AITS – 1 MATHEMATICS (MAIN)

PART (C): MATHEMATICS

SECTION I: (SINGLE CHOICE QUESTIONS)

This section contains 30 Multiple Choice Questions. Each question has four choices (1), (2), (3)
and (4) out of which ONLY ONE is correct.

61. The median of the set of the observations 1,3,5,7,11,13,17 is

(1) 1 (2) 7 (3) 9 (4) 17
61. (2)
Conceptual
z z 1 i 3
62. The complex numbers z1, z2, z3 are such that 1 3  form
z 2  z3 2
(1) equilateral triangle (2) isosceles triangle
(3) right angled triangle (4) collinear points
62. (1)
  
z1  z3   z 2  z 3  cis  
 3 

63. If a = 7, b = 8, c = 9 then the length of the line joining B to the midpoint of AC is

(1) 5 (2) 8 (3) 7 (4) 10
63. (3)
1
BE  2a 2  2c2  b 2
2

1 1 1
64. If x > 1, y > 1, z > 1 are in GP, then , , are in
1  ln x 1  ln y 1  ln z
(1) AP (2) HP (3) GP (4) AGP
64. (2)
Since x, y, z are in GP,
Hence, y2 = xz
 2log y  log x  log z
2  log y  1  1  log x   1  log z 
1 + log x, 1 + log y, 1 + log z are in AP.
1 1 1
, , are in HP.
1  log x 1  log y 1  log z

65. If f (x + 1) + f (x – 1) = 2 f (x) and f (0) = 0 then f  n  , n  N is

x
(1) n f (1) (2) f 1  (3) 0 (4) f (1)
65. (1)
Put x = 1 then f  2   f  0   2f 1  f  2   2f 1
Put x = 2 , f (3) = 3 f (1)….

66. If c  a  b and b  c  a then

2 2
(1) a .b  c (2) c.a  b (3) a  b (4) a b  c

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66. (4)
Conceptual

67. If the coefficient of x in the quadratic equation x2 + px + q = 0 was taken as 17 in place of 13, its
roots were found to be –2 and –15 the roots of the original equation are
(1) 4, 9 (2) –4, –9 (3) 3, 10 (4) –3, –10
67. (4)
x2 + 13x + 30 = 0

15
68. Number of irrational terms in the expansion of  2 3  are
(1) 9 (2) 7 (3) 16 (4) 10
68. (3)
15 r 15 r

 2 3    15Cr 2 2.3 2

n n2 n3
M C
69. Let D  n 2 n3 n 5 , then lim 11 233 is equal to
1 2 3
n 
 M13 
(where M ij is the minor & Cij is the cofactor of i th row & jth column’s elements)
(1) 0 (2) –1 (3) –2 (4) 3
69. (1)
n n2 n3
D  n2 n3 n5
1 2 3
 M11  3n 3  2n 5 , M13  2n 2  n 3 & C33  0
M11  C33 3n 3  2n 5
 lim 2
 2
n 
 M13   2n 2
 n3 
 3   3 
n 5  2  2  1 2  2 
n   n
  2
 0
 2   2 
n 6  4  1 n  4  1
n  n 

70. If x 2  y 2  z 2  0 , x = cy + bz, y = az + cx and z = bx + ay, then a 2 + b2 + c2 + 2abc =

(1) 2 (2) a + b + c (3) 1 (4) ab + bc + ca
70. (3)
det A = 0

71. Let A = {1, 2, 3} and R = {(1, 1), (1, 3), (3, 1), (2, 2), (2, 1), (3, 3)}then the relation R on A is
(1) reflexive (2) symmetric (3) transitive (4) equivalence
71. (1)
Reflexive but not Symmetric

72. The number of integral solutions to the system of equations x1 + x2 + x3 + x4 + x5 = 20 and x1 + x2 = 15

when x k  0 is
(1) 300 (2) 350 (3) 336 (4) 316

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72. (3)
x1  x 2  15; x k  0, r  2, n  15
No. of non negative inetegral solutions =  x  y1Cy1  16C1  16
x1  x 2  x 3  x 4  x 5  20
x 3  x 4  x 5  5  5 31C31  7 C 2  21
Total no. of solutions  16  21  336

