sr jee mains maths question bank - Sri Gayatri

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Page | 1 MATHS QUESTION BANK 2020

SRIGAYATRI EDUCATIONAL INSTITUTIONS

INDIA

SR MPC JEE MAINS

1. Number of integral values of x satisfying the equation 2

15sgn 1 2

1x

x

is:

(where [.] is greatest integer function,{ . } fractional part function and sgn(x) is signum function.

1) 5 2) 7 3) 15 4) 16

2. Range of the function 2

2 2log 2 log 16sin 1f x x is

1) 0,1 2) ( ,1] 3) 1,1 4) ,

3. If 2f x x bx c and 2 2f t f t for all real numbers t , then which of the

following is true ?

1) 1 2 4f f f 2) 2 1 4f f f

3) 2 4 1f f f 4) 4 2 1f f f

4. If , are two distinct real roots of the equation 3 1 0, 1,0ax x a a , none

of which is equal to unity, then the value of

3 2

1/ 1

1lim

1 1x x

a x x a

e x

is

aL k

then the value of KL is :

1) 1 2) 2 3) 3 4) 4

5. Let f x = max. sin :0 , min. sin :0t t x g x t t x a h x f x g x

where [ ] denotes greatest integer function, then the range of h x is:

1) 0,1 2) 1,2 3) 0,1,2 4) 3, 2, 1,0,1,2,3

6. If

11

27exp 3 log 27 9; 3

3 27

1 cos 3; 3

3 tan 3

x

x

exx

f x

xx

x x

If 3x

Lim f x

exist, then equals to ( where [ .] is greatest integer function)

1) 9 / 2 2) 2 / 9 3) 2 / 34) None of these

7. Number of ordered pair (a,b) from the set 1,2,3,4,5A so, that the function

3

2 103 2

x af x x bx is an injective mapping x R

1) 13 2) 14 3) 15 4) 16

MATHS QUESTION BANK



8. If 2 2 2.......

1 1 2 1 99

x x xf x x x x x

x x x

then value of

3f

is:

(where k and k denote greatest integer and fractional part functions of k

respectively).

1) 5050 2) 4950 3) 17 4) 73

9. Number of elements in the range set of 15

15

xf x

x

0,90x are ( where [ .] is

greatest integer function)

1) 5 2) 6 3) 7 4) infinite

10. If a point ,p x y lies on the curve y f x such that

1 1 1

1

, 1,2

1tan tan tan 3

lim sin 21 2x y

xy

yx y

exists, then

1

1/3lim

3 1x

f x

x

is equal to

1) 3 2) 3

5 3)

3

8 4)

3

10

11. The domain of f x is 0,1 therefore the domain of nxy f e f l x is:

1) 1

,1e

2) , 1e 3) 1

1,e

4) , 1 1,e e

12.

2

1x

x a

e xLim

x

equals to (where . is fractional part function and [ .] is greatest

integer function )

1) 2

I 2) 2e 3) I 4) Doesn’t exist

13. Let 23 22sin cosf x x x and 111 tan

2g x x , then the number of values of x

in interval 10 , 20 satisfying the equation sgnf x g x , is ( where sgn(x)

is signum function)

1) 6 2) 10 3) 15 4) 20

14. Let :f I I be a function ( I is set of integers) such that

0 1,f f f n 2 2f f n n . Then

1) 3 0f 2) 2 0f 3) 3 2f

4) f is many-one function.

15. Let a sequence of number is as follows:

1

3 5

7 9 11

13 15 17 19

21 23 25 27 29

_ ____ ____ ____ ____ ____ ___ ___

___ ____ ____ ____ ____ ____ ___ ___

If tn is the first term of thn row then limn

t nn

is equal to

1) 1/ 2 2) 1/ 2 3) 1 4) 1



16. The sum of all possible values of n where , 0n N x and 10 100n such that the

equation 22 0x x n has a solution, is equal to :(where x denotes largest

integer less than or equal to x )

1) 150 2) 175 3) 190 4) 210

17. The value of ‘a’ and ‘b’ for which 2,x b

e a has four distinct solutions, are

1) 3, , 0a b 2) 2, , 0a b

3) 3, ,a b R 4) 2, ,a b a

18. Let :f R R be defined as 3 3 sgn 2x x xf x e

(where sgn ( x ) denotes

signum function of x ). Then which one of the following is correct ?

1) f is injective but not surjective

2) f is surjective but not injective

3) f is injective as well as surjective

4) f is neither injective nor surjective

19. If is a root of the equation sin 1x x then min sin , { }

lim1x

x x

x

is

(Where [ . ] denotes greatest integer function {x} fractional part of x).

1) 1 2) 0 3) -1 4) does notexist

20. If 7 4

9

xf x

x

, then the range of function sin 2y f x is:

1) 0,1 2) 1

0,2

3) 1 1

0, ,12 2

4) 0,1

21. If the range of the function 2

1

1

xf x

p x

does not contain any values belonging to

the interval 1

1,3

then the true set of values of p , is:

1) , 1 2) 1

,4

3) 0, 4) ,0

22. The value of

2

0 2

1

lim

1x

xx x

x

e

x

x

is equal to

1) 1 2) 1/8 3) 3/ 2 4) 1/ 4

23.

40

cos tan coslimx

x x

x

1) 1/ 6 2) 1/ 3 3) 1/ 6 4) 1/ 3

24. If the equation 2 2 3 2 224 9 4 5 6 2 1 0

2

pp p x x p p p x p

is satisfied by all values of x in (0, 3] then sum of all possible integral values of ' 'p

is (where [.] is greatest integer function and { . } fractional part function)

1) 0 2) 5 3) 9 4) 10

25. If the equation 2 1x x p has exactly one solution, the number of integral values

of p , is:

1) 3 2) 4 3) 5 4) 7



26. If the functions 2 2 23 2 1f x k k x k x R and

2 3 2 26 5 2 1g x k k x k k x k k x R have the same graph, then the

number of real values of k is:

1) 0 2) 1 3) 2 4) 3

27. 2

20

sin cos tan sinlimx

x

x

1) 2) / 4 3) / 2 4) /8

28. Let

221 1

3

cos 1 . cos 12

2

x x

f xx x

. If 0f p and 0f q ,

then value of p

q is: ( . denotes fractional part function)

1) 4

2)

8

3) 4 4) 8

29. If [ . ] denotes greatest integer function then value of 1

lim 1 1 1x

x x x

is

1) 0 2) 1 3) -1 4) Doesn’t exist

30. Let x denotes the greatest integer less than or equal to x . If all the values of x such

that the product1 1

2 2x x

is prime, belongs to the set 1 2 3 4, ,x x x x , then

the value of 2 2 2 2

1 2 3 4x x x x is equal to

1)16 2) 7 3) 100 4) 11

31. If

1/ 1/

1/ 1/, 0

0, 0

x x

x x

e ex x

f x e e

x

then at 0,x f x is

1) Differentiable 2) Not differentiable

3) Differentiable and continuous 4) None of these

32. If f x y z f x f y f z with 1 1, 2 2 and , , ,f f x y z R then

1

3

4 . 3n

r

n

r f r

Ltn

is equal to

1) 4 2) 3 3) 6 4) 8

33. If [x] denotes the integral part of x and

sin sin 11

1

x xx

f xx

, then the number

of points where f(x) is discontinuous in [0, 5] is

1) 6 2) 5 3) 4 4) None of these



34. Let

2 2

f x f yx yf

for real values of x and y. If ' 0f exists and equals 1 and

0 1, then ' 2f f is equal to

1) 0 2) 2 3) 1 4) 3

35. If tan andf x x x g x is the inverse of f(x), then 'g x is equal to

1)

2

1

1 g x x

2)

2

1

2 g x x

3)

2

1

2 g x x

4) None of these

36. If y f x be a real valued function continuous and differentiable every where such that

' 0 and 1 1,f x x R f then range of 2 2

1

1,

xdx

xx y

is

1) 0,2

2) 0,4

3) ,

4 2

4) None of these

37. Let f(x) be a differentiable function wherever it is continuous and

1 2 1 2 1 2 1 2' ' 0, '' . '' 0, 5, 0 andf C f C f C f C f C f C C C . If f(x) is

continuous 1 2 1 2, and '' '' 0,C C f C f C then minimum number of roots of

1 2' 0 in 1, 1 ,f x C C is

1) 5 2) 4 3) 3 4) 2

38. Let f be a polynomial function such that

2 , 0f x f y f x f y f xy x y R and f(x) is one-one x R

with 0 1, 1 =2 and ' 1 2f f f . If

22min , , 1 ,h x x x

f x

then the

number of points of non-differentiability of h(x) is

1) 3 2) 4 3) 5 4) 6

39. Let :f R R be a function defined by 3max , ,f x x x then the set of all points where

f(x) is not differentiable, is

1) 1,1 2) 1,0 3) 0,1 4) 1,0,1

40. The number of points in (1, 3) where 2

, 1x

f x a a

is not differentiable, is ([.] is G.I.F)

1) 4 2) 5 3) 6 4) 7

41. Let sin , 0, , ,f x n P x x n I ‘P’ is a prime number and [.] is G.I.F, then the

number of points at which f(x) is not differentiable, is

1) P 2) 2P + 1 3) 2P 1 4) None of these



42. Number of points of non-differentiablity of sinf x x x in , ,2 2

is ([.] is

G.I.F)

1) 5 2) 4 3) 3 4) 2

43. If 2

1 1

2and exists, then

2

x x df x f x f x

dxx x

is equal to.

1)

2

3

1 x

2)

2

3

1 x 3)

2

1

1 x 4) None of these

44 If :f R R is an invertible function such that 1andf x f x are symmetric about the

line ,y x then

1) 'f x is even 2) f x is even

3) f x is neither odd nor even 4) None of these

45. If 1

21

3tan ,1

23

x

f x

x

then the value of ' 0f is

1) 3 2) 2 3) 0 4) None of these

46. If

2

2

ln 2

, thenx

x

d xy x e

dy

, is

1) 2

27

2)

2

27 3)

1

9

4)

1

9

47. If 2 1 1 ,f x x x x then 5 1 1 3

' ' ' '2 2 2 2

f f f f

is equal to

1) 1 2) 2 3) 1 4) 2

48. Number of points of discontinuity of 5 2

x xf x

, for 0,100 ,x is ([.] is G.I.F and

{.} is fractional part function).

1) 50 2) 51 3) 52 4) None of these

49. If f(x) is a thrice differentiable function such that

30

4 3 3 3 212,

x

f x f x f x f xLt

x

then the value of ''' 0f is equal to

1) 0 2) 12 3) 2 4) None of these



50. If , 0 andf xy f x f y x y f x is differentiable at 1,x then

1) 2x

f f x f yy

2)

0

11

x

xLt f f e

x

3) f(x) is differentiable for all x > 0 4) None of these

51. If a function : 2 ,2f a a is an odd function such that 2f x f a x for

,2x a a and the left hand derivative at x a is 0, then the left hand derivative at ,x a is

1) 0 2) 1 3) 1 4) Does not exist

52. Let 2

2

2, 3

3, 3

ax bx xf x

bx x

, then the values of ‘a’ and ‘b’ so that f(x) is differentiable

everywhere, are

1) 35 10

,3 9

2) 35 10

,9 3

3) 3 10

,35 9

4) 35 3

,9 10

53. If 2 2f x x x and

min : 2 , 2 0,

max :0 ,0 3

f t t x xg x

f t t x x

then

1) f(x) is everywhere continuous and differentiable

2) f(x) is differentiable {0}x R

3) f(x) is differentiable x R only 4) g x is differentiable 2,3x

54. Let 2tan ,f x x

([.] is G.I.F), then

1) ' 0 1f 2) 0x

Lt f x

does not exist

3) f(x) is not differentiable at 0x 4) f(x) is continuous at 0x

55. Let , andf x x g x x f g x h x where [.] is G.I.F, then ' 1 ,h is

1) 0 2) 1 3) 1 4) Does not exist

56. Let 2 , 1,1 ,f x x x x x then the number of points at which f(x) is discontinuous, is

1) 2 2) 1 3) 0 4) None of these

57. Let

4

tan , then4 x

dyy x

dx

, is

1) 1 2) 1 3) 0 4) Does not exist

58. If

, 0x

f x xx

, ([.] is G.I.F), then ' 1f , is

1) 1 2) 1 3) 0 4) Does not exist



59. Let cos sin ,0 2f x x x x , ([.] is G.I.F), then the number of points of

discontinuity of f(x) is

1) 6 2) 5 3) 4 4) 3

60. Let 0

1sin ,

x

f x t dtt

then the number of points of discontinuity of the function f(x) in (0,

), is

1) 0 2) 1 3) 2 4) Infinite

61. 2 2sin tan

dx

x x

1) 11 1 tan

cot tan2 2 2 2

xx c

2) 11 1

cot tan2 2 2 2

xx c

3) 1 1

cot tan2 2

x x c 4) 1 tan

cot tan2

xx c

62. 3 sin 2

cos 1

2 x

x xdx

x e x

1) sinln 2 1xxe c 2) sinln 2 1xxe c

3) sin

sin

2 1ln

2 1

x

x

xec

xe

4)

sin

sin

2 1 1ln

2 1 1

x

x

xec

xe

63

2 3

1

1

x dx

x x x x

1) 1 1tan x c

x

2) 1 1

2 tan 1x cx

3) 1 12 tan 1x c

x

4) 1 1tan 1x c

x

64. 2 24 3 3 4

dx

x x

1) 1

2

1 2tan

5 3 4

xc

x

2) 1

2

1 5tan

10 2 3 4

xc

x

3) 1

2

1 5tan

5 2 3 4

xc

x

4) 1

2

1 5tan

10 3 4

xc

x

65

2

2

2

1 1

xe xdx

x x

1) 2

2

x xe c

x

2) 2

2

x xe c

x

3) 1

1

x xe c

x

4)

1

1

x xe c

x



66.

2

sin cos

xdx

x x x

1) sec

cotsin cos

x xx c

x x x

2)

sectan

sin cos

x xx c

x x x

3) sec

cotsin cos

x xx c

x x x

4)

sectan

sin cos

x xx c

x x x

67. 3cos 2sin 2

dx

x x

1)

3

2tantan

3

xx c 2)

3

2tantan

3

xx c

3)

5

2tantan

5

xx c 4)

5

2tantan

5

xx c

68 If

2

3

8

3

2 3 2 3

4 54 5

bx x

dx a cx

x

then ab =

1) 1

5 2)

1

10 3)

1

11 4)

1

22

69. 21 sec

dx

x

1) 11 cos

cos2 2

xc

2) 1 cos

cos2

xc

3) 11 sin

sin2 2

xc

4) 1 sin

sin2

xc

70. 1 sec xdx

1) 1cos cos x c 2)

1sin cos x c

3) 12cos cos x c 4)

12sin cos x c

71. If 1

1 1

x

x x x

e dxf x C

e e e

, where C is arbitrary constant then

xLt f x

1) 4

2)

3

3)

6

4)

72 If

3 3

2 2

3 3

sin cos

cos tan sin cos sin cotsin cos sin

x x dx

a x b x cx x x

, then ab =

1) 4cos 2ec 2) 8cos 2ec 3) 8sec 4) 4sec

73 2

21ln 1

4

1.

1

xx x

e dxx

1) 2

2

1ln x c

x

2)

2

1ln 1 c

x

3) 2

2

1 1ln

2x c

x

4)

2

1 1ln 1

2c

x



74. 9sin 2 tan

dx

x x

1) 91tan

9x c 2) 91

tan18

x c 3) 91cot

18x c 4) 91

cot18

x c

75. 2 2 1

dx

x x x

1) 1 1cos

2

xc

x

2) 1 1

cos2

xc

x

3) 1 1sin

2

xc

x

4) 1 1

sin2

xc

x

76. If 2

3 1 2 4 2 4

2

1cos 1 1

1

xx dx A x B x Cx x D

x

, then A+B+C=

1) 0 2) 1

2 3)

1

2 4) -1

77. If 5 4 6

2 2 21 1 1x x dx A x x B x x C

, then A+B =

1) 5

6 2)

5

12 3)

5

24 4)

5

3

78 If 1 1 21sin 2 tan sin 1

1

xdx A x Bx x C

x

, then A + B =

1) 0 2) 1

2 3) 1 4)

1

2

79. If 2 22 tan ln tan 2 tanxdx x x f x c , where f(x) =

1) 1 sin

sin2

x

2) 1 sin

cos2

x

3) 1 cos

cos2

x

4) 1 cos

sin2

x

80.

2

21

1 ln

1 ln lnx x

x dx

x x

1) ln 1 ln x c 2) ln 1 ln x c

3) ln 1 lnx x c 4) ln 1 lnx x c

81. If

1 7 5 313 5 5 5 52 2 2 22 2 2 2 21 1 1 1x x dx a x b x c x d

, then b =

1) 2

5 2)

4

5 3)

4

25 4)

8

25

82. If

3

2 3 3 34 3

1ln

11

dx x b ca d

x x xx x

then a b c

1) 0 2) 1

3 3)

2

3 4)

4

3



83. If

1 1

4 44 41

4 141

4 4

1 11 ln tan

1

x x xx dx A B c

xx x

, then A + B =

1) 1

4 2)

1

2 3)

1

4 4)

1

2

84. cos5 5cos3 10cos

cos6 6cos4 15cos2 10

x x xdx

x x x

1) ln tan2 4

xc

2) ln tan

2 4

xc

3) 1

ln tan2 2 4

xc

4)

1ln tan

2 2 4

xc

85.

1

3 5 4sin cos

dx

x x

1) 1

42 tan x c 2) 1

42cot x c 3)1

44 tan x c 4) 1

44cot x c

86. If 1 1

1 ln1 1 1

x xx

x x

xe edx Ax B e C D

e e

, then A + B + C =

1) 2 2) -2 3) 4 4) 0

87. If

11 3 14

14 4 44 tan1

x dxAx Bx x C

x

, then A + B =

1) 2

3 2)

4

3 3)

4

3 4)

8

3

88. 1

.1

x dx

xx

1) 11 1ln sin

xx c

x

2) 11 12ln cos

xx c

x

3) 11 12ln 2sin

xx c

x

4) 11 12ln 2cos

xx c

x

89. 2

3 4 2

1

2 2 1

x dx

x x x

1) 4 2

2

2 2 1x xc

x

2)

4 2

3

2 2 1x xc

x

3) 4 22 2 1x x

cx

4)

4 22 2 1

2

x xc

x

90. 3 5

2 4

cos cos

sin sin

x xdx

x x

1) 1sin 6tan sinx x c 2) sin cosecx x c

3) 1sin 2cosec 6tan sinx x x c 4) 1sin 2cosec 5tan sinx x x c



91.

2 2 11027 27

1 2 1 2

sin sin

n n

n n n

xdx xdx

1) 54 2) 0 3) -54 4) 27 /2

92. ,n N nI 1

0

1 16 6nxe x dx e then n

1) 3 2) 4 2) 2 4) 5

93. Let f be integrable over 0, , R and /2

2

0

cos sin cosf d

/2

2

0

sin 2 sin cosk f d

then k

1) 2 2) -1 3) 1 4) 1/ 2

94. 64

1/3

0

, .x dx represent fractional part of x

1) 36 2 ) 30 3) 39 4) 33

95. 01 sin

xdx

x

1) 2) 2 3) 0 4)

96. 0

3 2sin cos

dx

x x

1) / 2 2) / 4 3) / 3 4) / 6

97. ln3

2 3

ln 2

2 3 .......x x xe e e dx

1) 1/2 2) 2/3 3) 3/2 4) 1/3

98. 0

sin 2 sin cos2

2

x x x

dxx

1) 2

4

2)

2

2

3)

2

8

4)

2

1

99

/2

0

cos3 1

2cos 1

xdx

x

1) 2

2) 1 3)

4

4)

1

2

100 If

/2

0

/2

0

sin o oc s c sn n nx xdx xdx

then

1) 1

1

2n 2)

1

1

2n 3)

1

2n 4)

2

1

2n

101.

1 1 2

2

0

sin

1 3

xdx

x x K

then K

1) 2 2) 3 3) 6 4) 12

102. 5 /4

/4

3 /4

cos sin

1 x

x x

e

1) 0 2) 2

3)

4

4)



103. If 1 1

100 10150 50

1 2

0 0

1 , 1I x dx I x dx and 1

2 5050

I K

I then K

1) 5049 2) 5051 3) 1 4) 5050

104.

2

7/22

0

______1

x dx

x

1) 2/15 2) 1/3 3) 4/15 4) 1/5

105.

1 1

1 1

sin secx xa

x a x x

e e dxa R

Tan e Tan e e e

1) 0 2) ln 24

3) ln 2 4) ln 2

2

106 /2

2

cos6 6cos 4 15cos 2 10

cos5 5cos3 10cos

x x xdx

x x x

1) 0 2) 1 3) 2 4) 4

107. Let 2 2 2

/ 4 / 4 / 4 / 4

1 2 3 4

0 0 0 0

, , cos , sinx x x xI e dx I e dx I e xdx I e xdx

then

1) 1 2 3 4I I I I 2) 1 2 3 4I I I I 4) 2 1 4 3I I I I 4) 1 2 4 3I I I I

108.

/2

2 1

0

/2

2 1

0

sin

sin

xdx

xdx

1) 1

2 1 2)

2

2 1 3)

2

2 1 4)

1

2

109. If /2 /2

0 0

cos 2 logcos tan . sin 2nx xdx k x nx dx

then k

1)1

n 2)

2

n 3)

1

2n 4)

1

4n

110 2 2

0

cos cos 2 cos3 sin sin 2 sin3x x x x x x dx

1) 32

2)

22 3

3

3)

23

3

4) 2 3

3

111 1/

2

1/

2019sin 2018cos

n

nn

Lt n x x x dx

1) 2018 2) 2019 3) 4036 4) 4037

112 If 1

0

x

x

f t dt x t f t dt then 1f

1) 0 2) 1

2 3) 1 4)

1

2

113.

2

2

2

2

2 tan

sec

2 tan

sec

3

3

x

x

x

x

t f t t dt

f y y dy

1) 3 2) 2/3 3) 3/2 4) 1/2



114.

3

1 1

1

3

1 1

1

1

Tan x Cot x dx

Tan x Tan dxx

1) 1 2) 1

2 3) 2 4) 4

115. Let f x be a continuous functin such that 1

0

12

3f x x f x dx then

1 _____f

1) 1 2) 1

3 3) 3 4) 0

116

2

2 1

n

nK r

nLt

r K n r

1) 1

2 2) 1 3) 2 4)

3

2

117.

2018

1

0

cot x dx where . in G.I.F.,

1) 0 2) cot 1 3) cot 2018 4) 2018-cot1

118. If

2

0

xdt

f xf t

and

1/3

2

0

6

xdt

f t then 9 _____f

1) 1

3 2) 3 3) 9 4) 1/39

119. 1

110

1_____

n n

Kr

x r dxx K

1) !n 2) 1 !n 3) 1 !n n 4) !n n

120.

40

cos 4 4cos 3 6cos 2 4cos cosx x x x xf x Lt

then

/2

/2

f x dx

1) 0 2) 1 3) 4 4) 2

121. If the area bounded by 2 2 3y x x and the line 1y kx is least, then the least area in

square units is

1) 31

3 2)

32

3 3)

34

3 4)

35

3

122. The area defined by 1 2 1 2x y in square units is

1) 2 2) 4 3) 6 4) 8

123. If ,

1,

x x zf x

x z

and 2

g x x , (where . denotes fractional part of x), then area

bounded by f x and g x for 0,10x is

1) 5/3 2) 5 3) 10/3 4) 20/3



124. Let 2 , cosf x x g x x and , be the roots of the equation 2 218 9 0x x .

Then the area bounded by the curves y fog x , the ordinates ,x x and the x-axis in

square units is

1) 1

32 2)

3

3)

4

4)

12

125. If 1

max sin ,cos ,2

f x x x

then the area of the region bounded by the curves y= f(x), x

axis, y axis and the line 5

3x

is

1) 5

312

square units 2)

5 3

12 2square units

3) 5

3 212

square units 4)

5 32

12 2square units

126. The area between the curves 4 22 ,y x x the x-axis and the ordinates of two minima of the

curve is in square units.

1)7

120 2)

5

127 3)

7

127 4)

7

60

127. A square ABCD is inscribed in a circle of radius 4. A point P moves inside the circle such that

,d P AB min {d(P, BC),)d(P, DA)} where d(P, AB) is the distance of a point P from line

AB. The area of region covered by the moving point P in square units is 1) 4

2) 8 3) 8 16 4) 4 4

128. The area bounded by the curves y = sin–1

|sin x| and y = (sin–1

|sin x|)2, 0 < x < 2 , is

1) 21

3 4 2)

31

6 8 3) 2 4)

24 3

3 6

129. Area bounded between the curves 2 2y 4 x and y 3 x in square units is

1) 1

3 2)

2 1

3 3 3)

2 1

3 4)

2 3

3 3

130. Let ‘f’ be a differentiable function such that 2

0

,

xtf x x e f x t dt then f(1) + f(2) + f(3) +

...... + f(9) =

1) 960 2) 1000 3) 1024 4) 1126

131. The area enclosed between the curves 1

ln , lny x e xy

and x-axis in square units is

1) 1 2) 2 3) 4 4) 6

132. The area bounded by the curve 32 2 2a x y a y in square units is

1) 22 a 2) a 3) 23 a 4) 2a

133. The area contained by ellipse 2 22 6 5 1x xy y in square units is

1) 2) 2 3) 3 4) 4



134. A point P moves in the xy plane in such a way that 1x y , where [.] denotes the

greatest integer function. Then the area of the region representing all possible positions of the

point P in square units is

1) 2 2) 5 3) 7 4) 8

135. The area of region represented by 5x y x y , for , 0, 0x y x y in square units is

1) 1

2 2)

3

2 3)

5

2 4)

7

2

136. The orthogonal trajectories of the family of curves are given by

1) constant 2) constant

3) constant 4)

137. The general solution of 1

1 log2log

xdyx x y x

dx

is

1) 1 1

1 log log2 2

x x

y x cx

2) 1 1

log log2 2.

x x

y x x c

3)

2log

2

x

y e x c 4)

21 1 1log 1 log log

2 2 2x x x

y e x x c

138. If the population of a country doubles in 50 years in how many years will it become thrice the

original, assume the rate of increase is proportional to the number of inhabitants

1) 75 2) 250log 3 3) 350log 2 4) 100

139. A curve is such that the mid point of the portion of the tangent intercepted between the point

where the tangent is drawn and the point where the tangent meets y-axis, lies on the line y = x.

