Higher Maths Booklet # 1

Higher Maths Booklet
This special booklet on higher maths aims to provide you practice on few of
the higher Maths topics asked at some of the management entrance tests.
In tests like JMET and XAT, some questions on the following mentioned
topics are covered.
Calculus
Coordinate Geometry
Trigonometry
Modern and discrete Algebra
3-D-Vector
Statistics
This booklet will provide answers to all your queries of higher maths. This
booklet contains 250 questions with detail solutions to each provided at the
end.
You are therefore required to solve all the questions provided in this booklet
to grasp all the Higher Maths Topics & their short-cuts, tricks etc.
If you have any queries, please write to us at [email protected]
Best Wishes !

PT Education, All rights reserved.
Some of the questions covered in this booklet have been taken from the
previous year tests and hence will give you a fair idea regarding the efforts
and practice needed to crack these questions.
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IC : PTpnrhm01
(1) of (48)
DIRECTIONS: For the following questions, four options are given. Choose the correct option.
1.
If the distances of 2 points P and Q from the focus of a parabola y 2 = 4ax are 4 and 9 respectively, then the distance of the point
of intersection of tangents at P and Q from the focus is
(1)
2.
x
1 + x2
1
log (1 + x + x 2 )
2
(2)
6
(2)
3
(2)
z
e
The value of the integral
1
(1)
6.
3
1
log
2
+
FG
H
IJ
K
+ xI
J
− xK
1
x
5 1 + x2
F1 + x
GH 1 + x
2
2
(4)
13
(3)
log (1 – x + x 2)
(4)
None of these
5
+... is
5
(3)
9
(4)
None of these
F
GG
H
z
1
0
I
JJ
K
xα − 1
dx , α ≥ 0
log x
log (1 + α)
6
(3)
1
(4)
2
(3)
log 3
(4)
– log 3
(3)
log (1/1 + α)
(4)
None of these
(3)
Symmetric
(4)
None of these
3
(4)
1/3
dx
is
x(1 + log x )
(2)
LM
MM
N
– log 2
and f(0) = 0, then the value of f(1) is
(2)
log (1 – α)
OP
PP
Q
Unitary
(2)
Orthogonal
The greatest value of f(x) = (x + 1) 1/3 – (x – 1) 1/3 in [0, 1] is
(1)
9.
IJ
K
5
−2 / 3 1 / 3 2 / 3
If A = 2 / 3 2 / 3 1 / 3 , then A is
1 / 3 −2 / 3 2 / 3
(1)
8.
log 2
If f (α ) =
(1)
7.
FG
H
1
x
3 1 + x2
(3)
If the system of equations, (λ – 1)x + (3λ + 1)y + 2λz = 0, (λ – 1)x + (4λ – 2)y + (λ + 3)z = 0, 2x + (3λ + 1)y + 3(λ – 1)z
= 0, has infinite solutions, then the value of λ is equal to
(1)
5.
+
6
If the coefficients of the (2r + 4)th and (r – 2)th terms in the expansion of (1 + x) 18 are equal, then r is equal to
(1)
4.
(2)
The sum of the series
(1)
3.
8
1
(2)
2
(3)
The number of solutions of the equation a f(x) + g(x) = 0 where a > 0, g(x) ≠ 0 and g(x) has minimum value 1/4, is
(1)
one
(2)
z
two
(3)
infinitely many
(4)
zero
2
10. The value of the integral
x 2 dx , where [x], denotes the greatest integral function, equals to
0
(1)
5 + (√2 + √3)
(2)
5 – (√2 + √3)
(3)
11. If m is a variable, the locus of the point of intersection of the line
(1)
(2) of (48)
parabola
(2)
ellipse
(3)
0
(4)
None of these
x y
x y
1
is a/an
− = m and
+ =
3 2
3 2 m
hyperbola
(4)
None of these
IC : PTpnrhm01
12. If
ex
= A 0 + A 1 x + A 2x 2 + ... (–1 < x < 1), then which of the following is true?
1− x
(1)
A 1 = 5/2
(2)
A 2 = 3/2
An =
(3)
n
(n + 1)!
FG 3 x
H2
13. The coefficient of the term independent of x in the expansion of (1 + x + 2 x 3 )
(1)
1
3
(2)
19
54
(4)
2
−
1
3x
IJ
K
A n +1 − A n =
1
(n + 1)!
9
is
(3)
17
54
(4)
1
4
14. The derivative of sin–1 x with respect to cos−1 1 − x 2 is
1
(1)
1 − x2
(2)
cos –1 x
(3)
1
(4)
(2)
5
2
(3)
5
18
(4)
None of these
z
π /2
15.
1 + cos x
dx is equal to
(
1
−
cos x )5 / 2
π /3
(1)
3
2
16. The determinant ∆ =
a
b
ax + b
b
c
bx + c
ax + b bx + c
0
2
9
is equal to zero, if
(1)
a, b, c, are in A.P.
(2)
a, b, c, are in G.P.
(3)
x is a root of ax 2 + 2b x + c = 0
(4)
Both (2) and (3)
(3)
–2r √pq
17. The minimum value of px + qy when xy = r2 is
(1)
2r √pq
(2)
2 pq √r
(4)
None of these
18. The distance of that point on y = x4 + 3x 2 + 2x which is nearest to the line y = 2x – 1 is
4
(1)
5
(2)
3
2
(3)
5
(4)
z
x
19. The points of extremum of the function φ( x ) = e − t
2
/2
1
5
5
(1 − t2 )dt are
1
(1)
x=±3
(2)
x=±2
(3)
x = ± 1/2
(4)
x=±1
(4)
None of these
20. If the function f(x) = x 2 + a/x has a local minimum at x = 2, then the value of ‘a’ is
(1)
8
IC : PTpnrhm01
(2)
16
(3)
18
(3) of (48)
21. A curve is given by the equation x = a cos θ +
by
(1)
θ = sin−1
F 2a + b I
GH 5ab JK
2
2
(2)
θ = tan−1
1
d2 y
1
= 0 is given
b cos2θ, y = a sin θ + bsin2θ. Then the value of θ for which
2
2
d x2
F 3a + 2b I
GH 4ab JK
2
2
(3)
θ = cos−1
F a + 2b I
GH 3ab JK
2
2
(4)
None of these
(4)
8
22. The area bounded by the curve y = 2x, x-axis and ordinates x = –2, x = 2 is equal to
(1)
2
(2)
3
(3)
4
23. If the plane x – 3y + 5z = d passes through the point (1, 2, 4), then the lengths of intercepts cut by it on the axes of x, y, z
are respectively
(1)
15, – 5, 3
(2)
1, – 5, 3
(3)
– 15, 5, – 3
(4)
1, – 6, 20
24. The direction ratios of the diagonals of a cube which joins the origin to the opposite corners are (when the 3 concurrent edges
of the cube are coordinates axes)
2
(1)
3
,
2
3
,
2
(2)
3
1, 1, 1
(3)
2, – 2, 1
(4)
1, 2, 3
(4)
2
(4)
None of these
25. Distance between parallel planes 2x – 2y + z + 3 = 0 and 4x – 4y + 2z + 5 = 0 is
(1)
2
3
(2)
1
3
(3)
1
6
26. The number of terms common to the two A.P.’s 3, 7, 11, ... 407 and 2, 9, 16, ... 709 are
(1)
10
(2)
12
(3)
14
27. If the complex numbers z 1 = a + i, z2 = 1 + ib, z3 = 0 form an equilateral triangle (a, b are real numbers between 0 and 1), then
(1)
a = 3 ± 1, b = −
3
2
(2)
a = 2 − 3, b = 2 − 3
(3)
a=
1
3
,b=
3
4
28. If z 1 and z2 both satisfy the relation z + z = 2 | z − 1| and arg (z1 − z2 ) =
(1)
0
(2)
1
(3)
(4)
None of these
π
, then Im (z 1 + z 2 ) is equal to
4
2
(4)
3
29. For all complex numbers z 1 , z2 satisfying |z 1 | = 12 and |z 2 – 3 – 4i| = 5, the minimum value of |z 1 – z2 | is
(1)
0
(2)
2
(3)
7
(4)
0.7
30. The minimum and maximum value of ab sin x + b (1 − a2 ) cos x + c; (|a| ≤ 1, b > 0) respectively are
(1)
{b – c, b + c}
(2)
{b + c, b – c}
(3)
{c – b, b + c}
31. If x = X cos θ – Y sin θ, y = X sin θ + Y cos θ and x 2 + 4xy + y 2 = AX 2 + BY 2, 0 ≤ θ ≤
(1)
(4) of (48)
θ=
π
6
(2)
θ=
π
4
(3)
A=–3
(4)
None of these
π
, then
2
(4)
B=1
IC : PTpnrhm01
32. The number of solutions of the equation cos (π
(1)
None
(2)
33. If sin –1 x + sin –1 y =
(1)
x − 4 ) cos (π x ) = 1 is
One
(3)
Two
(4)
More than two
(3)
π
6
(4)
p
2π
, then cos –1 x + cos –1 y =
3
2π
3
(2)
π
3
cos A
cos B
cos C
is equal to
+
+
a
b
c
34. In a ∆ABC, the sides a, b, c are the roots of the equation x 3 – 11x2 + 38x – 40 = 0. Then
(1)
1
(2)
FG
H
35. If f ( x ) = sin2 x + sin2 x +
3
4
(3)
IJ
K
FG
H
π
π
+ cos x . cos x +
3
3
IJ
K
9
16
(4)
None of these
(4)
not defined
and g(5/4) = 1, then (gof ) x is
(1)
a polynomial of the first degree in sin x, cos x
(2)
a constant function
(3)
a polynomial of the second degree in sin x, cos x
(4)
None of these
36. If the function f : [1, ∞) → [1, ∞) is defined by f(x) = 2 x(x–1) , then f–1 (x) is
(1)
FG 1 IJ
H 2K
x ( x −1)
(2)
1
(1 + 1 + 4 log2 x )
2
(3)
1
(1 − 1 + 4 log2 x )
2
37. The greatest value of the function f(x) = cos [x e [x] + 2x 2 – x], x ∈ (– 1, ∞) where [x] denotes the greatest integer function
(less than or equal to x) is
(1)
0
(2)
1
(3)
2
(4)
3
38. The line joining (5, 0) to (10 cos θ, 10 sin θ) is divided internally in the ratio 2 : 3 at P. If θ varies, then the locus of P is
(1)
a pair of st. lines
(2)
a circle
(3)
a st. line
(4)
None of these
(4)
None of these
39. Two vertices of an equilateral triangle are (–1, 0) and (1, 0). Its circumcircle is
(1)
F
GH
x2 + y −
I
J
3K
1
2
=
4
3
(2)
F
GH
x2 − y +
I
J
3K
1
2
=
4
3
(3)
F
GH
x2 + y −
I
J
3K
1
2
=−
4
3
40. The triangle formed by the tangent to the parabola y = x 2 at the point whose abscissa is x 0 [x0 ∈ [2, 3]], the y-axis and the
straight line y = x0 2 has the greatest area if x 0 =
(1)
2
(2)
1
(3)
3
(4)
4
41. If the lines 2 (sin a + sin b) x – 2 sin (a – b) y = 3 and 2 (cos a + cos b) x + 2 cos (a – b) y = 5 are perpendicular, then
sin 2a + sin 2b is equal to
(1)
sin (a – b) – 2 sin (a + b)
(2)
sin 2 (a – b) – 2 sin (a + b)
(3)
2 sin (a – b) – sin 2 (a + b)
(4)
sin 2 (a – b) – sin (a + b)
IC : PTpnrhm01
(5) of (48)
42. Let P be a point on the hyperbola x 2 – y 2 = a 2 where a is a parameter such that P is nearest to the line y = 2x. The locus of P
is
(1)
2x – y = 0
(2)
2y – x = 0
(3)
2y + x = 0
(4)
2x + y = 0
43. A ball is thrown from the top of the Qutab Minar 200 ft high with a velocity 80 ft per second at an elevation of 30° above the
horizon. The horizontal distance from the foot of the minar to the point where it hits the ground is (g = 32 ft/sec 2 )
(1)
200 ft
(2)
200√3 ft
(3)
300√3 ft
(4)
200/√3 ft
44. A body is in equilibrium on a rough inclined plane of which the coefficient of friction is (1/√3). The angle of inclination of the plane
is gradually increased. The body will be on the point of sliding downwards when the inclination of the plane reaches
(1)
15°
(2)
30°
(3)
45°
(4)
60°
(4)
None of these
45. If z = x + iy, then the inequality |z – 4| < |z – 2| represents the region given by
(1)
Re (z) > 0
(2)
Re (z) < 0
(3)
Re (z) > 2
46. Let z 1 and z2 be two complex numbers such that z1 ≠ z2 and |z1 | = |z2 |. If z 1 has positive real part and z 2 has negative imaginary
part, then
(1)
z1 + z2
may be
z1 − z2
zero
(2)
purely imaginary
47. If z is a complex number such that
(3)
real and negative
x – axis
(2)
straight line y = 5
(3)
a circle passing through the origin
(4)
None of these
(1)
both (1) and (2)
(4)
xy +
z − 5i
= 1, then the locus of z is
z + 5i
(1)
48. If 2 cos θ = x +
(4)
1
1
and 2 cos φ = y +
, then the value of cos (θ + φ) will be
y
x
x y
+
y x
(2)
2
LM x + y OP
Ny xQ
(3)
LM
N
1
1
xy +
xy
2
OP
Q
1
xy
49. If the function f(x) = x 3 – 6ax 2 + 5x satisfies the conditions of Lagrange’s mean value theorem for the interval [1, 2] and the
7
is parallel to the chord that joins the points of intersection of the curve with the
4
ordinates x = 1 and x = 2, then the value of a is
tangent to the curve y = f(x) at x =
(1)
35
16
50. If f ( x ) =
35
48
(2)
(3)
7
16
(4)
5
16
x
x
and g( x ) =
, where 0 < x ≤ 1, then in this interval
sin x
tan x
(1)
both f(x) and g(x) are increasing functions
(2)
both f(x) and g(x) are decreasing functions
(3)
f(x) is an increasing function
(4)
g(x) is an increasing function
z
sin2 θ
51. A function g (θ) =
z
cos 2 θ
f ( x ) dx +
0
f ( x ) dx is defined in the interval
0
FG − π , π IJ
H 2 2K
where f(x) is an increasing function, then g(θ) is
increasing in the interval
(1)
(6) of (48)
FG − π ,0IJ
H 2 K
(2)
FG − π ,− π IJ
H 2 4K
(3)
FG 0, π IJ
H 4K
(4)
FG − π ,0IJ
H 4 K
IC : PTpnrhm01
52. A box is constructed from a rectangular metal sheet of 21 cm. by 16 cm., by cutting equal squares of sides x from the corners
of the sheet and then turning up the projected portions. For what values of x the volume of the box will be maximum?
(1)
1
(2)
2
(3)
3
(4)
–1
(3)
198
(4)
199
(3)
2n
n +1
(4)
None of these
(3)
1
2
(4)
53. The greatest integer less than or equal to (√2 + 1) 6 is
(1)
54.