73. A and B alternately throw a pair of symmetrical dice. A wins if he throws 6 before B throws 7 and B
wins if he throws 7 before A throws 6. If A begins the game, the probability of his winning is
11 30 31 1
(1) (2) (3) (4)
36 61 61 36
73. (2)
a = p(A getting 6)
b = p(B getting 7)
  1  a 1  b 
5
a 30
p A   36 
1   1  31 5 61
36 6

3CK
74. The range of a random variable X is {1, 2, 3, 4, ….} and the probabilities are p  X  K   ;K=
K!
1, 2, 3, 4, …., then the value of C is
(1) loge 3 (2) loge 2 (3) log 3  loge 2  (4) log 2  log e 3
74. (3)
 P X  x   1
i

75. The minimum value of the quantity

a 2
 3a  1 b 2  4b  1 c2  5c  1
, where a, b and c are
abc
positive inters, is
11.13.15
(1) 125 (2) 210 (3) 60 (4)
23
75. (2)
 a 2  3a  1 b2  4b  1 c2  5c  1
abc
 1  1  1 
  a   3  b   4  c   5 
 a  b  c 
1
x   2; x > 0
x
So minimum value of expression
= (5) (6) (7) = 210

76. The slope of the straight line which is both tangent and normal to the curve 4x3 = 27y2 is
1 1
(1)  1 (2)  (3)  (4)  2
2 2

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76. (4)
dy
x  3t 2 , y  2t 3 , t
dx
The tangent at t, (y – 2t3) = t (x – 3t2) tx – y = t3 …. (1)
The normal at t1
t1 y  x  2t14  3t12 …. (2)
t 1 t3
(1), (2) are identical   4
1 t1 2t1  3t12
1
 t 3  2t13  3t1 , t1  
t
Eliminating t1, we get t4 = 2 + 3t2  t2 = 2, t   2

n
 r3  1 
77. lim   3 
r2  r 1 
n 

1 3
(1) 1 (2) (3) (4) 2
2 2
77. (3)
n 2
 r  1  n  r  r 1 
lim     2
. 
r  2  r  1  r 2  r  r  1 
n 

3 4 5 n 1 3 7 n2  n  1
 lim . . .... . . ....
n  1 2 3 n  1 7 13 n 2  n  1
1 3
 lim n  n  1 2
n  1.2 n  n 1
1
1
3 n 3
 lim
n  2 1 1
1  2 2
n n

78. Let f (x) = maximum x  x , x   x  , where [x] is the greatest integer less than or equal to x. Then
2

 f  x  dx 
2
(1) 1 (2) 3 (3) 5 (4) 7
78. (3)
2
0 2
 f  x  dx    x   x  dx    x  x  dx  5
2
2 0

x 1
  r 
79. lim  sin   is
r 1 n  n 
x 

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

79. (4)
n 1
  r  1
lim  sin     sin x dx  2
r 1 n n 
n  0

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1
80. Let g(x) be the inverse of the function f(x) and f 1  x   3
. Then g1  x  =
1 x
1 1 3 3
(1) 3
(2) 3
(3) 1   g  x   (4) 1   f  x  
1  g  x  1   f  x 
80. (3)
Given that g  x   f 1  x 
 f g  x    x
Differentiable with respective x,
1 3
f 1 g  x   g1  x   1  g1  x    1  g  x 
f  g  x  
1

1
81.  1  2 tan x  tan x  sec   2
dx 
(1) log sec x  sec x  tan x   C (2) log cosec x  sec x  tan x   C
(3) log sec x  sec x  tan x   C (4) log  sec x  tan x   C
81. (3)
12 12
 1  2 tan x  2 tan x sec x  dx    sec2 x  tan 2 x  2sec x tan x  dx
2

 log  sec x  tan x  sec x  c

2

82. If [x] denotes the greatest integer less than or equal to x then  dx is equal to
0  ex 
 
(1) ln2 (2) e2 (3) 0 (4) 2
e
82. (1)
 2 ln 2 
0  e x  dx  0 1dx  ln 2 0  ln 2
 

 p 1
 x sin   , x  0
83. Let f  x    x , Then f(x) is continuous but not differentiable at x = 0 if
 0, x0

