( sec )( cos )( )

1 1

1 1 cosec(c) h(x) = max(sin x, cos x)

(d) p x x x x x( ) [ ] ,

1

323

3 10

where

[.] denotes the greatest integer function.Comprehension Type

Let A be the set of all 3 × 3 symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0.

14. �e number of matrices in A is(a) 12 (b) 6 (c) 9 (d) 3

15. �e number of matrices A in A for which the system

of equations Axyz

100

is inconsistent, is

(a) 0 (b) more than 2(c) 2 (d) 1

Matrix Match Type16.

Column I Column IIP. A line from the origin meets the lines

x y z

2

11

21

1,

x y z

8 3

23

11

1/

at P and Q respectively. PQ2 is

1. 7

Q. �e values of x satisfying

tan–1(x + 3) – tan–1(x – 3) sin 1 35are

2. 6

R. Non-zero vectors

a b c, , satisfy

a b b a b c 0 0, ( ) ( ) ,2| | | | .

b c b a If

a b c 4 , then the values of are

3. 5

S. L e t f b e t h e f u n c t i o n o n [–, ] given by f (0) = 9 and

f xx

x( )

sinsin

9 2

2/

/ for x 0.

�e value of 2

f x dx( )

is

4. 4

P Q R S(a) 2 4 3 4(b) 3 4 2 1(c) 3 2 1 4(d) 1 2 3 4

Integer Answer Type17. �e distance between the skew lines

x y z4

73

72

and x y z

42

63

3410

is

18. �e resultant of vectors

a b cand is . If c trisects the angle between

a b a band and | | ,| | , 6 4 then | |c is19. Let A = {1, 2, 3, 4} and B = {0, 1, 2, 3, 4, 5}. If m is the

number of increasing functions from A to B and n is the number of onto functions from B to A, then the last digit of n – m is

20. �e area bounded by the y-axis, the tangent and normal to the curve x + y = xy at the point where the curve cuts the x-axis is

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< 60% NOT SATISFACTORY! Revise thoroughly and strengthen your concepts.

No. of questions attempted ……

No. of questions correct ……

Marks scored in percentage ……


SECTION-I

SINGLE OPTION CORRECT

1. If p1, p2, ..., pn be probabilities of n mutually exclusive and exhaustive events E1, E2, E3, ..., En

then p rr rr

nl o g ( )( )

∈ 2

11

(a) (0, 1) (b) (0, n) (c) (1, 2) (d) (1, n)

2. Sum of squares of all trigonometrical ratios has minimum value as (a) 6 (b) 7 (c) 3 (d) None of these

3. �e area bounded by y x x x 2 43 , is

(a) 160 sq. units (b) 1 2 85

s q . u n i t s

(c) 45

s q . u n i t s (d) None of these

4. If a1, a2 be the roots of x2 – 32x + 16 = 0 and a3, a4

be the roots of x2 + 32x + 1 = 0 then aii

2

1

4

(a) 0 (b) 32 (c) 2013 (d) None of these

5. If e1 and e2 be the eccentricities of two concentric ellipses such that foci of one lie on the other and they have same length of major axes. If e1

2 + e22 = p

then (a) p < 1 (b) p > 1 (c) p 1 (d) p 1

6. If f (x) f (y) = f (x) + y f (y), where f (x) 0 x ∈ N then fofofo f. . . . ( )1

2 0 1 3 t i m e s

(a) 2 (b) 2013 (c) 2014 (d) None of these

7. If f x xx

x( ) s i n

3 6 12 2

2 1 and g xf xR

( ) ( )

be continuous x ∈ R then least positive integral

value of k is (where [·] denotes greatest integer function)(a) 6 (b) 7 (c) 8 (d) None of these

8. In ABC, if x = (a)2, y = a2, z = c o t A2

and w = cotA then (a) xy = zw (b) xw = yz(c) xz = yw (d) None of these

SECTION-II

MORE THAN ONE OPTION CORRECT

9. If ar r

br r

cr rr r r

12

12

142

12

13

1, ,

then which of the following is (are) correct?(a) a + b = 2 (b) b + c = 1(c) a – c = 1 (d) None of these

10. If A = (1 + )(1 + 2)(1 + 4) ... (1 + 2n – 1) ; n ∈ N,

then A = (a) 1, if n be even (b) – 2, if n be even(c) 1, for all n ∈ N (d) – 2, if n be odd