196
(2)
197
C 0 C2 C4 C6
cn
+
+
+
+... +
, is equal to
1
3
5
7
n+1
2n +1
n +1
(1)
55. The sum to infinity of 1 +
(2)
n =0
(1)
∞
x 3n
(3)
x 3n −1
5e
(4)
5e + 1
, then the value of a 3 + b 3 + c 3 – 3 abc is
n =1
1
(2)
FG 1 + 1 . 1
H5 3 5
log 2 + 2
59. If S =
(1)
∞
x 3n − 2
n =1
58. The value of log 2 + 2
(1)
6e–1
∑ (3n)! , b = ∑ (3n − 2)! and c = ∑ (3n − 1)!
57. If a =
2
4 11 22 37 56
+
+
+
+
+ ..... is :
1! 2! 3! 4 ! 5!
6e
∞
FG 1 IJ
H 13 K
(2)
56. The sum of the series
(1)
1 1 1. 3 1
. +
.
+... is
2 2 2 . 4 22
FG 2 IJ
H 3K
(1)
2 n +1 − 1
n +1
(2)
3
(2)
0
+
(3)
–1
(4)
–2
(3)
1
log 2
2
(4)
log 3
(3)
loge
FG e IJ
H 4K
(4)
IJ
K
1 1
. +.....+∞ is
5 55
log 2 + 2
1
1
1
1
−
+
−
+....+ ∞ , then e S equals
1. 2 2 . 3 3 . 4 4 . 5
loge
FG 4 IJ
H eK
(2)
4
e
e
4
60. The equation of common tangent of parabolas x 2 = 108y and y 2 = 32x, is
(1)
4x + 5y = 36
(2)
2x + 3y + 36 = 0
(3)
3x + 2y = 56
61. The equation of ellipse whose focus is (1, 2), directrix is 3x + 4y = 5 and eccentricity is
(1)
(x + 1) 2 + (y + 2) 2 = (3x + 4y + 5) 2
(3)
(x – 1) 2 + (y – 2) 2 =
IC : PTpnrhm01
FG 3x + 4y − 5 IJ
H 10 K
(4)
None of these
1
is
2
(2)
(x – 1) 2 + (y – 2) 2 = (3x + 4y – 5)2
(4)
None of these
2
(7) of (48)
x2 y2
+
= 1 with foci at S and S’. If A be the area of triangle PSS’, then the maximum
25 16
62. Let P be a variable point on the ellipse
value of A is
(1)
24 sq. units
(2)
12 sq. units
(3)
36 sq. units
(4)
None of these
(4)
D ≠ 0, h 2 = 0.
63. The equation ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 represents a hyperbola if
D ≠ 0, h2 < ab.
(1)
(2)
D ≠ 0, h 2 > ab.
(3)
D ≠ 0, h 2 = ab.
64. Which of the following is false?
(1)
The three co-ordinate plane divides the whole system in eight octant.
(2)
The perpendicular distance of any point P (3, 4, 5) from the z–axis is 5.
(3)
In a tetrahedron, each of the four vertices is the intersection of four lines.
(4)
The yz–plane divides the line joining the points (2, 4, 5), (3, 5, –4) in the ratio –2 : 3.
65. If a line makes the angles α, β, γ with the axes, the value of sin²α + sin² β + sin² γ is equal to
(1)
1
(2)
5
4
(3)
3
2
(4)
2
66. Which of the following is false?
(1)
The intercepts made on the axes by the plane x + 2y – 2z = 9 are 9,
9
9
and – .
2
2
(2)
The angle between two planes is equal to the angles between the normals to them from any point.
(3)
The distance of the point (2, 3, 4) from the plane 3x – 6y + 2z + 11 = 0 is 1.
(4)
The two planes 2x – y + z = 16 and – x + 2y + 4z = 1 are parallel.
67. The equation of the plane passing through (–1, 3, 2) and perpendicular to each of the two planes x + 2y + 2z = 5 and
3x + 3y + 2z = 8 is
(1)
2x – 4y + 3z + 8 = 0
(2)
x – 3y + 2z + 6 = 0
(3)
3x + 2y – z = 1
(4)
2x – 3y + 5z + 6 = 0
(4)
2x + 3y – 3z = 5
68. The equation of the plane passing through points (1, 1, 0), (1, 2, 1), (–2, 2, –1) is
(1)
x–y+z=1
(2)
2x + 3y + 3z = 5
(3)
4x – 3y + 3z = 1
69. The equation of the plane passing through the line of intersection of planes 2x + 3y – 4z = 1, 3x – y + z + 2 = 0 and the point
(0, 1, 1) is
(1)
70. If
3x + 2y – 3z + 1 = 0
(2)
x + 2y – 3z + 1 = 0
(3)
5x + 2y – 4z + 2 = 0
(4)
5x + 2y – 3z + 1 = 0
(3)
E = D/2
(4)
None of these
(3)
ab + ca + ca
(4)
None of these
b+c c+a a+b
a b c
q + r r + p p + q = D and p q r = E , then
y+z z+x x+y
x y z
(1)
D=E
(2)
D = E2
a2
a2 − (b − c)2
bc
2
b2 − (c − a)2
ca ?
c − (a − b)
ab
71. Which of the following is a factor of b
c2
(1)
(8) of (48)
a+b+c
(2)
2
a2 + b 2 + c 2
2
IC : PTpnrhm01
72. In the determinant
2
1
3
1
4
2
5
2
−7 15 8 11
1
(1)
b2
c2 + a2
c
74.
2
c
b2
LM
MM
N0
77.
–62
(4)
None of these
2
(2)
4abc
(3)
4 a2 b 2 c2
(4)
a 2b 2c 2
(2)
1
abc
(3)
ab + bc + ca
(4)
0
(3)
1, 3, 8
(4)
None of these
(3)
635 I
(4)
None of these
(4)
2
bc
ca is equal to
ab
0
3
2
OP
PP
Q
0
2 , then the characteristic roots of A are
2
2, 3, 4
(2)
1, 3, 4
76. Using Cayley-Hamilton theorem, the value of A 8, if A =
(1)
(3)
is equal to
a +b
2
abc
3
75. If A = 0
(1)
2
41
a2
abc
1 / a a2
1 / b b2
1 / c c2
(1)
3
(2)
a2
(1)
1
–18
b 2 + c2
73.
2
, cofactor of element ‘8’ is
626 I
LM l
0
The matrix M
MM n
MN−m
(2)
m
n
625 I
LM1
N2
2
OP is
Q
−1
OP
−1P
is orthogonal when
0P
P
0 PQ
0
0
0
l
−m
n
−l
(1)
l = 2/7, m = 3/7, n = 6/7
(2)
l = 3/7, m = 3/7, n = 6/7
(3)
l = 3/7, m = 3/7, n = 3/7
(4)
l = 2/7, m = 5/7, n = 8/7
(3)
0
LM
MM
N −3
OP
P
2 PQ
3 −1 2
78. The rank of the matrix −6 2 4 is
(1)
3
1
(2)
1
79. Let a and b be respectively the degree and order of the differential Equation of the family of circles touching the lines
y 2 – x2 = 0 and lying in the first and second quadrant then
(1)
a = 1, b = 2
IC : PTpnrhm01
(2)
a = 1, b =1
(3)
a = 2, b = 1
(4)
a = 2, b = 2
(9) of (48)
80. A curve passes through the point (1, π/4) and its slope at any point (x, y) on it is given by dy/dx = y/x – cos2 (y/x). The equation
of the curve is
(1)
tan
FG y IJ + log x = c
H xK
(2)
tan
FG y IJ + log x = 1
H xK
(3)
tan
FG x IJ + log x = 1
H yK
(4)
y = e tan(y+x) + 1
81. An equation of the curve passing through the point (2, 1) and for which the sum of the subtangent and the x coordinate is equal
to 1 is
(1)
(x – 1) 2 y = 1
(2)
(x – 1) y = 1
(3)
(x – 1) y 2 = 1
(4)
xy = y – 1
(3)
(cos x + 54 cos 5 x )
2
(4)
2 4 cos 2x – 5 4 cos 5x
(3)
ny
(4)
n 2y
82. The fourth differential coefficient of sin 2x sin 3x is
(1)
cos x – 5 4 cos 5x
83. If y = ax n+1 + bx –n, then x 2
(1)
(cos x − 54 cos 5 x )
2
(2)
d2 y
is equal to
dx 2
n(n – 1)y
(2)
n(n + 1)y
84. Let f(x) and g(x) be two functions having finite non-zero 3rd order derivatives f'''(x) and g'''(x) for all x ∈ R. If f(x) g(x) = 1 for
all x ∈ R, then
(1)
3
f"' g"'
−
is equal to
f'
g'
FG f" − g" IJ
Hg fK
(2)
3
FG f" − g" IJ
H f gK
FG g" − f" IJ
H g gK
(3)
3
(3)
f” (ex)
FG f" − g" IJ
Hf fK
(4)
3
(4)
f”(e x) e 2x + f’(e x ) ex
(4)
1
f x
2
85. Let f be a polynomial. Then the second derivative of f(e x ) is
(1)
f”(e x) e x + f’ (e x )
(2)
f” (e x) e 2x + f” (e x ) e x
[ f ( x + h)]2 − [ f (x )]2
is equal to
h→ 0
2h
86. For a differentiable function f, the value of lim
(1)
[f’ (x)]2
(2)
f(x) f’(x)
87. Let f : R → R be a function such that f
(1)
(3)
(3)
f(x) is continuous in R.
π
2
2
2
bg
− f x
≤x≤
(2)
f(x) is continuous but not differentiable in R.
(4)
f(x) is bounded in R.
2
π
, then
2
(1)
f(x) is not differentiable at '0'.
(2)
f(x) is differentiable at p/2.
(3)
f(x) has local maxima at '0'.
(4)
None of these
(3)
1
20
R| sinx – x + x U|
|
6 | is equal to
lim S
V|
x
||
|W
T 1
bg
FG x + y IJ = f (x ) + f (y) , f(0) = 3 and f'(0) = 3, then
H 3 K
3
f (x)
is differentiable in R.
x
88. f(x) = min. {1, cos x, 1 – sin x}, –
bg
1
f x
2
3
89.
5
x →0
(1)
(10) of (48)
120
(2)
−
1
120
(4)
None of these
IC : PTpnrhm01
xf (2) − 2f ( x )
is
x−2
90. Let f(2) = 4 and f' (2) = 4, then lim
x →2
(1)
2
(2)
1
91. Let f ( x ) =
(1)
(18 −
x2 )
–2
(3)
, then the value of lim
x →3
0
−
(2)
FG f (x ) − f (3) IJ
H x−3 K
1
9
F 1 + 4 + 9 +....+ n I
GH n + 1 n + 1 n + 1 n + 1JK
–4
(4)
3
(4)
1
9
0
is
(3)
−
1
3
2
92.
lim
n→ ∞
(1)
93.
94.
95.
z
3
3
3
3
1
is equal to
(2)
2
3
(3)
1
3
(4)
(2)
π
x2
x−
+c
2
2
(3)
x2
π
x+
+c
2
2
(4)
π x2
+
+c
2 2
(3)
– sin 2x + c
(4)
sin 2x + c
sin−1 (cos x) dx =
(1)
π
− x2 + c
2
z
sin 4x
dx =
cos 2x
(1)
– cos 2x + c
(2)
z
3x + 4 − 3x + 1
1
dx =
{
}
(1)
2
(3 x + 4 )3 / 2 + (3x + 1)3 / 2
9
(3)
2
(3 x + 4 )3 / 2 + (3x + 1)3 / 2
27
96. If
(1)
{
z
2 + 3 cos x
sin2 x
–3
ze
0
0
z
0
π log 2
IC : PTpnrhm01
(4)
2
(3x + 4 )3 / 2 − (3x + 1)3 / 2
27
(3)
0
(4)
1
7
(3)
5 log 13
(4)
2 log 5
log 2
(3)
0
(4)
None of these
–2
x log x
1 + x2
j
2
(2)
∞
(1)
}
2
(3 x + 4)1 / 2 + (3x + 1)1/ 2
9
}
(2)
∞
98. The value of
{
(2)
{
}
dx = –2.cotx + a cosec x + c, then a =
97. The value of the integral
(1)
cos 2x + c
FG
H
log x +
dx is
IJ
K
1
dx
⋅
is
x 1 + x2
(2)
(11) of (48)
z
1
99. The value of
0
(1)
ln x
1 − x2
dx is
π ln 2
(2)
−
π
ln 2
2
(3)
π
l n2
2
(2)
log2 1
+
2
2
(3)
−
(4)
None of these
(4)
None of these
(4)
69760
z
a
100. The value of
(1)
x 5 dx
is
(2a2 − x 2 )3
0
log 2 1
−
2
4
log2 1
−
2
2
101. The number of five-digit telephone numbers having at least one of their digits repeated is
(1)
90000
(2)
100000
(3)
30240
102. A class has 30 students. The following prizes are to be awarded to the students of this class. First and second in Mathematics;
first and second in Physics first in Chemistry and first in Biology. If N denotes the number of ways in which this can be done, then
N is divisible by
(1)
400
(2)
600
(3)
8100
(4)
All of them
103. The sum of all the five digit numbers that can be formed using the digits 1, 2, 3, 4 and 5 (repetition of digits not allowed) is
(1)
360000
(2)
660000
(3)
366000
(4)
None of these
104. Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs
from amongst the chairs 1 to 4 and then the men select from the remaining chairs. The numbers of possible arrangements is
(1)
6C
3
× 4 C2
(2)
4C
2
× 4P 3
(3)
4P
× 6P 3
(4)
None of these
(3)
0 ≤ P(E) ≤ 1
(4)
–1 ≤ P(E) ≤ 1
(4)
7
5
2
105. If P(E) denotes the probability of an event E, then
(1)
P(E) ≤ 0
(2)
P(E) ≥ 1
106. A bag contains 5 white and 7 red balls. The probability of drawing a red ball is
(1)
8
12
(2)
7
12
(3)
5
7
107. A bag contains 10 white and 15 black balls. One ball is drawn randomly. the probability that it is black or white is
(1)
1
(2)
6
26
(3)
1
5
(4)
0
108. One ball is drawn from a bag containing 5 white and 7 black balls. The probability that the ball drawn white is
(1)
1
2
(2)
5
12
(3)
7
12
(4)
1
4
109. If | a | < 1 and | b | < 1, then the sum of the series 1 + (1 + a) b + (1 + a + a 2 ) b 2 + (1 + a + a 2 + a3 ) b 3 + ...is
(1)
(12) of (48)
1
(1 − a) (1 − b)
(2)
1
(1 − a) (1 − ab)
(3)
1
(1 − b) (1 − ab)
(4)
1
(1 − a) (1 − b ) (1 − ab )
IC : PTpnrhm01
1
110. The sum to n terms of the series
2n + 1
(1)
1+ 3
(2)
+
1
3+ 5
+
1
5+ 7
1
2n + 1
2
+ ... is
2n + 1 − 1
(3)
(4)
1
( 2n + 1 − 1)
2
111. Let S 1, S 2... be squares such that for each n ≥ 1, the length of a side of Sn equals the length of a diagonal of S n+1. If the length
of a side of S1 is 10 cm, then for which of the following values of n is the area of S n less than 1 sq. cm?