If the curve passes through (1, 0), then the curve is

1) 2y = x2 – x 2) y = x

2 – x 3) y = x – x

2 4) y = 2(x – x

2)

140 The solution of the differential equation2 2 2

2 2

x dx y dy a - x - y

x dy - y dx x y

is

1) 2 2 1 yx y a cos c tan

x

2) 2 2 1 y

x y a sin c tanx

3) 2 2 1 x

x y a sin c tany

4)

2 2 1 xx y a cos c tan

y

141. The solution of differential equation 3 2 2 2 22x ydy (1 y )(x y y 1)dx 0 is

1) 2 2x y (cx 1)(1 y ) 2)

2 2x y (cx 1)(1 y )

3) 2 2x y (cx 1)(1 y ) 4) None of these

142. The differential equation of all conics whose centre lies at origin, is given by

1) 2 2

2 3 1 2 1 23 3xy x y y xy xy y xy x y

2) 2 2

1 2 1 3 1 2 33 3xy x y y xy xy y xy x y

3) 2 2

2 3 1 1 1 23 3xy x y y xy xy y xy x y

4) 1 1 0xy yx

1n na y x

2nx n y 2 2ny x

2 nn x y 2 constantnn x y



143. Solution of 2 2dy

x y adx

is

1) 1tanyx

y a ca

2) 1cot

yxy a c

a

3) 1sinyx

y a ca

4) 1cos

yxy a c

a

144. The equation of curve passing through (1, 0) and satisfying

2 2

2 22 2 1dy dy

y x y xdx dx

, is given by

1)

1 2 2

22

2y y x

xx

2)

2 2

2 22

y y xx

x

3)

1 2 2

22

2y x y

yx

4)

1 2 2

22

2y x y

yx

145. Solution of 4

2 2

22

dyx y

ydx x ydy x

y xdx

is

1) 2 2

1yC

x x y

2)

2 2

1yC

x x y

3)

2 2

1yC

x x y

4)

2 2

1yC

x x y

146. A curve y f x passes through the origin. Though any point ,x y on the curve, lines are

drawn parallel to the coordinate axes. If the curve divides the area formed by these lines and

coordinate axes in the ratio m:n. Then the equation of curve is

1) /m ny Cx 2)

2 /m nmy Cx 3) 3 /m ny Cx 4) y Cx

147. The family of curves, the sub tangent at any point of which is the arithmetic mean of the

coordinates of the point of tangency, is given by

1) 2

x y Cy 2) 2

y x Cx 3) 2

x y Cxy 4) 2 4x ay

148. Water is drained from a vertical cylindrical tank by opening a valve at the base of the tank. It is

known that the rate at which the water level drops is proportional to the square root of water

depth y, where the constant of proportionality k > 0 depends on the acceleration due to gravity

and the geometry of the hole. If t is measured in minutes and 1

15k , then the time to drain the

tank, if the water is 4 m deep to start with is

1) 30 min 2) 45 min 3) 60 min 4) 80 min

149. The differential equation 21 ydy

dx y

determines a family of circles with

1) variable radii and a fixed centre at (0 , 1)

2) variable radii and fixed centre at (0, -1)

3) fixed radius 1 and variable centres along the x-axis

4) fixed radius 1 and variable centres along the y-axis

150. The solution of the differential equation 2 0ydx xdy xy dx , is

1) 2x

xy

2) 2

2

x x

y 3)

2

22 4

x x

y 4)

2xy

y



151. The function

ln

ln

xf x

e x

is

1) increasing on 0,

2) decreasing on 0,

3) increasing on 0, / e ,decreasing on / ,e

4) decreasing on 0, / e ,increasing on / ,e

152. The point (s) on the curve 3 23 12y x y where the tangent is vertical , is (are)

1) 4

, 23

2) 11

,13

3) 0,0 4) 4

,23

153. Tangent is drawn to ellipse 2

2 1 3 3 cos ,sin 0, / 2 .27

xy at where

Then the value of such that sum of intercepts on axes made by this tangent is minimum , is

1) 3

2)

6

3)

8

4)

4

154. If 3 2 20f x x bx cx d and b c then in ,

1) f x is a strictly increasing function

2) f x has local maxima

3) f x is a strictly decreasing function

4) f x is bounded

155. Tangent to the curve xy e drawn at the point ( , )cc e intersects the line joining the points

11, cc e and 11, cc e

1) on the left of x=c 2) on the right of x=c

3) at no point 4) at all points

156. The total number of local maxima and local minima of the function

3

2/3

2 , 3 1( )

, 1 2

x xf x

x x

is

1) 0 2) 1 3) 2 4) 3

157. Let the function g : , ,2 2

be given by 1( ) 2 tan

2

ug u e . Then g is

1) even and is strictly increasing in 0,

2) odd and is strictly decreasing in ,

3) odd and is strictly increasing in ,

4) neither even nor odd ,but is strictly increasing in ,

158. Let 2 4 2

0 1 2 .......... n

nP x a a x a x a x be a polynomial in a real variable x with

0 1 20 ......... .na a a a The function P x has

1) Neither a maximum nor a minimum

2)Only one maximum

3) Only one maximum and only one minimum

4) Only one minimum



159. If f x is differentiable function such that '' 5f x for each 0,4x and f

assumes its largest value at an interior point of this interval then ' '0 4f f

1) can assume 20 2) can assume 21

3) can assume 22 4) can assume 23

160. Let , , &P Q R S be four points in order on the parabola 2y ax bx c with

2,3 , 1,1 & 2,7P Q S . Then the sum of the co-ordinates of R so that quadrilateral

PQRS has maximum area is

1) 9

4 2)

1

2 3)

3

4 4)

5

4

161. Consider 2 2019

1 ......1! 2! 2019!

x x xf x Then

1 ) 0f x has exactly one real root

2) 0f x has at least 2017 real roots

3) 0f x has 1009 real roots

4) 0f x has only 3 real roots

162. let f x be a function defined on 0, such that 0 0,f x x and differentiable on

0, such that ' cos sin 0f x x f x x x Then 5

3f

1) Can be a prime 2) Can be 1 3 ) Can be -1 4) can’t be positive

163. The maximum distance from origin of a point on the curve

sin sin , cos cos ,at at

x a t b y a t bb b

both a,b 0 is

1) a b 2) a b 3) 2 2a b 4) 2 2a b

164. If 2 3 6 0, , ,a b c a b c R then the quadratic equation 2 0ax bx c has

1) at least one root [0,1] 2) at least one root [2,3]

3) at least one root [4,5] 4) none of these

165. If the function 3 2 22 9 12 1f x x ax a x ,where a>0, attains its maximum and minimum at

p and q respectively such that 2p q ,then a equals

1) 1/2 2) 3 3) 1 4) 2

166. A point on the parabola 2 18y x at which the ordinate increases at twice the rate of the

abscissa is

1) 9 9

,8 2

2) 2, 4 3) 9 9

,8 2

4) 2,4

167. A function y f x has a second order derivative " 6 1f x x . If its graph passes

through the point 2,1 and at that point the tangent to the graph is 3 5y x ,then the

function is

1) 2

1x 2) 3

1x 3) 3

1x 4) 2

1x

168. The normal to the curve 1 cos , sinx a y a at ' ' always passes through the fixed

point

1) ,a a 2) 0,a 3) 0,0 4) ,0a



169. Area of the greatest rectangle that can be inscribed in the ellipse 2 2

2 21

x y

a b is

1) 2ab 2) ab 3) ab 4) a

b

170 A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts

at a rate of 350 / mincm . When the thickness of ice is 5 cm , then the rate at which the

thickness of ice decreases is

1) 1

/ min36

cm

2) 1

/ min18

cm

3) 1

/ min54

cm

4) 5

/ min6

cm

171. If the equation 1

1 1......... 0n n

n na x a x a x

1 0, 2a n , has a positive root x , then the

equation 1 2

1 11 ....... 0n n

n nna x n a x a

has a positive root, which is

1) greater than 2) smaller than

3) Greater than or equal to 4) equal to

172. The function 2

2

xf x

x has a local minimum at

1) x=2 2) x=-2 3)x=0 4) x=1

173. A value of c for which conclusion of mean value theorem holds for the function logef x x

on the interval 1,3 is

1) 3log e 2) log 3e 3) 32log e 4) 3

1log

2e

174. The function 1tan sin cosf x x x is an increasing function in

1) 0,2

2) ,2 2

3) ,4 2

4) ,2 4

175. Suppose the cubic 3x px q has three distinct real roots where 0p and 0q . Then which

one of the following holds ?

1) The cubic has minima at 3

p and maxima at -

3

p

2) the cubic has minima at -3

pand maxima at

3

p

3) The cubic has minima at both 3

pand -

3

p

4) The cubic has maxima at both 3

p and

3

p

176. How many real solutions does the equation 7 5 314 16 30 560 0x x x x have ?

1) 7 2) 1 3) 3 4) 5

177. Given 3 3 2P x x ax bx cx d such that 0x is the only real root of ' 0P x . If

1 1P P , then in the interval 1,1

1) 1P is not minimum but P(1) is the maximum of P

2) 1P is the minimum but P(1) is not the maximum of P

3) Neither 1P is the minimum nor P(1) is the maximum of P

4) 1P is the minimum and P(1) is the maximum of P



178. The equation of the tangent to the curve 2

4y x

x , that is parallel to the x – axis, is

1) 1y 2) 2y 3) 3y 4) 0y

179. Let :f R R be defined by 2 , 1

2 3, 1

k x if xf x

x if x

. If f has local minimum at 1x ,

then a possible value of k is

1) 0 2) 1

2 3) -1 4) 1

180 The shortest distance between line 1y x and curve 2x y is

1) 3 2

8 2)

8

3 2 3)

4

3 4)

3

4

181. Let L1 be the line 1ˆ ˆˆ ˆ ˆ2 2r i j k i k and let L2 be the line

2

ˆˆ ˆ ˆ ˆ3r i j i j k . Let . be the plane which contain the line L1 and is

parallel to L2. The distance of the plane . from the origin is

1) 2

7 2)

1

7 3) 6 4)

1

6

182. Let P be the point of interesction of the three planes 1 2 3ˆ ˆ ˆ. 0, . 1 . 2r n r n and r n where 1 2

ˆ ˆ.n n and

3n are unit vectors along ˆ ˆˆ ˆ ˆ ˆ2 ,5 12 3 4j k i j and i k respectively then the projection of OP on z-

axis (O being origin) is

1) 3

2 2)

5

2 3)

7

2 4)

11

2

183. Let , ,a b c be unit vectors with ,[ ] 2r b c c a a b ab c and the angle between r

and a b c is 4

with 2r then the max value of

1) 2 2 2) 6 3) 1

2 4)

3

2

184. If 2 3 , 2a i j k b i j k and u is a vector satisfying a u a b and . 0a u then 2

2 u is

equal to

1) 5 2)2 3)3 4) 1

185. If A, B, C, D are four points in space satisfing 2 2 2 2

AB.CD K AD BC AC BD

then the

value of K is

1) 2 2) 1

3 3)

1

2 4) 1

186. Let , ,a b c be unit vectors, equally inclined to each other at an angle ,3 2

.If these are

the position vectors of the vertices of a triangle and g is the position vector of the centriod of the

triangle , then

1) 1g 2) 3

2g 3)

3

2g 4)

2

3g

187. If , , ,a b c d are non-zero coplanar vectors and 1 2 32 2 3 0x a x b c x d , then

1) 2 2 2

1 2 3

1

2x x x 2) 2 2 2

1 2 3

1

17x x x 3) 2 2 2

1 2 3

1

20x x x 4) 2 2 2

1 2 3

1

34x x x



188. G is the centroid of triangle ABC and 1A and 1B are the midpoints of sides AB and AC,

respectively. If 1 be the area of quadrilateral 1 1GA AB and be the area of triangle ABC, then

1

is equal to

1) 3

2 2) 3 3)

1

3 4)

2

3

189. Let A a and B b be points on two skew lines r a p and r b uq and the shortest

distance between the skew lines is 1, where p and q are unit vectors forming adjacent sides of a

parallelogram enclosing an area of 1

2 units. If an angle between AB and the line of shortest

distance is 600

then AB

1) 1

2 2) 2 3) 1 4) 0R

190. If . 0,a b where a and b are unit vectors and the unit vector c is inclined at an angle to both

a and b . If c = m a +n b + ,p a b , ,m n p R then

1) 4 4

2)

3

4 4

3) 0

4

4)

30

4

191.. Let two non collinear unit vectors a and b form an acute angle. A point P moves so that at any

point t, the position vector OP (where O is the origin) is given by cos sin .a t b t When P is

farthest from the origin O, let M be the length of OP and u be unit vector along OP . Then

1) 1/2

1 .a b

u and M a ba b

2)

1/2

1 .a b

u and M a ba b

3) 1/2

1 2 .a b

u and M a ba b

4)

1/2

1 2 .a b

u and M a ba b

192.. In a quadrilateral ABCD, AC is the bisector of AB and AD , angle between AB and AD is

2,15 3 5

3AC AB AD

. Then the angle between BA and CD is

1) 1 14

cos7 2

2)

1 21cos

7 3

3)

1 2cos

7

4) 1 2 7

cos14

193.. Let P x y y z y z z x z x x y and

8( ) . ,P x y y z z x for positive numbers , and , then value of 2

x y z is

1) 4 2) 5 3) 8 4) 9

194.. Let , ,a b c are three vectors having magnitudes 1,2,3 respectively satisfy the relation 6a b c .

If d is a unit vector coplanar with b and c such that . 1b d then evaluate

2 2

.a c d a c d

1) 9 2) 3 3) 12 4) 15

195.. If is the acute angle between the medians drawn through the acute angle of an isosceles right

angled triangle then the value of 4sec .

1) 3 2) 10 3) 8 4) 5



196.. Suppose that a, b, c do not lie in the same plane and are nonzero vectors such that

1, 2, 2, . 1, . 2a b c a b b c and the angle between a b and b c is 6

. If d is any

vector such that 2 2

. . .d a d b d c and d k for any scalar , then k is equal to

1) 3 2) 1 3) 0 4) 2

197. If , ,a b c are unit vectors such that . 0 .a b a c and the angle between b and c is / 3 , then the

value of a b a c is

1) 1/ 2 2) 1 3) 2 4) 3

198. Points , ,a b c and d are coplanar and sin 2sin 2 3sin3 0a b c d . Then the least

value of 2 2 2sin sin 2 sin 3 is

1) 1/14 2) 14 3) 6 4) 1/ 6

199. If a and b are any two vectors of magnitudes 1 and 2, respectively, and

22

1 3 . 2 3 47a b a b a b , then the angle between a and b is

1) / 3 2) 1cos 1/ 4 3) 2

3

4) 1cos 1/ 4

200. ,a b and c are unit vectors such that 3 4a b c . Angle between a and b is 1 , between b

and c is 2 and between a and c varies / 6,2 / 3 . Then the maximum value of

1 2cos 3cos is

1) 3 2) 4 3) 2 / 2 4) 6

201. The position vectors of the vertices A. B and C of a triangle are three unit vectors ˆˆ,a b and c ,

respectively. A vector d is such that ˆˆ ˆ. . .d a d b d c and ˆ ˆd b c . Then triangle ABC is

1) Acute angled 2) obtuse angled 3) right angled 4) Equilateral

202. If ,a b and c are such that 1, ,abc c a b

2, 3a b and 1

3c , then the angle

between a and b is

1) 6

2)

4

3)

3

4)

2

203. If O (origin) is a point inside the triangle PQR such that 1 2 0OP k OQ k OR where 1 2,k k are

constants such that

4

Area PQR

Area OQR

, then the value of 1 2k k is

1) 2 2) 3 3) 4 4) 5

204. If u and v are two non-collinear unit vectors such that2

u vu v

, then the value of

2

u u v is equal to

1) 1

4 2)

1

2 3)

2

3 4)

3

4

205. If a and b are two vectors such that 1, 4, . 2,a b a b If 2 3c a b b then angle between

b and c is

1) 6

2)

3

3)

2

3

4)

5

6



206. Let 4 4 4 4ˆ ˆˆ 4 cos3 3 sin3 4 sin3 3 cos3t t t tu e t e t i e t e t j and f t u . The ratio of the

maximum value of f to its minimum value given that 0 5t is

1) 20e 2) 10e 3) 10 4) 5

207. Suppose in a tetrahedron ABCD, AB = 1; 3CD ; the distance and angle between the lines AB

and CD are 2and 3

respectively. If the volume of the tetrahedron is V then the value of (60V)

1) 29 2) 30 3) 39 4) 40

208. If a and b are non-zero and non-collinear vectors , then the value of for which the vectors

1 2v a b and 2 2 3 3v a b are collinear is

1) 3

2 2)

2

3

3)

2

3 4)

3

2

209. Let , ,a b c be three vector of magnitude 1.1

, 22

respectively, satisfying 1.abc If

13

2 .a b c a c a c bk

, then the value of k.

1) 2 2) 3 3) 4 4) 5

210. Let 3 , 1 , 2a b c and 3 0a a c b then 2

a c equals

1) 3 2) 2 3) 1

2 4)

3

4

211. Let , ,a b c be distinct nonnegative numbers. If 2 5 3x y z

a a c

and

1 5 4x y z

c c b

are

coplanar then c is

1) The geometric mean of a and b 2) The artithmetric mean of a and b

3) Equal to zero 4) The harmonic mean of a and b

212. The angle between the lines 2 3x y z and 6 4x y z is

1) 00 2) 030 3) 045 4) 090

213. The image of 1,3,4 in the plane 2 0x y is

1) 12 1

, ,45 5

2) 12 1

, ,05 5

3) 9 13

, ,45 5

4)9 13

, ,05 5

214. The equation of the plane containing the line 2 5 3, 4 5x y z x y z and parallel to the

plane 3 6 1x y z is

1) 3 6 7 0x y z 2) 3 6 7 0x y z 3) 3 6 4x y z 4) 3 6 9x y z

215. The distance between the parallel planes 2 3 1x y cz and 3 6 13ax y z is

1) 12

7 2)

14

7 3)

2

7 4)

12

49

216. 0,7,10 , 1,6,6 , 4,9,6A B C be vertices of a triangle. If , , are the angles of the

traingle then possible value of is 0, 0, 0

1)6

2)

4

3)

3

4)

2

217. If 1, 2, 8 , 5, , , 11,3,7A B a b C are collinear then B divides AC in the ratio is___

1) 2 :3 2) 3: 2 3)3: 2 4) 2 :3

218. The distance between the lines 1 1 1

1 2 3

x y z and

5 2 1

1 2 3

x y z is____

1) 21 2) 45 3) 18 4) 41



219. If the vector equation of the plane passing through the intersection of the planes . 6r i j k

and . 2 3 4 5r i j k

, and the point 1,1,1 is . 26r a i b j k

⋋ then ⋋=____

1) 20 2) 23 3) 69 4) 59

220 If is the acute angle between the line 3r i k

⋋ 2 3 6i j k

and the normal to the

plane . 10 2 11 3r i j k

then

1)8

sin21

2) 8

cos21

3) 10

sin21

4) 10

cos21

221. Shortert distance between the lines 6 2 2r i j k

⋋ 2 2i j k

and

4 3 2 2r i k i j k

is____

1) 3 2) 81 3) 9 4) 7

222. If length of projection of the line segment joining , ,a b c and 1,1,1 on the plane

2 3 6x y z ⋋ is maximum then the possible value of 2 3 6a b c is 0, 0, 0a b c

1) 3 2) 10 3) 7 4) 11

223 If the lines x a d y a z a d

and

x b c y b z b c

k

are coplanar then k

______

1) 2) 3) 4)

224. The angle between one of the diagonals of a cube and one of its edge is

1) 1 1cos

3

2) 1 1

cos3

3_) 1 2

cos3

4) 1 1

cos2

225. If a plane has the intercepts , ,a b c on the axes and is at a distance of ‘p’ units from the origin

then 2 2 2 2 2 2 2a b b c c a p

1) 2 2 2a b c 2) 2 2 2a b c 3) 2 2 2

1 1 1

a b c 4) 6 6 6a b c

226. The vector equation of the line passing through (1,2,3) and perpendicular to the plane

. 2 5 9 0r i j k

is

1) 2 3r i j k

⋋ 3 i j k

2) 2 5r i j k

⋋ 2 3i j k

3) 2 3r i j k

⋋ 8k

4) 2 3r i j k

⋋ 2 5i j k

227. The point of intersection of the two lines 2 3r i j k

⋋ 3 3 9i j k

and

7 4 3r i j k i j

is_____

1) 3 2i j k

2) 2 3i j k

3) 2 3i j k

4) 3 2i j k



228. In a tetrahedron ABCD, 1, 1,2A and 3,1,1G is the centroid o fhte tetrahedron. 1G is the

centroid of the the traingle BCD then 1 ___AG

1)9

4 2) 4 3) 2 4)

9

2

229. If (0,0,0) is the orthocenter of a triangle formed by

cos ,sin ,0 , cos ,sin ,0 , cos ,sin ,0 then cos 2 =______

1) 0 2) 3 3) 4 4) 3

2

230. If a line makes acute angles , , with the coordianate axes such that

2cos cos cos cos

9 and

4cos cos

9 then cos cos cos

1) 5

4 2)

3

5 3)

4

5 4)

5

3

231. If the d.c’s of two lines are connected by 2 2 0l m n and 0mn nl lm then angle

between the lines is

1) 1 1cos

6

2) 0 3) 2

4)

3

232. The ratio in which the plane . 2 3 17r i j k

divides the line joining the points

2 4 7i j k

and 3 5 8i j k

is

1) 3:10 ext 2) 3:5 ext 3)3:5 int 4) 3:10 int

233. The image the pooint (1,6,3) in the line 1 2

1 2 3

x y z is

1) 1,3,5 2) 1,0,7 3) 1,9,1 4) 0,3, 2

234. 1,2,3 , 3,1,2P Q . If P’Q’ is the reflection of the line PQ in the plane 9x y z then the

point which doest not line on P’Q’ is____

1) 3,4,2 2) 5,3,4 3) 7,2,3 4) 1,5,6

235. The distance between the line 2 2 3r i j k

⋋ 4i j k

and the plane . 5 5r i j k

is

1)10

27 2)

10

3 3 3)

20

27 4)

20

3 3

236. The area of the traingle formed by the three lines ;1 2 3 2 1 3

x y z x y z and

1 2, 3

1 1

x yz

is

1) 3 3

2 2)

3

2 3)

3 3

4 4) 3

237. 1 1 1, ,l m n and 2 2 2, ,l m n are d.c’s of two lines and anlge between them is 6

. Then d.c’s of a

line perpendicular to both of these lines are 1 2 2 1, 1 2 2 1 1 2 2 1,k m n m n n l n l l m l m then _____k

1)1

2 2) 2 3)

3

2 4)

2

3



238. The minimum distance of the point 1,1,1 from the plane 1x y z measured perpendicular

to the line 1 1 1

1 2 3

x x y y z z is

1)28

3 2)

28

3 3)

27

3 4)

4 7

3

239. The plane 3 4 0y z is rotated about its line of intersection with the plane 0x through an

angle 3

. The equation of the plane in its new position is 3 4 0y z ax then a

1) 2 3 2) 5 3 3) 3 3 4) 3

240. The value of ‘⋋’ for which the lines 3 2 5 0 2 3x y z x y z and

2x y ⋋z 0 7 10 8x y z are perpendicular to each other is___

1) 1 2) 2 3) 1 4) 2

241.If 3 7z i p iq where p, q {0},I is purely imaginary, then minimum value of 2

z

1) 0 2) 58 3) 3364

3 4) 3364

242. Sum of common roots of the equations 3 22 2 1 0z z z and 97 29 1 0z z is

1) 0 2) -1 3) 1 4) 2

243. The diagram shows several numbers in the complex plane. The circle is the unit circle centred

at origin. One of these numbers is the reciprocal of F, which is

1) A 2) B 3) C 4) D

244 Principal argument of

251 1 3

2 3

i i

i i

is

1) 1912

2)

7

12

3)

5

12

4)

5

12

245. Define a sequence of complex numbers 2

1 10, 1n nz z z i for n . In the complex plane,

how far from origin is 111 ?z

1) 1 2) 2 3) 3 4) 110

246. The continued product of all the values of 1/49

2 3i is

1) 2+3i 2)-2+3i 3) -2-3i 4) 2-3i

247 For 1

6 6 62 3

1 1 1; ,

1 3 3 3

i i iz z z

i i i

which of the following holds good?

1) 2

1

3

2z 2)

4 4 8

1 2 3z z z

3) 3 3 6

1 2 3z z z

4)4 4 8

1 2 3z z z



248 If is non real root of 6 1,x then 5 3

2

1

1

1) 2 2) 0 3) 2 4)

249 If 3

2

iz

then

2018iz

1) 1 3

2 2

i 2)

1 3

2 2

i 3)

1

2 2

i 4) 1

250. There is only one way to chose real numbers A and B such that when the polynomial 4 3 25 4 3x x x Ax B is divided by 2 1,x the remainder is 0. If A and B assume these

unique values, then A B

1) -6 2) -2 3) 6 4) 2

251. One of the values of

8

3

1 sin cos8 8

1 sin cos8 8

i

i

is =

1) 0 2) 8 3) 3 4) 1

252. The locus represented by the equation 1 1 2z z is

1) An ellipse with foci (1, 0) (-1, 0)

2) One of the family of circles passing through the points of intersection of the circles

1 1z and 1 1z

3) Radical axis of the circles 1 1z and 1 1z

4) The portion of the real axis between the points (1, 0) ; (-1, 0) including both

253. Imaginary part of 2018

!

0

1n

n

i is i

1) 1 2) 2 3) 2018 4) 2i

254. The thn roots of 12 ,n N are

1) in A.P and outside the unit circle with centre at origin

2) in H.P and inside the unit circle with centre at origin

3) in G.P and outside the unit circle with centre at origin

4) in G.P and inside the unit circle with centre at origin

255. Identify the incorrect statement

1) No non zero complex number ‘z’ satisfies the equation 4z z

2) z z implies that z is purely real

3) z z implies that z is purely imaginary

4) If 1 2,z z are the roots of the quadratic equation 2 0az bz c such that Im 1 2( )z z 0 then

a, b, c must be real numbers

256. If x is a rational number then the solutions for x of 1 2x xi are

1) 0 only 2) 0, 1 3) 2

4) there may exist many solutions



257. If 1 2 1 2z z z z then absolute value of the difference in the amplitudes of 1z and 2z is

1) 4

2) 0 3) 4)

2

258. Number of solutions of the equation

2

33

0z

zz

where Z is a complex number is

1) 2 2) 3 3) 6 4) 5

259. If m and n are the smallest positive integers satisfying the relation 2cis 4cis6 4

m n

then

m n

1) 120 2) 96 3) 72 4) 60

260. A value of for which 2 3 sin

1 2 sin

i

i

is purely imaginary, is

1) 3

2)

6

3) 1 3

sin4

4)1 1

sin3

261. The least value of 2 2 2 2

1 2 2 3 3 4 2z i z i z i z i occurs when z =

1) 1+i 2) -1+i 3) -1-2i 4) 1+2i

262. Let 1 2,z z be non zero complex numbers satisfying the equation 2 2

1 1 2 22 2 0.z z z z The

geometrical nature of the triangle whose vertices are the origin and the points representing 1z

and 2z is

1) An isosceles right angled triangle 2) A right angled triangle which is not isosceles

3) An equilateral triangle 4) An isosceles triangle which is not right angled

263. It 1 2 81, , ,....., are 9th

root of unity taken in counter clock wise sequence then

1 3 5 72 2 2 2

1) 255 2) 511 3) 1023 4) 15

264. Let ‘p’ is not multiple of ‘n’ and let 2 11, , ,.... n be thn roots of unity then

2 11 ....p p

p n

1) 0 2) 2n 3) 2n 4) n

265. Let ω be a complex number such that 2ω+1=z where 3z .If

2 2

2 7

1 1 1

1 1 3

1

k

, then k =

1) z 2) 1 3) 1 4) z

266. If 1 2 91, , ,.... are 10th

root of unity then 1 2 9

1 1 1......

1 1 1

1) 0.9 2) 10.23 3) 4.5 4) 9

267. If 1

2

,z

i Rz

then

2 2

1 2 1 2

2 2

1 2

3z z z z

z z

1) 2 2) 4 3) 8 4) 16



268. Number of ordered pairs ,a b of real numbers such that 2018

( )a ib a ib holds good is

1) 2018 2) 2020 3) 2019 4) 1

269. If 4 1

14 1

i iz i

i i

then

z

amp z=

1) 1 2) 3) 3 4) 4

270. The complex number z having least positive argument which satisfy the condition

25 15z i is

1) 25i 2) 12+5i 3) 16+12i 4) 12+16i

271. If

2

1 2

2

sin 0 0

0 sin 0

0 0 sin

A

and

2

1 2

2

cos 0 0

0 cos 0

0 0 cos

B

where α, β, γ are any real

numbers and 5 5 1 1 3 3 2 2 1 15 10C A B A B A B A B A B then find C .

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

272. If

3 3 4

2 3 4

0 1 1

A

; then 1A

A) A B) 2A C) 3A D) 4A

273. Let matrix

3 2

1 4

2 2

x

A y

z

; if 2xyz and 8 4 3 28x y z , then (adj A) A equals;

A)

1 0 0

0 1 0

0 0 1

B)

0 0

0 0

0 0

C)

2

2

2

0 0

0 0

0 0

D)

2 0 0

0 2 0

0 0 2

274. A square matrix P satisfies 2P I P , where I is identity matrix. If 5 8nP I P , then n is:

A) 4 B) 5 C) 6 D) 7

275. Let matrix 1 2 3

1 1 2

x y z

A

where , ,x y z N . If det.(adj. (adj.A)) = 8 42 .3 then the number of

such matrices A is: [Note: adj. A denotes adjoint of square matrix A.]

A) 220 B) 45 C) 55 D) 110

276. Let A be a square matrix satisfying 2A +5A+5I=0 . The inverse of A+2I is equal to:

A) A-2I B) A+3I C) A-3I D) non-existent

277. If M be a square matrix of order 3 such that 2M , then2

Madj

equals to:

A) 1

2 B)

1

4 C)

1

8 D)

1

16



278. Let the following system of equations

2

1kx y z

x ky z k

x y kz k

has no solution. Find k .