(1) p < 0 (2) p = 0 (3) 0  p  1 (4) p  1
83. (3)
For continuity, lim f  x   f  0   0
x 0

1
 lim x p sin  0
x 0 x
This is possible only when p > 0
1 
 0  h  sin 
p
0
 0h  1
f  x   lim
  lim h p 2 sin  
h  h h  0
h
’
For f (x) to exist p – 1 > 0  p > 1
f(x) will not be differentiable if p  1
0  p  1

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84. The range of values of ‘a’ such that the angle  between the pair of tangents drawn from (a, 0) to the

circle x2 + y2 = 1 satisfies     , is
2
(1) (1, 2) 
(2) 1, 2  
(3)  2, 1  
(4)  2, 1  1, 2   
84. (4)
Equation of pair of tangents,  a 2  1 y 2  x 2  2ax  1  0 angle between them,
2 h 2  ab 2 a 2  1
tan  
ab
 2
a 2
  
 0  a   2, 1  1, 2 

x 2 y2
85. If ,  are the eccentric angles of the ends of a focal chord of the ellipse   1 , then the
a 2 b2
eccentricity of the ellipse is
sin   sin  sin   sin  cos   cos  cos   cos 
(1) (2) (3) (4)
sin      sin      cos      cos     
85. (2)
x     y        
cos    sin    cos   passing through (ae, 0)
a  2  b  2   2 

1 1 
86. An ellipse has eccentricity and one focus at S  ,1  . If one directrix is the common tangent,
2 2 
2 2 2 2
(nearer to S) to the circle x + y = 1 and x – y = 1. The equation of the ellipse is
2 2
 1 2  1 2
(1) 9  x    12  y  1  1 (2) 12  x    9  y  1  1
 3   3 
2
 1
 x    y  1 2 1
2
2 
(3) 
2
 1 (4) 3  x    4  y  1  1
12 9  2
86. (1)
Common tangents to x2 + y2 = 1 and x2 – y2 = 1 are x  1
2 2
 1 2 1 2
 Equation of ellipse is  x     y  1    1  x 
 2 2

87. T is a point on the tangent is drawn at the point P = (16, 16) of the parabola y2 = 16x. If S be the
focus of the parabola, then  TPS can be equal to
3 1 1 1 
(1) tan 1   (2) tan 1   (3) tan 1   (4)
4 2 4 2 4
87. (3)
Slope of tangent = 1/2 and slope of PS = 4/3
4 1
  
1  3 2
TPS  tan
4 1
1 .
3 2

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1
 tan 1  
2

i  2j 2i  j  3k
88. If a 
5
and b 
14
    
then value of 2a  b . a  b  a  2b  __________ 
(1) 2 (2) 4 (3) 5 (4) 0
88. (3)
   
a  b  a  2b  2a  b (after simplification)

89. If t1, t2 and t3 are distinct, the points  t1 , 2at1  at13  ,  t 2 , 2at 2  at 32  and  t 3 , 2at 3  at 33  are collinear,
then t1 + t2 + t3 =
(1) 0 (2) –1 (3) 1 (4) 2

89. (1)
Let the three given points lie on the line lx + my + n = 0, where l, m and n are constants, them
l  t   m  2at  at 3   n  0
 am t 3   2am  l  t  n  0
For t = t1, t2, t3
i.e t1, t2, t3 are roots
Then t1 + t2 + t3 = 0

90. Consider the family of lines 5x  3y  2  1  3x  y  4   0 and x  y  1   2  2x  y  2   0

equation of a straight line that belonging to the both families is
(1) 25x – 62y + 86 = 0 (2) 62x – 25y + 86 = 0
(3) 25x – 62y = 86 (4) 5x – 2y – 7 = 0
90. (4)
5x + 3y – 2 = 0
3x – y – 4 = 0
(x, y) = (1, –1)
x– y+ 1 =0
2x – y – 2 = 0
(x, y) = (3, 4)
Required line passing through (1, –1) and (3, 4)

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