(1)
7
(2)
112. If the expression e {(1 + |cos x | + cos
y2
2
8
x + | cos 3 x | + |cos 4 x|+ ... ∞) loge 4 }
(3)
9
(4)
10
(4)
{π/3, π/2, 2π/3}
satisfies the equation
– 20y + 64 = 0 for 0 < x < π, then the set of values of x is
(1)
{π/3, 2π/3}
(2)
{π/2, π/3}
(3)
{π/2, 0, 2π/3}
113. The points (α, β), (γ, δ), (α, δ) and (γ, β), where α, β, γ, δ are different real numbers, are
(1)
collinear
(2)
vertices of a square
(3)
vertices of a rhombus
(4)
concyclic
114. If O be the origin and if the coordinates of any two points Q1 and Q2 be (x1, y1) and x 2, y2) respectively, then OQ1 × OQ2 cos Q1OQ 2
is equal to
(1)
x 1 x 2 – y 1y 2
(2)
x 1 y 1 – x 2y 2
(3)
x 1 x 2 + y 1y 2
(4)
x 1 y 1 + x 2y 2
115. To what point must be the origin be transferred so as to remove the terms of the first degree in x2 + xy + 2y2 – 7x – 5y + 12 = 0?
(1)
FG 23 , 3 IJ
H 7 7K
(2)
FG −23 , 3 IJ
H 7 7K
(3)
FG 23 , −3 IJ
H7 7K
(4)
None of these
116. Area of a triangle with vertices (a, b),( x 1 , y 1 ) and (x 2 , y 2 ), where a, x1 and x2 are in GP with common ratio r, and b, y 1 and y 2
are in GP with common ratio s, is given by
(1)
ab (r–1) (s – r)
(2)
1
ab(r + 1) (s + 1) (s – r)
2
(3)
1
ab (r – 1) (s –1) (s – r)
2
(4)
ab (r +1) (s +1) (r – s)
117. If each of the points (x 1 , 4), (–2, y 1 ) lies on the line joining the points (2, –1), (5, –3), then the point P(x 1 , y 1) lies on
(1)
6(x + y) – 25 = 0
(2)
2x + 6y + 1 = 0
(3)
2x + 3y – 6 = 0
(4)
6(x + y) + 25 = 0
118. Line L has intercepts a and b on the coordinate axes. When the axes are rotated through a given angle; keeping the origin fixed,
the same line has intercepts p and q, then
(1)
a 2 + b2 = p2 + q2
(2)
1
1
1
1
+
=
+
a2 b2 p2 q2
(3)
a 2 + p2 = b2 + q2
(4)
1
1
1
1
+
=
+
a2 p2 b2 q2
119. The point A (2, 1) is translated parallel to the line x – y = 3 by a distance 4 units. If the new position A’ is in third quadrant, then
the coordinates of A’ are
(1)
(2 + 2 2 , 1 + 2 2 )
IC : PTpnrhm01
(2)
( −2 + 2 , − 1 − 2 )
(3)
(2 − 2 2 ,1 − 2 2 )
(4)
None of these
(13) of (48)
120. A tangent is drawn to each of the circles x² + y² = a² and x²+ y² = b². If two tangents are mutually perpendicular, then locus
of their point of intersection is
(1)
x 2 + y2 = a 2 – b 2
(2)
x2 + y2 = a2
(3)
x2 + y2 = a2 + b2
(4)
x 2 + y 2 = b2
121. The area of the triangle formed by the tangent at (3, 4) to the circle x² + y² = 25 and the coordinate axes is
(1)
625
sq. units
624
(2)
24
sq. units
625
(3)
625
sq. units
24
(4)
624
sq. units
625
122. The condition that the circle x² + y² + 2gx + 2fy + c = 0 bisects the circumference of the circle x² + y² + 2g 1x + 2f 1y + c1 = 0 is
(1)
g 2 + f2 = c
(2)
gg 1 + ff 1 =cc 1
(3)
2g 1 (g + g1 ) + 2f (f + f 1 ) =c + c 1
(4)
2g 1 (g – g 1 ) + 2f 1 (f – f 1 ) = c – c1
123. If the square of the radii of three concentric circles are in A.P., then the square of the lengths of the tangents from any point
to these circles are in
(1)
G.P.
(2)
A.P.
(3)
H.P.
(4)
None of these
(3)
± 2i, 1
(4)
± 2i, – 1
124. The roots of the equation x 3 + x2 + x + 1 = 0 are
(1)
± 1, i
(2)
± i, – 1
125. If two roots of the equation x 3 + mx2 + 11x – n = 0 are 2 and 3, then value of m + n is
(1)
–1
(2)
–2
(3)
–3
(4)
None of these
(3)
two solutions.
(4)
no solution.
126. The equation |x – x 2 – 1| = |2x – 3 – x2 | has
(1)
infinite solutions.
(2)
one solution.
127. The cubic equation which has the roots 2, 3, 4 is
(1)
x 3 – 26x 2 – 9x + 9 = 0
(2)
x 3 + 9x 2 + 26x + 24 = 0
(3)
x 3 – 9x2 + 26x – 24 = 0
(4)
x 3 + 5x 2 + 24 = 0
128. The number of real roots of the equation 2x 2 + 3|x| + 1 = 0 is
(1)
1
(2)
4
(3)
1
(4)
None of these
(3)
13
(4)
None of these
(4)
68°
129. The maximum value of 5 cosθ + 12 sinθ is
(1)
17
(2)
7
130. If the three angles of a quadrilateral are 60°, 60 g and
(1)
78°
(2)
88°
5π
, then the fourth angle is
6
(3)
85°
131. 3 (sin x – cos x) 4 + 4 (sin 6 x + cos6 x) + 6 (sin x + cos x) 2 is equal to
(1)
11
(2)
12
(3)
13
(4)
14
(3)
2
(4)
3
132. sin 6 θ + cos6 θ + 3 sin 2 θ cos 2 θ is equal to
(1)
0
(2)
1
133. The area between the curve y = 2x 4 – x 2 , the x-axis and the ordinates of two minima of the curve is
(1)
(14) of (48)
7
120
(2)
9
120
(3)
11
120
(4)
13
120
IC : PTpnrhm01
134. In the binomial expansion of (a – b) n, n ≥ 5, the sum of the 5th and 6th terms is zero. Then a/b equals
(1)
n−5
6
(2)
n−4
5
(3)
LM
MM
N1
2
135. The value of x for which the matrix product 0
(1)
1
2
(2)
5
n−4
OPLM
PPMM
QN
n
(2)
137. Let (1 + x 2 ) 2 (1 + x) n =
n +4
∑a x
K
K
6
n−5
OP
PP
Q
7 − x 14 x 7 x
0 0
1
0 equals an identity matrix is
–2 1 x −4 x −2x
0
1
1
3
(3)
1
4
n!
(n + 1)! (n + 2)!
136. For a fixed positive integer n, if D = (n + 1)! (n + 2)! (n + 3)! , then
(n + 2)! (n + 3)! (n + 4)!
(1)
(4)
n+1
(3)
(4)
LM D
N (n!)
3
−4
OP
Q
Can't say
1
5
is divisible by
(4)
None of these
(4)
2, 3, 4
. If a 1, a 2, a 3 are in A.P., then n is equal to
K =0
(1)
6
(2)
5
(3)
7
138. The slope of a curve at any point is the reciprocal of twice the ordinate at the point and it passes through the point (4, 3). The
equation of the curve is
(1)
x2 = y + 5
(2)
y2 = x – 5
(3)
y2 = x + 5
(4)
x2 = y – 5
139. The equation of a circle with origin as centre, passing through the vertices of an equilateral triangle whose median is of length
3a is
(1)
x 2 + y 2 = 9a 2
z
(2)
x 2 + y 2 = 16a2
(3)
x 2 + y 2 = a2
(4)
None of these
(3)
H.P.
(4)
None of these
(4)
None of these
π/4
140. Let an =
tann x dx. Then a 2 + a 4 , a 3 + a 5 , a 4 + a 6 are in
0
(1)
141. If I =
(1)
A.P.
z
(2)
G.P.
sec 2 x cosec 4 x dx = K cot 3 x + L tan x + M cot x + C, then
K=
1
3
(2)
L=2
(3)
M=–2
142. If a line passing through origin touches the circle (x – 4) 2 + (y + 5) 2 = 25, then its slope should be
(1)
±
3
4
(2)
0
(3)
±3
(4)
±1
143. The acute angle between the line joining the points (2, 1, – 3), (– 3, 1, 7) and a line parallel to x − 1 = y = z + 3 through the
3
4
5
point (– 1, 0, 4) is
(1)
cos −1
IC : PTpnrhm01
F 7 I
GH 5 10 JK
(2)
cos −1
F
GH
I
J
10 K
1
(3)
cos−1
F 3 I
GH 5 10 JK
(4)
cos −1
F 1 I
GH 5 10 JK
(15) of (48)
FG
H
144. 1 + cos
π
8
IJ FG1 + cos 3π IJ FG1 + cos 5π IJ FG1 + cos 7 π IJ
KH
8K H
8K H
8K
1
2
(1)
(2)
cos
is equal to
π
8
1
8
(3)
145. The sum of first n terms of the series 1 2 + 2.2 2 + 32 + 2.4 2 + 5 2 + 2.6 2 + ... is
(4)
1+ 2
2 2
n (n + 1)2
where n is even. When n is odd, the
2
sum is
(1)
(n + 1) n2
2
146. If z r = cos
(1)
(2)
n (n + 1)2
2
1
(2)
of area zero
148. If u = tan−1
(1)
(4)
None of these
(4)
–i
2r π
z + z1
=
where r = 0, 1, 2, 3, then 0
4
z2 + z 3
–1
147. The complex numbers z 1 , z 2 and z 3 satisfying
(1)
n2 (n − 1)
2
(3)
(2)
(3)
z1 − z3
1− 3i
are the vertices of a triangle which is
=
z2 − z3
2
right angled isosceles
F x + y I , the value of x ∂u + y ∂u
GH x − y JK
∂x
∂y
3
i
(3)
equilateral
(4)
obtuse angled isosceles
(3)
sin 2u
(4)
sec2 2u
3
tan 2u
(2)
is
cos 2u
149. From the bottom of a pole of height h, the angle of elevation of the top of a tower is α and the pole subtends angle β at the top
of the tower. The height of the tower is
(1)
h tan(α − β)
tan(α − β) − tan α
(2)
h cot(α − β)
cot(α − β) − cot α
(3)
cot(α − β)
cot(α − β) − cot α
(4)
None of these
150. The volume of the solid formed by the revolution of x = a(θ – sin θ), y = a(1 – cos θ) about its base will be
(1)
3π 2 a 3
151. If 0 < x < 1000 and
(2)
4π 2 a 3
(3)
5π 2a 3
(4)
None of these
LM x OP + LM x OP + LM x OP = 11x , where [x] is the greatest integer less than or equal to x, the number of possible
N10 Q N 20 Q N 30 Q 60
values of x is
(1)
15
(2)
16
(3)
17
(4)
33
152. If the circle C 1 : x2 + y2 = 16 intersect another circle C2 of radius 5 in such a manner that the common chord is of maximum length
and has a slope equal to 3/4, then coordinates of the centre of C 2 are
(1)
FG − 9 , 12 IJ , FG 9 ,− 12 IJ
H 5 5 K H5 5 K
(2)
FG − 9 ,− 12 IJ , FG 9 , 12 IJ
H 5 5 K H5 5 K
(3)
FG 12 , − 9 IJ , FG − 12 , − 9 IJ
H 5 5K H 5 5K
(4)
None of these
153. An isosceles triangle of vertical angle 2θ is inscribed in a circle of radius a. The area of the triangle is maximum when θ =
(1)
(16) of (48)
π
6
(2)
π
4
(3)
π
3
(4)
π
2
IC : PTpnrhm01
154. I f t h e c o e f f i c i e n t s o f p t h , ( p + 1 ) t h a n d ( p + 2)th terms in the expansion of (1 + x) n are in A.P., then
155.
(1)
n 2 – 2np + 4p 2 = 0
(3)
n2
– n (4p + 1) +
4p 2
=0
(2)
n 2 – n(4p + 1) + 4p 2 – 2 = 0
(4)
None of these
(3)
n (n + 1)
2
C1 2C2 3C3
nCn
+
+
+ ... +
=
C0
C1
C2
Cn −1
(1)
n (n − 1)
2
n (n + 2)
2
(2)
(4)
(n − 1)(n − 2)
2
156. If the three linear equations x + 4ay + az = 0, x + 3by + bz = 0 and x + 2cy + cz = 0 have a non-trivial solution, then a, b, c
are in
(1)
A.P.
(2)
10
157. The value of ∆ =
(1)
C4
10
11
C6
11
12
C8
12
C5
11
G.P.
H.P.
(4)
None of these
(4)
None of these
Cm
C7
12
Cm + 2
C9
13
Cm + 4
6
(3)
(2)
is equal to zero, where m is equal to
4
(3)
5
158. The locus of the points representing the complex number z for which |z| – 2=|z – i| – |z+5i| = 0 is
(1)
a circle with centre at the origin
(2)
a straight line passing through the origin
(3)
the single point (0, – 2)
(4)
None of these
159. Given that |z – 1| = 1, where z is a point on the Argand plane. Then
(1)
2−z
= i tan(arg z)
z
(2)
z−2
= i tan (arg z)
z
(3)
z+2
= i tan (arg z)
z
(4)
None of these
160. If α, β are the roots of x2 – 2ax + b 2 = 0 and γ, δ are the roots of x2 – 2bx + a 2 = 0, then
(1)
A.M. of α, β = G.M. of γ, δ
(2)
α, β, γ, are in A.P.
(3)
α, β, γ, δ are in G.P.