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

279. If the system of linear equations

2 0

3 0

4 0

x ay az

x by bz

x cy cz

has a non-zero solution, then a, b, c:

A) Are in A. P. B) are in G. P. C) are in H. P. D) satisfy 2 3 0a b c

280. The value of the determinant

1 0 1

1 1

1

a a

b a a b

depends on:

A) only a B) only b C) neither a nor b D) both a and b

281. The value of the determinant

2 2

2 2

2 2

x y z x x

y y z x y

z z z x y

A) 2

xyz x y z B) 2

x y z x y z C) 3

x y z D) 2

x y z

282. The determinant

2

2 0

2

a b c d ab cd

a b c d a b c d ab c d cd a b

ab cd ab c d cd a b abcd

for

A) 0a b c d B) 0ab cd

C) 0ab c d cd a b D) any a, b, c, d

283. Let ab=1,

2 2

2 2

2 2

1 2 2

2 1 2

2 2 1

a b ab b

ab a b a

b a a b

then the minimum value of is:

A) 3 B) 9 C) 27 D) 81

284. If

2 3

2 3

2 3

1 1 1

2 2 2

3 3 3

x x x

x x x

x x x

is expressed as a polynomial in x, then the term independent of x is:

A) 0 B) 2 C) 12 D) 16

285. If A is matrix of order 3 such that 5A and B = adjA, then the value of 1 TA AB

is equal

to (where A denotes determinant of matrix A. TA denotes transpose of matrixA, 1A denotes

inverse of matrix A. Adj A denotes adjoint of matrix A)

A) 5 B) 1 C) 25 D) 1

25

286. If 1

cos sin,

sin cosA A

is given by:

A) –A b) TA C) TA D) A



287. If

1 2 2

2 1 2

2

A

a b

is a matrix satisfying the equation 9TAA I , where I is a 3 3 identity matrix,

then ordered pair (a, b) is equal to:

A) 2,1 B) 2,1 C) 2, 1 D) 2, 1

288. If

1 3

1 3 3

2 4 4

P

is the adjoint of a 3 3 matrix A and 4A , then is equal to:

A) 11 B) 5 C) 0 D) 4

289. If A is an 3 3 non-singular matrix such that T TAA A A and 1 TB A A , then TBB equals:

A) I B B) I C) 1B D) 1T

B

290. Let

1 2 0

2 6 3 3

5 3 1

A B

and

2 1 5

2 2 1 6

0 1 2

A B

then r rT A T B has the value equal to:

A) 0 B) 1 C) 2 D) none of these

291. Let sin 0

0 sinA

. If TA A is a null matrix, then the number of values of in 0,6 , is

A) 4 B) 3 C) 2 D) 1

292. If A, B and C are n n matrices and det(A) = 2, det(B)= 3 and det(C) = 5, then the value of the

2 1det A BC is equal to:

A) 6

5 B)

12

5 C)

18

5 D)

24

5

293. Let

a b c

A p q r

x y z

and

4 2

4 2

4 2

x a p

B y b q

z c r

. If det(A) = 2, then the value of det(B) is equal to:

A) -8 B) 8 C) -16 D) 16

294. Let ijA a be a 3 3 matrix defined as

2 2

,,

,ij

i j i ja

i j i j

then det . .adj A is equal to:

A) 16 B) 25 C) 64 D) 0

295. Let

3 1

1 12 2,

0 11 3

2 2

A B

and TC AB A , then 3TA C A is equal to:

A)

3 1

2 2

1 0

B)

1 0

31

2

C)

31

2

0 3

D) 1 3

0 1

296. Let A and B are two square matrices matrices of order 3 such that det(A) = 3 and det(B) =2,

then the value of 1

1 1det .adj B A

is equal to:

[Note: adj M denotes the adjoint of a square matrix M.]

A) 6 B) 9 C) 18 D) 36



297. The determinant

cos sin cos 2

sin cos sin

cos sin cos

is:

A) 0 B) independent of θ C) independent of D) independent of θ and both

298. Given 1 3 1 0

,2 2 0 1

A I

. If A I is a singular matrix then:

A) B) 2 3 4 0 C)

2 3 4 0 D) 2 3 6 0

299. If 1 , is the complex cube root of unity and matrix 0

0H

, then 70H is equal to:

A) H B) 0 C) – H D) 2H

300. If , 0 and n nf n and

2 2 2

3 1 1 1 2

1 1 1 2 1 3 1 1

1 2 1 3 1 4

f f

f f f K

f f f

then K is equal to:

A) B) 1

C) 1 D) -1

301. The number of values of k for which the lines 2 2 0kx y , 2 3 0x ky ,

3 3 0x y k are concurrent is

1)1 2) 2 3) 3 4)4

302. If the ends of the base of an isosceles triangle are at (2, 0) and (0, 1), and the equation of one

side is x=2, then the orthocenter of the triangle is

1) 3 3

,2 2

2) 5

,14

3) 3

,14

4) 4 7

,3 12

303. If the equation of base of an equilateral triangle is 2 1x y and the vertex is 1,2 , then

the length of the sides of the triangle is

1) 20

3 2)

2

15 3)

8

15 4)

15

2

304. The area of a triangle, two of whose vertices are (2,1) and (3,-2) is 5. The coordinates of the

third vertex cannot be

1) (6, -1) 2) (4, 5) 3) (-1, 20) 4) (2, 9)

305. A straight line 1l with equation 2 10 0x y meets the circle with equation 2 2 100x y

at B in the first quadrant. A line through B perpendicular to 1l cuts the y-axis at 0,P t . The

value of t is

1) 12 2) 15 3)20 4) 25

306. A straight line L through the point (3, -2) is inclined at an angle 600 to the line 3 1x y . If

L also intersects the x-axis, then the equation of L is

1) 3 2 3 3 0y x 2) 3 2 3 3 0y x

3) 3 3 2 3 0y x 4) 3 3 2 3 0y x



307. A straight line with slope 2 and y-intercept 5 touches the circle 2 2 16 12 0x y x y c at

a point Q. Then the coordinates of Q are

1) (-6, 11) 2) (-9, -13) 3) (-10, -15) 4) (-6, -7)

308. If , is a point on the circle whose center is on the x-axis and which touches the line

0x y at (2, -2), then the greatest value of is

1) 4 2 2) 6 3) 4 2 2 4) 4 2

309. Given 1x y

a b and 1ax by are two variable lines, ‘a’ and ‘b’ being the parameters

connected by the relation 2 2a b ab . The locus of the point of intersection of the lines has

the equation

1) 2 2 1 0x y xy 2) 2 2 1 0x y xy

3) 2 2 1 0x y xy 4) 2 2 1 0x y xy

310. The locus of the midpoint of the chords of contacts of 2 2 2x y from the points on the line

3 4 10x y is a circle with center at P. If O is the origin, then OP is equal to

1) 2 2) 3 3) 1

2 4)

1

3

311. A line which makes an acute angle with the positive direction of x-axis is drawn through the

point P(3,4) to meet the lines x=6 and y=8 at R and S respectively then RS= 8 2 3 , if

1)

3

2)

4

3) 6

4)

12

312. The line 2 1 0x y is tangent to the circle at the point (2,5) and the centre of the circle lies

on 2 4x y . Then the radius of the circle is

1) 3 5 2) 5 3 3) 2 5 4) 5 2

313. Equation of tangent to the circle, at the point (1, -1), whose centre is the point of intersection of

the straight lines 1x y and 2 3x y is

1) 3 4 0x y 2) 4 3 0x y 3) 3 4 0x y 4) 4 3 0x y

314. The circle with equation 2 2 1x y intersects the line 7 5y x at two distinct points A

and B. Let C be the point at which the positive x-axis intersects the circle. ThenACB is

1)

1 4tan

3 2)

1 3tan

4 3)

4 4)

1 3tan

2

315. The image of the line 3 2x y in the line 1y x is

1) 3 2x y 2) 3 2x y 3) 3 2x y 4) 2x y

316. A circle of radius 5 is tangent to the line 4 3 18x y at M(3,-2) and lies above the line (in

the increasing direction of y). The equation of the circle is

1) 2 2 6 4 12 0x y x y 2) 2 2 2 2 3 0x y x y

3) 2 2 2 2 23 0x y x y 4) 2 2 6 4 12 0x y x y

317. Two circles intersect at the point P(2,3) and the line joining the other extremities of

theirrespective diameters through P makes an angle

6 with x-axis, then the equation of the

common chord of the two circles is

1) 3 2 3 3 0x y 2) 3 2 3 2 0x y

3) 3 2 3 3 0x y 4) 3 2 3 3 0x y



318. A variable point P lies inside the triangle formed by the lines 0y , 4 3 0x y and

3 4 9 0x y . If sum of the distances of the point P to the sidesof the triangle is 3,then the

locus of P satisfies the equation

1) 7 6 24 0x y 2) 2 24 0x y

3) 7 4 6 0x y 4) 2 6 0x y 319. Through a point on the hypotenuse of right angled triangle, lines are drawn parallel to the legs

of the triangle so that the triangle is divided into a square and two right triangles. The area of

one of the two small triangles is ‘m’ times the area of the square. The ratio of the area of the

other small right triangle to the area of the square, is

1) 1

4m 2)

1

2 1m 3)

2

1

8m 4)

1

m

320. The straight line passing through the point (8, 4) cuts y-axis at B and x-axis at A. The locus of

mid point of AB is

1) 2 4 64xy x y 2) 2 4 0xy x y

3) 4 2 8 0xy x y 4) 4 2 72xy x y

321. A line 2 6 0x y cuts coordinate axes at points A and B. Two lines 1L and 2L are drawn

from origin which divide AB in three equal parts, then sum of slopes of lines 1L and 2L is

1) 0 2) 10

2 3)

7

2 4)

9

4

322. If the chord of the circle 2 2 4 0x y y along the line 1x y subtends an angle at a

point on the major arc of the circle then cos

1) 1

2 2)

1

2 3)

3

2 4)

1

2 2

323. A square ABCD has 1sq.unit area. A circle touches the sides AB & AD of the square and passes

through the vertex C. Then, the radius of the circle is

1) 2 2 2) 2 1 3) 2 2 4) 1

2 2

324. Tangent to the curve 2 6y x at the point 1,7P touches the circle

2 2 16 12 0x y x y c at the point Q. Then, coordinates of Q are

1) 6, 11 2) 9, 13 3) 10, 15 4) 6, 7

325. A chord of the circle 2 2 32x y makes equal intercepts of length ‘l ’ on x-axis and y-axis

then the maximum value of ‘l ’ is

1) 4 2) 4 2 3) 16 4) 8

326. A circle passes through A(0,4) and B(8,0) has its centre on x-axis. If point C lies on the

circumference of the circle and m is the greatest area of triangle ABC then m is equal to

1) 10 5 1 2) 10 5 1 3) 20 5 1 4) 20 5 1

327. Tangents are drawn from external point P(6,8) to the circle 2 2 2x y r . The radius r of the

circle such that area of triangle formed by the tangents and chord of contactof P with respect to

the given circle ismaximum is

1)25 2) 15 3)5 4)20



328. From points on the line 4 3 7 0x y tangents are drawn to the circle

2 2 6 4 12 0x y x y which include an angle 1 24tan

7

between them. If S denotes the

sum of the distances of all such points from (-1, -1) then 3S equals to

1)64 2) 20 3) 40 4) 32

329. P is a point on 2 2 25x y and Q is a point on the line 3 3 0x y and the line

6 0x y is perpendicular bisector of PQ. Then, which of the following may be the

coordinates of P

1)11 2

,5 5

2)2 11

,5 5

3) 3,4 4) 0, 5

330. Let PQ and RS be tangents at the extremities of a diameter PR of a circle of radius R such that

PS and RQ intersect at a point X on the circumference of the circle. If PQ=3 and radius of the

circle =2, then XS is

1) 64

3 2)

32

5 3)

16

5 4)

64

15

331. If a circle passes through the point ,a b and cuts the circle 2 2 2x y k orthogonally, then the

equation of the locus of its center is

1) 2 2 22 2 0ax by a b k 2) 2 2 22 2 0ax by a b k

3) 2 2 2 2 23 4 0x y ax by a b k 4) 2 2 2 2 22 3 0x y ax by a b k

332. If the circles 2 2 2 0x y ax cy a and 2 2 3 1 0x y ax dy intersect at two distinct

points P and Q , then the line 5 0x by a passes through P and Q for

1) exactly one value of a 2) no value of a

3) infinitely many value of a 4) exactly two values of a

333. Consider a family of circles which are passing through the point 1,1 and are tangent to the

x axis. If ,h k are the coordinates of the centre of the circles, then set of value of k is given

by the interval

1) 0 1 2k 2) 1 2k 3) 1 2 1 2k 4) 1 2k

334. Consider circles 0p

2 2

1 : 2 2 0C x y x y p

2 2

2 : 2 2 0C x y x y p

2 2 2

3 :C x y p

Statement –I: If the circle 3C intersects 1C orthogonally then 2C does not represent a circle

Statement –II: If the circle 3C intersects 2C orthogonally then 2C and 3C have equal radii

Then which of the following is true?

1) Statement II is false and statement I is true

2) Statement I is false and statement II is true

3) Both the statements are false

4) Both the statements are true

335. Equation of the straight line meeting the circle with centre at origin and radius equal to 5 in two

points at equal distances of 3 units from the point 3,4A is

1) 6 8 41x y 2) 6 8 41 0x y 3) 8 6 41 0x y 4) 8 6 41 0x y



336. If two circles, each of radius 5 units, touch each other at 1,2 and the equation of their common

tangent is 4 3 10x y , then equation of one of these two circles, a portion of which lies in all

the quadrants is

1) 2 2 10 10 25 0x y x y 2) 2 2 6 2 15 0x y x y

3) 2 2 2 6 15 0x y x y 4) 2 2 10 10 25 0x y x y

337. The circles 2 2 2

12 0x y g x a and 2 2 2

22 0x y g x a cut each other orthogonally. If

1p and 2p are perpendiculars from 0,a and 0, a on a common tangent of these circles,

then 1 2p p is equal to

1) 2

2

a 2)

2a 3) 22a 4)

2 2a

338. If the circle 2 2

1 : 16S x y intersects another circle 2S of radius 5 in such a manner that the

common chord is of maximum length and has a slope equal to 3

4, the coordinates of the centre

of 2S are

1) 9 12 9 12

, , ,5 5 5 5

2) 9 12 9 12

, , ,5 5 5 5

3) 12 9 12 9

, , ,5 5 5 5

4)

12 9 12 9; , ,

5 5 5 5

339. Consider the following statements

I. Circle 2 2 1 0x y x y is completely lies inside the circle 2 2 2 2 7 0x y x y

II. Number of common tangents of the circles 2 2 14 12 21 0x y x y and 2 2 2 4 4 0x y x y is 4.

Which of these is/are correct

1) Only I 2) Only II 3) Both I and II 4) Neither I nor II

340. The radical centre of three circles described on the three sides 4 7 10 0x y , 5 0x y

and 7 4 15 0x y of a triangle as diameters is.

1) 1,2 2) 1, 2 3) 1,2 4) 1, 2

341. If the line 3 3 0y x cuts the parabola 2 2y x at A and B, then .PA PB is equal to

[where, 3,0P ]

1) 4 3 2

3

2)

4 2 3

3

3)

4 3

3 4)

2 3 2

3

342. The locus of the point through which three normals to the parabola 2 4y ax passes, such that

two of them make angles and respectively with the axis of the parabola such that

tan tan 2 is

1) 2 4 0x ay 2) 2 4 0y ax 3) 2 4 0x ay 4) 2 4 0y ax

343. The equation of the parabola, the extremities of whose latusrectum are 1,2 and 1, 4 is

1) 2

1 3 2 5x y 2) 2

1 3 2 5y x

3) 2

1 3 2 5y x 4) 2

1 3 2 1y x



344. If 9 ,6a a is a point bounded in region formed by parabola 2 16y x and 9x then

1) 0 1a 2) 1

4a 3) 1a 4) 0 4a

345. The length of the chord of the parabola 2 4x y passing through the vertex and having slope

cot is

1) 24cos cosec 2) 4tan sec 3)

24sin sec 4) 24cot cosec

346. The triangle PQR of area A is inscribed in the parabola 2 4y ax , 0a such that P lies at the

vertex of the parabola and base QR is a focal chord. The numerical difference of the ordinates

of the points Q and R is

1) 2

A

a 2)

A

a 3)

2A

a 4)

4A

a

347. The tangent drawn at any point P to the parabola 2 4y ax meets the directrix at the point K,

then the angle which KP subtends at its focus is

1) 030 2)

045 3) 060 4)

090

348. Let 2,3 be the focus of a parabola and 0x y and 0x y be its two tangents. Then

equation of its directrix will be

1) 2 3 0x y 2) 3 4 0x y 3) 5x y 4) 12 5 1 0x y

349. AB is a chord of the parabola 2 4y ax with vertex A,BC is drawn perpendicular to AB

meeting the axis of the parabola at C. The projection of BC on the axis of the parabola is

1) a 2) 2a 3) 4a 4) 8a

350. The parabola 2 8y x and the circle 2 2 2x y

1) have only two common tangents which are mutually perpendicular

2) have only two common tangents which are parallel to each other

3) have infinitely many common tangents

4) does not have any common tangent

351. The shortest distance between the parabola 2 4y x and the circle 2 2 6 12 20 0x y x y is

1) 4 2 5 2) 0 3) 3 2 5 4) 1

352. The ends of a line segment are 1,3P and 1,1Q , R is a point on the line segment PQ such

that : 1:PR QR . If R is an interior point of the parabola 2 4y x , then

1) 0,1 2) 3

,15

3) 1 3

,2 5

4) 1,

353. If angle between two focal chords of a parabola 2

5 8 1y x which are tangents to the

circle 2 2 9x y is 1tana

b

where a and b are relatively prime, then a b

1) 1 2) 7 3) 4 4) 2

354. The locus of centre of a circle which cuts orthogonally the parabola 2 4y x at 1,2 is

1) 1 0x y 2) 2 0x y 3) 3 0x y 4) 5 0x y

355. The length of the normal chord of the parabola which subtends a right angle at the focus is

1) 6 3a 2)5a 3) 5 5a 4) 5a



356. Area of the triangle formed by the tangents at 2 2

1 1 2 2,2 ,2P at at and Q at at and the chord PQ

of the parabola 2 4y ax is

1)2

2

a 2)

2

1 22

at t 3)

3

1 22

at t 4)

23

1 22

at t

357. Let A,B and C be three distinct points on 2 8y x such that normals at these points are

concurrent at P. The slope of AB is 2 and abscissa of centroid of ABC is 4

3. Which of the

following is not true.

1) Area of ABC is 8 sq. units

2) Coordinates of 6,0P

3) Circumcenter of ABC is 2,0

4) Angle between normalsare 0 0 045 ,45 ,90 358. Which of the following is not true

1) The tangents at the extrimities of focal chord of a parabola intersect at right angles

2) The locus of point of intersection of perpendicular tangents to the parabola is its directrix.

3) the circle having focal chord of a parabola as diameter touches the directrix of the parabola

4) If is angle between pair of tangents drawn from 1 1,x y to the parabola 2 4y ax then

11

1

tans

x a

359. If 2,5 and 3,7 are the points of intersection of the tangent and normal at a point on a

parabola with the axis of the parabola, then the focal distance of that point is

1) 29

2 2)

5

2 3) 29 4)

2

5

360. If the parabola 2y a b x b c x c a touches the x-axis then the line 0ax by c

1) always passes through a fixed point

2) represents the family of parallel lines

3) is always perpendicular to x-axis

4) always has negative slope

361. The locus of the foot of the perpendicular drawn from the centre of the ellipse 2 23 6x y on

any tangent to it is

1. 2

2 2 2 26 2x y x y 2. 2

2 2 2 26 2x y x y

3. 2

2 2 2 26 2x y x y 4. 2

2 2 2 26 2x y x y

362. The area (in sq.units) of the quadrilateral formed by the tangents at the end points of the latus

rectum to the ellipse

2 2

19 5

x y , is

1. 27

2 2. 27 3. 27

4

4. 18

363. If OB is the semi-minor axis of an ellipse 1F and 2F are its focii and the angle between 1F B

and 2F B is a right angle, then the square of the eccentricity of the ellipse is:

1. 1

2 2. 1

2 3.

1

2 2 4. 1

4



364. If 1P and 2P are two points on the ellipse

22 1

4

xy at which the tangents are parallel to

the chord joining the points 0,1 and 2,0 , then the distance between 1P and 2P is

1. 2 2 2. 5 3. 2 3 4. 10

365. The ellipse 2 24 4x y is inscribed in a rectangle aligned with the coordinate axes, which is

in turn inscribed in another ellipse that passes through the point 4,0 . Then the equation of

the ellipse is:

1. 2 212 16x y 2.

2 24 48 48x y

3. 2 24 64 48x y 4.

2 216 16x y

366. An ellipse passes through the focii of the hyperbola, 2 29 4 36x y and its major and minor

axes lie along the transverse and conjugate axes of the hyperbola respectively. If the product of

eccentricities of the two conics is 1

2, then which of the following points does not lie on the

ellipse?

1. 13

, 62

2. 39

, 32

3. 13 3

,2 2

4. 13,0

367. Let 3sec ,2tanP and 3sec ,2tanQ where 2

, be two distinct points on

the hyperbola

2 2

19 4

x y . Then the ordinate of the point of intersection of the normals at P

and Q is

1. 11

3 2.

11

3 3.

13

2 4.

13

2

368. If the focii of the ellipse 2 2

21

16

x y

b coincide with the focii of the hyperbola

2 2 1

144 81 25

x y ,

then 2b is equal to

1. 8 2. 10 3. 7 4. 9

369. If the product of perpendicular drawn from any point on the hyperbola 22 2 0x y to its

asymptotes is 3

K where K is

1. 2 2. 4 3. 6 4. 1

370. If the chord of hyperbola 2 2 2x y a touches the parabola

2 4y ax , then the locus of the

middle point of this chord is

1. 3 2x x a y 2. 2 2x x a y 3. 3 2y x a x 4. 3x x a y

371. If the latusrectum of a hyperbola through one focus subtends 060 at the other focus, then its

eccentricity is

1. 2 2. 3 3. 5 4. 6



372. The locus of the middle point of the normal chords of the hyperbola 2 2 2x y a is

1. 3

2 2 2 2 24y x a x y 2. 3

2 2 2 24y x ax y

3. 3

2 2 2 2 24y x a x y 4. 3

2 2 2 24y x a xy

373. A ray emanating from the point 3,0 is incident on the ellipse 2 216 25 400x y at the

point P with ordinate 4. Then the equation of the reflected ray is

1. 4 3 12 0x y 2. 4 3 12x y

3. 3 4 16x y 4. 3 4 9 0x y

374. Two sets A and B are as under:

, : 5 1, 5 1A a b R R a b ;

2 2, : 4 6 9 5 36B a b R R a b , then

1. A B 2. A B (an empty set)

3. neither A B nor B A 4. B A

375. If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its

nearest vertex on the major axis is 3

2 units, then its eccentricity is?

1. 1

2 2.

2

3 3.

1

9 4.

1

3

376. A sequence of concentric ellipses 1 2, ,....., nE E E are drawn such that nE touches the

extremities of the major axis of 1nE and the focii of coincide with the extremities of

minor axis of . If the eccentricity of the ellipse is independent of n, then the value of the

eccentricity, is

1. 2. 3. 4.

377. For all real values of , the line touches a fixed ellipse. The

eccentricity of the ellipse is

1. 2. 2 3. 3 4. 4

378. Coordinates of the vertices B and C of a triangle ABC are and respectively. The

vertex A is varying in such a way that and locus of A is

, then is

1. 1 2. 2 3. 3 4. 4

nE

1nE

5

3

5 1

2

5 1

2

1

5

1,1p 22 1 1px y p

3

2

2,0 8,0

4tan tan 12 2

B C

2 2

2

51

25

x y

k

k



379. If a tangent of slope 2 of the ellipse is normal to the circle ,

then the maximum value of is

1. 16 2. 8 3. 4 4.

380. If A and B are focii of ellipse image of the focus of A

w.r.to any tangent to the given ellipse is then is

1. 2 2. 4 3. 4.

381. If is the chord of contact of the hyperbola , then the equation of the

corresponding pair of tangents is

1. 2.

3. 4.

382. Consider a branch of the hyperbola with vertex at the point

A. Let B be one of the endpoints of its latus rectum. If C is the focus of the hyperbola nearst to

the point A, then the area of triangle ABC is

1. 2. 3. 4.

383. The two common tangents to the parabola and the ellipse meet at M.

Of the two tangents, one meets the parabola and the ellipse at and respectively, and the

other meets them at and respectively. Then the area of the quadrilateral is

1. 30 2. 45 3. 15 4. 20

384. Let be a point on the hyperbola . If the normal at the point P intersects

the x-axis at , then the eccentricity of the hyperbola is

1. 2. 3. 4.

385. A ray emanating from the point is incident on the hyperbola at the

point P with abscissa 8, then the equation of the reflected ray after first reflection is

1. 2.

3. 4.

386. If a normal to a rectangular hyperbola at P meets the transverse axis at Q and focii of the

hyperbola are S and , then is equal to

1. 2 2. 6 3. 9 4. 3

387. If the angle between the asymptotes of hyperbola is . Then the eccentricity of

conjugate hyperbola is

1. 4 2. 3 3. 1 4. 2

2 2

2 21

x y

a b

2 2 4 1 0x y x

ab

5

2 2

2 3 8 4 4 20x y x y

1A 1BA

2 2 2

9x 2 2 9x y

2 29 8 18 9 0x y x 2 29 8 18 9 0x y x

2 29 8 18 9 0x y x 2 29 8 18 9 0x y x

2 22 2 2 4 2 6 0x y x y

21

3

31

2

21

3

31

2

2 4y x 2 24 8x y

1P 1E

2P 2E 1 1 2 2P E E P

6,3P2 2

2 21

x y

a b

9,0

3

2 2

3 2 3

5,0 2 29 16 144x y

3 7 0x y 3 3 13 15 3 0x y

3 3 13 15 3 0x y 3 14 0x y

1S

2

6SP

SQ

2 2

2 21

x y

a b

3



388. Consider the hyperbola and a circle S with centre . Suppose that H

and S touch each other at a point with and . The common tangent to

H and S at P intersects the x-axis at point M. If is the centroid of the triangle PMN, then

the wrong expressions is

1. 2.

3. 4.

389. Tangents are to drawn to the hyperbola at the points P and Q. If these tangents

intersect at the point then the area (in sq. units) of the is:

1. 2. 3. 4.

390. The locus of the point of intersection of the lines, and

( is any non- zero real parameter) is.

1. A hyperbola with length of its transverse axis

2. An ellipse with length of its major axis

3. An ellipse whose eccentricity is

4. A hyperbola whose eccentricity is

391. The number of solutions of the equation is

1) 4 2) 6 3) 8 4) 10

392. Let the smallest positive value of x for which the function achieves

its maximum value be . Express in degrees i.e., . Then the sum of the digits in is

1) 15 2) 17 3) 16 4 ) 18

393. true for

1) 2) 3) 4)

394. The set of values of ‘a’ for which the equation does not have any real solution

is

1) 2) 3) 4)

395. The number of solutions in satisfying the equation

is

1) 0 2) 2 3) 3 4) > 4

2 2: 1H x y 2,0N x

1 1,P x y 1 1x 1 0y

,l m

121 1

11 1

3

dlfor x

dx x 1

1211

1

3 1

dm xfor x

dx x

121 1

11 1

3

dlfor x

dx x 1

1

10

3

dmfor x

dy

2 24 36x y

0,3T PTQ

54 3 60 3 36 5 45 5

2 4 2 0x y k

2 4 2 0kx ky k

8 2

8 2

1

3

3

3

1 1 12 2

esin x cos x log x

3 11

x xf x sin sin , x R

0x 0x0

0x

2 2 2 2 2 2x xx sin x cos xe ln x x sin x cos xe ln x x

0, 02

,

2

,

2 2 1n , n ,n N

2 0sin x x a

3

2 6,

3

2 6,

3

2 6,

3

2 6,

0 2,

3 33 3 3 1tan xlog tan x log log



396. Given that where do not differ by an even multiple of

then,

1) 0 2) 2 3) 1 4)

397. The value of is equal to

1) 2) 3) 4)

398. Let the value of where b > a and , then the value of

1) 2) 3) 4)

399.

1) 2) 3) 4)

400. If are the smallest positive angles in ascending order of magnitude which have their

sines equal to the positive quantity K the value of

1) 2) 2 3) 4)

401. The value of the expression is

1) 0 2) 1 3) 2 4) 3

402. The equation will have a solution if b belongs to

1) 2) 3) 4)

403. Let . The sum of all distinct solutions of the equation

in the set S is equal to

1) 2) 3) 0 4)

404. If then

1) 2)

3) 4)

405. The number of points inside the curve satisfying is

1) 4 2) 6 3) 8 4) 10

406. If are two distinct solutions lying between and of the equation

then

1) 0 2) 1 3) 4/3 4) 8/3

1cos sin cos sin

cos A sin A cos A sin A

,

2 2

sin sin cos cos

sin A cos A

1

2

100

1

101k

sin kx cos k x

101

1012

sin x 99 101sin x 50 101sin x 100 101sin x

19

24Tan a a b ab

a,b N

3

0

1 2 18r

cos r .....

