03_Maths sol(1)

7/30/2019 03_Maths sol(1)

http://slidepdf.com/reader/full/03maths-sol1 1/8

MATHEMATICS1. A 2. C 3. C 4. B

5. B 6. B 7. C 8. B

9. D 10. B 11. A 12. D

13. B 14. D 15. B 16. D

17. B 18. D 19. D 20. D

21. C 22. B 23. A 24. B

25. B 26. C 27. A 28. D

29. B 30. B 31. B 32. A

33. A 34. A 35. C 36. C

37. B 38. C 39. C 40. B

41. B 42. B 43. B 44. C

45. A 46. C 47. B 48. B

49. D 50. D 51. D 52. D

53. C 54. C 55. A 56. A

57. B 58. C 59. A 60. C

61. D 62. D 63. B 64. B

65. A 66. B 67. A 68. D

69. B 70. C 71. B 72. A

73. B 74. A 75 C 76. (a)

77. (c) 78. (c) 79. (d) 80. (d)

81. (c) 82. (c) 83. (a) 84. (b)

85. (c) 86. (d) 87. (a) 88. (c)

89. (a)

90. (c)

91. (a)

92. (b)

93. (a)

7/30/2019 03_Maths sol(1)


94. (d)

95. (c)

96. (c)

97. (a) 98. (a)

99. (b)

100. (d)

HINTS/SOLUTIONS

2. Given equation is

cos222

cossin22

cos

sin + cos = 4 2n2 = /4

& sin - cos = 4 2 n - 2 = - /4

3. Solve taking z = x + iy and iyxz

4. Rewrite as (1 + i)2i + (-2i)

3

5. AM GM. 2 + a 2 a2 , 2 + b 2 b2

2 + c 2 c2

(2 + a) (2 + b)(2 + c) 8 abc8 64

6. S19 = ]393[2

19]tat[

2

19191 = 399

8.

03x5x2positivepositive

2

positive

4 No real solution

9. Substitute x = x2 + x + 1 = 0

13.!4

P

!2!.3!.5

!12 413

17.n

0r

r nC)1r 2(

=

n

0r r

n

1r

1nn

0r

CCn2

7/30/2019 03_Maths sol(1)


= 2n 2n-1 + 2n = (n + 1) 2n

18.n

n

81

1012

= 1

24. For unique solution 0

52

321

111

0 8

25. A is lower triangular matrix if all entries above the diagonal vanish.

26. Since A2 = 0 A is nilpotent matrix

30. x = 1 is one solution01tansec

sec = 1

sec is the solution.

31. Apply AM GM

a + b > 2 ab

b + c > 2 bc

c + a > 2 ca

a 8

33. x3sinx5sinxsin

2

1x2cosor 0x3sin0]1x2cos2[x3sin

,3

2,

3,0x or

6

5,

6

number of solution = 6

38. a12s

r

24a4abcR

39.2

Acot

c2s2b2s2

a2s2s2

cbabac

acbcba 2

40. AsinCsinc

aCcos. Acos

Ccosc

Csin

a

Acos

Ccos. Acos Asin)CBsin(

41. AB = 20m

AQ = h cot6

= h 3 A

B

P

Q/3/6

7/30/2019 03_Maths sol(1)


Q and BQ = h3

=3

h

AB = AQ – BQ = h 3 -3

h = 20

2h = 20 3 h = 10 3 m

42. tan =2

1= Ap

AB

tan( - ) =4

1

AP

AB2

1

AP

AC

Now tan = tan{ - ( - )}

=9

2

4

2.

2

11

41

21

)tan(tan1

)tan(tan = tan-1 2/9

43. AD2 = AB2 – BD2 = a2 -4

a2

AD =2

a3

AD = h cot

Therefore 2

a3

= h cot

a =3

2(h cot )

48. Diagonals of a parallelogram bisect each other

02

1

2

54,1

2

2

2

63 y

y x

x

)0,1(

52.

5h = 20 cos + 15 cos =4

3h

5k = 24 sin sin =24

k5

124

k5

4

3h22

1)415(

k

42

)3h(2

22

55. Origin lies in the directrix of the given parabola angle between the tangents = 900

B

A

C

5, 0 (10 cos , 12sin )

32

P(h, k)

7/30/2019 03_Maths sol(1)


59. log2 sin x – log2cos x – log2(1 – tan x) – log2(1 + tan x) = -1

2

1

)xtan1(xcos

xsin2

tan2x = 1

least = /8

60. log10xy 2 xy 100(x + y)2 = (x – y)2 + 4xy

(x – y)2 + 400 400

possible smallest value x + y = 20

61. log42 – log82 + log162 + . . . . . .