(4)
None of these
161. The angle of elevation of the top of a T.V. tower from three points A, B, C in a straight line, (in the horizontal plane) through
the foot of the tower are α, 2α, 3α respectively. If AB = a, the height of the tower is
(1)
a tan α
(2)
a sin α
(3)
a sin 2α
(4)
a sin 3α
(4)
None of these
162. Expansion of the polynomial x5 – 2x 4 + x 3 – x 2 + 2x – 1 in powers of (x – 1) is
(1)
2(x – 1) 5 + (x – 1)4 + 3(x – 1) 3 + 2(x – 1)2 – (x – 1) – 17
(2)
(x – 1) 5 + 3(x – 1) 4 + 3(x – 1) 3 + 2(x – 1) 2 – 17
(3)
(x – 1) 5 + 3 (x – 1) 4 + 3 (x – 1) 3
(4)
(x – 1) 5 + 3 (x – 1) 4 + 3(x – 1) 2
163. If u = sin–1
(1)
F x + 2y + 3z I ,
GH x + y + z JK
7 tan u
IC : PTpnrhm01
8
8
8
then x
(2)
∂u
∂u
∂u
+y
+z
is equal to
∂x
∂y
∂z
–7 tan u
(3)
–(1/7) tan u
(17) of (48)
164. The area of the loop of the curve ay 2 = x2 (a - x) is
(1)
1 2
a
5
8 2
a
15
(2)
(3)
7 2
a
15
(4)
None of these
165. A person standing at the foot of a tower walks a distance 3a away from the tower and observes that the angle of elevation of the
top of the tower is α. He then walks a distance 4a perpendicular to the previous direction and observes the angle of elevation to be
β. The height of the tower is
(1)
3a tan α
5a tan β
(2)
(3)
4a tan β
(4)
7a tan β
166. If A and B are square matrices of order 3 such that |A| = – 1, |B| = 3, then |3AB| is equal to
(1)
–9
(2)
LMcos α
167. Let F(α ) = M sin α
MN 0
(1)
–81
(3)
–27
(4)
81
F(2α)
(4)
None of these
( A −1 )T1 ( A T )−1
(4)
|A| ≠ 0
OP
PP
Q
− sin α 0
cos α 0 , where α ∈ R, then [F(α) –1 ] is equal to
0
1
F(– α)
F(α–1)
(2)
(3)
168. If A is an invertible matrix, then which of the following is correct?
(1)
A–1 is multivalued
A–1 is singular
(2)
(3)
169. The system of equations –2x + y + z = a, x – 2y + z = b, x + y – 2z = c has no solution, if
(1)
a+b+c≠0
(2)
a+b+c=0
LM
MM
N
(3)
a = 1, b = 1, c = –2
(4)
None of these
(3)
4
(4)
1
(4)
(2, 2)
OP
PP
Q
1 1 1 −1
170. The rank of matrix A = 1 2 3 4 is
3 4 5 2
(1)
2
(2)
3
171. The point on the curve x 2 = 2y which is closest to the point (0, 5) is
(1)
(1, 1/2)
(2)
(–1, 1/2)
(3)
(2√2, 4)
172. In a certain factory turning out razors blades, there is a small chance of 0.002 for any blade to be defective. The blades are
supplied in packets of 10. Using Poisson distribution, the approximate number of packets containing two defective blades
respectively in a consignment of 10,000 packets is
(1)
1
(2)
2
(3)
4
(4)
6
173. Three machines M1, M 2 and M 3 produce identical items. Of their respective output 5%, 4% and 3% items are faulty. On a certain
day, M1 has produced 25 % of the total output, M2 has produced 30% and M3 the remainder. An item selected at random is found
to be faulty. What are the chances that it was produced by the machine with the highest output?
(1)
0.355
(2)
0.435
(3)
0.625
(4)
0.1
(3)
1
30
(4)
13
30
174. The probability density function of variate X is
X:
0
1
2
4
5
6
Y:
k
3k
5k
9k
11k
13k
What will be the minimum value of k so that P(X ≤ 2) > 0.3?
(1)
(18) of (48)
1
49
(2)
24
49
IC : PTpnrhm01
175. One hundred identical coins, each with probability p of showing up heads, are tossed. If 0 < p < 1 and the probability of heads
showing on 50 coins is equal to that of heads showing on 51 coins, the value of p is
(1)
1
2
49
101
(2)
(3)
50
101
51
101
(4)
176. In sampling a large number of parts manufactured by a machine, the mean number of defective in a small sample of 20 is 2. Out
of 1000 such samples, how many would be expected to contain at least 3 defective parts?
(1)
292
(2)
323
(3)
312
(4)
268
177. α, β, γ and δ are the roots of the equation x 4 + ax3 + 2x 2 + bx + 1 = 0, then (1 + α 2 ). (1 + β 2 ) (1+ γ 2).(1 + δ 2 ) equals
(1)
a 2 + b2
(2)
(a – b) 2 + 1
(3)
(a – b)2
(4)
2(a2 + b2) +1
178. Given a random variable whose range is set {1, 2} and whose probability density is f(1) = 1/4 and f(2) = 3/4. The mean and
variance of this distribution are, respectively
(1)
7 13
,
2 4
LM1
179. If S = M1
MN1
1
ω2
LM1
MM1
N1
ω2
(1)
ω
1
ω
7 13
,
4 4
(2)
(3)
7 3
,
2 16
(4)
7 3
,
4 16
(4)
1 1
1
1 ω
3
1 ω2
OP
P
ω PQ
1
ω and ω is the cube root of unity, then S –1 is equal to
2
OP
P
ω PQ
LM
MM
N
1
ω
1 1
1
1 ω2
2
1 ω
(2)
2
OP
P
ω PQ
1
ω
(3)
2
LM
MM
N
1 1
1
1
1 ω ω2
4
1 ω2 ω
OP
PP
Q
LM
MM
N
1
ω2
ω
OP
PP
Q
180. To a man running at a speed of 20 km/hr, the rain drops appear to be falling at an angle of 30° from the vertical. If the rain drops
are actually falling vertically downwards, their velocity in km/hr is
(1)
10√3
(2)
10
(3)
20√3
(4)
40
181. The end of a heavy uniform rod AB can slide along a rough horizontal rod AC to which it is attached by a ring. B and C are joined
by a string. If ∠ABC be a right angle, when the rod is on the point of sliding, µ the coefficient of friction and α the angle between
AB and the vertical, then
(1)
µ = 2 tan α/(2 + tan 2 α) (2)
µ = tan α/(2 + tan 2 α)
(3)
µ = 2 cot α/(1 + cot 2 α) (4)
µ = cot α/(2 + cot 2 α)
182. The centres of a set of circles, each of radius 2, lie on the circle x 2 + y2 = 36. The locus of any point in the set is
(1)
4 ≤ x 2 + y 2 ≤ 64
(2)
4 ≤ x 2 + y 2 ≤ 25
(3)
9 ≤ x 2 + y 2 ≤ 25
(4)
16 ≤ x 2 + y 2 ≤ 64
183. The equation of circle passing through (3, –6) and touching both the axes is
(1)
x 2 + y 2 – 6x + 4y – 3 = 0
(2)
x 2 + y 2 + 6x – 6y + 9 = 0
(3)
x2
(4)
x 2 + y 2 – 30x + 30y + 225 = 0
+
y2
+ 30x – 30y + 225 = 0
184. The length of the chord of the circle x2 + y 2 = 25 joining the points, tangents at which intersect at an angle of 120° is
(1)
5
2
(2)
5
(3)
10
(4)
None of these
185. The equations of four circles are (x ± a)2 + (y ± a) 2 = a 2 . The radius of a circle touching all the four circles is
(1)
( 2 − 2) a
IC : PTpnrhm01
(2)
2 2a
(3)
( 2 + 1) a
(4)
(2 + 2 ) a
(19) of (48)
186. A square is inscribed in the circle x 2 + y 2 – 10x – 6y + 30 = 0. One side of the square is parallel to y = x + 3, then one vertex
of the square is
(1)
(4, 3)
(2)
(7, 3)
(3)
(6, 3 – 3 )
(4)
(6, 3 +
3 )
187. Equation of a circle through the origin and belonging to the co-axial system, of which the limiting points are (1, 2), (4, 3) is
(1)
x 2 + y 2 – 2x + 47 = 0
(2)
x 2 + y2 + 8x + 6y = 0
(3)
2x 2 + 2y 2 – x – 7y = 0
(4)
x 2 + y 2 – 6x – 10y = 0
188. The number of common tangents that can be drawn to x 2 + y 2 + 3x + 7y – 10 = 0 and 2x 2 + 2y 2 + 6x + 7y + 5 = 0 are
(1)
0
(2)
1
(3)
2
(4)
4
189. A bag contains an assortment of blue and red balls. If two balls are drawn at random, the probability of drawing two red balls
is five times the probability of drawing two blue balls. Furthermore, the probability of drawing one ball of each color is six times
the probability of drawing two blue balls. The number of red and blue balls respectively in the bag is
(1)
6 and 3
(2)
3 and 6
190. The probability of a man hitting a target is
least once is greater than
(1)
(3)
2 and 3
(4)
None of these
1
. How many times must he fire so that the probability of his hitting the target at
4
2
?
3
2
(2)
3
(3)
4
(4)
5
(4)
101
121
191. Two integers x and y are chosen, without replacement, at random from the set
{x:0 ≤ x, y ≤ 10, yx is an integer} the probability that |x – y| ≤ 5 is
(1)
87
121
(2)
89
121
(3)
91
121
192. A second order determinant is written down at random using the numbers 1, – 1 as elements. The probability that the value of
the determinant is non zero is
(1)
1
3
(2)
3
8
(3)
5
8
(4)
1
2
193. In a family of n children, let A be the event that the family has children of both sexes and let B the event that there is at most
one girl in the family. Then the value of n for which the event A and B are independent is (assuming that each child has
probability
(1)
3
1
of being a boy)
2
(2)
4
(3)
5
(4)
6
194. A book contains 1000 pages. A page is chosen at random. The probabilities that the sum of the digits of the marked number on
the page is equal to 9 is:
(1)
23
500
(2)
11
200
(3)
7
100
(4)
None of these
195. An unbiased die with faces marked 1, 2, 3, 4, 5 and 6 is rolled four times. Out of the four face values obtained, the probability
that the minimum face value is not less then 2 and the maximum face value is not greater than 5, is
(1)
1
81
(2)
16
81
(3)
65
81
(4)
80
81
196. If X and Y are independent binomial variates B(5, 1/2) and B(7, 1/2), then P(X + Y = 3) is
(1)
(20) of (48)
55
1024
(2)
55
4098
(3)
55
2048
(4)
None of these
IC : PTpnrhm01
197. The probability that at least one of the events A and B occur is 0.6. If A and B occur simultaneously with probability 0.2, then
P ( A) + P (B ) , where A and B are complements of A and B respectively, is equal to
(1)
0.4
(2)
198. If u = log(x3 + y 3 + z 3 – 3xyz) and
(1)
199. If
9
At 4
(1)
Bt 3
+
+
Ct 2
+ Dt + E =
33
200. If u =
(1)
(2)
x 2 ( x 2 − y 2 )2
( x 2 + y 2 )2
u
(3)
LM ∂ + ∂ + ∂ OP
N∂ x ∂ y ∂ zQ
2
–9
u=
(4)
1.4
3
(4)
–3
(3)
27
(4)
24
(3)
1/2u
(4)
3u
(3)
t −1
t−3
t +1
2−t
t −3
t+4
t − 3 , then E equals
3t
–39
1.2
k
, then k is equal to
( x + y + z)2
t 2 + 3t
(2)
, then x .
0.8
∂u
∂u
+ y.
is equal to
∂x
∂y
(2)
2u
201. In a series of 5 observations, the values of mean and variance are 4.4 and 8.24, if three observations are 1, 2 and 6, then the
other two are
(1)
4, 3
(2)
6, 7
(3)
4, 9
(4)
9, 6
202. In a surprise check on passengers in a local bus, 20 ticketless passengers were caught. The sum of squares and the standard
deviation of the amount found in their pockets were Rs.2000 and Rs.6, respectively. If the total fine imposed on these
passengers is equal to the total amount found in their pockets and the fine imposed is uniform, what is the amount that each one
of them will have to pay as fine?
(1)
Rs.10
(2)
Rs.8
(3)
Rs.12
(4)
Rs.16
203. A company has three establishments E 1, E2 and E 3 in three cities. Analysis of the monthly salaries paid to the employees in the
three establishment is given below:
E1
E2
E3
Number of employees
20
25
40
Average monthly salary (Rs.)
305
300
340
Standard deviation of monthly salary (Rs.)
50
40
45
The standard deviation of the monthly salaries of all the 85 employees in the company
(1)
48.69
(2)
39.48
(3)
43.46
(4)
58.95
204. The standard deviation of the combined group of 500 items from the following data.
Group I
Group II
Group III
No. of items :
100
150
250
Arithmetic mean :
50
55
60
Variance :
(1)
8.6
100
(2)
9.0
121
(3)
7.2
144
(4)
11.98
205. For a group containing 100 items, the arithmetic mean and standard deviation are 8 and 10.5 . For 50 observations selected
from these 100 observations, the mean and standard deviation are 10 and 2 respectively. The mean and standard deviation of
the remaining 50 observations is
(1)
2
IC : PTpnrhm01
(2)
3
(3)
4
(4)
5
(21) of (48)
206. The S.D. from the following data is
Group I
Group II
Group III
50
60
90
200
6
7
?
7.746
113
?
115
116
Number
Standard deviation
Mean
(1)
8
(2)
5
(3)
4
(4)
Combined
6
207. The mean, median and the coefficient of variation of 100 observations are found to be 90, 84 and 80 respectively. The
coefficient of skewness for this system is
(1)
0.15
(2)
0.25
(3)
0.4
(4)
0.6
208. Karl Pearson's coefficient of skewness of a distribution is + 0.32. Its standard deviation is 6.5 and mean is 29.6. The mode of
the distribution is
(1)
24.3
(2)
35.8
(3)
20.2
(4)
27.5
209. A coin is tossed 2n times. The chance that the number of times one gets head is not equal to the number of times one gets tail is
FG IJ
H K
(2n) ! 1
(n !)2 2
(1)
2n
(2)
1−
(2n) !
(n !)2
(3)
1−
(2n) ! 1
⋅
(n !)2 4n
(4)
None of these
210. The mean of a certain distribution is 50, its standard deviation is 15 and coefficient of skewness is –1. The median of the
distribution is
(1)
40
(2)
45
(3)
55
(4)
48
211. The coefficient of correlation between x and y, when n = 10, Σx = 60, Σy = 60, Σx 2 = 400, Σy 2 = 580 and Σxy = 305 is
(1)
0.68
(2)
–0.68
(3)
0.58
(4)
–0.58
212. The coefficient of correlation between two variables X and Y is 0.3. Their covariance is 9. The variance of X is 16. The standard
deviation of Y series is
(1)
7.5
(2)
3.2
(3)
8.9
(4)
6.4
213. If in bivariate distribution cov(x,y) = 40, var(x) = 50 and var(y) = 72, then coefficient of correlation in x and y is
(1)
0.67
(2)
0.33
(3)
0.001
(4)
27
214. In two sets of variables x and y with 50 observations each the following data were observed:
Mean x Series = 10, S.D. x Series = 3. Mean y Series = 6, S.D. y Series = 2.
Coefficient of correlation between x and y = +0.3. However, it was later on found that one value of X (= 10) and one value of
Y (= 6) were inaccurate and hence weeded out. With the remaining 49 pairs of values how is the coefficient of correlation
affected?
(1)
Increase by 20%
(2)
Decrease by 20%
(3)
Remains unchanged
(4)
None of these
(4)
3, 2/7
(4)
y = –0.4x + 7
215. You are given the following data.
Series
X
Y
Mean
6
8
Standard deviation
4
12
Coefficient of Correlation = 2/3.
The regression coefficients b yx and b xy are
(1)
2, 2/9
(2)
2, 9/2
(3)
4, 3/8
216. The equation of the line of regression of y on x for the following data is
x:
5
2
y:
5
8
(1)
y = 0.4x + 7
(22) of (48)
(2)
1
4
4
2
y = 0.4x – 7
(3)
3
10
y = –0.4x – 7
IC : PTpnrhm01
217. Which of the statement/s is/are correct?
i.
The regression coefficient of y on x is 3.2 and that of x on y is 0.8.
ii.
The two regression coefficients are 0.4 and –0.2.
iii.
The two regression coefficient are given to be 0.8 and 0.2 and the coefficient of correlation is 0.4.
iv.
40x – 18y = 5 and 8x – 10y + 6 = 0 are respectively the regression equations of y on x and x on y.