3

1

b

1

2a b

1

2b a

b

b a

14 8 16

cos ec cosec cos ec

8cot

16cot

32cot

2

16cos ec

, , ,

4 3 22 2 2 2

sin sin sin sin

2 1 K 1 K 2 K 1K 6 0 4 0 2 020 33 20 27 20tan tan tan

8 4 1 0cos x bcos x

2, 2, 2, 1,

02

S x , : x ,

3 2 0sec x cosecx tan x cot x

7

9

2

9

5

9

3 3

3 3cos sinm,

cos sin

4 2

2

2 9 82

m mcos

m

23 mcos

m

4 2

2

2 9 82

m mcos

m

22 mcos

m

2 2 4x y 4 4 21 3tan x cot x sin y

, 2

2

2 2tan x sec x

tan tan



407. If then the value of

1) 2 2) -2 3) 3 4) -3

408. If then is equal to

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

409. If then the minimum value of is equal to (where

x is variable)

1) 2) 3) 4)

410. The number of solutions of is

1) 0 2) 6 3) 4 4) 8

411. The greatest possible value of the expression on the

interval is

1) 2) 3) 4)

412. The number of non similar isosceles possible triangle’s such that is

1) 1 2) 2 3) 3 4) 4

413. If then the possible

value of sec x is

1) 1 2) -1 3) 4)

414. If then the roots of f(x) = 0 are

1) 2) 3) 4)

415. Fundamental periodicity of is

1) 2) 3) does not exist 4) 1

416. Let .

If f(x) vanishes for x = 0 and (where ) then

1)

2)

3) has only two solutions 0,

4) f(x) is identically zero

3

4 2 4 2

y xtan tan

0x

sin yLt

x

2 3cos x cos x cos xcos x

a b c d

a c

b d

a sin x bcos x bcos x d cos

2 21

2d a

b 2 21

2d a

a 2 21

2d a

d 2 21

d ad

esin x log x

2

3 6 6tan x tan x cos x

5

12 3,

122

5

112

6

123

5

113

6

100tan A tan B tanC

2 24

cos x cos x sin x sin x sec x

2 2

4cos x sin x sec x cos x cos x

2 2

2 2

2 2

2 2

02

sin cos x

f x cos x sin , ,

x sin cos

11

2,

11 0

2, ,

11 0

2, ,

11 0

2, ,

2 2f x sec x tan x

2

1 1 2 2 n nf x a cos x a cos x .... a cos x

1x x 1x k ,k Z

1 1 2 2

1

2n na cos a cos ....... a cos

1 1 2 2

1

2n na sin a sin ....... a sin

0f x 1x

x



417. If has no solution then the complete set of values of ‘a’ is

1) 2)

3) 4)

418. The number of integral values of ‘b’ for which the equation

1) 2 2) 6 3) 4 4) 8

419. The value of expression

1) 3 2) -1 3) 1/2 4) 0

420. Let such that, and if the maximum value of the expression

equals p < q then the minimum value of q-p =

421. and are the length of medians of triangle ABC drawn through the angular points A,

B and C respectively. value of is equal to

(A) (B) (C) (D) None of these

422. In the given figure, ‘P’ is any interior point of the equilateral triangle ABC of side length 2

units. If, and represents the distance of P from the sides BC, CA and AB respectively,

then is equal to

(A) 6 (B) (C) (C)

423. In triangle ABC the value of the expression is always

equal to

(A) (B) (C) (D) none of these

424. If in a triangle ABC, a, b and are given and are the possible values of the remaining

side, then is equal to

(A) (B) (C) (D)

425. In triangle ABC, a = 5, b = 4, c = 3. ‘G’ is the centroid of triangle. circumradius of triangle

GAB is equal to

(A) (B) (C) (D)

426. In triangle . The value of is equal to

(A) 3 (B) 2 (C) 9 (D) None of these

427. In any triangle, is always equal to

(A) (B) (C) (D)

31

2cos x x a

a R3

2 3a

30

2 3a ,

3 3

2 2 3a ,

2 2 3 23 4 5 3 4 3 2 6 0sin x cos x b b sin x cos x b b b

0 0 0 0

1 1 1 1

6 24 48 12cos sin sin sin

a,b,c R 2 2 2 100a b c

2 2 02

a b sin x c sin x, x

p q, p,q N

,a bm m cm2 2 2

a b cm m m

2 2 21

2a b c 2 2 23

2a b c 2 2 23

4a b c

,a bx x cx

a b cx x x

3 3 / 2 2 3

2 2cos2 cos2 2 cosa B b A ab A B

2a 2b 2c

A1 2,c c

1 2| |c c

2 2 2sinb a A 2 2 22 sinb a A 2 2 2sina b A 2 2 22 sina b A

2 135

1312

513

3

313

2

cos cos cos,

A B CABC

a b c 1 2 3r r r

r

cos cos cosA B C

R

r1

R

r 1

r

R

r

R



428. In triangle ABC, line joining circumcentre and incentre is parallel to side AC, then

is equal to

(A) (B) 1 (C) (D) 2

429. If in triangle ABC, D is a interior point of side BC such that BD : DC = 1 :

3, then the value of is equal to

(A) (B) (C) (D)2

430. In a triangle ABC, 2B = A + C and . Then the value of is equal to

(A) 1/2 (B) 1 (C) 2 (D) none of these

431. In triangle ABC it is given that . Then is equal to

(A) (B) (C) (D)

432. In any right angled triangle ABC, the value of is always equal to

(A) 2 (B) 4 (C) 6 (D) 8

433. If in the triangle ABC, , then is equal to

(A) (B) (C) (D)

434. If in triangle ABC, . Then


435. In triangle ABC, where . If , then x is equal to

(A) (B) (C) (D)

436. In any triangle ABC, the value of is always equal to

(A) 2r (B) (C) (D)

437. In the adjacent figure, ‘P’ is any arbitrary interior point of the triangle ABC. and

are the length of altitudes drawn from vertices A, B and C respectively. If and

represent the distance of ‘P’ from sides BC, AC and AB respectively to the opposite sides then

is always equal to

(A) 3 (B) 2 (C) 1 (D) none of these

cos cosA C

1

2

3

4

, .3 4

B C

sin

sin

CAD

BAD

3 2 3 6

2b ac 2

3

a a b c

abc

2cos cos 2cosA B C a b

a b c bc ac A

3

2

6

4

2 2 2

2

a b c

R

2B

2 2 2 2

1 2 3

1 1 1 1

r r r r

2

2 2

4b

a c

2

2 2

2b

a c

2

2 2

8b

a c

2

2 2

b

a c2 2 2cos cos cos 1A B C

2A

2B

2C

: : 1 :1: 1a b c x x 0,1x2

A C

1

6

1

2 3

1

7

1

2 7

1 2

1 cos

r r

C

2R22r

R

22R

r

,a bH H cH

,a bx x cx

a b c

a b c

x x x

H H H



438. If , then the value of is

(A) (B) (C) (D)

439. is equal to

(A) (B) (C) (D)

440. is equal to


441. The maximum value of is equal to

(A) (B) (C) (D) none of these

442. In triangle ABC, then the value of where are the

sides of the triangle, is equal to


443. If , then value of the expression ,

is equal to

(A) (B) (C) (D)

444. The number of real solutions of is equal to

(A) 1 (B) 2 (C) 3 (D) none of these

445. Total number of positive integral values of ‘n’ such that the equations

and are consistent, is equal to

(A) 1 (B) 4 (C) 3 (D) 2

446. The angle of elevation of the top of a tower from a point on the ground is 30° and it is 60°

when it is viewed from a point located 40 m away from the initial point towards the tower. The

height of the tower is

(A) (B) (C) (D)

447. A tower of x metres high has a flagstaff at its top. The tower and the flagstaff subtend equal

angles at a point distant y metres from the foot of the tower. Then the length of the flagstaff (in

metres) is

(A) (B) (C) (D)

448. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P

be a point on the ground such that AP = 2 AB. If , then is equal to:

(A) (B) (C) (D)

,2 2

x

1 1tan 3sin 2tan tan

4 5 3cos 2

x x

x

/ 2x 2x 3x x1

1

2 11

2tan

1 2

rn

rr

1tan 2n 1tan 24

n 1 1tan 2n 1 1tan 24

n

1 1cos cos 2cot 2 1

2 1 / 4 3 / 4

2 2

1 1sec cosx ec x

2

2

2

4

2

,2

A

1 1tan tan ,

c b

a b a c

, ,a b c

2

3

4

,2

x

1 1 1sin cos cos cos sin sinx x

2

2

1 1 2tan 1 sin 12

x x x x

2

21 1cos sin

4

nx y

2

21 1sin cos

16y x

10 3 m3

m20

3 m

1020 3 m

2 2

2 2

y x y

x y

2 2

2 2

x y x

y x

2 2

2 2

x x y

x y

2 2

2 2

x x y

x y

BPC tan

1

4

2

9

4

9

6

7



449. Let h be the height of the peak of the hill from a station to be . He walks c metres along a

slope inclined at the angle and finds the angle of elevation of the peak of the hill to be .

Then the height of the peak above the ground is

(1) 2) 3) (4)

450. From the top of a ‘h’ meters high cliff the angle depression of the top and bottom of a tower are

observed to be 300 and 60

0 respectively the height of the tower is

(1) (B) (3) (4)

451. If a, b, c are positive real numbers such that a > b > c and the quadratic equation

(a + b – 2c) x2 + (b + c – 2a)x + (c + a – 2b) = 0 has a root in the interval (– 1, 0) then the

roots of the equation x2 + (a + c) x + 4b2 = 0 are

(1) imaginary (2) real and unequal.

(3) real and equal. (4) less than – 1.

452. If the equation x2 + 2(l + 1)x + l2 + l+ 7 = 0 has only negative roots, then the least

value of l equals

(1) 4 (2) 7 (3) 1 (4) 6

453. SB Let a > 1 be a fixed real number. If S is the set of real numbers x that are solutions to the

equation then

(1) S is f. (2) S is an infinite set.

(3) S is a doubleton. (4) S is a singleton.

454. Let P(x) = , where a1, a2, ......., a12 are positive reals and

where b1, b2, ......., b13 are non-positive reals, then which one of the following is

always correct?

(1) A > 0, B > 0 (2) A > 0, B < 0 (3) A < 0, B > 0 (4) A < 0, B < 0

455. The set of all real values of x for which both and are meaningless, is

equal to

(1) [– 4, – 3] (2) (–3, – 2) (3) (–3, 2] (4) (– 3, 1)

456. Infinite number of triangles are formed as shown in figure.

If total area of these triangles is A then 8A is equal to

(1) 3 (2) 4 (3) 1 (4) 2

sin sin

sin

c

sin

sin

c

sinc sinc

3

h 2

3

h

2

h

4

h

2 22log log5 4 ,

x aa x

2

10

4log (4.9)

3

xx

12

1

( )i

i

A P a

13

1

( )j

j

B P b

2

2

3

log ( 1)x

x

x x

2 9x

x

y

1

2

3

3

1

9

1

27

1O



457. Let a1, a2, a3, a4 and b be real numbers such that and .

The sum of all possible integral value of 'b' will be

(1) 6 (2) 3 (3) 10 (4) 15

458. If two roots of the equation (x – 1) (2x2 – 3x + 4) = 0 coincide with roots of the equation

x3 + (a + 1) x2 + (a + b) x + b = 0 where a, b Î R then 2(a + b) equals

(1) 4 (2) 2 (3) 1 (4) 0

459. If p1, p2 are the roots of the quadratic equation ax2 + bx + c = 0 and q1, q2 are the roots of the

quadratic equation cx2 + bx + a = 0 (a, b, c Î R) such that p1, q1, p2, q2 are in A.P. of distinct

terms, then

equals

(1) – 1 (2) 1 (3) (4) 2

460. If the equation 2x + 2–2 = 2k has exactly one real solution, then sum of all integral values of

k in [– 100, 100] is equal to

(1) 5050 (2) 10100 (3) 0 (4) – 5050

461. If H1, H2, H3 , ............, H101 are in H.P., then is equal to

(1) 1/101 (2) 101 (3) 100 (4) 1/100

462. If and a, b, c are not in A.P., then

(1) a, b, c are in G.P. (2) a, , c are in A.P.

(3) a, , c are in H.P. (4) a, 2b, c are in H.P.

463. Let a, b, c be real numbers such that a + b + c = 6 and ab + bc + ca = 9. If exactly one root

of the equation x2 – (m + 2)x + 5m = 0 lies between minimum and maximum value of c,

then find the number of integral values of m.

(1) 9 (2) 8 (3) 10 (4) 7

464. Suppose that the temperature T at every point (x, y) in the cartesian plane is given by the

formula

T = 1 – x2 + 2y2

The correct statement about the maximum and minimum temperature along the line x + y = 1, is

(1) Minimum is – 1. There is no maximum.

(2) Maximum is – 1. There is no minimum.

(3) Maximum is 0. Minimum is – 1

(4) There is neither a maximum nor a minimum along the line

465. Given a and b are the roots of the quadratic equation x2 – 4x + k = 0 (k ¹ 0). If ab, ab2 +a2b,

a3 + b3 are in geometric progression then the value of 'k' equals

(1) 4 (2) (3)3/7 (4) 12

466. If 'a' and 'b' are the first and the last terms of an A.P. as well as of an H.P. both having n terms.

The product of the rth term of the first series and (n – r + 1)th term of the second series, is

(1) independent of both n and r and equals ab

(2) independent on both n and r and equals a2b

(3) independent on both n and r and equals ab2

(4) dependent on n and r.

4

1

8K

K

b a

4

2 2

1

16K

K

b a

1

2

1001

1 1

( 1)i i i

i i i

H H

H H

1 1 1 10

2 2a a b c c b

2

b

2

b



467. If the values of y satisfying the equation x2 – 2x sin (xy) + 1 = 0 is expressed in the form of

kp (k Î R), then the sum of all possible values of k in (0, 48) is

(1) 568 (2) 562 (3) 560 (4) 564

468. Let a, b, c, d are positive integers such that logab = and logcd = . If (a – c) = 9, then the

value of (b – d) is

(1) 7 (2) 697 (3) 124 (4) 93

469. 21st term in the sequence of numbers x1 , x2 , x3 , x4 , ....... , x2005

which satisfy = = = .............. =

Also x1 + x2 + x3 + ........ + x1005 = 2010 , is ..

(1) (2) (3) (4)

470. The smallest integral value of a such that |x + a – 3| + |x – 2a| = |2x – a – 3| is true , " x R

, is

(1) 1 (2) 0 (3) -1 (4) 2

471. If a, b are the roots of the quadratic equation x2 – ,

then the value of a2 + ab + b2 is equal to

(1) 3 (2) 5 (3) 7 (4) 11

472. Let f (x) = 3ax2 – 4bx + c (a, b, c Î R, a ¹ 0) where a, b, c are in A.P. Then the equation f (x) = 0

has

(1) no real solution. (2) two unequal real roots.

(3) sum of roots always negative. (4) product of roots always positive.

473. Let a, b, g are the roots of the cubic equation a0x3 +3a1x2 + 3a2x + a3 = 0 (a0¹ 0).

Then the value of (a – b)2 + (b – g)2 + (g – a)2 equals

(1) (2) (3) (4)

474. The solution set of is (0, a) È (b, c) then (abc) has the value equal to

(1) 12 (2) 15 (3) 20 (4) None

475. The value of equals

(1) (2) (3) (4)

476. Find the sum of all positive integral value(s) of 'n', n Î [1, 300] for which the quadratic equation

x2 – 3x – n = 0 has integral roots.

(1) 1668 (2) 1600 (3) 1664 (4)1632

477. If the range of real values of b for which the equation

(x4 + 4x2 + 4) – (b + 4) (x4 + 6x2 + 8) – (b + 5) (x4 + 8x2 + 16) = 0

has atleast one real solution is [a, b) , then the value of (2a – 5b) is

(1) 11 (2) 6 (3) 9 (4) 15

478. If the equation (x2 + a|x| + a + 1)(x2 + (a + 1) | x| + a) = 0 has no real root,

then the range of values of 'a' is

(1) (0, 1) (2) (– ¥, 0) (3) (0, ¥) (4) (–1, 0)

3

2

5

4

1

1 1

x

x

2

2 3

x

x

3

3 5

x

x

1005

1005 2009

x

x

86

1005

83

1005

82

1005

79

1005

32 3 2log 2log 3 log 2 log 3

3 2 3 2 3 2 0x

2

2 0 1

2

0

18( )a a a

a

2

2 0 1

2

0

18( )a a a

a

2

0 1 2

2

0

18( )a a a

a

2

1 0 2

2

0

18( )a a a

a

2 2log (4 ) log (1 )1 1

x xx x

1

1

( 1)5

n

nn

n

5

12

5

24

5

36

5

16



479. If 1 + log5(x2 + 1) ³ log5(ax2 + 4x + a), then 'a' can be equal to

(1) (– , 3] [7, ) (2) a (2, 3]

(3) (– , – 2) (2, ) (4) (– , 3]

480. If the solution set of the inequality logx is (a, b) È (c, d) then find the value of

where (a < b < c < d).

(1) 2/5 (2) 4/5 (3) 10 (4) 5/4

481. The value of the expression is equal to

1) 2) 3) 4)

482. If r, s, t are prime numbers and p, q are the positive integers such that the LCM of p, q

is , then the number of ordered pairs (p, q) is

1) 252 2) 254 3) 225 4) 224

483. Let be positive integers such that . Then

the number of such distinct arrangements is

1) 8 2) 7 3) 6 4) 5

484. Let denote the number of triangles, which can be formed using the vertices of a

regular polygon of ‘n’ sides. If , then n equals

1) 5 2) 7 3) 6 4) 4

485. Let . Find the total number of divisors of ( denotes )

1) 5 2) 7 3) 6 4) 8

486. Number of four – digit numbers of the form which satisfy the following three

conditions

(i) (ii) N is a multiple of 5

(iii)

is equal to n, then the value of n/3 is ________________

1) 8 2) 6 3) 7 4) 5

487. Number of permutations of 1,2,3,4,5,6,7,8 and 9 taken all at a time such that the digit

1 appearing somewhere to the left of 2

3 appearing to the left of 4 and

5 somewhere to the left of 6, is Then the value of is __________

1) 6 2) 8 3) 9 4) 7

x R

5 11

2 x

cd

ab

547 52

4 3

1

j

j

C C

47

5C 52

5C 52

4C 47

4C

2 4 2r t s

1 2 3 4 5n n n n n 1 2 3 4 5 20n n n n n

1 2 3 4 5, , , ,n n n n n

nT

1 21n nT T

0

( )n n

r k r

kf n

r

(9)f

n

r

rnC

N abcd

4000 6000N

3 6b c

7!k k



488. is selected from the set and the number is formed. Total number of

ways of selecting so that the formed number is divisible by 4 is equal to

1) 50 2) 49 3) 48 4) 47

489. The total number of ways in which number of identical balls can be put in n

numbered boxes such that box contains at least number of balls is

1) 2) 3) 4)

490. Let . The total number of functions that

are onto and there are exactly three elements in A such that is equal to

1) 490 2) 510 3) 630 4) 530

491. Two players play a series of games. Each game can result in either a win

or a loss for . The total number of ways in which can win the series of these games

is equal to

1) 2)

3) 4)

492. Total number of numbers less than that can be formed using the digits 1, 2, 3 is

equal to

1) 2) 3) 4)

493. The number of integral solutions of with is

1) 134 2) 136 3) 138 4) 140

494. Let there be circles in a plane. The value of for which the number of radical

centers is equal to the number of radical axes is (assume that all radical axes and radical

centers exist and are different)

1) 7 2) 6 3) 5 4) 3

495. Let A be a set of distinct elements. The number of triplets of the A

elements in which at least two coordinates of equal to

1) 2) 3) 4)

496.

1) 2) 3) 4)

n {1,2,3,.....,100} 2 3 5n n n

n

2n

(1,2,3,.... )n thi i

2

1

n

nC

2 1

1

n

nC

2 2

21

n n

nC

2

1

n

nC

1 2 3 7 1 2 3, , ,..., , , ,A x x x x B y y y :f A B

x2( )f x y

1 2P and P 2n

1P 1P

2 212

2

n n

nC 2 212 2

2

n n

nC

212

2

n n

nC 212 2

2

n n

nC

83 10

9 813 4 3

2 91

3 32

817 3 3

2 9 81

3 3 32

0x y z 5, 5, 5x y z

3n n

( 3)n , ,x y z

3

n P 3

3

nn P 23 2n n 23 1n n

21

1

2 2 1

.r

m

r C

r r m

m r m

1m

m

1m

m

12m

m

2m

m



497. If , then

1) 2) 3) 4)

498. The coefficient of the term independent of in the expansion of

is

1) 210 2) 105 3) 70 4) 112

499. The value of is

1) 2) 3) 4)

500. The number of distinct terms in the expansion of is/are (with respect

to different power of )

1) 225 2) 61 3) 127 4) 60

501. The value is equal to

1) 2) 3) 4)

502. If the coefficients of the , , terms in the expansion of are in

A.P, then the largest value of is

1) 10 2) 9 3) 8 4) 6

503. Let be the coefficient of in and

respectively. Then the value of is

1) 7 2) 6 3) 4 4) 5

504. The value of is equal to.

1) 6 2) 5 3) 2 4) 1

505. Let and for all , let

. If the value of

where , then the value of is.

1) 9 2) 10 3) 5 4) 7

12n m m N

62 4

0 2 4 6......

2 3 2 3 2 3

nn nn CC CC

2 2

11 3

n

m

2 2

13 1

n

m

2 2

3 1

n

2 2

3 1

n

x

10

2/3 1/3 1/2

1 1

1

x x

x x x x

4040 30

0

r r

r

r C C

69

2940 C 70

3040 C 69

29C 70

30C

15

2

2

1 1x x

x x

x

220

20

0

20 r

r

r r C

39

20400 C 40

19400 C 39

19400 C 38

20400 C

thr 1th

r 2th

r 14

1 x

r

a and b 3x 4

2 31 2 3x x x 4

2 3 41 2 3 4x x x x

4 /a b

1

1 0

1lim . . .3

5

n rn r t

r tnnr t

C C

1

2233 1a 3n

11 2 3 0

0 1 2 1( ) . . . ..... 1 . .nn n n n n n n

nf n C a C a C a C a

(2007) (2008) 3kf f k N k



506. The coefficient of in the expansion of is

1) 7 2) 6 3) 8 4) 9

507. Let be the smallest positive integer such that the coefficient of in the expansion

of for some positive

integer ‘n’. Then the value of ‘n’ is

1) 4 2) 5 3) 6 4) 7

508. If . Then the value of

is

1) 2) 1 3) 4) 2

509. The last two digits of the number are

1) 01 2) 03 3) 09 4) 05

510. If , then

1) 2) 3) 4)

511. Three dice are thrown if (a, b, c) is a sample point, given the condition that the chance

that equals

1) 2) 3) 4)

512. When 6 distinct toys are distributed at random to 3 persons the conditional probability that each

gets at least 2 toys, given that each person has got at least one toy is

1) 2) 3) 4)

513. A matrix is formed taking elements from the set . Given the condition that its trace is

2, then the probability that it is a diagonal matrix or symmetric matrix is

1) 2) 3) 4)

514. Two cards are drawn at random from a pack of cards. Given the condition that there is atleast

one ace, the chance that there is atleast one face card is _____

1) 2) 3) 4)

515. A coin is tossed until a head appears or it has been tossed 3 times. Given that head doesn’t

appear on the first toss, the probability that coin is tossed 3 times is _________

1) 2) 3) 4)

516. A fair die is thrown 20 times. The probability that on the 8th

throw, the fourth six appears is ___

then is _____

1) 5 2) 6 3) 7 4) 8

9x 2 3 1001 1 1 ..... 1x x x x

m2x

2 3 49 50 51

31 1 .... 1 1 3 1x x x mx is n C

2010

2008 2009 2

0 1 23 n

nx x a a x a x a x

0 1 2 3 4 5 6

1 1 1 1

2 2 2 2a a a a a a a

2010320102

14

23

2

0 1 21 ....n n

nx C C x C x C x 0 2 1 3 2 4 2..... n nC C C C C C C C

2

2 !

!

n

n

2 !

1 ! 1 !

n

n n

2 !

2 ! 2 !

n

n n

2 !

2 ! 2 !

n

n n

a b c

a b c

2

7

5

7

6

7

5

14

1

5

2

3

1

36

1

6

3 3 0,1

1

4

2

3

1

12

1

8

8

33

3

11

4

33

1

2

1

4

3

8

1

2

1

8

5

8

5

6K K



517. . A subset A is selected keeping back elements of A in S, again a subset B is

selected. Given that the probability that is ____

1) 2) 3) 4)

518. A draws two cards at random from a pack of 52 cards. After returning them to the pack and

shuffling it, B draws two cards at random. The probability that their draws contain exactly one

common card is _______

2) 3) 4)

519. In constructing a problem on vectors, the three components of a vector are randomly chosen

from the digits 0 to 5 with replacements. Given that the condition that the magnitude of the

vector is 5, the chance that no two components of the vector are equal is ____

1) 2) 3) 4)

520. 5 letters a, b, c, d, e are kept in 5 addressed envelopes A, B, C, D, E (one in each) at random.

Given the condition that no letter has gone into its own envelope, the probability that ‘a’ goes to

C and b goes to D is

1) 2) 3) 4)

521. A function is selected at random and if

and then

equals

1) 2) 3) 4)

522. Let then

1) 2) 3) 4)

523. Four persons independently solve a certain problem correctly with probabilities

then the odds in favour of the problem is solved by atleast one of them is

1) 235 : 21 2) 221 : 35 3) 123 : 5 4) 51 : 13

524. Bag A contains 3 white, 2 red ballsBag B contains 1 white ball. A fair coin is tossed.

If head appears one ball is transferred from A to B. If tail appears 2 balls are

transferred from A to B. Now one ball is drawn from B, the chance that it is white is

1) 2) 3) 4)

525. If and then the value of _______

1) 2) 1 3) 4)

1,2,3,4,5,6S

1,2,3,4A B 1,2A B

4

81

1

27

1

9

2

27

25

663

50

663

25

546

1

13

1

2

1

3

2

5

2

3

1

3

2

3

1

11

2

11

1,2,3,4,5

1,2,3,4,5,6,7

A

B

:f A B

1 ,f i jE f i f j 2f i jE f i f j

752

101 5

CEP

E C

3

11

4

11

5

11

6

11

1 2, 1,2,3,4,......100n n 1 2

1 2

1600

100

n nP

n n

20

33

58

99

13

33

59

99

1 3 1 1, , , ,

2 4 4 8

13

30

23

30

11

30

19

30

0.3P A 0.8, 0.1P A B P A B A B

P PB A

22

21

21

22

3

4



526. In an examination a student will be qualified if he passes at least two of the three tests. The

probability that he will pass the first test is . If he fails in one of the tests then the probability

of his passing in the next test is , other wise it is . The probability that he gets qualified is

1) 2) 3) 4)

527. 3 persons A, B, C in order drawn a card at random from a pack containing 52 cards, replacing it

after each draw. The first who draws a diamond or an ace is declared as winner. The probability

that B wins the game is ______

1) 2) 3) 4)

528. In a game of chess, the chance of A’s win is three times the chance of a draw. The chance B’s

win is two times the chance of a draw. If they agree to play 3 games, then the odds in favour of

they win alternately is ____ (No match is draw)

1) 5 : 31 2) 7 : 29 3) 11 : 25 4) 1 : 1

529. An urn contains 16 different white and 10 different black balls. Balls are drawn one by one until

only 4 white balls remain in the urn. The probability that the last ball drawn is black is ______

1) 2) 3) 4)

530. If head means one and tail means two, and if the coefficients of quadratic expression

are chosen by tossing three fair coins. Given that has imaginary

roots, the chance that the minimum value of is greatest is _______

1) 2) 3) 4)

531. Let let is its power set. If two elements A, B of selected at

random, give that the chance that is

1) 2) 3) 4) None

532. A bag contains 10 fair coins, and 25 coins having heads on both sides. A coin selected at

random and tossed. If it gives head, the probability that it was a fair coin is _____

1) 2) 3) 4)

533. is equal to

1) 2) 3) 4)

534. If p : It rains today, q : I go to school, r : I shall meet any friends and s : I shall go for a movie,

then which of the following is the proposition : If it does not rain or if I do not go to school,

then I shall meet my friend and go for a movie.