11. If 03

a

i i n( )1 , 2 , 3 , . . . . , and

( s i n )zrr

r

n

1

13a , then |z| ∈

(a) 12

,

(b) 3

445

,

(c) 34

1,

(d) None of these

15 CHALLENGINGPROBLEMS

For Entrance Exams


12. If y = ax + a be the common tangent to y2 = 4ax and x2 = 4by (where a, b > 0) then (a) a= 1 (b) a< 0, < 0(c) common tangent never passes through 1st quadrant(d) None of these

13. I f f xx x

x x

a b cd

a b cd

( ) ,

∈

2

24 3

1( w h e r e

a, b, d are integers and c prime) then which of the following are prime?(a) a + c (b) a + d (c) b + d (d) c + d

14. If A

5 31 1 1 3 3 6

and

det (–3A2013 + A2014) = aa2 (1 + + 2) then(a) a = 2013 (b) = 3(c) = 10 (d) None of these

15. From a point A on x-axis, 2 tangents are drawn to x2 + y2 = 16 meeting y-axis at P and Q then (a) minimum value of AP2 + AQ2 is 128(b) minimum area of APQ is 32 sq. units(c) minimum value of OA2 + OP2 is 64(d) for minimum area of APQ, the point A is

( , ) ( , )4 2 0 4 2 0o r

SOLUTIONS

1. (a) : For mutually exclusive and exhaustive events, we have p1 + p2 + p3 + ... + pn = 1 ...(i)

p r p p p nr rr

n

n nl o g ( ) l o g l o g . . . l o g ( )( ) ( )21

1 3 2 4 21 2 3 1

p r p p p nr rr

n

n nl o g ( ) l o g l o g . . . l o g ( )( ) ( )21

1 3 2 4 21 2 3 1 | l o g | | | | l o g | | | . . . | l o g ( ) | | |( )3 1 4 2 22 3 1p p n pn n

| l o g | | | | l o g | | | . . . | l o g ( ) | | |( )3 1 4 2 22 3 1p p n pn n

< |p1| + |p2| + |p3| + ... + |pn| [Q log32 < 1 etc.]< p1 + p2 + p3 + ... + pn< 1 [using (i)]

∈ p rr rr

nl o g ( ) ( , )( )2

11 0 1

2. (b) : ... sin2 + cos2 + sec2 + cosec2 + tan2 + cot2= 1 + (1 + tan2) + (1 + cot2) + tan2 + cot2

= 3 + 2(tan2 + cot2)

3 4

2

2 2t a n c o t

3 4 2 2t a n c o t [ . ] A . M G . M . 7 Minimum value = 7.

3. (b) : y x x 2 3 ...(i) y x x 2 3 ...(ii) x = 4 ...(iii)(i) meets x-axis at x = 0 i.e. at (0, 0) only and then increases with x.

On solving (ii) with x-axis, we get

x x x( ) ,2 0 0 4

Moreover, y – if x

Required area = ( )y y dx1 20

4

(where y1 and y2 are values of y from (i) and (ii) respectively)

{ ( ) ( ) }/ /2 23 2 3 2

0

4x x x x dx

2 2 25

3 2

0

45 2

04

x dx x/ /[ ]

45

3 2 1 2 85

( ) s q . u n i t s

4. (d) : a1 + a2 = 32, a1a2 = 16 a3 + a4 = –32 ; a3a4 = 1

a a a a aii

2

1

4

12

22

32

42

= (a1 + a2)2 – 2a1a2 + (a3 + a4)2 – 2a3a4= 322 – 2(16) + (–32)2 – 2 · 1 = 2(1024 – 16 – 1) = 2014

5. (c) : Let be the angle of inclination of major axes and equation of one of the ellipses be

x

a

y

b

2

2

2

2 1

...(i)

whose foci are S and S′ and eccentricity = e1 b2 = a2(1 – e1

2) ...(ii)

Let P and P′ be foci of other ellipse whose eccentricity = e2Let PP′ = 2R OP = R = ae2... Diagonals PP′ and SS′ bisect each other PS′P′S is a parallelogram.Here, co-ordinates of P are (Rcos, Rsin)

From (i), R

a

R

b

2 2

2

2 2

2 1c o s s i n

a e

a

a e

a e

222 2

2

222 2

212

11

1( s i n ) s i n

( )