(1)
i and iii
(2)
ii, iii and iv
(3)
only iv
(4)
only iii
218. The equations of two regression lines obtained in a correlation analysis are 3x + 12y = 19 and 3y + 9x = 46. Then the
correlation coefficient is
(1)
0.35
(2)
–0.35
(3)
0.43
(4)
–0.28
219. If the regression coefficients be given by b yx = 1.6 and b xy = 0.4 and θ be the angle between two regression lines, then the
value of tan θ
(1)
0.6
(2)
0.2
(3)
0.1
(4)
0.9
220. A department in a works has 10 machines which may need adjustment from time to time during the day. Three of these machines
are old, each having a probability of 1/11 of needing adjustment during the day, and 7 are new having corresponding probabilities
1/21. Assuming that no machine needs adjustment twice on the same day, the probability that on a particular day just 2 old and
no new machine need adjustment is
(1)
0.016
(2)
0.12
(3)
0.09
(4)
0.08
221. While walking in a forest an ecologist got 300 insect bites in two and half hours. For how many one-minute intervals was he free
from insect bites? (e –2 = .1353)
(1)
10 minute
(2)
15 minute
(3)
20 minute
(4)
None of these
(4)
0.18
(4)
0.53
222. For a certain normal variate X, the mean is 12 and the S.D. is 4. Then P(4 ≤ X ≤ 20)
(1)
0.15
(2)
0.35
(3)
0.95
223. X is a normal variate with mean 30 and S.D. 5. The probability that |X – 30| > 5
(1)
0.73
(2)
0.85
(3)
0.31
224. A coin is tossed n times. The probability of getting atleast one head is greater than that of getting at least two tails by
Then n is
(1) 5
(2)
10
(3)
15
(4)
5
.
32
none of these
225. If mean of a Binomial Distribution is 20 and standard deviation is 4, then number of events is
(1)
50
(2)
25
(3)
100
(4)
80
226. A normal distribution with x = 50 and σ = 10 is given. The value of x that has 16% of the area to its left is
(1)
30
→
(2)
40
→
→
227. If a = $i + 2 $j + 3k$ , b = − $i + 2$j + k$ and c = 3 $i + $j, then
(1)
3
(2)
4
(3)
FG a + t b IJ
H K
→
→
(3)
35
(4)
25
→
is perpendicular to c , if the value of t is
5
(4)
6
228. The work done by the forces P= +2$i − 3$j + k$ and Q = $i + 5$j − 3k$ to displace the particle from point A to point B whose position
vectors are −2 $i + 5$j + 7k$ and 3 $i + 7 $j + 2k$ , is
(1)
31 units
IC : PTpnrhm01
(2)
29 units
(3)
9 units
(4)
None of these
(23) of (48)
→
→
229. The unit vector which is perpendicular to each of the vectors a = 3 $i + 2 $j − k$ , and b = 12 $i + 5 $j − 5k$ is
5 $i + 3$j − 9k$
(1)
−5$i + 3$j + 9k$
(2)
115
(3)
115
−5$i + 3$j − 9k$
115
→
(4)
−5$i − 3 $j + 9k$
115
→
230. The area of triangle whose two adjacent sides are a = $i + 4 $j − k$ and b = $i + $j + 2k$ is
(1)
3√7
2/5 √61
(2)
(3)
3/2√11
→
(4)
None of these
→
231. The unit vector which is perpendicular to the vectors a = 2 $i + $j + k$ and b = 3$i + 4 $j − k$ . is
$i + $j + k$
(1)
− $i + $j − k$
(2)
3
(3)
3
$i − $j + k$
(4)
3
− $i + $j + k$
3
232. The moment of the force 3 $i + $j − 2k$ acting through the point $i − 2$j + 2k$ about the point 2 $i − $j + k$ is
$i + $j + k$
(1)
1
$i + $j + 2k$
(2)
(3)
6
e$i − $j − 2k$ j
(4)
None of these
(4)
4
(4)
None of these
(4)
zero
(4)
None of these
(4)
None of these
233. The vectors 2 $i − $j + λk$ , $i + 2 $j − 3k$ and 3$i − 4 $j + 5k$ are coplanar if λ is equal to
(1)
1
(2)
→ → →
FG
H
→
2
→
IJ
K
(3)
→
→
→
3
→
234. For non–zero vectors a , b , c a × b ⋅ c = a ⋅ b ⋅ c is true only when
→
LM
N
→
→
→
a=b= c .
(1)
→ →
→ →
→
235. a × b b × c c × a
L
2M a
N
→
(1)
→
→
b c
(2)
OP
Q
→ →
→ →
→ →
a⋅ b = b ⋅ c = c ⋅ a = 0 .
(3)
→ →
a⋅ b = 0
is equal to
OP
Q
LM
N
→
(2)
→
→
3 a b c
OP
Q
LM a
N
→ → →
(3)
b c
OP
Q
2
→
→ → →
236. If a , b , c are three non-coplanar non-zero vectors and r is any vector in space, then
→
→
→
→
→
→
→
→
→
→
→
→
( a × b ) × ( r × c )) + ( b × c ) × ( r × a ) + ( c × a ) × ( r × b ) is equal to
→→→ →
(1)
(2)
2[ a b c ] r
→
→
→
→→→ →
3[ a b c ] r
→→→ →
(3)
[a b c] r
→
237. If AB = b and AC = c , then the length of the perpendicular from A to the line BC is
→
→
→
→
| b × c|
(1)
(24) of (48)
| b + c|
→
→
→
→
→
| b × c|
(2)
| b − c|
(3)
→
1 | b × c|
2 → →
| b − c|
IC : PTpnrhm01
→
→
→
→
→
→
→
238. The projection of the vector i + j + k on the line whose vector equation is r = (3 + t) i + (2t − 1) j + 3t k , t being a scalar, is
1
(1)
6
(2)
14
6
(3)
→
→
→
→
(4)
14
None of these
→ →
239. A vector r satisfies the equations r × a = b and r . a = 0 . Then which of the following is true?
(1)
→
→
a× b
→
r =
(2)
→ →
→
→
r =
→
a× b
a. b
→
(3)
→ →
→
a× b
→
r =
(4)
→ →
None of these
b. b
a. a
→
→ → →
→
→
→
→
→
→
→
→
240. If the vectors a , b , c are non-coplanar and l, m, n, are distinct scalars, then [l a + m b + n c l b + m c + n a l c + m a + n b ] = 0 , if
(1)
lm + nm + nl = 0
(2)
l+m+n=0
(3)
→
→
→
→
l2 + m2 + n2 = 0
→
→
→
→
l3 + m3 + n3 = 0
(4)
→
→
→
→
→
→
241. The vector x is perpendicular to the vectors a = 2 i + 3 j − k and b = i − 2 j + 3 k . If x .(2 i − j + k ) = −6 , then x is equal to
(1)
→
→
→
→
(2)
−3 i + 3 j + 3 k
→
→
(3)
3 i − 3 j+ 3k
→
→
→
(4)
3 i + 3 j− 3k
None of these
242. The direction cosines of a line connected by the relations l + m + n = 0, mn – 2nl – 2lm = 0 are
(1)
–
1
6
1
,
6
1
,
6
(2)
1
–
,
−2
6
6
,
2
(3)
6
1
6
,
2
6
,
−2
6
1
(4) −
6
243. The vector equation of a line passing through (2, –1, 1) and parallel to the line whose equation is
(1)
r = 2i + j – k + λ (2i – 7j + 3k)
(2)
r = 2i – j + k + λ (2i + 7j – 3k)
(3)
r = 2i – j – k + λ (2i + 7j – 3k)
(4)
None of these
,
2
6
,
−1
6
x −3 y +1 z −2
is
=
=
2
7
−3
244. The direction cosines of three concurrent lines are proportional to (l, –2, 2) (0, 2, –1), (1, 2, 0) are
(1)
coplanar
245. Let f ( x ) =
(1)
(2)
1
(18 − x )
2
collinear
, the value of Lim
0
x →3
(2)
FG f (x) − f(3) IJ
H x −3 K
(3)
parallel
(4)
None of these
(3)
–1/3
(4)
1/9
is
– 1/9
z
f (x )
246. Let f : R → R be a differentiable function and f(1) = 4. Then the value of Lim
x→ 1
(1)
8f’ (1)
(2)
(1)
π
2
IC : PTpnrhm01
(3)
1
247. The value of lim
n→∞
4f’ (1)
1 1 1
.
+
2 2 2
1 1 1
.
+
2 2 2
(2)
π
4
2f’ (1)
4
2t
dt is
x −1
(4)
f’ (1)
is
1
+
2
1
... n terms
2
(3)
π
3
(4)
π
6
(25) of (48)
248. Let f(x) be strictly increasing and differentiable then lim
x →0
(1)
1
(2)
249. The series Un =
(1)
n+1 − n
np
divergent
–1
FG p > 1 IJ
H 2K
(2)
f ( x )2 − f ( x )
is
f( x ) − f (0)
(3)
0
(4)
2
(3)
oscillatory
(4)
None of these
(4)
None of these
is
convergent
250. The value of Lim [ x 2 + x + sin x] where [.] denotes the greatest integer function
x→ 0
(1)
does not exist
(2)
is equal to zero
(3)
–1
(26) of (48)
IC : PTpnrhm01
Detailed Solutions
1.
If S is the focus of the parabola and T is the point of intersection of tangents
at P and Q, then
= SP × SQ ⇒
ST 2
2.
ST 2
7.
= 4 × 9 ⇒ ST = 6. Ans.(2)
We know that, for |y| < 1
1
1
1
y + y 3 + y 5 +... = [log (1 + y) – log (1–y)]
3
5
2
x
Thus,
+
1+ x 2
FG
H
1
x
3 1+ x 2
IJ
K
3
+
FG
H
1
x
5 1+ x 2
IJ
K
5
+...
We have, f (x) = (x + 1) 1/3 – (x – 1) 1/3
LM
MN
1
1
1
−
3 ( x + 1)2 / 3 ( x − 1) 2 /3
2r+3
= 18 C
r–3
.
9.
5.
6λ − 2
2
=
z
1 α
0
x dx =
z
1
0
F
GH
I
JK
2
x 2 dx +
z
2
x 2 dx
3
2
3
2
dx + 2 dx + 3 dx
1
2
3
12.
z
1 x α log x
0
log x
dx
0
ex
= e x (1− x )−1
1− x
We have
F
GH
I
JK
x2 x 3
+
+... (1+ x + x 2 + x 3 +...)
2 ! 3!
= A o + A 1 x + A 2x 2 + .... where A n = 1+
LMQ d (n ) = n log nOP
N dt
Q
t
t
Thus, A 2 = 1 + 1 +
13.
From (i), when α = 0, f(0) = 0
∴ from (ii), f(0) = log (1) + c, i.e., c = 0.
1 5
= , and A n+1 – A n = 1/(n+1)!. Ans.(4)
2 2
[(3/2)x 2 – (1/3x)] 9 is given by
Tr +1 = 9 Cr
FG 3 x IJ FG − 1 IJ
H 2 K H 3x K
Now integrating both sides w.r.t. α,
f(α) = log (1 + α) + c ....(ii)
1 1
1
+ +...+ ∀n ≥ 0
n!
1! 2 !
The (r + 1)th term in the expansion of
1
x α +1
1
=
α +1
1+ α
2
This is clearly a hyperbola. Ans.(3)
.... (i),
∂ x −1
dx =
∂α log x
3
FG x − y IJ FG x + y IJ = mFG 1 IJ ⇒ x − y = 1
H 3 2 KH 3 2 K H mK 9 4
= 0, λ = 0, 3 . Ans.(1)
−1
dx
log x
then f '(α ) =
1
z
The required locus is obtained by eliminating the variable m from the given
equations of the lines. We then have
= 1+ x +
α
x 2 dx +
2
1 xα
0
2
= √2 – 1 + 2 √3 – 2 √2 + 6 – 3 √3 = 5 – (√2 + √3). Ans.(2)
11.
e
z
z
x 2 dx +
= 0 + √2 – 1 + 2 (√3 – √2) + 3(2 – √3)
z
f (α ) =
1
0
0
= log(1+ log x) 1 = log 2 − log1 = log 2 . Ans.(1)
6.
z
z z z z
1
e
e
dx
x dx
=
I=
x(1+ log x ) 1 1+ log x
1
z
x 2 dx =
= 0 dx +
3 λ + 1 6λ − 2
λ − 1 5λ + 1
Let I =
0
λ − 1 3 λ + 1 5λ + 1
or if, 0
λ−3
0 =0
⇒ (λ − 3 )
z
2
10.
− λ + 3 = 0 (operate C 3 + C 2 )
3λ + 1 3 λ − 3
2
af(x) = – g (x) ≤ – 1/4. Since g(x) ≥ 1/4 for all x.
1
2
( x − 1)2 / 3 − ( x + 1)2 / 3
3( x 2 − 1)2 / 3
Thus the number of solution is zero. Ans.(4)
2λ
λ−3
=
But this is not possible as a f(x) > 0 for all x.
λ − 1 3λ + 1
2λ
λ − 1 4 λ − 2 λ + 3 = 0 (operate R 2 − R 1 )
2
3λ + 1 3λ − 3
0
OP
PQ
Clearly, f ’(x) ≠ 0 for any other value of x ∈ [0,1]. The value of f(x) at
x = 0 is 2. Hence, the greatest value of f(x) is 2. Ans.(2)
The given equation will be consistent, if
or if
1/ 3
Now, f’(x) = 0 ⇒ (x–1) 2/3 = (x + 1) 2/3 ⇒ x = 0
We know that the coefficient of the (r + 1)th term in the expansion of
(1 + x) 18 is 18 C r .
λ − 1 3λ + 1
OP
P
2 / 3 PQ
1/ 3
2 / 3 −2 / 3
Clearly, f’(x) does not exist at x = ± 1.
2r + 3 + r – 3 = 18 ⇒ r = 6. Ans.(1)
4.
−2 / 3
2/3
OP
PP
Q
∴ f '( x ) =
2
18 C
1/ 3 × 1/ 3
Hence the matrix is orthogonal. Ans.(2)
8.
1
x
x
− log 1−
log 1+
2
1+ x 2
1+ x 2
According to the given condition
LM−2 / 3
MM
N 2/3
2/3
LM
MM
N
2
3.
OP
P
2 / 3 PQ
1/ 3
2/3
4 / 9 + 1/ 9 + 4 / 9 −4 / 9 + 2 / 9 + 2 / 9 −2 / 9 − 2 / 9 + 4 / 9
= −4 / 9 + 2 / 9 + 2 / 9 4 / 9 + 4 / 9 + 1/ 9
2/9−4/9+2/9 =I
−2 / 9 − 2 / 9 + 4 / 9 2 / 9 − 4 / 9 + 2 / 9
1/ 9 + 4 / 9 + 4 / 9
LM FG
IJ FG
IJ OP
K H
K PQ
MN H
1 F 1+ x + x I
= logG
J . Ans.(2)
2 H 1+ x − x K
=
LM−2 / 3
MM
N
A' A = 2 / 3
1/ 3
= 9 Cr ( −1)r
2
3 9 −2r
2 9 −r
9 −r
x 18 −3 r
r
.... (1)
Since we are looking for the coefficient of the term independent of x in the
Hence (ii) gives, f(α) = log (1 + α). Ans.(1)
IC : PTpnrhm01
(27) of (48)
FG 3 x
H2
expansion of (1+ x + 2 x 3 )
2
−
1
3x
IJ
K
.... (2)
19.