1) 2)

3) 4) None of these

535. For two data sets each of size 5, the variance are given to be 4 and 5 and the corresponding

means are given to be 2 and 4 respectively. The variance of the combined data set is

1) 6 2) 3) 4)

1

3

1

6

1

3

1

2

5

27

1

3

1

9

81

367

117

367

169

367

33

122

1

2

5

11

1

10

5

13

2f x ax bx c 0f x

f x

2

7

1

7

1

4

3

4

1,2,3,4,5,6S f S f S

B A n A n B

1

4

1

2

5

16

1

2

1

3

2

3

1

6

~ ( (~ ))p q

~ p q (~ )p q ~ ~p p ~ ~p q

~ ( ) ( )p q r s ~ ( ~ ) ( )p q r s

~ ( ) ( )p q r s

13

2

5

2

11

2



536. In an experiment with 15 observations the results were available .

One observation that 20 was found wrong and was replaced by the correct value 30. The

corrected variance is

1) 8.33 2) 78.00 3) 188.66 4) 177.33

537. If the standard deviation of 0,1,2,3,…..,9 is K, then the standard deviation of 10,11,12,13…..19

is

1) 2) 3) 4)

538. If a variable takes values 0,1,2,3,….. with frequencies proportional to

then the mean of the distribution is

1) 2) 3) 4)

539. The average marks of boys in a class is 52 and that of girls is 42. The average marks of boys

and girls combined is 50. The percentage of boys in the class is

1) 60 2) 40 3) 20 4) 80

540. If and then the standard deviation of is

1) 4/9 2) 9/4 3) 3/2 4) 2/3

******

2 2830, 170x x

10K K 10 K 10 K

2 3

, , , ,....2! 3!

e ee e

e e 21

2e

18

1

8 9i

i

x

18

2

1

8 45i

i

x

1 2 18, ,.....,x x x



SRIGAYATRI EDUCATIONAL INSTITUTIONS

INDIA

SR MPC JEE MAINS

MATHS

1) 2 2) 2 3) 2 4) 1 5) 3 6) 2 7) 3 8) 4 9) 2 10) 4

11) 2 12) 4 13) 3 14) 3 15) 2 16) 3 17) 3 18) 4 19) 4 20) 3

21) 2 22) 2 23) 2 24) 2 25) 3 26) 2 27) 1 28) 1 29) 4 30) 4

31) 2 32) 1 33) 2 34) 3 35) 3 36) 2 37) 2 38) 4 39) 4 40) 4

41) 3 42) 3 43) 2 44) 1 45) 2 46) 1 47) 2 48) 1 49) 2 50) 3

51) 1 52) 2 53) 2 54) 4 55) 4 56) 3 57) 4 58) 4 59) 2 60) 1

61) 1 62) 4 63) 3 64) 2 65) 4 66) 4 67) 4 68) 4 69) 4 70) 3

71) 4 72) 2 73) 3 74) 4 75) 1 76) 4 77) 3 78) 3 79) 1 80) 4

81) 4 82) 1 83) 3 84) 4 85) 3 86) 4 87) 4 88) 3 89) 4 90) 3

91) 2 92) 1 93) 3 94) 1 95) 4 96) 2 97) 1 98) 3 99) 2 100) 3

101) 3 102) 1 103) 2 104) 1 105) 4 106) 4 107) 1 108) 2 109) 3 110) 4

111) 1 112) 4 113) 3 114) 3 115) 1 116) 3 117) 2 118) 2 119) 4 120) 4

121) 2 122) 3 123) 3 124) 4 125) 4 126) 1 127) 1 128) 4 129) 4 130) 1

131) 2 132) 4 133) 1 134) 4 135) 2 136) 2 137) 1 138) 2 139) 3 140) 2

141) 3 142) 1 143) 1 144) 1 145) 1 146) 1 147) 1 148) 3 149) 3 150) 2

151) 2 152) 4 153) 2 154) 1 155) 1 156) 3 157) 3 158) 4 159) 1 160) 1

161) 1 162) 4 163) 2 164) 1 165) 4 166) 1 167) 2 168) 4 169) 1 170) 2

171) 2 172) 1 173) 3 174) 4 175) 1 176) 2 177) 1 178) 3 179) 3 180) 1

181) 1 182) 4 183) 4 184) 1 185) 3 186) 4 187) 2 188) 2 189) 2 190) 2

191) 1 192) 3 193) 3 194) 1 195) 4 196) 1 197) 2 198) 1 199) 3 200) 2

201) 3 202) 2 203) 2 204) 4 205) 4 206) 1 207) 2 208) 3 209) 1 210) 1

211) 1 212) 4 213) 3 214) 2 215) 2 216) 4 217) 4 218) 1 219) 3 220) 2

221) 3 222) 4 223) 2 224) 2 225) 2 226) 4 227) 2 228) 2 229) 2 230) 4

231) 3 232) 4 233) 2 234) 1 235) 2 236) 1 237) 2 238) 1 239) 2 240) 3

241) 4 242) 2 243) 3 244) 3 245) 2 246) 1 247) 2 248) 1 249) 2 250) 3

251) 4 252) 4 253) 2 254) 4 255) 4 256) 1 257) 4 258) 4 259) 3 260) 4

261) 4 262) 1 263) 2 264) 1 265) 4 266) 3 267) 2 268) 2 269) 4 270) 4

271) 2 272) 3 273) 2 274) 3 275) 3 276) 2 277) 4 278) 3 279) 3 280) 3

281) 3 282) 4 283) 3 284) 3 285) 2 286) 2

287) 3 288) 1 289) 2 290) 3

291) 4 292) 2 293) 3 294) 4 295) 4 296) 4 297) 2 298) 2 299) 1 300) 3

301) 2 302) 2 303) 1 304) 4 305) 3 306) 2 307) 4 308) 3 309) 1 310) 3

311) 3 312) 1 313) 2 314) 3 315) 3 316) 3 317) 3 318) 4 319) 1 320) 2

321) 2 322) 4 323) 3 324) 4 325) 2 326) 2 327) 3 328) 3 329) 3 330) 4

331) 1 332) 2 333) 2 334) 2 335) 1 336) 2 337) 2 338) 1 339) 1 340) 1

341) 1 342) 2 343) 3 344) 1 345) 1 346) 3 347) 4 348) 1 349) 3 350) 1

351) 1 352) 1 353) 2 354) 1 355) 3 356) 4 357) 3 358) 4 359) 1 360) 1

361) 1 362) 2 363) 1 364) 4 365) 1 366) 3 367) 4 368) 3 369) 1 370) 1

371) 2 372) 1 373) 2 374) 1 375) 4 376) 2 377) 1 378) 4 379) 3 380) 2

381) 2 382) 2 383) 1 384) 2 385) 2 386) 4 387) 4 388) 3 389) 4 390) 1

391) 2 392) 4 393) 1 394) 3 395) 2 396) 3 397) 3 398) 3 399) 3 400) 2

MATHS QUESTION BANK



401) 4 402) 3 403) 3 404) 1 405) 1 406) 4 407) 3 408) 3 409) 1 410) 2

411) 4 412) 2 413) 3 414) 1 415) 1 416) 4 417) 1 418) 2 419) 4 420) 1

421) 3 422) 2 423) 3 424) 4 425) 2 426) 3 427) 3 428) 2 429) 3 430) 2

431) 2 432) 4 433) 3 434) 3 435) 3 436) 2 437) 3 438) 4 439) 2 440) 3

441) 2 442) 3 443) 4 444) 2 445) 1 446) 4 447) 2 448) 2 449) 1 450) 2

451) 1 452) 4 453) 4 454) 1 455) 3 456) 1 457) 1 458) 3 459) 1 460) 1

461) 3 462) 4 463) 2 464) 1 465) 2 466) 1 467) 4 468) 4 469) 3 470) 1

471) 3 472) 2 473) 4 474) 1 475) 3 476) 2 477) 1 478) 3 479) 2 480) 3

481) 3 482) 3 483) 2 484) 2 485) 4 486) 1 487) 3 488) 2 489) 3 490) 1

491) 2 492) 3 493) 2 494) 3 495) 2 496) 2 497) 1 498) 1 499) 1 500) 2

501) 4 502) 2 503) 3 504) 4 505) 1 506) 3 507) 2 508) 3 509) 3 510) 3

511) 4 512) 4 513) 4 514) 1 515) 3 516) 3 517) 1 518) 2 519) 4 520) 3

521) 4 522) 4 523) 1 524) 2 525) 1 526) 2 527) 2 528) 1 529) 2 530) 2

531) 3 532) 4 533) 2 534) 1 535) 4 536) 2 537) 2 538) 2 539) 4 540) 3

MATHS

HINTS & SOLUTIONS

1. 2

2

15sgn 1 14

1x

x

2. 2

220 log 16sin 1 log 17x

2

2 22 log 17 2 log 16sin 1 2x

2

20 2 log 16sin 1 2 1x f x

3. Since 2 2f t f t

Function is symmetric about line 2x . Also, 2 0x bx c is symmetric about / 2x b / 2 2 4b b

2 4f x x x c

3 groups are possible In (1) and (2) c is positive and (3) ‘c’ is negative.

12 3

2

0f c

Let ‘c’ is positive

1 3f c

2 4f c

4f c

3c

then 1 0; 2 1, 4 3f f f

2 1 4f f f

Again ' 'c is negative, Let 3c



1 6, 2 7, 4 3f f f

2 1 4f f f

4. 3 1 1 0a x x

21 1 0, , 1x ax ax a so , are roots of 2 1 0ax ax a

1

1,a

a

3 2 33 2

1 11 1

11lim lim

1 1 1 1x xx x

x x a xa x x a

e x e x

2 2 2

111 1

1 1lim lim

111

1

xxx x

x a x x a x ax a

eex

x

a

5. sin ; 0 x< /2

1 ; / 2 2

xf x

x

0 ; 0 x<

sin ; 3 / 2

1 ; 3 / 2 2

g x x x

x

0 ; 0 x< / 2

1 ; / 2 3 / 2

2 ; 3 / 2

h x x

x

Hence range of h x is 0,1,2

6. R.H.L

0 0

1 cos h 1 cos h 1 cos hlim lim

sin h.tan h. 1 cos hcos h

h hhh

2

L.H.L

4 6

27 3

3 30 0

exp 6 log 27 9 3 9lim lim

3 27 3 3 1

h h

h hh h

eh

62

3 3

0 0

9 3 1 9 3 1 1lim lim .

/ 327 27 33 1

3 1

h h

hh h

h

h

h

1

9

7. 2'f x x ax b is injective if 0D

2 4 0a b 1a , 1,2,3,4,5b No.of pair = 5

2a , 1,2,3,4,5b No.of pair = 5

3a , 3,4,5b No.of pair = 3

4a , 4,5b No.of pair = 2

5a ‘b’ has no value



8. 100f x x

9. 15

; 0,9015

xf x x

x

0 15x 0f x

75 90 ; 5x f x

15 30x 1f x Total Integes

30 45x 2f x

0, 1, 2, 3, 4, 5f x

45 60x 3f x

60 75x 4f x

10.

1 1 11

, 1,2

1tan tan tan 3

sin 2lim .

1 2x y

xyy

Lx y

1 1

1

, 1,2

1

tan tan 3

1sin 2

lim .1 2x y

xy

x

yy

x y

1 1

1

, 1,2

1tan tan 3

sin 2lim .

1 2x y

xy

yy x

x y

1 11 3 1tan tan 3

3

xy yx

y x y

1 3 1

3

xf x

x

11. 0,1fD

0 1 0xe x ………………(i)

0 ln 1 , 1 1,x x e e ………………(ii)

i ii

, 1x e

12. Let

{ }

2[ ]

{ } 1lim

{ }

x

x a

e xP

x

Put [ ]x a h , 0h

Then

{[ ] }

20

{[ ] } 1lim

{[ ] }

a h

h

e a hP

a h

20

1lim

h

h

e hP

h



0

1 1lim

2 2

h

h

e

h

[Using L Hospital Rule]

Next put [ ] , 0x a h h

then

{[ ] }

20

{[ ] } 1lim

{[ ] }

a h

h

e a hP

a h

1 1

2 20 0

(1 ) 1 2lim lim 2

(1 ) (1 )

h h

h h

e h e he

h h

Limit does not exist

13. 11tan 1 sgn 1

2g x x g x

23 22sin cos 1x x

23 22sin 1 cosx x which is possible if sin 1x and cos 0x sin 1, 2 / 2x x n

10 2 / 2 20n

21 39

5 94 4

n n

No.of values of 15x

14. f f n n ……..(i) and 0 1f

Put 0,n we get 0 0f f or 1 0f

Also 2 2f f n n ……….(ii)

Put 1n , we get

1 2 1f f or 0 2 1f or 2 1f

for 3f , put 2n in (ii)

0 2 2f f of 1 2 2f or 3 2f

15. 1 3 7 ......n nS t

11 3 7 ......n n nS t t

10 1 2 4 6 ...... n n nt t t

11 2 4 6......n n nt t t

1

1 2 2 2 22

n

nt n

21 1 1nt n n n n

2 2

2

2

1 1lim 1 lim

21n n

n n nn n n

n n n

16. We have 22 0x x n

x has to be an integer.

22 2 1n x x x x

n can be 21, 36, 55, 78 corresponding to 3,4,5,6x .

Hence sum of all values of ' 'n is 190.



17.

18. 3 ; 0

3 3 3 ; 0x x

xf x

x

as sgn 1xe x R

Clearly f x is many-one.

for 0,x f x is decreasing, hence range of f x for 0x is ,3 .

Hence Range of f x is ,3

19. LHL :

min(sin , [ ]

lim( 1)x

x x x

x

When 1 x

{ } 1 sinx x x

min{sin , 1} 1x x x

Required limit = 1

lim 11x

x

x

x

sin 1x x

RHL :

sin

lim 01x

x

x

sin

11

x

x

Hence LHL RHL sin

01

x

x

Limit does not exist

20. 7 4

9

xf x

x

[ 7, ) 9fD

Now

7 16

7 49 7 4

xf x

xx x

So range of f x is (0, / 4] /8

Hence range of sin 2y f x is 1

(0,1]2



21. 2

2

11

1

xy py x y y x

p x

2 1 1 0x y x y p

As x R . So 0 1 4 1 1 0D y y p

24 1 4 1 0y p y

Since, 1

1,3

y

So 2 14 1 4 1 0 1,

3y p y y

2 22 1 4 0y y p

2

2 1 11 1,

2 3

yp y

y

Hence

2

2 1/ min

2

yp

y

1

4p

22. 0

lim 1x

x

x

0

lim 1 0x

xx

So,

2

1

2 2 20 02

11 1 .....

2!

lim1 11

x

x x

xxx

x x xx x

xx x

e xlim

x xx

2 32

2 20

1 1......

1 1 12! 3!lim2 4 81 1

x

x xx

x x

x x

23. 0

tan tan2sin sin

2 2lim

tan tan

2 2

x

x x x x

x x x x

3

tan tan 1

4

x x x x

x x

1 1 1

22 3 3

24. 2 24 9 0 2, 3p p p

2

0 2 42

pp

2 1 0 2 1 ,p p n n I

1

2

np

; possible values of p are 2 and 3.

25. 3,3p

26. 2 2 2 23 2 0, 1 0, 6 5 0, 2 1 0k x k k k k k and 2 0k k must satisfied

simultaneously.



27. 2

20

sin cos tan sinlimx

x

x

22

20

sin sin tan sin sin tan sinlim

sin tan sinx

x x

xx

28.

21 2 1

20

/ 2 cos 1 . cos 10 lim

2 . 1h

h hf

h h

2

0

/ 4 lim 21 cos 4 2

02

f p

221 1

20

sin 1 1 cos 1 10 lim

2 1 1 1h

h hf

h h

1 22 2

20

sin 1 1lim

8 81 1h

hq

h

29. R.H.L 0 0

lim 1 lim 0h h

f h h h h

L.H.L 0 0

lim 1 lim 0 1h h

f h h h h

30. 3 1 3 5

, ,2 2 2 2

x

31. LHD RHD.

32. f(r) = r for r N.

33. sin 1 0,x x is discontinuous at integral points.

34. Use

0'

h

f x h f xf x Lt

h

35. 1 tanf f x g x g x

2

1'

2g x

x g x

36. Since, ' 0 , 1 1f x x R f

1 1f x x

2 2 21 1

1

x xdx dx

x f x x

37. Since 1 2 1 2'' . '' 0 and '' '' 0f C f C f C f C .

1 2'' 0 and '' 0f C f C

38. 21f x x



39. Use the graph.

40. Let 2,g x x clearly g x is discontinuous and hence not differentiable at

2, 3, 4, 5, 6, 7 and 8x .

41. sinf x n P x is discontinuous and non – differentiable at those points where

sinn P x is an integer.

42. Clearly f(x) is not differentiable at 1,0,1x .

43. 3

1 , 2,02

f x xx

1 32

1f x

x

44. If , lies on , then ,y f x lies on 1y f x .

, lies on y f x

y f x is odd.

45. 1 1tan 2 tan 3f x x

46.

2 2

2 3 2

1d x d y

dy dxdy

dx

47. 2 , 1f x x x

,0 1x x

, 1 0x x

2, 1x x

48. If x is a multiple of 10, then f(x) is continuous.

49. Apply L-Hospital’s rule successively.

50. ' 1 logxef x f



51.

0 0 0

20

h h h

f a a h f af a h f a f a h f aLt Lt Lt

h h h

52. ' 3 ' 3f f and f is continuous.

53. Use the graph.

54. 0

0 and ' 0 0xLt f x f f

55. h x x

RHD = , LHD = 0

56. , , 1x x x are continuous everywhere.

57. RHD = 1, LHD = 1.

58. RHD = 1, LHD = .

59. [x] is discontinuous at integral values of x.

60. f(x) is differentiable 0,x

61. Dividing by 2cos x in both numerator and denominator

2 2sin tan

dx

x x , put tanx = t

1

2 22 2

1 1 1 1 1tan

2 2 22 2 2 2

dt tdt c

t t tt t

11 1 tancot tan

2 2 2 2

xx c

62. 3 sin 2

cos 1

2 x

x xdx

x e x

put 2 sin2 1xt xe

1 cot ln sin4 4

x dx x x c

62.

2

2 32

11

1

111 1

x dxx dx x

Ix x x x

x xx

, put 2 11t x

x

1 1

2

12 2 tan 2 tan 1

1

dtt c x c

t x

64. put 2

3sin 2

2 9sin 16

dx I



2

2 2

2 2

1

cos2 usingsec 1 tan9tan 16sec

d

2

1 1

2 2

sec 1 5 1 52 tan tan tan

16 25tan 10 4 10 2 3 4

d xc c

x

65.

2

2 2

1 1 1 1

11 1 1 1

x

xe x dx x

I e dxxx x x x

1 1

1 1

x xd x xe dx e c

dx x x

66.

2

cos 1sec . sec

sin cossin cos

x xI x x dx x xd

x x xx x x

2secsec sec

sec tansin cos sin cos sin cos sin cos

d x xx x x x xscexxdx x c

x x x x x x x x x x x x

67.

73 4

2

1 1

2 2cos 2sin 2 cos tancos sin

dx dx dx

x x x xx x

21 1

, tan2 t

tdt t x

5 51 3 2 22 2

tan1tan

2 5 5

xtt t dt t c x c

68. Put 2 5

3 32 3 1 3

4 5 22 110

xt I t dt t c

x

5

33 2 3 3 5 1.

110 4 5 110 3 22

xc ab

x

69. 1

2 2

cos cos sinsin

2cos 1 2 sin

xdx xdx xI c

x x

70. cos 1

cos

xI dx

x

, Put 2cos x t

2

1 1

2 2

12 2 2cos 2cos cos

1 1

t dtdt t c x c

t t

71. 2 2

2 22 2 1

x x

x xx x

e e dxI

e e e e

Substitute 2 2

22

1

x xdt

t e e It t

1 1 2 2sec 2secx x

t c e e c

72. 3 3cos sin cos cos sin sin sin cos cos sin

dx dxI

x x x x x x



2 2sec cos

cos tan sin cos sin cot

xdx ec xdxI

x x

2 2

cos tan sin cos sin cotcos sin

x x c

2 2

. 8cos 2cos sin

ab ec

73. 2 4

2 4 4 2

11 11 .

1 1

x x xdxI x dx

x x x x

Put 2x t

2

22

11 1 2 1

2 2 11

t dt tdt

t tt t

2

2

2

1 1 1 1 1 1ln ln ln

2 2 2

tc t c x c

t t x

74. 9 8 2

9

1 cos 1cot cot .cos

sin 2 .tan 2 cos 2

dx ecxx dx x ec xdx

x x x

8 91 1cot . cot cot

2 18x d x x c

75. 22

2

1

2 12 11

dx dxI

xx x x

x x

, put

1t

x

1 1

2 2

1 1cos cos

2 21 2 2 1

dt dt t xc c

xt t t

76. 2

3

4

1

1

xI x dx

x

, put 2 sinx

2sin 1 sin cos 1sin sin

cos .2 2

x dd

x

1

1 1 1 1cos 1 cos2 cos sin cos

2 4 2 4 4d D

4 1 2 2 4 4 1 2 2 4

1

1 1 1 1 1 11 sin 1 1 cos 1

2 4 4 2 4 4x x x x D x x x x D

1 1 1

14 2 4

A B C

77. Put 2 2

21, 1

1

tdx dtt x x dt dx x

tx

2 21 1 12 1 1

2t x x t

t t

From (i) & (iii) 6 5 7 4 61 1 1 1 1

2 2 8 12I t t dt t t dt t t c

t

1 1 5

8 12 24A B

78. 1 1sin 2 tan ,

1

xdx

x

2

2tansin 2

1 tan



2 2 111 1 sin

2 2

xx dx x x c

1 1

12 2

A B

79. 2 2

2

2 2 2

2 tan sec2 tan

2 tan 2 tan 1 sec

x x dxxdx dx

x x x

2

2 2 2

tan sincosln tan 2 tan

2 tan cos 1 2 sin

d x d xxdxx x

x x x

2 1 1sin sinln tan 2 tan sin sin

2 2

x xx x c f x

80.

2 2

2 21

1 ln 1 ln

1 1 ln ln1 ln lnx x

x dx x dx

x x x xx x

21 ln 1 ln

ln 1 ln1 ln 1 ln 1 ln

x dx xdxx x c

x x x x x

81.

113 5 22 21 ,x x dx

put 5

2 21t x

7 3

2 2 54 4 21

5 5 7 5 3

t tt t dt t c

7 5 35 5 52 2 22 2 2

4 8 4 81 1 1

5 25 15 25x x x c b

82. Put 31t x

2 222

1 1 2 1 2 1

3 3 11 1

dtdt

t t tt t t

3

3 33

2 1 1 2 1 1 1ln ln

3 1 3 3 1 3 33 1

t xd d

t t t x xx

2 1 1

03 3 3

a b c

83. 1

4 41

4

4

1

11

dxx dx

xx

Put 2

4

4 4 2 2

1 1 1 11

1 2 1 1

tt dt dt

x t t t

1 1

4 44 4

1 1

14 4

1 11 1 1 1 1ln tan ln tan

4 1 2 4 21

x x xtt c c

t xx x

1 1 1

4 2 4A B

84. cos6 6cos4 15cos2 10 cos6 cos4 5 cos4 cos2 10 cos2 1x x x x x x x x

cos cos5 5cos3 10cosx x x x



1 1 1 1 sin 1

sec ln sec tan ln ln tan2 2 2 cos 2 2 4

x xxds x x c c c

x

85.

1 3 5

3 5 4 4 4sin cos sin cos

dx dx

x x x x

, Put tan x t

3 1 1

4 4 44 4tant dt t x c

86. Put 21 xe t

2

2 2 2

2 2

12 ln 1 2 ln 1 4 2 ln 1 4 1

1 1

tt dt t t dt t t dt

t t

2 1 1 12 ln 1 2 2ln 1 2 4 2ln

1 1 1

xx

x

t et t c e x c

t e

2 4 2 0A B C

87.

1

4

1

xdx

x , Put 4x t

4

2

2 2

14 4 1

1 1

tdt t dt

t t

3 1 1

3 1 14 4 44 4

4 4tan 4 4tan3 3

t t t c x x x c

4 8

43 3

A B

88. 1

.1

x dx

xx

, put 2sinx , 2sin cosdx d

1 sin

2 . 2 cos 1sin

d ec d

11 12ln cos cot 2 2ln 2

xec c sin x c

x

89.

1 1

3 52

3 4 2

2 4

1

2 12 2 12

x x dxx dx

x x x

x x

Put 2 4

2 1 1 12

4 2

dtt t c

x x t

4 2

2 4

1 2 1 2 2 12

2 2

x xc c

x x x

90.

22 4 2 23 5

2 4 2 4 2 4

cos cos cos 1 1cos cos, sin

sin sin sin sin

x x xdx t tx xdx dt t x

x x x x t t

4 2

2 4 2 22 2

2 3 2 4 2 61 1

11

t t tdt dt dt

t t t tt t

1 126tan sin 2cos 6tan sint t c x ecx x c

t



91.

2 3

27 27

3 2

sin sin ..... 0xdx xdx

92. 1

0 11, 1n

n nI e I n I

.

3 16 6I e

93. /2

2

0

cos sin 2 sin cos 0f d

/2 /2

2 2

0 0

cos sin cos sin 2 sin cosf d f d

94.

64 1 8 27 64

1/3

0 0 1 8 27

0 2 3 36x dx dx dx dx dx

95. 2

0 0

1 sin2 2 2

1 sin cos

xdx xI dx I

x x

96. Put

2

0

tan2 2 4 41 1

x drt

t

97.

2

22 .....

1

xx x

x

ee e

e

3

2

ln

ln

1 1

1 2xI

e

98. 0

2 sin 2 sin cos2

I x x dx

cos2

x t

2

162I

99. /2 /2

2

0 0

cos3 12cos cos 1 1 1

2cos 1 2 2

xx x dx

x

100. /2 /2

0 0 0

1 1 1 1sin 2 ,2 sin sin

2 2 2 2

n n n

n n nx dx x t t dt tdt

/2

0

1cos

2

n

ntdt

101.

1

220

22 .

2 2 331 3

2 2

dxI

x

2

6 3I

102. 5 /4

3 /4

2 sin cos 0I x x dx

103. 1 1

100 10050 50 50 49

2 1

0 0

1 1 1I x x dx I x x x dx



1 2

11

5050I I

104.

/2

2 3

0

2tan sin cos

15x d

105. 1 1 22 1

a x

x a x

e dxI

Tan e Tan e e

1 1

1

1

0 0

ln ln 22 2 2

a aTan e Tan e

a

a

dtt Tan e

t Tan e

106.

/2

2

2cos 2 2 4xdx

107. For 2 2 2

, / 4 , cos sinx x x xx o then e e e x e x

108.

/2

20

1sin ,n n

n

I nIn xdx

I n

109.

/2/2 /2

0 00

sin 2 1cos 2 logcos logcos sin 2 tan

2 2

nxnx xdx x nx xdx

n n

/ 2

0

1sin 2 tan

2nx xdx

n

110. 2 /3 2

2 0 2 /3

2cos 1 2cos 1 2cos 1x dx x dx x dx

2 33

111. 2 1 1 12 2018 sin cos 1

nLt n

n n n

1

2 2018 1 20182

112. 1 1 1 1f x xf x f f

113. 2 2

2 2

2 tan 2 tan

sec sec

3 3 3

x x

x x

I t f t t dt t f t t dt

2

2

2 tan

sec

2 3 3

x

x

I f t t dt

114. 3

1 1

1

cot 3 1 22

Tan x x dx

3 1 3

1 1 1 1 1 1

1 1 1

1 1 1Tan x Tan dx Tan x Tan dx Tan x Tan dx

x x x

22

115. 1 1

2 22

0 0

10

3x x f x dx x f x dx

f x x



116.

1

22 20

1 12 2

1K K

dx

K Kx K x

117.

cot1 2018

0 cot1

1 0 cot1dx dx

118.

1 3

2

13f x f x x c

f x

1/33 2 6 0 3 9 3f c c f x x f

119. 1

0

1 1 11 2 .... ....

1 2x x x n dx

x x x n

1 ! 1 !

! !

1 ...

ln 1 ... cos 1 ! !

1....

1

n n

n n

x x n t

x t dt t n n

dtdx

x t

120. cosf x x

/2 /2

/2 0

cos 2 cos 2x dx xdx

121. 1x and 2x are the roots of the equation

2 2 3 1x x kx

2 2 4 0x k x

1 2

1 2

2

4

x x k

x x

2

1

21 2 3x

xA kx x x dx

2

1

2

2 42

x

x

xk x

2 2

3 32 12 1 2 1

12 4

2 3

x xk x x x x

2

2

2 1 2 1 1 2

2 14

2 3

kx x x x x x

2

2 2

2 1 1 2

2 14 2 4 4

2 3

kx x x x k

2

22 16 1 162

6 6 3

kk



3/22

2 16

6

k

isminimum,if 2A k

32

Hence,3

minA

122. Put 2 , 1x x y y

Then, the required region is defined by 1 2x y .

Required area 2 2

2 2 2 6

Hence, (3) is the correct answer

123. As ,

1,

x x zf x

z z

and 2

g x x , where both f x and g x are [periodic

with period ‘1’ shown as;

Thus, required area 1 2

010 x x dx

1 1/2 2

010 x x dx

1

3/2 3

0

103 / 2 3

x x

2 1 10

10 sq units3 3 3

Hence, (3) is the correct answer

124. Here, 2 2cos cosy fog x f g x x x

Also, 2 218 9 0x x

3 6 0x x

, as ,6 3

x

Required area of curve is



/3 /3

2

/6 /6

1cos 1 cos2

2xdx x dx

/3

/6

1 sin 2 1 1 2 2sin sin

2 2 2 3 6 2 3 6

xx

1 1 3 3

2 6 2 2 2 12

Hence, (D) is the correct answer.

125.

cos , 04

5sin ,

4 6

1 5 5,

2 6 3

x x

f x x x

x

Required area

5 55

3 6 34

0 05

4 6

1cos sin

2f x dx xdx xdx dx

126. 4 22y x x has minimum 1

2x .

127. Shaded area is the required region

22 44

4 4

r sq. units.

128. The required area A is shown shaded in the figure.

21

2 2

0 1

4 x x dx+4 x x dx

3 24

43 24 8

3 24

3 6 2

24 3

3 6

129. Required area 1

2

0

2 4 x 3x dx

11/2

2 1

0

x 4 x 3.2x2 4 x sin

2 2 2 3

O

A B

CD



2 3

3 3.

130. 2 2

0

x

tf x x e f x t dt x 2

0 0

x x

x t x te f t dt x e e f t dt

0

' 2 0

x

x x x tf x x e e f x e e f t dt = 2x + f(x) – (f(x) – x2) = x2 + 2x

3 3

2 2

3 3

x xf x x k x

( f(0) = 0)

f(1) + f(2) + ....... + f(9) = 1/3

(13 + 2

3 + ....... 9

3) + (1

2 + 1

2 + ..... + 9

2)

1 81 100 9 10 19

9603 4 6

131. The given curves are lny x e and 1 1

ln x xx e y ey y

Using transformation of graph we can sketch the curves.

Hence, the required area

0

1 0ln x

ex e dx e dx

1 0

ln (putting )e

xtdt e dx x e t

1 0ln 1 1 2

e xt t t e

132. The given curve is 2 2 3 2 ..... 1a x y a y

It is symmetrical about y-axis and it cuts the y-axis at the points (0, 0) and (0, 2a).

The curve does not exist for y > 2a and y < 0.

The required area 2 areaOBA

2

02

a

xdy

O2-2

1, 3 B

2y 3 x

2y 4 x



3/2

2

0

22 dy,from(1).

a y a y

a

22 sin , ,Putting y a we get

/2 3/2 3

0

22 sin . 2 .cos .4 sin cosa a a d

a

2 3.1.132 . .