[Using (ii)]

s i n 2 2

2

12 2

222

11

e

ee e

s i n 2 1

222

12 2

2

11

e e

ee


( ) ( )

s i n1 1

112

22

12

22

2e e

e e

1 – e12 – e2

2 + e12 e2

2 e12 e2

2

e12 + e2

2 1 p 1 6. (c) : ... f (x) f (y) = f (x) + y f (y)

x = 1, y = 1 f (1) f (1) = f (1) + 1 · f (1) f (1) f (1) = 2 f (1) f (1) = 2 [Q f (x) 0 x ∈ N f (1) 0]x = 2, y = 2 f (2) f (2) = f (2) + 2 f (2) = 3 f (2) f (2) = 3. [By same reason, f (2) 0]x = 3, y = 3 f (3) f (3) = f (3) + 3 f (3) = 4 f (3) f (3) = 4 etc. fofofo f fofofo f. . . . ( ) . . . . ( )1 2

2 0 1 3 2 0 1 2 t i m e s t i m e s

fofofo f. . . . ( )32 0 1 1 t i m e s

= ...... = f (2013) = 2014

7. (c) : 3 6 2 x is de�ned if x2 36 –6 x 6

and, s i n s i n

1 1

212 2

12x

x xx

is de�ned if

12

12

xx

[taking x 0]

i.e., if 12

1 1 22

2

xx

x x| |

| |

1 2 0 0

1 2 0 0

2

2

x x x

x x x

,

,

i f

i f

( ) ,

( ) ,

1 0 0

1 0 0

2

2

x x

x x

i f

i f x = ± 1

Df = {–1, 1} f (–1) = 3 52

= 4.35 (approx.)

f ( ) . ( )1 3 52

7 4 9 a p p r o x .

Rf = {4.35, 7.49} (approx.)

Now, g xf x

k( ) ( )

will be continuous x ∈ R

if f xk( ) ( , )∈ 0 1 for which value of k must exceed the

greatest value of f (x) i.e., 7.49 least positive integral value of c = 8.

8. (b) :

c o tc o s

s i n c o s

c o ss i n

AA

A AA

A2

22

22 2

12

2 1

22

4

2 2 2bc Abc A

bc b c a( c o s )s i n

( )

c o t ( )A bc b c a

22

4

2 2 2

2

4

2 2 2( ) ( )bc ca ab a b c

( ) ( )a b c a x2 2

4 4 4 ...(i)

c o t c o ss i n

c o ss i n

AAA

bc Abc A

b c a

22 4

2 2 2

c o t ( )A

b c a a b c y2 2 2 2 2 2

4 4 4 ...(ii)

On dividing (i) by (ii), c o t

c o t

A

Axy

zw

xy

2

xw = yz

9. (a, b, c) : l o g ( ) . . . .e x x

x x x12 3 4

2 3 4 t o

Putting x = 1 on both sides, we getl o g . . .e 2 1 1

213

14

t o ...(A)

1

1 21

3 41

5 6. . . . t o

12 1 2

12

12 21

21( )r r r r

a

r r a = 2loge2 ...(i)

From (i), l o g . . . . .e 2 1 12

13

14

15

t o

1 12 3

14 5

. . . t o ...(B)

1

2 31

4 51

6 71 2. . . l o gt o e

12 2 1

1 21 r rr

e( )l o g

12

2 2 2 2 2 221 r r

br

e el o g l o g ...(ii)

Adding (A) and (B),

2 2 1 11 2

12 3

13 4

14 5

l o g . . .e

t o

1 21 2 3

23 4 5

. . . . t o

2 2 1 22 1 2 2 1

141

31

l o g( ) ( )e

r rr r r r r

c = 2loge2 – 1 ... (iii) From (i), (ii) and (iii) a + b = 2, b + c = 1 and a – c = 1

10. (a, d) : A = (1 + )(1 + 2)(1 + 4)(1 + 8) ... (1 + 2n – 1

)

= (1 + )(1 + 21)(1 + 22

)(1 + 23)(1 + 24

) ... to n terms

= (1 + )(1 + 2)(1 + )(1 + 2) ... to n terms= (–2)(–)(–2)(–) ... to n terms

12

,

,

i f b e e v e n

i f b e o d d

n

n


11. (b, c) : zr

rr

n

1

13s i n a

sina1 + zsina2 + z2sina3 + ... + zn – 1 · sinan = 3 |sina1 + zsina2 + z2sina3 + ... + zn – 1 · sinan| = | |3 3 |sina1| + |zsina2| +|z2sina3| + ...

+ |zn – 1sinan| 3 |sina1| + |z||sina2| + ... + |zn – 1||sinan|

3

23

23

22| | | | . . . .z z t o

03

0 32

a

ai is i n

3 32

1 2{ | | | | . . .z z t o }

2 11 | |z [Taking 0 < |z| < 1]

∈

1 12

12

12

1| | | | | | ,z z z

Options (b) and (c) are correct.12. (a, b, c) : y2 = 4ax .... (i) and x2 = ...(ii)

y m xa

m1 11

... (iii) and x my

bm

...(iv)

are any tangent to (i) and (ii) respectively.