⇒ φ'( x ) = e − x
[(3/2)x 2 – 1(3x)] 9 .
=
14.
3 9 −12
2 9−6
+ 2 . 9 C7 ( −1)7 .
y=
3 9 −14
and z =
Then y = z. Now
z
π/2
I=
x 2)
=
21.
2 sin 2 ( x / 2)
π/3
5 /2
We have x = a cos θ + (1/2) b cos 2θ,
z
∴
dy
dx
= −a sin θ − b sin 2θ, = a cos θ + b cos 2 θ
dθ
dθ
⇒
dy a cos θ + b cos 2θ
=
dx −a sin θ − b sin 2θ
1 cos( x / 2 )
dx
4 π/3 sin5 ( x / 2)
⇒
d2y
dx 2
Now put sin (x/2) = t and cos (x/2) dx = 2 dt
∴ I=
=
16.
1
2
z
1/ 2
1/2
dt
t
5
=
LM OP
MN PQ
1 t −4
2 −4
1/ 2
=
1/ 2
1
−1 −4
t
8
1
2
2
=
⇒
−1
( 2 )4 − (2 )4
8
dx 2
c h
a
b
ax + b
a b
ax + b
b
c
bx + c = b c
bx + c
d2y
∴
FG
H
IJ
K
d a cos θ + b cos 2θ dθ
dθ a sin θ + b sin 2θ dx
R| (a sin θ + b sin 2θ)(−a sin θ − 2b sin 2θ U|
| −(a cos θ + b cos 2θ)(a cos θ + 2b cos 2θ) |V
= −S
(a sin θ + b sin 2 θ)
||
||
T
W
2
dx 2
= 0 ⇒ (a sin θ + b sin 2θ) (–a sin θ – 2 b sin 2θ)
–(a cos θ + b cos 2 θ) (a cos θ + 2 b cos 2θ) = 0
0 0 −( ax 2 + 2bx + c )
0
=−
1
( −a sin θ − b sin 2θ)
Applying R 3 → R 3 – x R 1 – R 2 , we have
ax + b bx + c
⇒ – a 2 – 2 b 2 – 3 ab (cos 2 θ cos θ + sin θ sin 2 θ) = 0
= (b 2 – ac) (ax 2 + 2b x + c). Now ∆ = 0
⇒ a, b, c are in G.P. or x is a root of the equation
22.
ax 2 + 2b x + c = 0. Ans.(4)
The given equation is y = 2x.
Area OCD is below x-axis, therefore, we will take minus sign in the formula
for area OCD
Let z = px + qy. Then, z [∴ xy = r 2]
2
dz
qr 2
⇒ dz = p − qr . For max or min
=0⇒x =±
2
dx
p
dx
x
For x =
qr 2
p
, we have
Hence, z is min. for x =
2
d z
dx 2
qr 2
p
=
2 qr
2
x3
y
x’
>0
( –2
with the minimum value
zc
0
=
−2
= −2
y − 2x + 1
Distance of any point (x, y) from y = 2x – 1 is
y = x 4 + 3x 2 + 2x then this distance is S =
5
4
,–
C
O
B
2x
y=
(0, 0)
4)
D
(2, 4)
A
(2, 0)
x
y’
∴Required area = Area OCD + Area OAB
qr 2
qr 2
z=p
+
= 2r pq . Ans.(1)
p
qr 2
p
18.
a2 + 2 b2
. Ans.(3)
3 ab
⇒ a 2 + 2 b 2 = 3 a b cos θ ⇒ cos θ =
⇒ b 2 = ac or a x 2 + 2b x + c = 0
17.
d2y
−1
−1
3
4 − 16 =
−12 = . Ans.(1)
8
8
2
∆=
(1− x 2 ) .
f’(x) = 2x – a/x 2, f’(x) = 0 ⇒ 2x = a/x 2 ⇒ x = (a/2) 1/3 .
π /2
dx =
/2
y = a sin θ + (1/2) b sin 2 θ.
sin –1 x
dy
=1 .
dz
2 cos( x / 2 )
2
So 2 = (a/2) 1/3 ⇒ a = 16. Ans.(2)
Hence required differential coefficient = 1. Ans.(3)
15.
(1− t 2 )dt
Clearly for x = (a/2) 1/3 we have f”(x) > 0.
2 9−7
√(1 –
cos –1
/2
Hence, x = ± 1 are points of extremum of φ (x). Ans.(4)
20.
9.8.7 3 −3
9.8
3 −5
7
2 17
. Ans.(3)
. 3 + 2. ( −1). 2 =
−
=
12.3
12
18 27 54
.
.
2
2
sin –1 x
2
Now, φ’ (x) = 0 ⇒ 1 – x 2 = 0 ⇒ x = ± 1.
For x 0 , r must be 6 in (1); for x –1 , there is no value of r; and for x –3 , r must be
7 in (1). Therefore, the coefficient of the term independent of x in (2) is
C6 ( −1)6 .
−t
We have, φ( x ) = e
1
We must get the coefficients of x 0, x –1 and x –3 in
9
z
x
9
. If (x, y) is on
2
x + 3x + 1
5
dS 4 x 3 + 6 x
dS
=
⇒
= 0 ⇒ x = 0 . Also S’ (x) < 0 for x < 0 and S’ (x) > 0 for
dx
dx
5
23.
z
z z
2
h
0
2
− y dx + y dx = −2 x dx + 2 x dx
Fx I
GH 2 JK
2
−2
0
0
+2
−2
Fx I
GH 2 JK
2
0
2
= – (0 – 4) + (4 – 0) = 8. Ans.(4)
0
If plane x – 3y + 5z = d passes through (1, 2, 4) then 1 – 6 + 20 = d
or d = 15. ∴ Equation of plane x – 3y + 5z = 15
or
x
y
z
+
+ =1
15 −5 3
Hence lengths of intercepts are 15, – 5, 3. Ans.(1)
x > 0. Thus S is minimum when x = 0, and min. S is 1/√5. Ans.(4)
(28) of (48)
IC : PTpnrhm01
24.
Let length of side of cube = a. Then coordinates of corner (P) opposite to
origin are → (a, a, a).
∴
∴ 2 | x 1 − x 2 | = y12 − y22 = ( y1 + y2 ) ( y1 − y2 )
Direction ratio of diagonal
OP ⇒ a – 0, a – 0, a – 0, ⇒ a, a, a
or
25.
But arg (z 1 – z 2) =
D. Ratio are → 1, 1, 1. Ans.(2)
Distance between parallel planes d = |p 1 – p 2 |
∴ tan −1
where p 1, p 2 are lengths ⊥ ars from (0, 0, 0) to the plane. Length of ⊥ ar form
(0, 0, 0) to
3
2x – 2y + z + 3 = 0 is p 1 =
Similarly p 2 =
5
16 + 16 + 4
4+ 4+1
=
=
3
9
=1
26.
29.
5
6
|z 1 | = 12 gives a circle with centre at (0, 0) and radius 12.
B(0 + 12i)
lies on large circle]
ab sin x + b 1 − a 2 cos x
∴
= b (sin x cos α + 1− cos 2 α cos x)
Now p and q are + ive integers and hence from (1) we conclude that q is
multiple of 4 and so let q = 4s and as q lies between 1 and 102, therefore s
lies between 1 and 25.
= b (sin x cos α + sin α cos x) = b sin (x + α)
p+1 q
= = λ p + 1 = 7λ and q = 4λ
7
4
31.
Q
– 1 ≤ sin (x + α) ≤ 1
∴
c – b ≤ b sin (x + α) + c ≤ b + c
∴
b sin (x + α) + c ∈ [c – b, c + b]. Ans.(3)
x2
+ 4xy + y 2 = AX 2 + BY 2
∴ λ varies from 1 to 14 or from 1 to 25.
⇒
Hence we choose λ to vary from 1 to 14.
(X sin θ + Y cos θ) + (X sin θ + Y cos θ) 2 = AX 2 + BY 2
Thus there are only 14 common terms
On comparing coefficient of XY.
T p = 4p – 1 = 4(7λ – 1) – 1 = 28λ – 5
∴
– sin 2θ + 4 cos 2θ + sin 2θ = 0
∴
cos 2θ = 0 then 2 θ = nπ +
∴θ =
π
[for n = 0] (Q 0 ≤ θ ≤ π/2). Ans.(2)
4
Put λ = 1, 2, 3, ..., 14 and common terms are
23, 51, 79 .... . Ans.(3)
(X cos θ – Y sin θ) 2 + 4(X cos θ – Y sin θ)
For equilateral triangle
a2 + 1 = b 2 + 1 = (a – 1) 2 + (1 – b) 2
∴
a = b and (a – 1) 2 + (1 – b) 2 = a 2 + 1
⇒
a2 – 4a + 1 = 0
⇒
a=
4 ± 16 − 4
2
32.
cos ( π ( x − 4 ) cos ( π x ) = 1
Since x – 4 ≥ 0 and x ≥ 0
∴
z+z
=x
2
x ≥4
z+z
= | z − 1| = | x + iy − 1|
2
33.
∴
x = 0, 4 but x ≥ 4
∴
only one solution x = 4. Ans.(2)
cos −1 x + cos −1 y =
x 2 = (x – 1) 2 + y 2 = x 2 – 2x + 1 + y 2
If z 1 = x 1 + iy 1 , z 2 = x 2 + iy 2
Then 2x 1 = 1 + y12 ; 2 x 2 = 1 + y 22
IC : PTpnrhm01
....(2)
From (1), π x − 4 = 0 and π x = 0
Also from the given relation
⇒ 2x = 1 + y 2
....(1)
cos ( π x − 4 ) = 1 and cos ( π x ) = 1
Since 0 < a < 1, ∴ a = 2 – √3, b = 2 – √3. Ans.(2)
⇒
π
nπ π
⇒θ=
+
2
2
4
It is possible only when
= 2± 3
i.e. x = |(x – 1) + iy|
A’ (12 + 0i)
Let a = cos α then |a| ≤ 1
i.e. 4p – 1 = 7q – 5 or 4(p + 1) = 7q ...(1)
both p and q very from 1 to 102
O
(0, 0)
= OA – OP = 12 – 10 = 2. Ans.(2)
30.
be the general terms of the two series where both p and q lie between 1 and
102. We have to find the values of p and q for which T p = T q .
Let z = x + iy ∴
P A
1
5
5 + 4i
3
C
[Q z 2 lies on the small circle and z 1
and T q = 2 + 7(q – 1) = 7q – 5
28.
y1 − y2
π
= tan = 1
x1 − x2
4
of |z 1 – z 2| = PA
5 1
= . Ans.(3)
6 6
Let T p = 3 + 4(p – 1) = 4p – 1
27.
⇒
Minimum value
It is easy to observe that both the series consist of 102 terms.
∴
y1 − y2 π
=
x1 − x 2 4
∴ ± 2 = (y 1 + y 2) . 1 ⇒ y 1 + y 2 = ± 2 ∴ Im (z 1 + z 2) = ± 2. Ans.(3)
By d = p 1 – p 2
d = 1−
π
4
π
π
− sin −1 x + − sin −1 y
2
2
= π – (sin –1 x + sin –1 y) = π −
=
2π
(given)
3
π
. Ans.(2)
3
(29) of (48)
34.
Now centroid is
∴ a + b + c = 11, ab + bc + ca = 38 and abc = 40
cos A cos B cos C
a 2 + b 2 + c 2 ( a + b + c )2 − 2 ( ab + bc + ca)
+
+
=
=
a
b
c
2 abc
2 abc
∴
=
35.
FG 0, b IJ = FG 0, 3 IJ = FG 0, 1 IJ
H 3K H 3 K H 3 K
Since a, b, c are roots of x 3 – 11x 2 + 38x – 40 = 0
(11) 2 − 2 (38 ) 121 − 76 45 9 . Ans.(3)
=
=
=
2 (40 )
80
80 16
F
GH
( x − 0 )2 + y −
(g of) x = g{f(x)}
= g{sin 2 x + sin 2 (x + π/3)} + cos x cos (x + π/3)}
R|
= g Ssin
T|
=g
2
F1
x + G sin x +
H2
RS 5 sin
T4
2
x+
3
cos x
2
I
JK
2
+ cos x
F 1 cos x −
GH 2
3
sin x
2
F 0, 1 I
GH 3 JK
∴ circumcentre is
F
GH
1
F
GH
1
or x 2 + y −
I |U
JK V|
W
i.e. x 2 + y −
UV FG IJ
W H K
5
5
=1.
cos 2 x = g
4
4
40.
I
J
3K
1
I
J
3K
F
GH
2
= (1− 0)2 + 0 −
2
I
J
3K
∴ reqd. circle is
= 1+
2
=
I
J
3K
1
2
1 4
=
3 3
4
. Ans.(1)
3
Let A be the point (x 0, x 20) on the parabola y = x 2
Equation of the tangent at A to the parabola is
Hence (g of) x is a constant function. Ans.(2)
36.
f : [1, ∞) → [1, ∞)
Let
y = f(x) = 2 x(x–1) (Q x = f –1 (y))
⇒
log 2 y = x 2 – x ⇒ x 2 – x – log 2 y = 0
x=
∴
f −1 (y) =
37.
It meets the y-axis at
and the line y = x 02 meets the
2
(Q x ≥ 1)
2
1 + (1 + 4 log2 x )
2
PAQ =
+
∴
= 16
which represents a circle, as a = b and h = 0. Ans.(2)
43.
∴ The third vertex of the equilateral triangle must lie on y-axis.
and
C(0, b)
s = ut +
=4–1=3
∴ circumcentre is centroid of ∆ABC
80 sin 30° = 40 f t/sec.
If t is time from the top of minar to the point say P, where the ball strike the
ground, then using
1+ b 2 = 1+ b 2 = 2
Q triangle is equilateral
Horizontal and vertical upward components of velocity of projection at the
top of the minar are
80 cos 30° = 40√3 f t/sec.
Y
Now, AC = BC = AB
∴ b = √3 ( Q b > 0)
α
=2 ⇒ α=2β
β
∴ locus of P is x – 2y = 0, i.e., 2y – x = 0. Ans.(2)
Since two vertices are A(–1, 0) and B(1, 0).
∴
A(–1, 0)
dy
=0
dx
dy x
dy α
= . At (α, β),
=
dx y
dx β
∴
Let it be C(0, b).
2x – 2y
Also slope of the line y = 2x is 2.
or x 2 + y 2 – 6x – 7 = 0
b2
sin 2a + sin 2b = sin 2(a – b) – 2 sin (a + b). Ans.(2)
Point P (α, β) will be nearest to the line y = 2x if the tangent at P is parallel
to the line y = 2x.
Now x 2 – y 2 = a 2 ⇒
Squaring and adding, (h – 3) 2 + k 2 = 16
∴ Locus of P(h, k) is (x –
2 (cos a + cos b) 2 (sin a + sin b)
sin 2a + sin 2 b + 2 sin (a + b)
= −1 ⇒
=1
×
2 cos( a − b )
2 sin (a − b)
sin 2(a − b)
⇒
42.
1
∴ h – 3 = 4 cos θ and k =
(20 sin θ + 0) = 4 sin θ.
5
y2
If the given lines are perpendicular then,
−
θ + 15) = 4 cos θ + 3
3) 2
1
1
PQ × AQ = × 2 x 20 × x 0 = x 30 .