6.4.2 2a

by Wallis formula

Thus, the required area 2a .

133. The ellipse 2 25 6 2 1 0y xy x is centered at origin, with slanted principal

axes.

On solving the equation for y, we have

2 2

26 3 20 2 1 3 5

10 5

x x x x xy

2,5 0 5 5yis rea x x

If 5, 3 5x y

If 5, 3 5x y

The required area

5 2 2

5

3 5 3 5

5 5

x x x xdx

5 5

2 2

05

2 45 5

5 5x dx x dx

Put 5sin , 5 cosx dx d

When 0, 0x

When 0 5,2

x

Hence, the required area

2

2

0

45 5sin 5 cos

5d

2

2

0

14 cos 4 .

2 2d

134. 1and 0If x y then1 2,0 1x y

2, 1 1,2 , 1,1x y

If 0, 1x y , then 1,1 , 2, 1 1,2x y .



Area of the required region 4 2 1 1 1 8

135. 5x y x y For 0, 5x y x y

0 1,5 6x y x y

Similarly for 1 2,4 5x y x y and soon.

The required area

= area of rectangle (ABCD+DEGF+GHJI)

1 3

3 .1.12 2

square units

136. Differentiating, we have

Putting this value in the given equation, we have

Replacing by we have

constant. Which is the required family of orthogonal

trajectories.

137.

138. P – Population, y – population after ‘t’ years

1 1 1 1n n n ndy dxa nx a nx

dx dy

1n ndxnx y x

dy

dy

dx,

dx

dy

dxny x

dy

2 20nydy xdx ny x

1

2log

logdy x

y x xdx x

2loglog 1log

loglog2 22.

xx xdx x

xxI F e e e x

1log

2

1log

2

. .x

x

G Sisx y dx

y x x c

logdy dy dy

y ky kdt y kt cdt dt y

0& log 0t y p p c



139. The point on y-axis is .

According to given condition,

putting we get

x 1 – = x (as f (1) = 0).

140. Put x r cos , y rsin

2 2 2 yx y r , tan

x

xdx ydy r dr

2

2

x dy ydxsec d

x

given equation2 2

2 2

r dr a r

r d r

2 2

drd

a r

1 rsin C

a

r a sin C

141. 3 2 2 2 22x y dy (1 y )(x y y 1) dx 0

2

2 2 2 3

2y dy y 1 1

dx x(1 y ) 1 y x

Put 2

2

yu

1 y

2 2

2y dy du

dx(1 y ) dx

3

du u 1

dx x x

2 2 22

1u.x dx C x y (Cx 1)(1 y )

x

142. Equation of all conics whose centre lies at origin, is

2 22 1ax hxy by …… (i)

Differentiating Eq. (i) w.r.t.x, we get 1 12 2 2 2 0ax hxy hy byy

1 1 0ax h y xy byy

0& 2 log2 50 logt y p p k p

log 22log 50

50

pk k

p

?& 3 logt y p kt y c

log 2log3 log

50t p p

2

log350log 3

log 2 / 50t

dy0, y x

dx

x x dyy

2 2 dx

dy y2 1

dx x

yv

x

dvv 1

dx

yln 1 ln | x | c

x

y

x



Multiplying by x equation becomes,

2 2

1 1 0ax h xy x y bxyy …….(ii)

Subtracting Eqs. (i) and (ii), we get

2 2

1 1 1h xy x y b y xyy

1 1hx by y xy hx by

1 1hx by y xy

1

1hx by

y xy

…… (iii)

Again, differentiating w.r.t.x, we get

Or 1

1hx by

y xy

……. (iv)

From Eqs. (iii) and (iv), we get

2

1 21 2

1

y xy x yb y xy

y xy

2

1 2

3

1

y xy x yb

y xy

……. (v)

Again, differentiating both the sidesw.r.t.x, we get

221 2 21 1 2 3

3 4

1 1

330

y xy x y xyy y xy x y

y xy y xy

2 2

2 3 1 2 1 23 3xy x y y xy xy y xy x y

Hence, (1) is the correct answer.

143. 2 2dy

x y adx

Put 1dy dt

x y tdx dx

or 1dy dt

dx dx

Eq.(i) reduces to

2 2 2dtt a t

dx , separating the variable and integrating.

2 2

2 2 2 21

t adx dt dt

a t a t

1tant

x t a Ca

1, tanx y

ie x x y a Ca

1, tanx y

ie y a Ca

, is the required general solution.

144. The given differential equation can be written as

2 2

2 2 2 24 4 . 2 1dy dy dy

y x xy y xdx dx dx



2

11

2

dy y y

dx x x

…… (i)

Let dv dy

y vx v xdx dx

Eq.( )becomesi

211

2

dvv x v v

dx

or 21

12

dv dx

xv

22 log 2 logv v xC

2 22

2 log logy y x

xCx

, putting x = 1 and y = 0

2

2C

Curves are given by 12 2

22

2y y x

xx

Hence, (1) is the correct answer.

145. The given equation can be written as

2

2 2 22 2.

xdx ydy ydx xdy y

y xx y

2 2

2 2 22 2

12

/

d x y xd

x y yx y

Integrating both the sides, we get

2 22 2

1 1 1

/

yC C

x y x x yx y

146. Areaof OBPO m

Areaof OPAO n

0

0

x

x

xy ydx m

nydx

0

x

nxy m n ydx

Differentiating w.r.t.x, we get



dy

n x y m n ydx

dy

nx mydx

.m dx dy

n x y , integrating both the sides

/m ny Cx

147. Let the family of curves be y f x

'tan

'

l PP

l TP

'tan

'

l PP

l TP

l (subtangent)

'

f x

f x

' 2

y x y

y

(given)

2

'y

yx y

2dy y

dx x y

2

dy x y

dx xy

…… (i)

It is a homogeneous differential equation.

Put x vy

Differentiating w.r.t. y, we get

dx dv

v ydy dy

…… (ii)

In Eq. (u) replacing dx

dyby Eq. (ii), we get

1

2 2

dv vy y vv y

dy y

1 1 2 1

2 2 2

dv v v v vy v

dy

2

1

dydv

v y

Integrating, 1 1

2log 1log log 0

1

vy C C

12log 2log log logy x y y C



2

1log log logy x y C

2

1log log log ,wherelog logy x y C C C

2

log logy x yC

2

x y Cy , is the required equation of family of curves.

148. dy

k ydx

; when t=0 ; y = 4

0

4 0

tdyl dt

y

0

42

15

ty kt

0 415

t

60mint

149. 21 ydy

dx y

2

2

1.2 1

21

ydy y x C

y

22 21 1y x C y x C

2 2 1x C y

Therefore, the differential equation represents a circle of fixed radius 1 and

variable centres along the x-axis

150. Given equation is,

2 0ydx xdy xy dx

Which could be converted into exact from

ie, 2

0ydx xdy

xdxy

2

02

x xd d

y

Integrating both the sides, we get

2

constant2

x x

y

or 2

2

x x

y

151.

ln

ln

xf x

e x

2

2

1 1ln ln

'ln

ln ln

ln

e x xx e x

f xe x

e x e x x x

e x x e x



0 0,on since 1 e

f x decreases on 0,

152. The given curve is 3 23 12y x y

2

2

23 6 12

4

dy dy xy x

dx dx y

For vertical tangents 214 0

0

dyy

dx

2

2

2

24 8 162,

3 3

4 3

24 82,

3

y

For y x

x

Fory x ve not possible

Re . 4 / 3,3q ,

153

11

154. 3 2 2,0f x x bx cx d b c

2' 3 2f x x bx c

Discriminant 2 24 12 4 3 0b c b c

' 0f x x R

f x is strictly increasing x R

155.



156. The given function is

3

2/3

2 , 3 1( )

, 1 2

x xf x

x x

The graph of y f x is as shown in the figure .From graph clearly there is one local

maximum ( at 1x ) and one local minima (at x=0)

total number of local maxima of minima =2

157.

158.



159. let maximum of f occurs at C. so ' 0f c APPLY LMVT for 'f x

' '

''0

0,0

f c ff c

c

This implies ' 0 5f c --------(1) similarly apply LMVT FOR 'f x ON , 4c

To get ' 4 5 4f c ------(2)

Adding (1) & (2)

' '0 4 20f f

160. 2y ax bx c pass through , ,P Q S , so

1a b c for area to be maximum tangent at ,R s t should be parallel to QS

Then 1 7

,2 4

R

161. 0f x has strictly increasing

162. clearly cos 0d

f x xdx

.cosso g x f x x is non increasing

And has roots 2 1

,2

nn Z

g is non-increasing

0 0g f x 5

03

f

163.



164.

165.

166. CONCEPTUAL

167.

168.

169.



170. Given that

3 3

2

2

450 / min 50

3

4 50

50 1/ min

184 15

dv dcm r

dt dt

drr

dt

drcm

dt

171.

172.



173

174.

175. 3 23dy

Let y x px q x pdx

2

2

2

2 2

2 2

3 3

0 3 03

6

minimumat x maxima at x3 3

x

p px x

dy pFor x p x

dx

d yx

dx

d y d yve and ve

dx dx

p py has and



176 7 5 314 16 30 560Let f x x x x x

6 4 2' 7 70 48 30 0,f x x x x x R

f is an increasing function on R

Also lim limx x

f x and f x

The curve y f x crosses x- axis only once

0f x has exactly one real root

177.

178. Since tangent is parallel to x – axis ,

3

80 1 0 2 3

dyx y

dx x

Equation of tangent is y-3=0(x-2)=>y=3

179.

180. CONCEPTUAL

181. Equation of the plane containing 1A(x 2) B(y 1) C( 1) 0L z

Where 2 0; 0A C A B C

2 , 3 ,A C B C C C

Plane is 2(x 2) 3(y 1) z 1 0 or

2x 3y z 2 0

2 2

p | |714

Hence



182. The planes are 2y z 0,5x 12y 13 and 11

3 4 10 2

x z solving we get z

183. Taking dot product with a on both the sides, we have . . [ ]r a b c a abc

So 1 1 1

. . . .2 2 2

r a Similarly r band r a

1

( )2

Hence b c r a

1 1 1 3

| || | cos 2(| | | | | |)2 4 2 22

r a b c a b c

184. ( ) 0a u a b a u b

u b ta for some t

u b ta (2 ) (1 2 ) (1 3 )u t i t j t k

. 0 2 2(1 2 ) 3(1 3 ) 0Now a u t t t

21 3 1. ( ) ( ) 2 | | 5

2 2 2t Henceu i k and u .

185. Let A be the origin of reference and the position vector of B,C,D , , . . . . , be b c d wr t A So AB b CD d c,AD d,BC c b,AC c and

. . . . .( ). BD d b The L H S is equal to b d c The R.H.S. is

2 2 2 2K | d | | b | | c | | d b |c

K[d.d c.c b.b 2c.b .c.d d b.b 2d.b]c

1

2K[b.(d c)]. K2

Hence

186. . . . cosa b b c c a

1

( )3

g a b c

21 1| ( ) 3 6cos

3 3g a b c

1 2

1 2cos33

1

cos3 2

since

187. 1 2 32 2 3 1x x x

2 2 2

1 2 3 1 2 3| 2 2 3 | 4 4 9x x x x x x

2 2 2

1 2 3

1

17x x x

188. Let PV of A, B and C be 0,b and c , respectively

Therefore, 3

b cG

1 1,2 2

b cA B

1

1

1 1 1Δ

2 2 3 2 12AB G

b c cAG AB b c



1

1 1 | 1Δ

2 2 3 2 12AAG

b c bAG AA b c

1 1

1 1 1 1Δ Δ

6 3 2 3GA AB ABCb c b c

1

Δ3

Δ

189. ( )

1 ( )| |

p qb a

p q

| | cos60 1 2b a AB

190. c ma nb p(a b)

Taking dot product with , and b we ea hav m n cos

| | | cos cos p( ) | 1c a b a b

Squaring both sides, we get

2 2 2cos cos 1p

21 p

cos2

1 1

Now cos ( )2 2

for real value of

3

cos4 4

191. OP a cos sint b t

2 2 2 2 2| OP | | a | cos | | sin . sin 2t b t a b t

1 . sin 2a b t

max | | 1 . ,4

M OP a b t

a

2

bOP

u | |

a bis the unit vector along OP

a b

192. | | 0Let AC

1 5 | AC | 3| AB| 5 | | AD |Then from

| | 5AB

CDLet be the angle between BA and

BACD b (d c)

cos| BA | CD | | b || d c |

Now . . .b d c b c b d



2

| || | cos | || | cos3 3

b c b d

1 1

(5 )( ) (5 )(3 )2 2

2 2

25 1510

2

Denominator of (i) b d c

Now 2 2 2| d c | | d | | | 2c.dc

2 29 2( )(3 )(1/ 2)

2 2 210 3 7

Denominator of (i) 2(5)( 7 ) 5 7

2

2

10 2cos

5 7 7

193. P [ ] [ ] [ ]xyz y xyz z xyz x

2 P ( ) [ ]Now xyz x y ………(i)

2P ( ) [ ] y z xyz ………(ii)

2P ( ) [ ] x z xyz ……….(iii)

28( ) ( )[ ]xyz

2[ ] 8xyz

194. / Let the angle betweena and b is and a b and c is

[ ] 6 sin cos 1abc

So 0 0sin 1,cos 1 90 , 0

, , a b c are mutually perpendicular

4 0 1

1[ ] 0 0 9 0

.

1 . 1

again bcdc d

c d

3 3

.2

c d 0we havead

2

1 0 0

3 3 27 9| . | 0 9 9

2 4 4

3 30 1

2

a c d

2

( )a c d

2

2 3 3 27| ( . ) ( . )

2 4a d c c d a a

So 2 2 36| . | | ( ) | 9

4a c d a c d



195. Without loss of generality, let the right angled OAB be such that 1OA OB

unit .Along OA take unit vector as i and along OB take unit vector as j

So, OA iOˆ ˆB j

j

AD OD OA i2

i

BE OE OB jˆ

ˆ2

Let θ be the acute angle between the medians

. 1 4

cos5 5

4

AD BE

AD BE

4sec 5

196. d ( ) 0 .( ). We have a b and d b c Therefore d is

0. perpendicular to a b and b c So vector d is

( ) ( )scalar multiple of a b b c

2 2 2 2 2| d | | | | | | | sin6

b c

a b

2 2 2 2 2 1| | | | | | 2 )(| | | | 2

4a b a b b c b c

2 21| | .3.4 3 | |

4x

197. 2 2| | | ( ) |a b a c a b c

2 2| | | |a b c

2 2

. a b c b c

2 2| | | | 2 | | cos 13

b c b c

‖

198. , , . . Pointsa b c and d are coplanar Therefore

sin 2sin2 3sin3 1

| sin 2sin 2 3sin3 |Now

2 2 21 4 9 sin sin 2 sin 3

or 2 2 2 1sin sin 2 sin 3

14

199. 2 21 9( ) 6( ) 4 | |a b a b a

2 2| | 9 | | 4 47b a b a b

Or 1 4 4 36 4cos 47

Or 1

cos2

2

, 3

Hence the angle betweena and b is

.



200. 2| 3 | 16a b c

2 2 2

1 2| | | | 9 | | 2cos 6cosa b c

3 36cos 16, [ / 6,2 / 3]

Or 1 2 32cos 6cos 5 6cos

Or 1 2 max(cos 3cos ) 4

201. ˆˆ ˆd a d b d c

( ) (1ˆ ˆ ˆˆ ˆ ˆ ) (1 )ˆ â b a c b c b c

Or ˆ ˆˆ ˆ ˆ1 ˆb c a b a c

Or ˆ ˆ1 ˆ ˆ 0ˆ â b b c a c

Or ˆ ˆˆ ˆ ˆ1 ( ) 0a b b a c

Or ( ) (ˆ ˆˆ ˆ ˆ) 0â a b b a c

Or ( ) (ˆ ˆ ˆˆ ˆ ˆ ) 0 ˆ â c a b Hencea cis perpendi

( ) . ., ˆ .ˆcular to a b i e The triangle is right angled

202. ( )c a b

Or c c a b c

1

3

2 2 2 | | | |Also c a b

2 2 2 21 1 1( sin ) 2 3sin

3 9 9a b

Or 2 1sin

2 Or

4

203.Given that 1 2 0a k b k c Now,

1| ( ) ( ) |

( ) 2 4

1( )| ( ) |

2

b a c aArea PQR

Area OQRb c

1 2 1 2|{ 1 ) } { (1 ) }|4

| |

k b k c k b k c

b c

1 2 1 2(1 )(1 ) 4k k k k

Hence 1 2 3k k

204. Given 2

2| |2 2

u v u vu v u v

2 2sin sin2

, Where is the angle between u and v

2 2 24sin cos sin2 2 2



2sin 02

or 2 1

cos2 4

0 (is not possible) 2 1cos

2 4

1 2

1 cos co1

s2 2 3

But 0 as u and v are non-collinear. So

22 20 2 3

sin90 sin4

u u v u u v u v

205. Given 2 3c a b b

. . 2 3b c b a b b

2

. 3b c b

Now 22

2 2 .a b a b a b

16 4 12 and 2

2 2 3c a b b

2

2 26 . 2

4 9b a b

c a b bzero

48 144 192

8 3c

Now ,

2

3 3 3 4. 3cos

28 3

b bb c

cb c b c

Hence 5

6b c

206.

207. 1; 3b c d shortest distance between AB and CD is 2

Also 3

b c d

equation of AB, r b …….(1)

and equation of CD: r c c d …….(2)

n b c d

...

ˆ

c b c dc nS D projectionof c on n

n b c d



sin / 3

c b c b c d

b c d

02

1. 3. 3 / 2

b c d

3b c d

Volume 1 1 1

.36 6 2

b c d

208.

209. Given 1

| || || | 1( )( 2) 12

a b c

Also, | [ ] | 1 [ ] | | || || | , ,ab c ab c a b c a b c are

Mutually perpendicular vectors .

Also, (( ) ( ) ) ( ) ( )a c a c b a c a a c c b

(( ) 0) (( ) ( ) )a a c a c c c c a b

22 c c a b c 2a b a

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

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

1 1 13

4(1) 2 62 2 2

Alternatively: For objective,

Put ˆ

2

ˆˆ, , 2j

a i b c k .

210.

211. 20

1 0 1

a a c

c c b ab c

212. 1 2 1 2 1 2 0a a bb c c

213. 2 71 3 4

1 2 0 5

214. If 0y then 1, 1x z

215. 1 13

4 9 36Dist

216. ABC form a right angled isosceles triangle

Possible value of 2

or 0



217. 2

11 5 6 43

mm n m n m n

n

218. Given lives are parallel and 1 2 3 1 2 3 1 2 1 2 1 2, , 1,2,3 ; , , 4,1, 2 0a a a b b b a a bb c c

21Dist

219. Eqn.of plane is 2 3 4 5x y z ⋋ 6 0x y z 1,1,1 lies on the plane

⋋14

3

220. .

cos , 2 3 6 , 10 2 11a b

a i j k b i j k

a b

221. 2 2 , 3 2 2b i j k d i j k

6 2 2 , 4a i j k c i k

then

.. 9

a c b dS D

b d

222. The Drs 1, 1, 1a b c and 2,3,6 are perpendicular 2 3 6 2 3 6a b c

223.

2

0 2 0

2

K K K

a b c d a b a b c d a b a b a b c d

224. Drs of one diagonal is (1,1,1) and edge is (1,0,0) 1

cos3

225. 1x y z

a b c

2 2 2 2

1 1 1 1

p a b c

226. r a

⋋ b

, 1,2,3 , 1,2, 5a b

227. Given lines are

2 3 4 5 6 2 3r i j k i j k i j k

And 7 4 3 7 4 3 2 36

r i j k i j k i j k

Point of intersection is 2 3i j k

228. 3AG

1

44

3AG AG

229. Given points form an equilateral triangle

230. 2 2 2 4

cos cos tan 1 29 9 9

16 25

19 9



231. 1 1

1 02 2

232. 1 1 1

2 2 2

6 3

20 10

ax bx z dm

n ax by z d

233. Any point on the line is (⋋,2⋋+1, 3⋋+2) Foot of perpendicular is (1,3,5) image= (1,0,7)

234. ' 3,4,5 , ' 5,3,4P Q

Equation of P’Q’ is 3 4 5

2 1 1

x y z

(3,4,1) does not lie on the line

235. The plane and the line are parallel

Dist from 2, 2,3 5 5 0to x y z is 10

27

236. Vertices are (0,0,0),(1,2,3),(2,1,3)

Area is 3 3

2

237. 2 2 22

1 2 2 1 1 2 2 1 1 2 2 1 1k m n m n n l n l l m l m

2 2sin 16

k

2 11 2

4k k

238. 6

sin7

14

9AP

28

tan3

PBPB

AP

239. 2

1 9 16cos

2 5 25 a

2 75 5 3a a

240. ; 5,7,1Drs and -10⋋-8, 7⋋-16, -27

50⋋+40+49⋋-112-27=0 99⋋=99⋋=1

241. 3 7 7 3z p q i p q

z is purely img : 7

3

qp

p, q are integers, q = 3, p = 7

2 2

7 3 3364z p q

242. 3 22 2 1 0z z z 3 21 0 z z z 2

1 1 0 z z z

21 1 0 z z z , 21, ,z w w

Common roots are 2,w w



243. Conceptual

244.

525 2

2 c s .24 3

52 .2

2 6

i cis

cis cis

argument =

19

12

, principal argument

5

12

245. 1 2 3 4 5 60, , 1 , , 1 , ,......z z i z i z i z i z i

246. Continued product of all the values of 1/n

a ib is 1

1 ,n

a ib n N

247. 6 6 6

1 2 3

1 1 1, ,

2 2 2z z z ,

1/12

1 2 3

1

2

z z z

4 4 8

1 2 3z z z

248. 5 3 2 4

2

2 2

1

1 1

249. iz w , 2018 2018 2 1 3

2 2

iiz w w

250. 4 3 25 4 3f x x x x Ax B , 0 2 4f i Ai B i ,

0 2 4f i Ai B i 4, 2A B

251.

8/3

sin icos8 8

8/3

8/3i Cos isin8 8

1/3

cos isin

1/3

1 1

252. ' 2 & ' 2sp s p a ss a

253. 1 1 1 1 1...... 1i i 2013 2i

255. Conceptual

256. Conceptual

257. 1 2 1 2z z z z

Quadrilateral formed by 1 2 20, , ,z z z z is

rectangle or square , Adjacent angle is 90

259. 2 46 4

m n

cis cis

Equating modulii we get m = 2n

2

6 4

n n

cis cis

31

4

cis n

cis n

13 4

n ncis

2

12n

, 24 48n m

260. 2 3isin 1 2isin

1 2isin 1 2isin

2

2

(2 6sin ) i 7sin

1 4sin



22 6sin 0

1

sin3

261. Least value occurs when 1 2 3 4

4

z z z zz

262. 2 2

1 1 2 22 2 0z z z z ,

2

1 1

2 2

2 2 0z z

z z

11 2

2

45z

i i z ozz

(coney’s theorem)

1

2

1z

iz

1 21 2

2

90

z z

i z z oz

(coney’s theorem)

263. Observe that 8 1 7 2 6 3 5 4, , ,

9

1 2 81 1 ....... x x x x x , 9

1 2 8

1......

1

xx x x

x , put x = 2

1 2 82 2 ....... 2 511 , 2

1 3 5 72 2 2 2 511

1 3 5 72 2 2 2 511

264. Conceptual

265. z i 3

1 3

i2 2

Given determinant 23 3i 3

266. 10

1 2 91 1 ......x x x x x

10

1 2 9

1......

1

xx x x

x

10

1 2 9log 1 log 1 log log ...... logx x x x x

diff. 9

10

1 2 9

10 1 1 1 1.......

1 1

x

x x x x x

Take 1

limx

267. 1

2

zi

z ,

2z

o 1z

1 2z z

268. Let ia ib re ,

2018 2018i ir e re , Case 1 : r = 0 is one possibility(or)

Case 2 : 2017 2019 1ir e ,

2018

2017

1ier

2019th roots of a positive real number

There will be 2019 different values

Total no.of possibilities = 2020

269. z 2 i

270 Apply coney’s theorem



(0,25)15

Z

20

271. 5 51 1 1 1 1 1A B B A C A B I

272. 4 3A I A A I

273. adj A A A I

8 4 3 28 2A xyz x y z

274. 2P I P

or 3 2 2P P P P I

or 4 2 3P I P

or 5 3 5P I P

or 6 5 8P I P

275. 2

1nadj adj A A

12A x y z

1, 1, 1x y z

11

2 55C

276. 2 5 6A A I I

2 3A I A I I

2A I and 3A I are inverse of each other

277.

2 21

2 2 8

M Madj M

278. 2

1 1

1 1 0 1 2 0

1 1

k

D k k k

k

also 1

2

1 1 1

1 0 1 2

1

D k k k k

k k

279.

1 22 1 1

1 3 0

1 4

a a

D b bb a c

c c

280. 1 1 2 3

0 0 1

0 1 1 1

1 1

C C C C a

a a b

281. 1 1 2 3R R R R

1 1 2

1 1 1

2 2

2 2

x y z y y z x y C C C

z z z x y

and 2 2 3C C C



3 3

0 0 1

1 1 2

0 1

x y z y x y z

z x y

282.

2 1 1 0 1

2 0 1 0

0 0 0 02

a b c d ab cd a b ab

a b c d a b c d ab c d cd a b c d a b c d cd

cd abab cd ab c d cd a b abcd

283. 1 1 3 2 2 3,C C bC C C aC

2 3

2 2 2 2

2 2

1 0 2

1 0 1 2 1

1

b

a b a a b

b a a b

284.

2

2

2

1 1 1

1 2 3 1 2 2 2 1 2 3

1 3 3

x x

x x x x x x x x

x x

285. 1 1 1

5A

A

2 3. . 5 5 5

TAB AB A adj A A adj A

1

3

1 11

5 5

T TA AB AB AB

286. cos sin

sin cosA

cos sin

sin cosadj A

1cos sin1

sin cos1

TA A

Thus, A is orthogonal matrix.

287. Given, matrix,

1 2 2 1 2

2 1 2 ; 2 1 2

2 2 2

T

a

A A

a b b

1 2 2 1 2

2 1 2 2 1 2

2 2 2

T

a

AA

a b b

2 2

9 0 4 2 1 0 0

0 9 2 2 2 9 0 1 0

0 0 14 2 2 2 2 4

a b

a b

a b a b a b

4 2 0a b

2 2 2 0a b

2 24 9a b

2; 1a b

288. adj A P



adj A P

2

A P

16 12 12 4 6 3 4 6

16 2 6

2 22; 11

289. 1 TB A A

1 T TAB AA A A

TT T TABB A B BA

1

T TT T TABB A A A A A

Multiplying with 1A

on both sides, TBB I

.

290.

1 2 0

2 6 3 3

5 3 1

A B

and

2 1 5

2 2 1 6

0 1 2

A B

2 1r rT A T B

2 3r rT A T B

Thus, 1rT A

1rT B

291. sin 0

0 sinA

sin 0

0 sin

TA

0 02sin 0

0 00 2sin

TA A

(Null matrix)

2sin 0;sin 0; n

0,6 only one solution.

292. 2, 3, 5A B C

2 1 2 1 2 1det A BC A BC A B C

2

2 1 4 3 12

5 5

A BA B C

C

293.

4 2

; 4 2

4 2

a b c x a p

A p q r B y b q

x y z z c r

8 8

x a p x y z

B y b q a b c

z c r p q r



8 8 16

a b c

p q r A

x y z

294. 2 2

;

;ij

i j i ja

i j i j

11 12 13

21 22 23

31 32 33

0 1 2

; 1 0 1

2 1 0

a a a

A a a a A

a a a

0A (determinant of skew symmetric matrix is zero)

2

det 0adj A A

296. 3; 2A B

11 1

1 1

1det

detadj B A

adj B A

2 2

1 1 1

1 1

B A AB

2 2 2AB A B

2 2

3 2 36

297. Expanding the determinant along 1R .

2 2cos sin cos2

1 cos2

Which is independent of

298. 1 3 0

2 2 0A I

1 3

2 2

det 0A I

So, 1 2 6 0

4 1 0

4 and 1

299. Here , 0

0H

2

2

2

0 0 0

0 0 0H

0

0

k

k

kH

Then

1

1

1

0

0

k

k

kH



70

70

70

0 0

00H H

300.

3 1 1 1 2

. . 1 1 1 2 1 3

1 2 1 3 1 4

f f

L H S f f f

f f f

2 2

2 2 3 3

2 2 3 3 4 4

3 1 1

1 1 1

1 1 1

2

2 2 2 2 2

1 1 11 1 1 1 1 1

1 1 0 1 1

1 1 0 1 1

2 2 2

1 1

So, K = 1.

301. Three non-parallel lines are concurrent if 0

2 2

2 3 0 2,3, 5

3 3

k

k k

k

But for k=2, lines are parallel.

302. From the figure

22 2

1 12 1y y

2 2

1 1 14 1 2y y y

15 2y and 1

5

2y

Equation of the line from (2, 5/2) to the given base is

5

2 22

y x

Or 2 5 4 2y x

at 1y

3

24

x or5

4x

303.