=4

1

3

1

2

1+ . . . . . .

= -log(1 - 1) + 1 = -ln2 + 1

63. 1b

y

a

ta2

2

2

42

4

2

2

t1b

y

1 – t4 0 t4 1

65. (A) x y = -5 xdx

dy+ y = 0

dx

dy= -

x

y> 0 (as xy = - 5 < 0)

The Slope of the normal is negative

-b

a< 0

b

a> 0 a > 0 , b > 0 or a < 0 , b < 0

67. P q total no.1 1, 2, 3, 4, 5, 6 62 1, 3, 5 33 1, 2, 4, 5 44 1, 3,5 35 1, 2, 3, 4, 6 56 1, 5 2 = 23

70. ea(x – 1)2 = 1 x – 1 = e-a/2

x = 1 e-a/2

e-a/2

> 0; 1 –

e-a/2

can not lie in the interval (1, 2)x = 1 + e-a/2 will lie in (1, 2) if

1 < 1 + e-a/2 < 2 0 < e-a/2 < 1

a > 0

72. Let the chord be y = mx + cc

mxy= 1

3x2 – y2 – 2x 0c

mxyy4

c

mxy

coeff. of x2 + coeff of y2 = 3c – c + 2m + 4 = 0c + m + 2 + 0

y = mx –

m –

2 = 0(y + 2) + ((1 – x)m = 0this will always pass thrown (1, -2)

7/30/2019 03_Maths sol(1)


73. 7 = 1

1 + + . . . . . .

6

= 0 and | | = 1

74. Write22

2

)1r (r

1r 3r =

22

2

)1r (r

r )1r (=

22 )1r (r

1

r

1

75. P(x = r) =!r

e r

76. (a)

77. (c) Here, y x

y x

y x y x

y x y x x

cos.cos

sincos

)cos()cos(

)cos()cos(2cos

22

or y y x 22 sin)cos1(cos

or 2

cos2cos22 y

x or 22

seccos

2 y

x

78. (c)

x

x1

2

1cos

x = cos i sin . Similarly, y = cos i sin

)(sin)(cos i y

x.

79. (d)

80. (d) 10

9

4cos

10

9sin

10

9cos

2

1

=

20

17cos

20

232cos

20

23cos

the value =20

17

20

17coscos

1.

81. (c) 4254 2 x

114

5 x

114

5 xor 11

4

5 x

5

8 x or x < 0.

7/30/2019 03_Maths sol(1)


So, the solution set = ( – , 0) ,5

8.

82. (c) 2 s(2 s – 2c) = ab or 4

1)(

ab

c s s

or 4

1

2cos2

C or

2

1

2cos

C acute bemust

2

C

83. (a)

84. (b)

85. (c)

86. (d) Let the point be (t , t )

So, 4

3

1

4

1

134

22

t t

22

3

1

4

141

34

t t

87. (a)

88. (c)

89. (a)

90. (c) The tangents to the parabola ax y 42 at the points (a, 2a), (a, – 2a) are y = x + a

and y = – x –

a.

The third side of the triangle is x = a.

Clearly, these lines form a right-angled triangle whose two sides are equal.

91. (a) The distance between foci = 2ae = 4 and3

2e .

a = 3. So, 59

419)1( 222 eab .

92. (b) For the ellipse, )1(;16 2222 eaba

7/30/2019 03_Maths sol(1)


4

16 2be ae = 216 b .

For the hyperbola, )1(;25

81,

25

144 22222 eabba

4

5e ae = 3. 316 2b .

93. (a)

94. (d) x + n – [ x + n] has the period 1 and2

tanx

has the period ,

2

i.e., 2, LCM of 1, 2 is 2.

95. (c) The function ][ x x x f is same as x x f , which is many-one and, therefore,

inverse function is not defined.

96. (c) x ye x log x = y + x;

1

1

dx

dy

x .

97. (a) 2

1...

3

1

2

1

lim)1(log

lim2

232

02

2

0 x

x x x x x

x

x x x

x

e

x.

98. (a) f (1 + 0) = 1)1(lim])1[|11(|lim01

hhhh x

0)0(lim])1[|11(|lim)01(00

hhh f hh

99. (b) h( x) = min },{2

x x = x, x 0, 2 x , 0 < x < 1, x, x 1.

As x, 2 x are polynomial functions, they are continuous and differentiable in their

respective intervals of definition. So, the only doubtful points are 0 and 1. Check

continuity and differentiability at x = 0, 1.

100. (d)

03_Maths sol(1)

Documents