2
2
Ans.(3)
41.
Let P(h, k) be the point dividing the line joining given points in the ratio
2 : 3 internally, then
∴
2
Which increases in the interval [2, 3] and hence is greatest when x 0 = 3.
. Ans.(2)
therefore the greatest value of f(x) is 1. Ans.(2)
39.
x
Hence the area of the triangle
1 + 1 + 4 log2 y
FG 1 IJ (20 cos
H 5K
A(x0,x02)
P(0, –x 0 )
y-axis at Q(0, x 0 2).
The greatest value of cos θ is 1. For the point x = 0, the argument of the
given cosine function is
h=
(0, x02)
O
P(0, –x 2 0)
0 e [0] + 2.0 2 – 0 = 0
38.
Q
or 2x 0 x – y = x 20
1± 1 + 4 log2 y
Hence f −1 ( x ) =
y
1
xx o = (y + x 20 )
2
Q
0(0, 0)
B(1, 0)
X
1 2
ft , we get
2
– 200 = 40t –
1
× 32 t 2 ⇒ t = 5 sec .
2
∴ Horizontal distance of P from the foot of minar
= H.V. × time = 40√3 × 5 = 200√3 ft. Ans.(2)
(30) of (48)
IC : PTpnrhm01
44.
The body will be on the point of sliding down wards, when
α = λ, where α is the inclination of plane.
g' ( x ) =
⇒ tan α = tan λ = 1/√3 ⇒ α = 30°. Ans.(2)
45.
⇒
⇒
cx − 4h + y
x2
+
y2
2
<
c x − 2h + y
2
x2
– 8x + 16 <
+
y2
2
⇒(x –
4) 2
+
y2
< (x –
2) 2
+
We have
c
c
he
he
e j
=
| z | −2 Ree z z j + z
2
2
2
1 2
1
im
47.
2
=
ez z j = 0 , then
e j
Reez z j− | z |
1
2
z +z
∴ 1 2
z1 − z 2
∴ sin 2θ > cos 2θ
is purely imaginary. If
hence −
52.
∴ (x – 3)(3x – 28) = 0
∴ x = 3.
1
x
∴ I + F + G = (√2 + 1) 6 + (√2 – 1) 6 = 2[ 6C 0 2 3 + 6 C 2 2 2 + 6C 4 2 + 6C 6 ]
= 198. Now 0 < F + G < 2. But F + G = 198 – I is integer.
∴ y = cos φ + i sin φ
F + G = 1 ∴ I = 198 – 1 = 197. Ans.(2)
∴ xy = (cos θ + i sin θ) (cos φ + i sin φ)
= cos (θ + φ) + i sin (θ + φ)
54.
1
= cos(θ + φ) – i sin (θ + φ)
xy
55.
IJ
K
1
1
xy +
= cos (θ + φ) . Ans.(3)
2
xy
f ( 2) − f (1)
= f (2) − f (1) = 18 – 24a – 6 + 6a = 12 – 18a
2 −1
FG 7 IJ = 147 − 84a + 5
H 4 K 16 4
=
227
35
35
∴ 12 − 18 a =
− 21 a ⇒
= 3 a ⇒ a = . Ans.(2)
16
16
48
50.
f '( x ) =
R|S (sin x − x cos x) U|V = R|S cos x (tan x − x) U|V .
|T sin x |W |T sin x |W
2
2
227
− 21a.
16
∴ f(x) is an increasing function. Next,
1 4
C0 2 c 2
+
= 1+ = . Ans.(3)
1
3
3 3
/
−12
FG IJ
H K
2
+...
= 21/2 = 2 . Ans.(4)
Let t n be the nth term of the series 4 + 11 + 22 + 37 + 56 + .. Since the
differences of the successive terms in this series are in AP. So, let
tn = an 2 + bn + c
Putting n =1, 2, 3 we get
a + b + c = 4, 4a + 2b + c = 11 and 9a + 3b + c = 22.
Solving these equations, we obtain a = 2, b = 1 and c = 1.
∴ t n = 2n 2 + n + 1, n = 1, 2, ....
∞
So, sum of the series =
Now 0 < x ≤ 1 ⇒ x lies in Ist quadrant,
hence tan x > x and cos x > 0 ∴ f’(x) > 0 for 0 < x ≤ 1.
C0 1
= = 1.
1
1
FG IJ
H K
FG 1 IJ
H 2K
56.
2
1
1 3
⋅
1 1 1 3 1
1 1
1
1+ ⋅ + ⋅ ⋅ 2 +...= 1+
+ 2 2
2 2 2 4 2
2 2
2! 2
= 1−
From the given conditions
Now f'(x) = 3x 2 – 12ax + 5 ⇒ f'
Put n = 1, then S1 =
At n = 2, S 2 =
1
= 2 cos (θ + φ)
xy
f '( 7 / 4) =
(√2 + 1) 6 = I + F where I is integer and 0 ≤ F < 1 and
(√2 – 1) 6 = G. where 0 < G < 1
We get x = cos θ + i sin θ
49.
d2 V
= 6x – 37 = –19 = –ve for x = 3.
dx 2
Hence V is max. when x = 3. Ans.(3)
53.
Taking +ve sign
FG
H
The dimensions of the box after cutting equal squares of side x on the corner
will be 21 – 2x, 16 – 2x and height x.
dV
= 12x 2 – 148x + 336 = 0 or 3x 2 – 37x + 84 = 0
dx
∴ x 2 – 2x cos θ + 1 = 0 ⇒ x = cos θ ± i sin θ
∴
π
π
< x < − . Ans.(2)
2
4
V = x(21 – 2x)(16 – 2x) = x(336 – 74x + 4x 2) or V = 4x 3 – 74x 2 + 336x.
⇒ y = 0. Therefore locus of z is x-axis. Ans.(1)
∴ xy +
f( x) dx
0
Hence f(sin 2θ) > f(cos 2 θ)
⇒ x2 + (y – 5)2 = x 2 + (y + 5)2
Since 2 cos θ = x +
z
cos 2 θ
f ( x) dx +
Now for g(θ) to be increasing g´(θ) > 0
⇒ |x + iy – 5i | = |x + iy+ 5i |
48.
2
or g´(θ) = 2sinθ cosθ [f(sin 2 θ) – f(cos 2θ)]
z − 5i
= 1 ⇒ | z – 5i | = | z + 5i |
z + 5i
Let z = x + iy. Then
2
as f(x) is an increasing function
iIm z 1 z 2
z1 + z2
can be zero also. Ans.(4)
z1 − z2
1 2
U| = R| (sin x cos x − x) U| = R| (sin 2 x − 2 x) U| .
V| S|T sin x V|W S|T 2 sin x V|W
W
Then g´(θ) = f(sin 2θ) 2 sinθ cosθ – f(cos 2θ) 2sinθ cosθ
j
j
1 2
2
g(θ) =
0
z1 + z2 z1 − z2
z1 + z 2
z z1 − z1z 2 + z 2 z1 − z 2 z2
=
= 1
z1 − z 2
z1 z1 − z1z 2 − z2 z1 + z 2 z2
z1 − z2 z1 − z2
| z1| − 2i Im z1 z 2 − | z 2 |
z
sin 2 θ
51.
⇒ Re (z) < 3. Ans.(4)
46.
x)
∴ g’(x) < 0 ⇒ g(x) is decreasing. Ans.(3)
y2
– 4x + 4 ⇒ – 4x < –12 ⇒ x < 3
2
2
Again as 0 < x ≤ 1, sin 2x – 2x < 0.
Let z = x + iy ⇒ |z – 4| < |z – 2| ⇒ |(x – 4) + iy| < |(x – 2) + iy|
2
R| (tan x − x sec
S| tan x
T
=2
∞
n2
∞
n
∞
2
2n + n + 1
n!
n =1
∑
1
∑ n! + ∑ n ! + ∑ n !
n=1
n=1
n =1
= 2(2e) + e + (e – 1) = 6e – 1. Ans.(2)
IC : PTpnrhm01
(31) of (48)
57.
We have, a =
∞
∑
∞
⇒a+b+c=
= 1+ x +
65.
∞
∞
x 3n −1
x 3n
x 3 n−2
and c =
,b =
( 3n) !
(3 n − 2)!
(3n − 1)!
n=0
n=1
n=1
∑
∑
∞
x 3n
x 3 n−2
∞
∴ (1 – sin² α) + (1 – sin² β) + (1 – sin² γ) = 1 or,
sin² α + sin² β + sin² γ = 2. Ans.(4)
x 3 n−1
∑ ( 3n)! + ∑ (3 n − 2)! + ∑ (3 n − 1)!
n=0
n=1
Because l = cos α, m = cosβ, n = cos γ. Because cos²α + cos²β + cos²γ = 1,
66.
(1)
The given equation can be written as
n =1
∴Intercepts on axes by this plane are 9, 9/2 and –9/2.
x2 x3
+
+... = e x .
2! 3!
∴ Statement (1) is true.
a + b ω + c ω2 = 1 + ω x +
ω 2 x2 ω 3x3
+
+... = e ωx ,
2!
3!
(2)
Statement (2) is true.
(3)
The perpendicular distance from (2, 3, 4) to the plane
3x – 6y + 2z + 11 = 0 =
2
and, a + b ω2 + cω = e ω x , ω is an imaginary cube root of unity.
Now, a 3 + b 3 + c 3 – 3 abc
( 4)
∴ The statement (4) is not true. Ans.(4)
5
67.
F 1+ 1 I R
GG 5 JJ |SQlogFG 1+ x IJ = 2 FG x + x + x −.....IJ U|V
GH 1− 51 JK |T H 1 − x K H 3 5 K |W
3
= log 2 + log
5
Let
get
FG 3 IJ = log 3. Ans.(4)
H 2K
1
F 1 1 IJ
T =
. Then T = G −
ncn + 1h
H n n + 1K
n
∴ S=
68.
n
1
1
1
1
−
+
−
+....∞
1. 2 2 .3 3.4 4 .5
= 4ax and
x2
= 4by is
yb 1/3
+x
we get
a 1/3
+
(ab) 2/3
∴ – 2(x – 1) – 3(y – 1) + 3z = 0 or 2x + 3y – 3z = 5. Ans.(4)
69.
e
e = log e
FG 4 IJ . Ans.(2)
H eK
2
62.
2
1 ( 3 x + 4y − 5)
(3 x + 4 y − 5 )
=
. Ans.(3)
4
25
100
b2
16 3
We have, a 2 = 25, b 2 = 16, e = 1− 2 = 1−
= .
25 5
a
The equation of the plane passing through the planes 2x + 3y – 4z = 1 and
3x – y + z + 2 = 0 is given by (2x + 3y – 4z – 1) + λ (3x – y + z + 2) = 0 ....(A).
Since it passes through (0, 1, 1), ∴ (0 + 3 – 4 – 1) + λ (0 – 1 + 1 + 2) = 0
⇒ λ = 1.
Substituting the value of λ in (A). ∴ (2x + 3y – 4z – 1) + 1.(3x – y + z + 2)
= 0 or 5x +2y – 3z + 1 = 0. Ans.(4)
= 0.
Focus (1, 2), directrix is 3x + 4y = 5, so the equation of the ellipse
= (x–1) 2 + (y – 2) 2 =
a
b
c
=
=
or a/–2 = b/–3 = c/3.
−1− 1 −3 − 0 0 + 3
∴Substituting, the value of a, b, c in (A).
Here a = 8, b = 27, hence tangent is 2x + 3y + 36 = 0. Ans.(2)
61.
Let the equation of the plane passing through (1, 1, 0) is a(x – 1)
+ b(y – 1) + c(z – 0) = 0 ...(A). Since this plane also pass through the points
(1, 2, 1) and (–2, 2, –1).
∴ 0.a + 1.b + 1.c = 0 ....(B), – 3.a + 1.b – 1.c = 0 ....(C). Solving (B) and (C),
The equation of common tangent for parabolas
y2
a
b
c
=
=
or a/2 = b/–4 = c/3
4−6 6−2 3−6
∴ From (A) the equation of required plane is 2 (x + 1) – 4 (y – 3) + 3 (z – 2)
= 0 or 2x – 4y + 3z + 8 = 0. Ans.(1)
F 1 1 I F 1 1 I F 1 1 I F 1 1I
= G − J − G − J + G − J − G − J +.....∞
H 1 2K H 2 3K H 3 4K H 4 5K
F 1 1 1 I
= 2 G 1− + − +.....J –1 = 2 log ( 1 + 1) – log
H 2 3 4 K
60.
The equation of plane through (–1, 3, 2) is a (x + 1) + b (y – 3) + c (z – 2)
= 0 ....(A). This plane is ⊥ to planes x + 2y + 2z = 5 and
3x + 3y + 2z = 8.
∴ a.1 + b.2 + c. 2 = 0 ....(B) and a.3 + b.3 + c.2 = 0 .... (C). Solving these, we
= log 2 + log
59.
b+c c+a a+b
70.
q +r
r +p
p + q . Applying C 1 + C 2 + C 3 and taking 2 common from
y+z
z+x
x+y
a+b+c c +a a+b
r +p p+q .
x +y+z z+x x +y
C 1, D = 2 p + q + r
Therefore coordinates of foci S and S’ are (3, 0) and (–3, 0) respectively. Let
P(5 cos θ, 4 sin θ) be a variable point on the ellipse.
1
2
3
0
1
−3
0
1 =
5 cos θ 4 sin θ 1
1
2
3
0
1
−6
0
0 =
5 cos θ − 3 4 sin θ 0
b c+a a+b
Applying C 1 – C 2, D = 2 q r + p p + q .
y z+ x x +y
Then A = area of ∆PSS’
=
7
=1 .
7
∴ The given planes are ⊥ to each other.
FG 1 + 1 . 1 + 1 . 1 +.....∞IJ
H5 3 5 5 5
K
3
=
∴ 2. (–1) + (–1). 2 + 1.4 = – 2 – 2 + 4 = 0.
2
We have log 2 + 2
9 + 36 + 4
The direction ratios of given planes are 2, –1, 1 and –1, 2, 4.
= e x e ωx e ω x = e x ( 1+ ω + ω ) = e 0 x = e 0 = 1. Ans.(1)
58.
3.2 − 6.3 + 2.4 + 11
∴ Statement (3) is true.
= (a + b + c) (a + bω + cω2) (a + b ω2 + cω)
2
x y z
+ − =1 .
9 9 9
2 2
24
sin θ = 12 sin θ.
2
b c+a a
b c a
p or D = 2 q r p
y z+ x x
y z x
Applying C 3 – C 1 ⇒ D = 2 q r + p
Clearly, maximum value of A is 12 sq. units. Ans.(2)
63.
Given equation ax 2 + 2hxy + by 2 + 2gx + 2fy + c = 0 represents a hyperbola
if ∆ ≠ 0, h 2 > ab. Ans.(2)
64.
Ans.(3)
Interchanging C 2 with C 3 and then C 1 with C 2 and hence changing the sign
a b c
D
r = 2E or E = . Ans.(3)
2
x y z
twice. D = 2 p q
(32) of (48)
IC : PTpnrhm01
71.
a2
b2
c
a 2 − ( b − c )2
b 2 − ( c − a )2
2
2
c − (a − b)
a2
=– b 2
b2 + c2
c2 + a2
c2
a2 + b2
2
bc
a2
ca = − b 2
(b − c )2
(c − a )2
2
2
ab
c
(a − b)
bc
ca . [Applying C 2 → C 2 – C 1 ]
78.
ab
bc
ca [Applying C 2 → C 2 + 2C 3 ].