2 2

2 2 1 55

52 1AD

0tan60

AD

BD or

53

BD

Or 5

3BD

5 20

2 23 3

BC BD

304.

1

2 1 1 10 3 7 10

3 2 1

x y

x y

305. From the figure,

Equation of the line

2 10y x

306. Let the slope of L be m then

03tan60 3

1 3

m

m

3 3 1 3m m

Or 3 3 1 3m m

0 3m or m

But 0m as L intersects x-axis

Hence equation of L is

2 3 3y x

Or 3 2 3 3 0y x



307. From the figure,

We have

1 12 5y x and 1

1

62 1

8

y

x

Or 1 6x and 1 7y

308. (a,0) is the center C and P is (2, -2), then 045COP . Since the equation of

POP is x+y=0, we have

2 2OP CP

Hence, OC=4.

The point on the circle with the greatest x coordinate is B.

4 2 2OB OC CB

309. Let (h, k) be point of intersection then

1h k

a b ---(i)

and 1ah bk

also it is given that 2 2a b ab --- (ii)

multiplying (i) and (ii), we get 2 2 1b a

h k hka b

or 2 2 1h k hk

or2 2 1 0x y xy

310. Chord with midpoint (h, k) is 2 2hx ky h k --- (i)

Chord of contact of 1 1,x y is 1 2 2xx yy --- (ii)

Comparing (i) and (ii) we get

1 2 2

2hx

h k

and 1 2 2

2hy

h k

1 1,x y lies on 3 4 10x y or 2 26 8 10h k h k

Therefore, the locus of (h, k) is 2 2 3 4

05 5

x y x y

Which is a circle with center 3 4

,10 10

P

. Therefore, 1

2OP .



311. 3 4

cos sin

x yr

3

cosPR

,

4

sinPS

RS=PS-PR

312

12

32tan1 4

1 22

2 2tan

8 4

3 5

r

r

313. Centre of circle is 4 1

,3 3

. Equation of circle is 2 23 3 8 2 0x y x y

Find 1 0S .

314.

r distance from origin to the line = 1

2

090AOB and

045ACB

315. Use reflection property

316.

Let 3 2

5cos sin

x y

; where

3tan

4



5cos 3x

and 5sin 2x

we have, 4

cos5

and 3

sin5

1x and 1y

centre (-1, 1)

Hence, required equation of circle is

2 2

1 1 25x y

2 2 2 2 23 0x y x y

317. Let the other extremities of the diameter through P(2,3) be 1 1,A a b , 2 2,B a b

then equation of the two circles are

1 12 3 0x a x y b y

and 2 22 3 0x a x y b y

So equation of the common chord is

2 1 2 12 3 0x a a y b b

Since AB makes an angle 6

with x-axis 2 1

2 1

1

3

b b

a a

318.

3 4 9 4 33

5 5

9 3 4 4 33

5 5

2 6 2 6

h k h kk

h k h kk

h k x y

319. Given 2

2

xymy

2x my --- (1)

Now 2

2 2 2

x y y zxy yzy

2 22xy y yz xy xz y yz

2y xz --- (2)

2

2 2

y y x yz

x x m m

(using (1))



Hence required ratio

2

12

42

yy

m

my

320. Let the equation of 1 1x y

ABa b . Let ,p h k locus of midpoint of AB.

2 , 2a h b k substitution in (1) 1

12 2 2

x ykh hy hk

h k and passes

through the point , 8,4x y

1

8 4 2 4 2 42

k h hk h k hk x y xy . The locus of p is

2 4 0xy x y

321.

6 6, 2,2

3 6P

3 12

, 1,43 6

Q

1OPm and 4OQm

322. Centre is (0, 2), radius 2, length of the perpendicular from the centre on the

line is 1

2 so

1/ 2 1cos

2 2 2

323.

2 2 21 1r r r

2 2r



324. The tangent at (1,7) on 2 6 0x y is 1 0S .

7

6 02

yx

or2 5 0x y

Q is foot of perpendicular drawn from (-8, -6) to 2 5 0x y

325. 2 22l r

22 2 3l

326.

22 16 8h h

2 216 16 64h h h

3h

So centre is (3, 0) and r=5

Equation if circle 2 23 25x y

Let 3 5cos , 5sinC

0 4 11

8 0 1 10 2sin cos 12

3 5cos 5sin 1

max 10 5 1A

327. 2 21

10 cos sin22

A

A is maximum when 1

tan3

5r

328.

1 cos 3

tan2 sin 4

329. Conceptual

330. 3

cot4

SPQ



331. Let the equation of the circle through ,a b be

2 2 2 2 0x y gx fy c …(i)

So, 2 2 2 2 0a b ag bf c …(ii)

Since circle (i) cuts 2 2 2x y k orthogonally, we have

22 0 2 0g f c k or 2c k

Putting 2c k in (ii) we get

2 2 22 2 0ag bf a b k

So, the locus of the centre ,g f is

2 2 22 2 0ax by a b k

or 2 2 22 2 0ax by a b k

332. 2 2

1 2 0S x y ax cy a

2 2

2 3 1 0S x y ax dy

The equation of the radical axis (common chord) of 1S and 2S is

1 2 0S S

5 1 0ax c d y a

Given that 5 0x by a passes through P and Q. Therefore,

1

1

a c d a

b a

21a a

2 1 0a a

It has no real roots

333. The equation of the circle is

2 2 2x h y k k

It passes through 1,1 then

2 2 21 1h k k

2 2 2 2 0h h k

Since h is real 0D

4 4 2 2 0k

2 1 0 1 2k k

334. If 3C and 1C intersect orthogonally then 2 0p p

21,0p C represents a circle

statement I is false

If 3C and 2C intersect orthogonally then 1p

for this ‘p’ 2C and 3C have equal radii

statement –II is correct

335. Equation of circle whose centre is at 3,4 and radius is equal to 3 is



2 2

3 4 9x y

Given circle is 2 2 25x y

Now, the equation of straight line is the common chord of the two circles 6 8 41x y

336. The centres of the two circle will lie on the line through 1,2P perpendicular to the common

tangent 4 3 10x y . If 1C and 2C are the centres of these circles, then 1 15PC r ,

2 25PC r . Also 1 2,C C lie on the line 1 2

cos sin

x yr

,

Where 3

tan4

. When 1r r the coordinates of 1C are 5cos 1,5sin 2 and

4 3cos , sin

5 5

When 2r r , the coordinates of 2C are 3, 1 .

The circle with centre 1 5,5C and radius 5 touches both the coordinate axes and hence lies

completely in the first quadrant

Therefore, the required circle is with centre 3, 1 and radius 5, so its equation is

2 2 23 1 5x y

2 2 6 2 15 0x y x y

Since the origin lies inside the circle a portion of the circle lies in all the

quadrants

337. Since the given circles cut each other orthogonally.

2

1 2 0g g a ….(i)

If 1lx my is a common tangent of these circles, then

2 2112 2

lg 1g a

l m

2 2 2 2 2

1 1g 1l l m g a

2 2 2 2 2

1 12 g 1 0m g l a l m

Similarly 2 2 2 2 2

2 22 g 1 0m g l a l m

So that 1g and 2g are the roots of the equation

2 2 2 2 22 g 1 0m g l a l m

2 2 2

2

1 2 2

1a l mg g a

m

[from Eq. (i)]

2 2 2 2 21a l m a m ….(ii)

Now, 1 2 2 2 2 2

1 1,

ma map p

l m l m

2 2

2

2 2

1 m aa

l m

[from Eq. (ii)]

338. Let ,P h k be the coordinates of the centre of circle 2S , then its equation is

2 2 25x h y k



The equation of 1S is 2 2 24x y and so the equation of the common chord of 1S and 2S is

1 2 0S S

2 22 2 9hx ky h k …(i)

Let p be the length of the perpendicular from the centre 0,0 of 1S to common chord

2 2

2 2

9

4 4

h kp

h k

Now, the length of the common chord 2 22 4 p

It will be of maximum length if 0p

2 2 9 0h k …(ii)

The slope of Eq. (i) is 3

4

3 4

4 3

h hk

k …(iii)

On substituting the respective value of k in Eq. (ii) we have

9

5h and

12

5k [using Eq. (iii)

The centres of circle 2S are

9 12

,5 5

C

or9 12

,5 5

C

339. Statement –I: The centre and radius of circle 2 2 1 0x y x y

are1 1

,2 2

and 3

2 respectively and the centre and radius of circle 2 2 2 2 7 0x y x y

are 1, 1 and 3 respectively.

Distance between the centres is 5 3

32 2 1 2 1 2C C r r

First circle is completely inside the second circle

Statement –II 1 7, 6C 1 8r

2 1,2C 2 3r

1 2 1 2 1 2r r c c r r

340. Since, the radical centre of three circles described on the sides of a triangle as diameters is the

orthocenter of the triangle

Radical centre = orthocenter

Given sides are 4 7 10 0x y ….(i)

5 0x y ….(ii)

7 4 15 0x y ....(iii)

and

Since, lines (i) and (iii) are perpendiculars the point of intersection of (i) and (iii) is (1,2) the

orthocenter of the triangle. Hence, radical centre is (1,2).

341. Given, 3,0P

Equation of line AB is

0 0

3 0

cos60 sin60

x yr

(say)



32

rx

and3

2

ry

Point 3

3 ,2 2

r r

lie on 2 2y x

23

3 24 2

r r

23

2 3 04 2

r r

Let roots be 1r and 2r .

Product of roots

1 2

2 3.

3 4PA PB rr

4 2 3

3

342. Let ,h k be the point of intersection of three normals to the parabola 2 4y ax . The equation

of any normal to 2 4y ax is

32y mx am am

If it passes through ,h k , then 32k mh am am

3 2 0am m a h k ...(i)

Let roots of Eq (i) be 1 2 3, ,m m m then from Eq (i),

1 2 3

km m m

a ....(ii)

Also 1 2tan , tanm m and tan tan 2 ....(iii)

1 2 2m m

From Eqs. (ii) and (iii), 32

km

a

3

2

km

a

Which being a root of Eq. (i) must satisfy it

i,e., 3

3 3 2 0am m a h k

3

2 02 2

k ka a h k

a a

3

20

8 2

k khk k

a a

2 4 0k ah

Required locus of ,h k is 2 4 0y ax

343. Let the extremities be 1,2L and ' 1, 4L

Slope of ' LL is not defined

Hence, latusrectum 'LL is perpendicular to x axis. Therefore axis of the parabola will be

parallel to x axis and tangent at the vertex will be parallel to y axis.



Let the equation of the parabola be

2

4y a x

Now, ' 4LL a

6 4 4 6a a ...(i)

Hence from Eq. (i), equation of parabola becomes 2

6y x ...(ii)

Since, L and 'L lie on Eq. (ii), therefore

2

2 6 1 ....(iii)

and 2

4 6 1 ....(iv)

From Eqs. (iii) and (iv) , we get

2 2

2 4

12 12 1

From Eq. (iii)

1 5

9 6 1 ,2 2

On putting the values of and in Eq. (ii), we get

2 21

1 6 1 3 2 12

y x y x

and 2 25

1 6 1 3 2 52

y x y x

344. Since, the point 9 ,6a a is bounded in the region formed by the parabola 2 16y x and 9x ,

then

2 16 0y x , 9 0x

236 16.9 0a a , 9 9 0a

36 4 0, 1a a a

0 4,a 1 0 1a a

345. Let A be the vertex of the parabola and AP is chord of parabola such that slope of AP is cot

Let coordinates of P be 22 ,t t which is a point on the parabola.

Slope of AP2

t

cot2

t

2cott

2 44AP t t

24t t

22cot 4 1 cotAP

22cot 4cos 4cot cosec ec

2cos4 cos 4cos cos

sinec ec

346. 0,0P 2

1 1,2Q at at and 2

2 2,2R at at

2 2

1 2 1 2

12 2

2A at at at at



2

1 2a t t 1 2 1t t

Difference of ordinates 1 22a t t

2

22 .

A Aa

a a

347. Let 2,2P at at any point on the parabola and focus is ,0a

The equation of tangent at P is 2yt x at

Since, it meets the directrix x a at K

Then, the coordinate of K is 2

,at a

at

Slope of SP 1 2

2

1

atm

a t

Slope of SK 2

2

1

2

a tm

at

2

1 2 2

12. 1

21

a tatm m

ata t

090PSK

348. Mirror image of focus in the tangent of parabola lies on its directrix

Here mirror images of 2,3 in given lines are 3,2 3, 2and

Equation of directrix is 2 3 0x y

349. Given equation of parabola is 2 4y ax

Let the coordinates of B are 2,2at at

Slope of AB 2

t

Since BC is perpendicular to AB

So, slope of 2

tBC

Equation of BC is 222

ty at x at

This line meets the x-axis at point C

Put 20 4y x a at

So, distance CD= 2 24 4a at at a

350. Let the equation y mx c be the common tangent to the curve 2 8y x and 2 2 2x y



Then, 2

cm

and 2 22 1c m

If 2m t , then 242 1 2 0t t t

t

2 1 0 1, 2t t t

Thus, 1m 2t

Hence tangents are y x c and y x c which are perpendicular to each other.

351. Normal at point 2, 2m m on the parabola 2 4y x is given by 32y mx m m . If this is

normal to the circle also, then it will passes through centre 3,6 of the circle.

36 3 2 1m m m m

Since, shortest distance between parabola and circle will occurs along common normal

Shortest distance= distance between 2, 2m m and centre 3,6 -radius of circle 4 2 5

352. Let any point on the line segment PQ is ,R then

1 1

11

and

3 1

1

( 0 as R is on segment PQ)

A point is inside parabola 2 4y x , if

2 4 0y x

2

3 14 1 0

1

3 1 3 1

2 2 01 1

5 3 1 0

3

15

So, 0 1 (Since 0 )

353. Focus of parabola is 3,5 let be angle between focal chords then

11

3tan

2 5

r

S

15tan

8

354. tangent at 1,2 to the parabola is normal to the circle and thus passes through centre ( they

cut each other orthogonally)

Therefore the locus of centre of circle is tangent equation i.e. 2 2 1 0y x , 1 0x y

355. Normal chord drawn at the point P t to the parabola 2 4y ax subtends a right angle at focus

then 2t

P t and 1Q t are ends of normal chord, then 1 1

22, 3t t t t

t

2 ,2 4 ,4P at at a a

2

1 1,2 9 , 6Q at at a a

2 25 10 5 5PQ a a a



356. If R is point of intersection of tangents then 1 2 1 2R at t a t t

Area of triangle PQR

2

1 1

2

2 2

1 2 1 2

2 11

2 12

1

at at

at at

at t a t t

= 2

3

1 22

at t

357. Let 2 2

1 1 2 22 ,4 , 2 ,4A t t B t t and 2

3 22 ,4C t t

Slope of 1 22 1AB t t and 1 2 3 0t t t

So, 3 1t

Also, 2 2 2

1 2 3 2 2

1 2

2 41

3 3

t t tt t

1 21, 0t t

2,4 , 0,0A B and 2, 4C

Hence 6,0P

358. Conceptual

359. For parabola 2 4y ax , tangent and normal at point 2,2P at at meets x-axis ar 2,0T at and

22 ,0N a at

Thus, focus S is the mid-point of NT

Also, SP=ST=SN=a+at2

2SP TN

For given data 29

2TN

29

2SP

360. Solving equation of parabola with x-axis (y=0). We get 2y a b x b c x c a = 0,

which should have two equal values of x, as x-axis touches the parabola.

2

4 0b c a b c a

2

2 0 2 0b c a a b c

Thus 0ax by c always passes through 2,1

361. 2 2

16 2

x y

,

Now, equation of any variable tangent is 2 2 2 ......y mx a m b i

So, equation of perpendicular line drawn from centre to tangent is ........x

y iim

Eliminating m , we get 4 4 2 2 2 2 2 22x y x y a x b y

2

2 2 2 26 2x y x y



362. The given ellipse is

2 2 5 21 1

9 5 9 3

x ye

2,5 / 3L,Equation of tangent at L is

21

9 3

x y

It meets x-axis at 9 / 2,0A and y-axis at 0,3B

Area of 1 9 27

32 2 4

OAB

Area of quadrilateral =27

363. Slope of 1F B slope of 2 1F B

1b b

ae ae

2

22

1 11 1

2 2

be

a

364. Here 2, 1a b ,

1

2m ;

221

4 1 22

c

So, 1

22

y x

For ellipse; 2 2

1 ......... 14 1

x y

We put 1

22

y x in (1)

2 2 2 2 0x x 2 2x or

If 1 1

2, 2,2 2

x y and x y

Points are 1 1

2, , 2,2 2

1 2 10P P

365. The given ellipse is

2 2

14 1

x y , So, 2,0 , 0,1A andB

If PQRS is the rectangle in which it is inscribed, then P=(2,1)

0,B b

' 0,B b

' ,0A a ,0A a ,01F ae ,02F ae



Then ellipse passes through P(2,1) 2 2

4 11...... A

a b lso given that, it passes through

(4,0) 2

2

160 1 16a

a

The required ellipse is 2 212 16x y

366. Equation of hyperbola is

2 2

14 9

x y ; Its focii 13,0 ;

13

2e

If e, be the eccentricity of the ellipse, then 1 113 1 1

2 2 13e e

Since ellipse passes through the foci 13,0k of the hyperbola,

therefore 2 13a

Now 2 2

1a b ae 2 213 1 12b b

Hence, equation of ellipse is

2 2

113 12

x y

Hence the point 13 3

,2 2

does not lie on the ellipse. 11 0S

367. Let coordinate at point of intersection of normals at P and Q be (h,k)

Eq. of normal to the hyperbola

2 2

2 21

3 2

x y at point 3sec ,2tanP is

2 23 cos 2 cot 3 2 ......... 1x y

Similarly, equation of normal to the hyperbola

2 2

2 23 2

x y at point 3sec ,2tanQ

is

2 23 cos 2 cot 3 2 .......... 2x y

Given 2 2

and these passes through ,h k

2 23 sin 2 tan 3 2 ............... 3h k

And 2 23 cos 2 cot 3 2 ............. 4h k

Solving (3) and (4), we get

PQ

RS

O

0,1B

2,0A 4,0

2,1



39 cos sin

6 cos sinK

Hence, ordinate of point intersection of normals at P and Q is 13

2

368.

2 2

21

16

x y

b

2 222516 ,0 ,0 3,0 7

25b b

369.

2 2

2 2

a b

a b

370. Let ,h k be the midpoint of the chord

2 2

1 11S S hx ky h k

Is a tgt to the parabola 2 2 2ak h k h

371. 0tan30

LS

SS

2

2

1

3 2

b

a e 3e

372. If ,h k be the middle point of the chord, then its equation is 1 11S S

2 2 1hx ky h k

Since (1) is a normal, it should be same as normal at point P ,

i.e., cos cot 2 2x y a

Comparing (1) and (2) , we get

2 2

cos cot 2

h k h k

a

2 2using sec tan 1 to eliminate

Hence required locus is 3

2 2 2 2 24y x a x y

373. Verification: Draw a tangent at B now

2 2 31;

25 16 5

x ye

,4P h lies on 2 2

1 0,425 16

x yP

Required line SP is 4 3 12x y

374. , : 5 1, 5 1A a b R R a b

Let 5 , 5a x b y

Set A contains all points inside 1, 1x y

2 2, : 4 6 9 5 36B a b R R a b



set B contains all points inside or on

2 211

9 4

x y

1, 1 lies inside the ellipse

Hence A B .

375. Distance between focus and vertex 3

12

a e …..(i)

Length of latus rectum

224

b

a

2 2b a ………(ii)

1 1 2a e e 1

3e

376. Eccentricity is independent of n. Thus, we have

22 2

21 1 0

/

b a be e e e

a b e a gives

5 1

2e e must be ve

377. We know that any line of the form 2 2 2y mx a m b touches an

ellipse

2 2

2 21

x y

a b

The given line is 2

2 2

2 12 1 1

1 1

ppx y p y x

p p

Comparing this equation with 2 2 2y mx a m b , we get

2

2

1

pm

p

and 2 2 2

2

1

1a m b

p

X'X

'Y

Y

1,1

1,0

0,1

0, 1 1, 1 1, 1

1,0

1,1



On eliminating m, we get

2 22

2 2

4 1

1 1

a pb

p p

for all real values of 1,1p

2 2 2 24 1 0a b p b real values of 1,1p

2 2 2 2 21

4 0 1 0 14

a b and b a and b

The eq. of ellipse

2 2

11/ 4 1

x y

1 31

4 2e

378.

1

4tan tan 12 2 4

s c s a s a s bB C

s s b s s c

1 2 5

4 3

s a s a

s a

5

6 10 63

b c a BC

Therefore, equation of locus of A is

2 251

25 16

x y

379. Equation of tangent is 2 22 4y x a b

Since this is normal to the circle 2 2 4 1 0x y x

This tangent passes through 2,0

2 2 2 20 4 4 4 16a b a b

Using A.M.G.M., we get

2 22 24

4 . 42

a ba b ab

380.

2 22 3 2 1

5 51

4 1 / 4

x y x y

' 2 4BA a

381. Let a pair of tangents be drawn from the point 1 1,x y to the hyperbola

2 2 9x y

Then the chord of contact will be 1 1 9xx yy ………………(i)

But the given chord of contact is x=9 ……………….(ii)

As (i) and (ii) represent the same line,

Therefore, the equation of pair of tangents drawn from 1,0 to 2 2 9x y

is

22 2 9 8 9x y x

2

11 1sinu g SS S



Or 2 29 8 18 9 0x y x

382.

2 22 2

14 2

x y

Now B is one of the end points of its latus rectum and c is the focus of the hyperbola nearest to

the vertex A.

Clearly, area of ABC does not change if we consider similar hyperbola with center at 0,0

or hyperbola

2 2

14 2

x y

Area of 1 1 36 2 1 1

2 2 2ABC AB BC sq.units

383. The point M is found to be 4,0 the equation of 1 2E E is 2x the

equation of 1 2P P is 4x

1 1 2 22,1 , 4,4 , 4, 4 , 2, 1E P P E

2 2

1 2 6 5 61P E ;

2 21

2 21

2 11 / 4

8 4

ME

MP

The area of 1 1 2 21

8 8 1 / 2 2 2 302

P E E P

384. Normal at 6,3P is 2 2

2 2 2 2

1 1

a x b ya b a e

x y

It passes through 6,3

2 2

2 2

6 3

a x b ya e passes through 9,0

29 2 2

6

aa e

3

2e

M

1E

2E

2P

1P



385. Foci 5,0 ; 8 3 3x y

Equation of the require line is the line joining 1 5,0S and 8, 3 3P

386.

222

2 2 22 2

sec 1 1

sec 1

a eSQ

SP ea e e

2e

387. 1 1

2tan3 3

b b

a a

2 1 4

13 3

e

2

'2

1 11

ee

'

'2

1 1 12

4 ee

388. Hyperbola and circle touch at 1 1,P x y

Equation of tangent to H at P is 1 1 1xx yy

It meets the x-axis at

1

1,0

x

; Now, centroid of PMN is ,l m

So,

1 21

1

3

x xx

l

and

211 1

3 3

xym

Curves are touch each other 1 2m m 1 2 12 1

1 1

2x x x

x xy y

So,

11

12 21 1 11 1

1

3

1 11 , ,

33 3 3

l xx

xdl dm dm

dx dy dxx x

389. Now, PQ is the chord of contact is 12y

3 5, 12 3 12 15P TR

Area of 1

2PQT TR PQ

115 6 5 45 5

2 sq.units

OX

Y

M

P

N

X'X

'Y

Y

0,3

PQR



390. 2 22 1 2 2 1

,k k

x yk k

1

2

xk

k

1

2 2

yk

k

2 2

144 2

y x

Length of transverse axis = 2 2 4 2 8 2a

Hence, the locus is a hyperbola with length of its transverse axis equals to 8 2 . 391. On squaring both sides, we get

3

1 1 1sin x log x

3

sin x ln x

There are 6 solutions 3 right side of y-axis and 3 left side of y-axis.

392. 13

xsin and 1

11

xsin

393. a b a b if ab < 0.

394. Given equation is 2sin x x a

Consider

2 2 2dy

y sin x cos xdx

Now y = sin2x,y = x +a touch each other if we have that

1

2 2 1 22 6

cos x cos x x

The point of contact is 3

6 2,

and the tangent at is

3

2 6y x



395.

33

3

3

3

31

3tan x

logloglog tan x

log log

33

23log tan x

log tan x

Let 3

23 1 1 0tan xlog y, y y y

y

1

3tan x

7

6 6x ,

396. 1cos x sin x

cos A sin A

On simplifying we get a quadratic in sinx.

Use sum of the roots product of the roots.

397. 100

1

101k

S sin kx cos k x

100 2 99 100S sin xcos x sin xcos x .... sin xcos x …………. (i)

100 2 99 100S cos x sin x cos x sin x ..... sin xcos x ……….. (ii)

(on writing in reverse order)

Adding. we get …….

100

2 101 101 101times

S sin x sin x ..... sin x x

Hence 100

101 50 1012

S sin x sin x

398. 19

2 2 3 6 2 324

tan a ,b

399. Use 2

cos ec cot cot

400. 0 0 0180 360 540, , then

4 3 22 2 2 2

sin sin sin sin

4 3 22 2 2 2

sin cos sin cos

2 1 2 1sin k

401. 0 3 0

0 0

2 0

3 20 2060 3 20

1 3 20

tan tantan tan

tan

402. 8 4 1 0cos x bcos x

4

4

12b cos x , x R

cos x

2b ,

403. 3 2 2sin x cos x cos x

23

cos x cos x

2 23

x n x



2 23

x n x

5 7

3 9 9 9x , , ,

0 0i ix x

404. 33cos mcos

33sin msin

3 3cos cos

3 3 3 3cos cos sin sin

2 21 6m sin cos …….. (1)

Also 2 2 2 6 63 3cos sin m cos sin

2 2 21 1 3m sin cos ……….. (2)

From (1) and (2) 22 m

cosm

4 2

2

2

2 9 82 2 1

m mcos cos

m

405. 2

2 2 2 3 3 3LHS tan x cot x ;RHS

Thus, given equation have a solution, if

LHS = RHS =3

Or 2 2 2 1tan x cot x &sin y

3

4 4x ,

and

3

2 2y ,

Clearly, 2 2 4x y is satisfied by 4 2

x , y

Only four pair of solutions are possible

406. 22 2 2sec x tan x

23 8 3 0tan x tan x

tan ,tan are the roots of this equation.

407.

3

1 12 2

1 12 2

y xtan tan

y xtan tan tan

2

2

3

1 3

sin y sin x

sin x sin x

408.

2 2

3 2 2

cos x cos x cos x cosa c b

b d cos x cos x cos x cos c

409. 2a sin x bcos x bcos x d a sin x bcos xcos d

2 2 24d a b cos

2 2

2

24

d acos

b

2 2

2

d acos

b



410. 5 5 sin x cos xsin x cos x

sin x cos x

5 5

1sin x cos x

sin x cos xsin x cos x

Sin2x 4 3 2 2 3 4sin x sin xcos x sin xcos x sin xcos x cos x

2

2 2 2 2 2 0sin x sin x ; 2 2sin x no solution

411. Let 6

u x

then 6 4

u ,

and then 2

3 2u ,

2

3 6tan x cot x cot u

Now 2

3 6 6tan x tan x cos x

2

2cot u tanu cosu cosu

sin u

Both 2

2sin u and cosu monotonic decreasing on

6 4,

and thus the greatest value occurs at

6u

.

412. 100tan A tan B tanC

Triangle ABC is isosceles triangle

Let A C, then 2B C

2 2 100tan A tan A

3 250 50 0tan A tan A

X = tan A

3 250 50f x x x ; 23 100 0f ' x x x

Critical points x = 0 and 100

3x

413. 24 4

cos x cos x cos x

2 2cos x sin x sin x sin x sec x

2 2 2sin xcos x cosx sin x sin xcos x sec x

2 24

sin x cos x ;x sec x

414. 1 1 2 3C C C C

2

2

2 2

1

1 1

1

cos x

f x x x sin

sin cos

2 2 1 3 3 1R R R ,R R R

2

2 2

2 2 2

1

1 1

1

cos x

f x x x cos sin x

sin cos cos x



2 2 2

2 2 2 211

2x x sin x cos sin cos

f(x) = 0 if

x=-1 or 2 2

4x cos sin ,

1x or 1

2 417. Slope of 1y sin x is – cosx. Slope of

3 3

2 2 2 2y x a for x is

3 7 7 3

2 6 6 2cos x x p ,

If 3

2 2y x a

passes through ‘p’ then

3 3

2 23 3a a

.

418. Put 3 4t sin x cos x and 0 5t

Now, given equation is quadratic in t.

With roots 2 2t b 3t b

419. Let 0 06 6z cos i sin

2 4 8

15 15

2 4 8 16

2 2 2 20

1 1 1 1

z iz iz izGE z i z i z i

z z z z

420. Cauchy – Schwartz inequality.

421.

422.



423.

424.

425.

426.



427.

428.

429.

430.

431.



432.

433.

434.

435.



436.

437.

438.

439.



440.

441.

442.



443.

444.

445.

446.

