03_maths sol(1)
TRANSCRIPT
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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)
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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
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= 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
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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)
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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)
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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.
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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
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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)