From (iv) ym

xb

m

12 ...(v)

If (iii) and (v) be same then mm

am

b

m1

12

1

a n d

On eliminating m, we get mab

mab1

31

3

Common tangent is yab

x aba

3 3

But, common tangent is given as y = ax + a

a ab

ba

3 3a n d

a < 0, < 0 and a= 1 [Q a, b > 0]Further, common tangent can be written as

xa

ya

a

1 in which intercepts on the axes are –ve.

Common tangent will never pass through 1st quadrant.

13. (a, b, c) :

L e t y f xx x

x x

( )2

24 3

1 (y – 1)x2 + (y – 4)x + (y – 3) = 0Q x being real, D 0 [Q 1 + x + x2 > 0 x ∈ R] (y – 4)2 – 4(y – 1)(y – 3) 0 3y2 – 8y – 4 0

yy

y228

343

43

2 89

y y

43

2 73

43

2 73

0

∈

y4 2 7

34 2 7

3,

On comparing, we can write a = 4, b = 2, c = 7 and d = 3. a + c = 11, a + d = 7 and b + d = 5 are primes.

14. (a, b, c) : A

5 31 1 1 3 3 6

|A| = 1680 + 333 = 2013 det(–3A2013 + A2014) = |A2013(–3I + A)|

= |A2013||–3I + A|

2 0 1 32 3

1 1 1 3 3 32 0 1 3

= 20132013(666 + 333) = 20132013(111 × 9) = 20132013 · 32(1 + 10 + 102) a = 2013, = 3, = 10

15. (a, b, c, d) :

OROA

OA c o s s e c 4

And, OROP

OP s i n 4 c o s e c

OA2 + OP2 = 16(sec2 + cosec2)= 16 (2 + tan2 + cot2)

3 2 3 2

23 2 3 2

2 22 2t a n c o t t a n c o t

[... A.M. G.M.] Minimum value of OA2 + OP2 = 64Now, AP2 + AQ2 = (OA2 + OP2) + (OA2 + OQ2) = 2(OA2 + OP2) 2(64) [... OP = OQ] Minimum value of AP2 + AQ2 = 128

Also, APQ = 12

12

2 4 PQ OA OP( ) s e c

= 4cosec · 4sec 3 2

23 2

s i n [... 0 < sin2 1]

Minimum area of APQ = 32 sq. units when = 45°

�en, OA = 4 sec 45° = 4 2

and so point A may be ( , ) ( , )4 2 0 4 2 0o r


1. �ere are four seats numbered 1, 2, 3, 4 in a room and four persons having tickets corresponding to these seats (one person having one ticket). Now the person having the ticket number 1, enters into the room and sits on any of the seats at random. �en the person having the ticket number 2, enters in room. If his seat is empty then he sits on his seat otherwise he sits on any of the empty seat at random. Similarly the other persons sit. Find the probability that the person having ticket numbered 4 gets the seat number 4. – Pratiksha, Kolkata

Ans. Way1 : 1 take 1, 2 take 2, 3 take 3, and 4 take 4

p (way 1) = 14

× 1 × 1 × 1 = 14

Way 2 : 1 take 2, 2 take 1, 3 take 3, 4 take 4

p(way 2) = 14

13

1 1 112


p(way 3) = 14

13

12

1 124


p(way 4) = 14

1 12

1 18

Required probability = 14

112

124

18

12

.

2. P(x1, y1), Q(x2, y2) with y1 < 0 and y2 < 0 be ends

of latus rectum of the ellipse x y2

24

1 . Find the

equations of the parabola with P and Q as the ends of latus rectum. – Md. Rizwan, Lucknow

Ans. Ends of latus rectum are

3 12

3 12

3 12

3 12

, , , , ,and and

3 1

23 1

23 1

23 1

2, , , , ,and and

P Q32

3 12

, , and

Vertices are A A0 1 32

0 1 32

, ,or

Required equation = x y2 2 3 3 3

or x y2 2 3 3 3.