In

Tan 60° =

……..(i)

BCD

A B C

D

h

30° 60°

40x

CD h

BC x

3h x



And in ,

Tan 30° =

447. (B)Let BC be the height of the tower and CD be the height of the flagstaff

In , ….(i)

In ……(ii)

448.

ACD

40

CD h

AC x

40 3 40 3 20x h x x x

20 3h m

A B

C

D

h

y

x

BAC tanx

y

, tan 2x h

DABy

22

2

22 tan

1 tan1

x

yx h x h

xy y

y

2 2 3 2 22xy xy x y x h

2 2

2 2

x y xh

y x

tan

1

tan2

C

x

x

P

B

A4 x

tan tan 1

tan tan 2

1 tan 1tan

4 2

1 tan2 tan

2



449. (A)

450.

451. Let f(x) = (a + b – 2c) x2 + (b + c – 2a) x + (c + a – 2b)

As a > b > c a + b – 2c > 0

f (– 1) f(0) < 0

(2a – b – c) (c + a – 2b) < 0 ...........(1)

Also a > b a – b > 0 and a > c a – c > 0

2a – b – c > 0 ...........(2)

From (1) and (2), we get

c + a – 2b < 0 c + a < 2b ............(3)

Now, discriminant of given equation = (a + c)2 – 16b2 = (a + c)2 – 4b2 – 12b2< 0

(As a + c < 2b).

452.

Let f (x) = x2 + 2( + 1)x + + + 7

If both roots of f (x) = 0 are negative, then

D = b2 – 4ac = 4( + 1)2 – 4( + + 7) 0 – 6 0

[6, ) ....... (1)

Sum of roots = – 2( + 1) < 0

(–1, ) ....... (2)

and product of roots = + + 7 > 0 R ....... (3)

From (1), (2), (3) we get [6, ) (As (1), (2), (3) must be satisfied

simultaneously.)

Hence the least value of = 6

453.SBGiven 222log log

5 4x a

a x

2 22log log

5 4a a

x x ........(1)

Let 2log a

x = t, we get

t2 = 5 + 4t t2 – 4t – 5 = 0 (t – 5)(t + 1) = 0 t = 5 or – 1

2log a

x = 5 log 2

5 ax .

Also = – 1 (rejected)

1 9 tan, tan

2

tan30h x

y

3y h x

0tan60 3h

h yy

1 2

33

hy h x



Hence S contains exactly one real number.

454. We have P(x) = 2

10

4log (4.9)

3

xx =

2

10

4log 5

3

xx =

2

10

4(1 log 2)

3x x

Hence for P(x), a > 0 and D =

24

3

– 4(1) (1 – log 2) = 44

0.79

< 0

P(x) > 0 x R.

Hence A > 0 and B > 0

455. For to be meaningless, x2 – 9 < 0 x (– 3, 3) ....(1)

For 2

2

3

log ( 1)x

x

x x

to be meaningless, 2

3

x

x

0 x (– 3, 2] ....(2)

Now (1) (2) x (–3, 2]

Note that base = 1 is not possible

456. We have

A = 1 1 1 1 1 1

1 2 3 .......2 3 2 9 2 27

=

1 1 2 3......

2 3 9 27 2

S

s

Let S = , then

= 0 +

————————————

(Subtracting)

= = = = S =

A = = 8A = 3

457. Given b + (a1 + a2 + a3 + a4) = 8

or a1 + a2 + a3 + a4 = 8 – b ....(1)

Also b2 + = 16

or = 16 – b2 ....(2)

Using R.M.S. A.M. for a1, a2, a3, a4

or 2 2 2 2

1 2 3 4 1 2 3 4

4 4

a a a a a a a a

or 2 2 2 2 2

1 2 3 4 1 2 3 4( )

4 16

a a a a a a a a

2log12log10log

2

10log5logAs 1010101010

xGraph of P(x)

)9.4(log3

x4x 10

2 P(x) =

9x2

.......3

3

3

2

3

132

3

S .......

3

3

3

2

3

1432

3

S2 ..........

3

1

3

1

3

132 3

11

3

1

3

23

1

2

1

4

3

2

S

8

3

24

23

22

21 aaaa

24

23

22

21 aaaa



16 – b22(8 )

4

b [ using (1) and (2) ]

64 – 4b2 64 + b2 – 16 b

5b2 – 16b 0

b(5b – 16) 0

Thus 0 b Hence bmax = 16

5

Hence possible integral values are 0, 1, 2, 3 Sum of all is 6.

458. We have

x3 + 2x2 – 4x – 4 = 0 has roots, a, b and c. .........(1)

On replacing x by 1

x in equation (1), we get

4x3 + 4x2 – 2x –1 = 0 3 2 1 1

02 4

x x x , which has roots 1

a ,

1

b and

1

c . ........(2)

On comparing equation (2) with x3 + qx2 + rx + s = 0, we get

1 1

1, ,2 4

q r s

Hence (q + r + s) = 1

4

459. Given that p1, q1, p2 q2 are in A.P.

(p2 – p1)2 = (q2 – q1)

2

(p2 + p1)2 – 4p1p2 = (q2 + q1)

2 – 4q1q2

2 2

4 4b c b a

a a c c

2 2

2 2

4 4b ac b ac

a c

Since b2 – 4ac is the discriminant of both the equations and roots are different

b2 4ac

a2 = c2 a = c (Not possible because two quadratic equations become identical)

or a = – c = – 1

460. We have

2x = 2k –

For exactly one real solution,

2k – > 0 k > (As 2x> 0)

Hence sum = 1 + 2 + ......... + 100 = 5050

461. Let 1

1 1

1 1 i i

i i i i

H Hk k

H H H H

(Hi – Hi + 1) = k Hi Hi + 1

5

16

c

a

4

1

y

x

y = 2x

y = 2k – 14

O (0,0)

4

1

8

1



(where k is the common difference of corresponding A.P.)

= =

=

= = = 100

462. We have

+ = 0

(a + c – 2b)

As a + c – 2b 0, so ac – 2bc + ac – 2ab = 0

2ac = 2b(a + c) b =

Hence a, 2b, c are in H.P. 1 1 1

b a c .

463. [Sol.56anWe havea a + b = 6 – c and ab = 9 – c (6 – c) = (c – 3)2

t2 + (c – 6)t + (c – 3)2 = 0

Now D 0, (As t is real)

(c – 6)2 – 4 (c – 3)2 0 (c – 6 + 2 c–6) (c – 6 – 2c +6) 0

(3c – 12) (– c) 0 c [0, 4]

Now, let f(x) = x2 – (m + 2)x + 5m

Since, exactly one root of f(x) = 0 lies in [0, 4],

sof(0) f(4) < 0

5m (16 – 4m – 8 + 5m) < 0 m (m + 8) < 0 m (–8, 0)

Now, checking at end points,

For m = – 8 x2 + 6x – 4 = 0 (x + 10) (x – 4) x = – 10, 4.

Also, for m = 0 x2 – 2x = 0 x = 0, – 2

Hence, m (– 8, 0]

So, number of integral values of m = 8

464. T = 1 – x2 + 2y2 where x + y = 1 [12th,

T = 1 – x2 + 2(1 – x)2

= 1 – x2 + 2(1 + x2 – 2x)

T = x2 – 4x + 3

= (x – 2)2 – 1

Tmax = D.N.E.

Tmin = – 1]

465. Given ; ( + ); 3 + 3 are in G.P.

+ = 4; = k ; 2 + 2 = ( + ) = 4k

3 + 3 = ( + )3 – 3( + )

= 64 – 3k(4) = 4(16 – 3k)

k ; 4k ; 4(16 – 3k) are in G.P.

16k2 = 4k(16 – 3k)

4k(4k – 16 + 3k) = 0

k = 0; k = 16

7

100

1i 1ii

1iii

HH

HH)1(

100

1i

i

k

)1( 1ii

1ii

HH

HH

100

1i

i

k

)1(

i1i H

1

H

1

10010145342312H

1

H

1.........

H

1

H

1

H

1

H

1

H

1

H

1

H

1

H

1

k

1

1101 H

1

H

1

k

1

k

k100

0b2c

1

c

1

b2a

1

a

1

b2c

1

a

1

c

1

b2a

1

0)b2a(c

1

)b2c(a

1

0)b2a(c

1

)b2c(a

1

ca

ac



466. a, (a + d), a + 2d), .................... thn term

b [TN, S & P] done

b = a + (n – 1)d d = 1

b a

n

Tr or A.P. = a + (r – 1)d

= ( 1)( ) ( 1)( ) ( )

1 ( 1) 1

r b a an a r b a an r b a ba

n n n

H.P. 1 1 1

, ,........( 1)A A D A n D

where =1

A a &

1

( 1)b

A n D

A + (n – 1) D = 1

D ;

1

a + (n – 1) D =

1

b ; D =

( 1)

a b

ab n

Tn – r + 1 = 1

( )A n r D =

1

1 ( )( )

( 1)

n r a b

a ab n

= ( 1)

( )

ab n

an r b a b

.............(2)

product = ab

467. [Sol. We have x2 – 2x sin(xy) + 1 = 0

x2 + 1 = 2x sin (xy) 2 1x

x

= 2 sin (xy) x +

1

x = 2 sin (xy)

If x > 0, then x + 1

x 2 and 2 sin (xy) 2,

equality can hold good for x = 1

Hence L.H.S. = R.H.S. is possible only if x = 1 and sin y = 1

|||ly If x < 0, then x + 1

x – 2

L.H.S. = R.H.S. is possible only if x = – 1 and sin y = 1

So, we conclude that sin y = 1 y = (4n + 1)2

, n I

(4n + 1) =2

k (Given) k = , n I

Sum of all values of k in (0, 48) is

S = 1

2(1 + 5 + 9 + 13 + ........ + 93) =

1

2 ×

24

2 (1 + 93) = 564.

468. b = a3/2 and d = c5/4

let a = x2 and c = y4, x, y N

b = x3 ; d = y5

given a – c = 9

x2 – y4 = 9

(x – y2)(x + y2) = 9; Hence x – y2 = 1 and x + y2 = 9

(no other combination in the set of + ve integers will be possible)

x = 5 and y = 2

b – d = x3 – y5 = 125 – 32 = 93

469. [Sol. Let 1

1 1

x

x = x1 =

1



Similarly, x2 = 3

1

, x3 =

5

1

and so on.

Hence x1, x2, x3,......., x2005 are in arithmetic progression ,

Given 1

1 1

x

x = 2

2 3

x

x = 3

3 5

x

x = .......= 21

21 41

x

x =....... = 1005

1005 2009

x

x and

1005

1

2010i

i

x

or 1

1

1x

x

= 2

2

3x

x

= 3

3

5x

x

= .......= 21

21

41x

x

=....... = 1005

1005

2009x

x

= k (say)

k = 1 2 3 21 1005

1 2 3 1005

( 1) ( 3) ( 5) ....... ( 41) ....... ( 2009)

...........

x x x x x

x x x x

k = 1 2 3 1005

1 2 3 1005

( ...... ) (1 3 5 ....... 2009)

( ........ )

x x x x

x x x x

[13th quiz]

k = 1 + 2(1005)

2010 k = 1 +

1005

2 =

1007

2

10051

2k

21

21

41x

x

= k x21 + 41 = k x21 x21 =

41

1k =

41 2

1005

=

82

1005

470. If | x | + | y | = | x + y |, then xy 0

Hence |x + a – 3| + |x – 2a| = |2x – a – 3| (x + a – 3) (x – 2a) 0

x2 – (a + 3)x – 2a(a – 3) 0 x R

D 0

(a + 3)2 – 8a(a – 3) 9a2 – 18a + 9 0 9a2 – 18a + 9 0 a2 – 2a + 1 0

(a – 1)2 0 a = 1

471. We know tha 2

2 32 22

11(log 3)

log 3 log 2log 3 log 3 log 32 2 2 3

(Using base changing

formula)

The given equation becomes x2 – 3x + 2 = 0

= 1, = 2

Hence 2 + + 2 = 7

472. Let three terms be (r – d), (r), (r + d)

a = r – d, b = r , c = r + d

Discriminant = D = 16b2 – (4)(3a)(c) = 4(4b2 – 3ac) = 4 [4r2 – 3(r2 – d2)] = 4 [r2 + 3d2]

D > 0 (As 'r' and 'd' cannot be simultaneously zero.)

Hence the equation f(x) = 0 has two distinct solutions. ]

473. We have

=

=

Now, –––= 2) – 2 = 2

= 2– 6 = 2

1

0

3a

a

2

0

3a

a

2( ) 2 2

222

1 0 21 2 1 2

2 2

0 0 0 0 0

18( )3 3 18 186

a a aa a a a

a a a a a

a x +3a x + 3a x + a =00 1 2 33 2



474. 4 – x > 0, 1 + x > 0 x (–1, 4) ....(1)

Let | x – 1 | < 1 x (0, 2) ....(2)

The inequality implies

log2(4 – x) < log2(1 + x) 4 – x > 1 + x x < ....(3)

(1) – (3) x

Let | x – 1 | > 1 x (– , 0) (2, ) ....(4)

The inequality implies

log2(4 – x) < log2(1 + x) 4 – x < 1 + x x > ....(5)

(1), (4), (5) x (2, 4)

Finally, we have x (2, 4)

475. Conceptual.

476. Discriminant = 9 + 4n = (2k + 1)2

n = k(k + 1) – 2

Now for n [1, 300]

k {2, 3, 4, ............,14, 15, 16}

Sum of all positive integral value(s) of n =

= (88 × 17) + (8 × 17) – 32 = 1496 + 136 – 32 = 1600.

477. We have

(x2 + 2)2 – (b + 4) (x2 + 2) (x2 + 4) – (b + 5) (x2 + 4)2 = 0

dividing by (x2 + y2)2,

– (b + 5) = 0

Let = t, so we get

t2 – (b + 4) t – (b + 5) = 0 , where t ...(1)

As roots of (1) are – 1 and b + 5, so

b + 5 < 1 b < – 4

478. We have (x2 + a|x| + a + 1) (x2 + (a + 1) |x| + a) = 0

Clearly for a > 0,

x2 + a |x| + a + 1 > 0 x R and x2 + (a + 1) | x | + a > 0 x R

Hence the given equation has no real root if a (0, )

479. We have log55(x2 + 1) log5(ax2 + 4x + a)

5(x2 + 1) ax2 + 4x + a x R

x2(5 – a) – 4x + (5 – a) 0 x R

5 > a ....(1)

and D 0

16 – 4(5 – a)2 0

3

2

30,

2

3

2

30,

2

16 16

2 1

( 1) 30 ( 1) 32k k

k k k k

16 162

1 1

(16)(17)(33) (16)(17)32 32

6 2k k

k k

22 2

2 2

2 2( 4)

4 4

x xb

x x

2

2

2

4

x

x

1,1

2

1

2

9

2



(2 + 5 – a) 0

(a – 3)(a – 7) 0now

(– , 3] [7, ) ....(2)

(1) (2) a (– , 3] ....(3)

But for the domain of log5(ax2 + 4x + a)

We must have ax2 + 4x + a > 0 x R

a > 0 and 16 – 4a2< 0

(– , – 2) (2, )

domain of 'a' is (2, ) ....(4)

(3) (4) a (2, 3]

480. Domain : x > ; x 1

Case-I: If x > 1 x + <

2(x2 + 1) < 5x 2x2 – 5x + 2 < 0

2x2 – 4x – x + 2 < 0

(x – 2)(2x – 1) < 0

but x > 1, x (1, 2)

Case-II: < x < 1 then (x – 2)(2x – 1) > 0

intersection is

I II x (1, 2) = 10

481.

=

= [Using ]

=

=

=

=

482. If L.C.M. of p and q is , then distribution of factors ‘r’ is as follows:

2

5

5 1

2x

x

1

x

5

2

1/20 2

2

5

1/2

2/5

1/2

1

2 1,

5 2

2 1,

5 2

cd

ab5

47 52

4 3

1

j

j

c C

47 51 50 49 48 47

4 3 3 3 3 3c C C C C C

51 50 49 48 47 47

3 3 3 3 3 4C C C C C C 1

1 1

n n n

r r rC C C

51 50 49 48 48

3 3 3 3 4C C C C C

51 50 49 49

3 3 3 4C C C C

51 51

3 4C C

52

4C2 4 2r t s

p q



Thus, factor ‘r’ can be distributed in ways,

similarly, factors ‘t’ and ‘s’ can be distributed in ways, respectively.

Hence, number of ordered pairs are .

483. Possible solutions are

1,2,3,4,10

1,2,3,5,9

1,2,3,6,8

1,2,4,5,8

1,2,4,6,7

1,3,4,5,7

2,3,4,5,6

Hence, 7 solutions are there

484. A regular polygon of n sides has ‘n’ vertices, no two of which are collinear. Out of these

points, triangle can be formed.

Given,

or

or

or

or

or

or

485.

=

Now,

=

486. We have

2 3 1

2 5 1 2 3 1and

2 3 1 2 5 1 2 3 1 225

n3

n C

1

3 1 3;n n

n nT C T C

1 21n nT T

1

3 3 21n nC C

1 1 ( 1)( 2)21

3 2 1 3 2 1

n n n n n n

( 1)( 1 2) 126n n n n

( 1) 42n n

( 1) 7 6n n

7n

1 2 .....n

k r r r n

r r r r r

k r

C C C C C

1 2 3

1 2 31 ....r r r n

n rC C C C

1 1 2

0 1 2 ....r r r n

n rC C C C

2

1

r C

3

2

r C and so on finally 1n

n rC

1 1

1

n n

n r rC C

1 1 1 1 1

1 1 2 3 1

0

( ) ....n

n n n n n

r n

r

f n C C C C C

1 1 1 1

0 1 2 1.... 1n n n n

nC C C C

1( ) 2 1nf n

10(9) 2 1 1023 3.11.31f

0r 2r1r 2r2r 2r2r 0r2r 1r



First place can be filled in 2 ways, i.e. 4, 5

For and , total possibilities are

i.e., 34, 35, 36, 45, 46, 56

Last place ‘d’ can be filled in 2 ways, i.e., 0, 5 (N is a multiple of 5)

Hence, total numbers = , then .

487. Number of digits are 9

Select 2 places for the digit 1 and 2 in ways

From the remaining 7 places, select any two places for 3 and 4 in ways and from

the remaining 5 places, select any two for 5 and 6 in ways.

Now, the remaining 3 digits can be filled in 3! Ways

total ways =

=

=

488. If is odd,

is divisible by 4 if

Thus, , i.e., can take 49 different values.

If is even,

is not divisible by 4

As will be in the form of .

Thus, the total number of ways of selecting ‘n’ is equal to 49.

489. If we put minimum number of balls required in each box, balls left are which

can be put in ways without restriction.

490. Three elements from set A can be selected in ways. Their image has to be .

Remaining 2 images can be assigned to remaining 4 pre – images in ways. But the

function is onto, hence the number of ways is . Then the total number of

functions is

491. must win at least games. Let win games . Therefore,

corresponding number of ways is . The total number of ways is

=

=

492. Formed number can be utmost of nine digits. Total number of such numbers is

a b c dN=

a (4000 6000)N

b c 6(3 6)b c

2 6 2 24 N / 3 8N

9

2C

7

2C

5

2C

9 7 5

2 2 2. . .3!C C C

9! 7! 5!. . .3!

2!.7! 2!.5! 2!.3!

9! 9.8.7!. 9.7!

8 8

n

1 23 4 1, 5 4 1n n

2 3 5n n n 2n

3,5,7,9....99n n

n1 23 4 1,5 4 1n n

2 3 5n n n

2 3 5n n n 4 2

( 1) / 2n n2( 1)/2

1

n n

nC

7

3C2y

4242 2

7

3 14 490C

1P 1n1P n r ( 1,2,...., )r n

2n

n rC

2 2 2 2

1 2 2

1

....n

n n n n

n r n n n

r

C C C C

2

22

2

nn

nC

2 212 2

2

n n

nC

2 3 8 83 3 3 .... 3 2 3 8 9 8 8

83(3 1) 3 3 4 3 7 3 32 3

3 1 2 2



493. Let , where .

Then the given equation reduces to

(1)

Now, we have to find non – negative integral solution of Eq. (1).

The total number f such solutions is

494. For a radical centre, 3 circles are required. The total number of radical centres is .

The total number of radical axis is .

Now,

495. Total number of triplets without restriction is . The number of triplets with all

different coordinates is .

Therefore, the required number f triplets is .

496.

=

=

=

=

= where

=

=

=

497.

= Real part of

= Real part of

= Real part of

5, 5 5x p y q and z r , , 0p q r

15p q r

15 3 1 17

3 1 2 136C C

3

n C

2

n C

2 3 5n nC C n

n n n

3

n P

3 1 2n n n n

21

1

2 2 1m

mr r

r r m

m r C

21

1 2

1m

mr

r r m r

m r C

21

1

1m

m mr r r

r r

m r C C

1

1

1

1

m

mmr rr

r r

CCm r

r

1

1 1

1m

m mr r r

r r

C C

1

1

1

,m

r r

r

t t

r m

r

rt

C

1mt t

1

1m m

m

m

C C

1m

m

62 4

0 2 4 6

2 2.... ( 1)

1 32 3 2 3 2 3

nnn n

n mCC CC

12 3

n

i

1 2 3n

i

1 tan12

n

i



= Real part of

= Real part of

=

=

=

498. We have

=

=

=

Let be the general term in . Then,

For this term to be independent of , we must have

So, the required coefficient is

499.

=

=

=

=

cos sin12 12

cos12

n

n

i

cos sin12 12

cos12

n

n ni

cos12

cos12

n

n

cos

cos12

n

n

2 2

11 3

n

m

3 1cos

12 2 2

2/3 1/3 1/2

1 1

1

x x

x x x x

31/3 3

2/3 1/3 1/2 1/2

1 1

1 1

x x

x x x x

1/3 2/3 1/3 1/2

2/3 1/3 1/2

1 1 1

1

x x x x

x x x

1/3 1/2 1/3 1/21 1x x x x

10

101/3 1/2

2/3 1/3 1/2

1 1

1

x xx x

x x x x

1rT 10

1/3 1/2x x

10

19 1/3 1/2

1 1r rr

r rT C x x

x

100 20 2 3 4

3 2

r rro r r or r

410

4 1 210C

4040 30

0

r r

r

r C C

40

39 30

1

0

40 r r

r

C C

40

39 30

1 30

0

40 r r

r

C C

39 30

1 3040 r rC

69

2940 C



500.

=

=

501.

or

=

= 400 coefficient of in

=

=

502. According to the question

are A.P., so

or

or

or

or

or

503. We have b = coefficient of in

= coefficient of

= coefficient of

Hence,

504.

15

2

2

1 1x x

x x

15

3 4

2

1x x x

x

2 60

0 1 2 60

30

....a a x a x a x

x

20 20

220 20 20

20

0 0

(20 ) 20r r r

r r

r r C r C r C

20

19 19

1 19

0

20 20r r

r

C C

20

19 19

1 19

0

400 r r

r

C C

18x 19 19

1 1x x

38

18400 C

38

20400 C

14 14 14

1 1, ,r r rC C C 2

a cb

14 14 14

1 12 r r rC C C

2 14! 14! 14!

14 ! ! 14 1 ! 14 1 ! 1 !r r r r r

2 1 1

14 13 ! 1 ! 15 14 13 ! 1 ! 13 ! 1 1 !r r r r r r r r r r r r

2 1 1

14 15 14 1r r r r r r

2 1 1

14 1 15 14r r r r r r

3 12 1

1 15

r

r r r

5 9r or 3x

4

2 3 41 2 3 4x x x x

0 1

3 4 2 3 4 4 4 2 3 4

0 1(1 2 3 ) 4 1 2 3 4 ....x in C x x x x C x x x x

4

3 2 31 2 3x in x x x a

4 / 4a b

1

1 0

1 1lim . .3 lim . 4 3

5 5

n rn r t n r r

r t rn mn nr t

C C C

1 1

1 1lim 4 3 lim 5 4 1

5 5

n nn r n r n n

r rn nn nr r

C C



505.

=

=

=

=

=

506. can be formed in 8 ways.

i.e., and coefficient in each case is 1.

Coefficient of times = 8

507. Coefficient of in expansion

As given

From

1 2 3 1 0

0 1 2 1( ) .... ( 1)n n n n n n n n

nf n C a C a C a C a

2 1

0 1 2 1

1..... ( 1)n n n n n n n r

nC a C a C a C a aa

11 1

n n n

na Ca

223

1 11

3

n

a

223

1

223

3 1( )

3 1

nn

f x

2007

223

1

223

3 1(2007)

3 1

f

2008

223

1

223

3 1(2008)

3 1

f

2007 2008

223 223

1

223

3 3(2007) (2008)

3 1

f f

1

99 223

1

223

3 3

3 1

1

223

9 9

1

223

1 3

3 3

1 3

9x

9 1 8 2 7 3 6 4 5 1 2 6 1 3 5 2 3 4, , , , , , ,x x x x x x x x

9 1 1 1 ...8x 2x

3 4 5 49 50 2

2 2 2 2 21 ..... .C C C C C m 1

1[ ]n n n

r r ras C C C

2 3 4 5 49 50 2

2 2 2 2 2 2... .C C C C C C m

3 4 50 2

3 2 2....C C C m

5 49 50 2

3 2 2.....C C C m

50 50 2 50 50

3 2 2 2C C m C C

51 50 2

3 2 1 ......(1)C C m

51 50

3 2

513 1 3 1 . . .....(2)

3n C n C

51 50 2

3 2(1) (2),3 . 1and n C C m



Then the value of is 5.

508.

509.

=

=

Therefore, last two digits are 09

510.

Multiplying Eqs. (1) and (2) and equating the coefficient of , we get

= Coefficient of in

=

=

511.

512.

50 50 2

2 2

513 . 1

3n C C m

2 1

51

mn

n2 1

1 2 3 ....n n n n n

n nb C C a C a C a

2 3

1 2 3

1....n n n n n

na C C a C a C aa

2

0 1 2

1.... 1n n n n n

nC C a C a C aa

11 1

2

na

2006 2005

2006 2005

1 11 1 1 1b b a a

a a

2006 20051

1 1 1 1a aa

2051

1 1 1a aa

2005

1 a

20051

4011 4 1

2005

4014

14 7 7

23 529 530 1

7 6 27 7 7 7

0 1 5 6530 530 .... 530 530 1C C C C

7 67 7

0 1530 530 .... 3710 1 100 3709C C m

2 3 1

0 1 2 3 11 .... (1)n n n

n nx C C x C x C x C x C x

1 2

0 1 2 11 ..... (2)n n n n

n nx C x C x C x C x C

2nx

0 2 1 3 2 4 2.... n nC C C C C C C C

2nx 2

1n

x

2

2

n

nC

2 !

2 ! 2 !

n

n n

20 5

56 14

n A BBP

A n A

3

6

6 3 6 3 6 3 61 2 3

6!3!

2 3! 90 1

540 63 2 1 0

n A BBP

A n A C C C



513.

514.

515.

516.

517.

518.

519.

520.

521.

522.

525.

526.

527.

Where

528.

Odds in favour = 5 : 31

529. Conceptual

3

6

1 2 1 1

82

n A BBP

A n A

4 121 1

4 8 41 1 2

8

334

n A BB C CP

A n A C C C

1 1

18 8. .1 1 1 2

4 8 8

R P

3 4 4 57

3 8 8

1 5 1 5 7 5. . 1 1 1...... 35

6 6 6 6 6R P C

2 2

1 224

1 1

1 2 4

813

n E EEP

E n E

51

2

52 522 2

52 2 52 51 25 2 50.

26 51 26 51 663

CR P

C C

6 6 2

6 3 9 3

n A BBP

A n A

4 1.

44 11R P

7 72 5 5

2 1 11 101 6 5

6.

11

n E C CE CE G E

n E C C

59

. 99. 9999

R P n A n A B

0.1, 0.8 0.3 0.5, 0.8 0.1 0.7P A B P A B P B

0.1 0.5 1 5 22. .

0.3 0.7 3 7 21

P A B P B AG E

P B P A

1 1 1 1 1 2 1 2 1 1 5. . . . . . . .

3 3 3 3 3 3 3 3 6 3 27R P

2

117 117.

169 117 81 3671

qR P

q q

16 9 4

52 13 13P q

13 2 1

6a a a a

3 1

6 2P A

2 1

6 3P B

1 1 1 1 1 1 1 1 5.

2 3 2 3 2 3 12 18 36R P



530.

531.

533. .

534. Correct result is

So, .

535. The combined variance, is given by

.

536. Corrected variance .

537. Conceptual.

538. Mean

539. Number of boys = b, Number of girls = g.

540.

.

.

******

1

7

n A BBP

A n A

1 22

61 1

20 5

162

n E EEP

E n E

63

6 6 6 6 6 6 60 1 2 3 4 5 6

C

C C C C C C C

~ ( (~ )) ~ ~ (~ ) (~ )p q p q p q

(~ ~ ) ( )p q r s

~ ( ) ( )p q r s 2

2 2 2 2

1 1 1 2 22

1 2

n d n d

n n

5 4 1 5 5 1 11

10 2

23330 180

7815 5

e e

52 42 50 : 4 :1b g b g b g

2

18 1822

1 1

1 18 8

18 18i

i i

x xi

2

1 1 5 1 945 .9

18 18 2 4 4

3. .

2S D

sr jee mains maths question bank - Sri Gayatri

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