3. n^ is a unit vector perpendicular to the plane containing the points whose position vectors are

a b r, , and satisfying |[ ] | | ( ) |^

r a b a n r b a

|[ ] | | ( ) |^

r a b a n r b a a b then �nd the value of l – Arun Kumar, Patna

Ans. ( ) ( )

r a b a is parallel to n^ ( ) ( ) ^

r a b a t n ...(i)|( ) ( ) | | |^

r a b a n t or |[ ]| | |^

r a b a n t ...(ii)From (i),

r b a a b t n( ) ^ or | ( ) ( ) | | |

r b a a b t ...(iii)From (ii) and (iii) we get,

|[ ]| | ( ) ( ) |^

r a b a n r b a a b

l = 1.

4. Prove that the roots of 1 + z + z3 + z4 = 0 are represented by the vertices of an equilateral triangle. – Manish Sharma, Punjab

Ans. �e given equation is (1 + z)(1 + z3) = 0. �e distinct roots being – 1, , 2 which if be represented by points A, B and C in that order

AB 32

34

32

, BC 0 3 3

CA 32

34

32

The three points represent the vertices of an equilateral triangle.

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SOLUTION SET-1711. (b) : Using the polar coordinates, x = rcos, y = rsin,

xdx + ydy = x y dx2 2 becomesdr(1 – cos) + rsind = 0 r(1 – cos) = c, r – x = c, (where c = constant)x2 + y2 = r2 = (c + x)2 y2 = 2cx + c2, family of parabolas.2. (b) : cos cos cos cos cos cos cos

( )A Ca c

Bb

b A b C a B c Bb a c

( cos cos ) ( cos cos )( )

b A a B b C c Bb a c

a cb a c( )

[ c = bcosA + acosB and a = bcosC + ccosB]

1b

3. (b) : 4 6 6

1 3 21 4 3

0

l

ll

l3 – 4l2 – l + 4 = 0 A3 – 4A2 – A + 4I = 0; Multiplying by A–1, we get

A A A I 1 214

14

a 14

1 14

, , , a + + = 1

4. (c) : Let C 2

in the triangle ABC.

r ab

a b a b

2 2

r a b ab r a b2 2 Squaring both sides, we get(a –2r)(b – 2r) = 2r2

�e number of triangles is half of the number of divisors

of 2r2 = 2 32 112 612, which is 12 × 2 × 33 = 27

5. (d) : A = (0, –1) and P = (2cos, sin)AP2 = 4cos2 + (sin+ 1)2 = 5 – 3sin2 + 2sin

16

33 1

3163

2sin

APmax .

43

6. (d) : r i j k r j k i

^ ^ ^ ^ ^ ^a

r j k i r i k j ^ ^ ^ ^ ^ ^

�e lines

r a b a and

r c d are skew if [ ]

c a b d 0

Here, [ ]^ ^ ^ ^ ^i j k i j 2 2

�e lines are skew.7. (d) : A = (–1, 2), B = (2, 3) Line AB is x – 3y + 7 = 0�e circle is (x + 1)(x – 2) + (y – 2)(y – 3) + l(x – 3y + 7) = 0or S = x2 + y2 + (l – 1)x – (3l + 5)y + 7l + 4 = 0 …(i)

Radius = 5 1 3 54

7 4 52 2

( ) ( ) ( )l l l l2 = 1, l = 1, –1l = 1 S is x2 + y2 – 8y + 11 = 0. It does not intersect x-axis.l = –1 S is x2 + y2 – 2x – 2y – 3 = 0 y = 0 x = –1, 3Length of intercept is 3 + 1 = 4.8. (a) : Since, S = x2 + y2 + (l – 1)x – (3l + 5) y + 7l + 4 = 0 ...(i)Putting y = – x in eq. (i)2x2 + 4(l + 1)x + 4 + 7l = 0

It has equal roots l 2 12

, .

l = 2, y = 0 (i) becomes x2 + x + 18 = 0, S does not intersect x-axis.

l 12

, y = 0 (i) becomes 2x2 – 3x + 1 = 0

x 1 12

,

Length of x-intercept = 1 12

12

.

9. (1) : In the given equation, a + = p, a = –(p + q)

�e given expression is ( )

( )( )

( )a

a

1

1 11

1 1

2

2

2

2q q

aa

a

11 1

11 1

1( ) ( )

Since, q – 1 = –(a + a + + 1) = –(a + 1)( + 1)

10. (c) :

f x xx

f x xx1 2′ ′ ( ) cos

| cos |, ( ) sin

| sin |

f x xx

f x xx3 4′ ′ ( ) cos

| cos |, ( ) sin

| sin |

32

∈

, the derivaties are –1, –1, 1, 1

4 32

∈

, the derivatives are –1, 1, 1, –1

5 32

2∈

, the derivatives are 1, 1, –1, –1

7 2 52

∈

, the derivatives are 1, –1, –1, 1

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