LM 3
MM−6
N−3
~
ab
OP
P
2 PQ
−1 2
1
LM 1
MM−2
N −1
LM 1
MM
N −1
OP
P
1PQ
−1 1
4 ~ −2
2
2 .[Applying
2
1
1
1
( C1), (C3 ) ]
3
2
OP
P
0PQ
0 2
0 0 [Applying C 2 → C 1 + C 2, C 3 → C 1 + C 3 ]. Obviously the 3rd
0
order minor zero. But there exists a second order non-zero minor i.e.,
2
2
2
b +c
2
2
2
c +a
2
2
2
a +b
a +b +c
C1 + C2 = − b + c + a
c +a +b
2
2
bc
2
2
ca
2
2
ab
1 2
≠ 0 . Hence rank of given matrix is 2. Ans.(4)
−2 0
79.
2
2
1 b +c
2
2
bc
2
2
ca
2
2
ab
2
= −( a + b + c ) 1 c + a
1 a +b
2
Its cofactor = ( −1)
1 1
2
1 1
4 2 2 =2 2
1 2 3
1 1 =0 ,
13
1
∆= 4
1
12 0
10 4 = 0
9
9
∴ 2x + 2yy’ – 2ky’ = 0 ⇒ k =
x + yy'
y'
FG x + yy' IJ + (x + yy')
H y' K 2(y')
y=
2
2
–x
y=
x
=0
∴ order = 1, degree = 2 i.e., a = 2, b = 1. Ans.(3)
1
3 0
80.
1
6 4 =8 0 3 4
4
5
.... (1)
from (1) x 2 + y2 − 2y
Put a = 1, b = 2, c = 3 so that
(0, k)
k2
⇒ x 2 + y 2 – 2ky +
=0
2
1 2 3
because, R 1 and R 2 are identical. Ans.(4)
73.
2
Ans.(2)
Element ‘8’ occurs in 3rd row and 3rd column.
3+ 3
FkI
GH 2 JK
2
k/√
72.
∴ equation of circle (x – 0) 2 + (y – k) 2 =
4 5
Putting V = y/x in the given equation, we have
V+x
1 2 5
dV
dV
2
2
= V − cos V ⇒ x
= − cos V
dx
dx
⇒ –sec 2 V dV = dx/x ⇒ –tan V = log x + C
= 8 [21 – 3] = 144. Option (3) also gives the same value on substituting a
= 1, b = 2, c = 3. Ans.(3)
Put y = (π/4) when x = 1, we have
– tan (π/4) = log 1 + C, i.e., C = –1.
74.
1/ a a 2
1/ b b 2
c2
1/ c
75.
bc
1 a3
1
ca =
1 b3
abc
ab
1 c3
abc
1 a3
abc
abc =
1 b3
abc
abc
1 c3
1
1 = 0 . Ans.(4)
1
Therefore, –tan (y/x) = log x – 1. Ans.(2)
81.
The characteristic equation of A is given by |A – λΙ| = 0
3−λ
0
0
0
3−λ
2
0
2
2−λ
i.e.
the integrating factor e
1− λ
2
2
−1− λ
=0
82.
so that A 8 = (5Ι) 4 = 625Ι. Ans.(2)
0
0
l
n
−m
−l
LM l + m + n
0
= MM
nl + lm − mn
MM−ml + mn − ln
N
2
2
2
OP LM l
−1P Mm
0 P Mn
PM
0 PQ MN 0
0
0
n
0
l
OP
n P
−l P
P
0 PQ
83.
−m
0 −m
−1 0
OP L1
PP = MM0
−nm + nl + ml M0
P M
n + l + m PQ MN0
OP
P
0P
P
1PQ
nl + lm − mn
−lm + mn − ln
0 0 0
1
0
0
1 0 0
0
n 2 + l2 + m 2
2
2
2
1
[cos (x + 2π) – 5 4 cos (5x + 2π)]
2
y = ax n+1 + bx –n
⇒
0
0 − mn + ln+ lm
1
(cos x − cos 5 x ) .
2
1
= [cos x − 5 4 cos 5 x ] . Ans.(2)
2
A square matrix is orthogonal if
n
y=
∴ y4 =
A2 – 5Ι = 0 or A 2 = 5Ι,
m
= elog y = y we get
xy = y + 1 or (x – 1) y = 1. Ans.(2)
The characteristic equation of given matrix is
LM l
0
AA’ = Ι M
MM n
MN−m
1
dy
y
Putting x = 2 and y = 1, we have C = 2 – 1 = 1.
or λ 2 – 5 = 0. By Cayley’s Hamilton theorem.
77.
z
d
( xy) = 1 ⇒ xy = y + C .
dy
= 0 i.e. (3 – λ)(λ – 1)(λ – 4) = 0. Therefore the
characteristic roots of A are 1, 3, 4. Ans.(2)
76.
Since the subtangent is given by y(dx/dy), so we have y (dx/dy) + x = 1 or
dx/dy + (1/y) x = 1/y, which is linear equation in x. Multiplying both sides by
0
1
0 0
dy
d2y
= a(n + 1) x n + b ( −n)x −n−1 ⇒ 2 = an (n + 1) x n−1 − bn ( −n − 1)x −n− 2
dx
dx
⇒ x2
d 2y
= an(n + 1) x n+1 + b n(n + 1)x −n
dx 2
= n(n + 1) (ax n+1 + bx –n ) = n(n + 1) y. Ans.(2)
i.e. l 2 + m 2 + n 2 = 1 and nl + lm – mn = 0, which is clearly satisfied by the given
values of l, m, n. Ans.(1)
IC : PTpnrhm01
(33) of (48)
84.
We have f(x) g(x) = 1.
Differentiating with respect to x, we get f’g + fg’ = 0 ... (i)
90.
xf (2) − 2 f ( x )
xf(2) − 2 f (2 ) + 2 f (2 ) − 2 f ( x )
= lim
x →2
x−2
x−2
lim
x →2
Differentiating (i) w.r.t. x, we get f”g + 2f’g’ + fg” = 0 ... (iii)
= 4 – 2 x 4 = – 4. Ans.(3)
91.
F f "' 3g" IJ ( f ' g) = −FG g"' + 3 f" IJ( fg')
⇒G +
H f' g K
H g' f K
[using (i)]
FG
H
∴ lim
IJ
K
x →3
f "' 3g" g"' 3 f"
f"' g"'
f " g"
. Ans.(2)
+
=
+
⇒ −
=3
−
f'
g
g'
f
f ' g'
f g
⇒
dy
= f '( e x ) ⋅ e x
dx
2
∴
d y
dx
2
92.
86.
87.
n→ ∞
n→ ∞
and
FG x + y IJ = f (x ) + f (y) . Replacing x by 3x and y by zero, then
H 3K 3
= lim
h→ 0
h
93.
....(1)
F 3 x + 3h IJ − f(x)
fG
H 3 K
f ( x + h) − f ( x )
= lim
f '( x ) = lim
h→ 0
h
h→ 0
94.
95.
f (3 x ) + f ( 3h)
− f (x)
f(3h) − f ( 0)
3
[from (1)] = f'(0) = 3
= lim
h→ 0
h
3h
Hence f(x) is continuous and differentiable every where. Ans.(3)
h→0
R| cos x,
f (x) = S
||T1− sin x,
=
z FGH
z
z
=
π
≤x≤0
2
π
0<x≤
2
=
−
f ( 0 − h) − f ( 0)
cos h − 1
= lim
=0 .
h→ 0
h→ 0
h
h
89.
Applying L-Hospital rule, lim
sin x − x +
x →0
x5
6x
− sin x +
6 = lim − cos x + 1
= lim
x→0
x→0
20 x 3
60 x 2
= lim
sin x
x→0 120 x
(34) of (48)
= lim
x→0
3x2
x3
cos x − 1+
6 = lim
6
x→0
5x4
IJ
K
z z
x2
π
π
π
− x dx =
+ c . Ans.(2)
1.dx − x .dx = x −
2
2
2
2
z
z
sin 4x
2 sin 2x cos 2x
dx = 2 sin 2x dx = – cos 2x + c. Ans.(1)
dx =
cos 2x
cos 2x
1
3x + 4 − 3x + 1
ze
dx
3x + 4 + 3x + 1
je
3x + 4 + 3x + 1
zc
z
3x + 4 − 3 x + 1
3x + 4 + 3x + 1
1
dx =
3
3x + 4 − 3 x + 1
1
3
hc
h
3 x + 4 dx +
R| 3x + 4
S| c 3h
|T 3 × 2
3/2
1
3
z
ze
j
dx
j
3 x + 4 + 3 x + 1 dx
3 x + 1 dx
U| R| 3x + 1
V| + 31 S| c 3h
|W |T 3 × 2
3/2
U|
V| + c
|W
= (2/27) {(3x + 4) 3/2 + (3x + 1) 3/2 }+ c. Ans.(3)
and Lf ' ( 0) = lim
Hence Rf'(0) ≠ Lf'(0), i.e. f(x) is not differentiable at x = 0. Ans.(1)
z
1
=
3
f (0 + h) − f ( 0)
1− sin h − 1
= lim
= −1
h→ 0
h
h
R|S FG π − xIJ U|V dx
|T H 2 K |W
sin −1(cos x ) dx = sin −1 sin
=
We have f(x) = min {1, cos x, 1 – sin x}
Now, Rf ' (0) = lim
FG1+ 1 IJ FG 2 + 1 IJ
H n K H n K = 1× 2 = 1 . Ans.(3)
F 1 I 6×1 3
6 G 1+ J
H nK
z
∴ f(x) = 3x + c, Q f(0) = 0 + c = 3, ∴ c = 3, then f(x) = 3x + 3
∴ f(x) can be rewritten as
3
n→ ∞
3
f ( 3 x ) + f (0 )
⇒ f ( 3 x ) − 3 f ( x ) = − f (0 )
f (x ) =
3
88.
2
= lim
1 d
1
( f ( x )) 2 = . 2 f ( x ) . f '( x ) = f(x) f’(x). Ans.(2)
2 dx
2
Given f
1
. Ans.(4)
9
3
By definition the given limit
=
1
x
, f ' (3 ) =
9
(18 − x 2 ) 3 /2
F 1+ 4 + 9 +.....+n I = lim n(n + 1) (2n + 1)
GH n + 1 JK
6 (n + 1)
lim
x
= e . f” (e x ) . e x + f’ (e x ) . e x
= e 2x . f” (e x ) + f’(e x ) . e x . Ans.(4)
= (18 − x 2 ) −1/ 2 .
(18 − x 2 )
f ( x ) − f (3 )
f '( x) − 0
= lim
(By L' Hospital rule)
x → 3 1− 0
( x − 3)
= f'(3) =
Let y = f(e x )
Then
1
Since f ( x ) =
∴ f '( x) =
FG f"' + 3g" IJ (fg') = −FG g"' + 3f" IJ fg'
H f' g K
H g' g K
⇒−
85.
x →2
f"'
g"'
3 f"
3g"
( f ' g) + ( fg') + ( fg') +
( gf ') = 0
f'
g'
f'
g
⇒
(x − 2 ) f( 2)
f(x) − f (2)
− 2 lim
= f(2) – 2 f'(2)
x →2
x−2
x−2
= lim
Differentiating (ii) w.r.t. x, we get f”’ g + g”’ f + 3f”g’ + 3g”f’ = 0
96.
z
z
2 + 3 cos x
2
3 cos x
+
dx =
dx
sin 2 x
sin2 x sin2 x
z
= (2 cos ec 2 x + 3 cot x cos ec x ) dx
z
z
= 2 cos ec 2 x dx + 3 cos ec x . cot x dx = –2 cot x – 3 cosec x + c. Ans.(1)
1
cos x
=
. Ans.(1)
120 120
IC : PTpnrhm01
z
∞
97.
0
z
1
x log x
dx =
(1+ x 2 )2
0
z
∞
x log x
dx +
(1+ x 2 ) 2
1
x log x
(1+ x 2 )
104. Two women can be arranged in the four chairs in 4P 2 ways. Then, three men
can be arranged in the remaining six chairs in 6P 3 ways. Hence, the number
of ways in which two men and three men can be arranged is 4 P2 × 6P3 . Ans.(3)
.... (i)
dx
105. Ans.(3)
(Q function changes its nature on x = 1)
z
∞
Consider I =
1+ y
⇒
z
2 2
(1+ x )
z
98.
dx =
Let I = log x +
0
z
z
z
2
j
2
j
2
dx put it in (i)
dx = 0 . Ans.(1)
2
12 C
P ( E) =
7
. Ans.(2)
12
⇒ dx = sec 2θ dθ
log (tan θ + cot θ)
10 C
1
+
n(S) =
25 C
z
1
= 25
2
z
2
π/2
log (sin θ cos θ) dθ = −
0
z
π/2
log sin θ dθ −
0
log cos θ dθ
l n sin θ
⋅ cos θ dθ =
cos θ
z
π /2
z
π/4
=
a
0
x
2 3
(2 a − x )
dx =
tan 3 θ (sec 2 θ − 1) dθ =
4
tan θ
4
n(E) 25
= = 1. Ans.(1)
n (S) 25
1
= 12
n(E) 5
= . Ans.(2)
n (S) 12
∑(1+ a + a 2 +...+ an−1)
π/4
−
0
2
tan θ
2
z
∞
0
z
0
0
1 1
log 2 1
− + log 2 =
− . Ans.(1)
4 2
2
4
101. The number of five-digit telephone numbers which can be formed using the
digits 0, 1, 2, ..., 9 is 10 5. The number of five-digit telephone numbers which
have one of their digits repeated is 10 P5 = 30240. Thus, the required number
of telephone numbers is 10 5 – 30240 = 69760. Ans.(4)
102. First and second prizes in Mathematics (Physics) can be awarded in 30 P 2
( 30 P2 ) ways. First prize in Chemistry (Biology) can be awarded in 30 (30)
ways. Therefore, N = ( 30 P 2 ) 2 (30 2 ) = 30 4 29 2 = 2 4 . 3 4 . 5 4 . 29 2. Since 400
= 2 4 . 5 2, 600 = 2 3, 3.5 2 and 8100 = 2 2. 3 4 . 5 2 we get
N is divisible by each of 400, 600 and 8100. Also, N is divisible by four
distinct primes, viz. 2, 3, 5 and 29. Ans.(4)
103. The total number of 5 digit numbers that can be formed using the digits 1, 2,
3, 4 and 5 is 5! =120. If 5 occurs at the units place, then the remaining digits
can be arranged in 4P 4 = 4! = 24 ways. Similarly, 4 can occur at the units
place in 4P 4 = 24 ways, and so on. Thus, the sum due to the units place of all
the 120 numbers is 24(1 + 2 + 3 + 4 + 5) units = 24 × 15 units = 360 units.
Next, 5 can occur at the tens place in 24 ways. The same is true of the
remaining digits. Thus, the sum due to the tens place of all the 120 numbers
is remaining digits. Thus, the sum due to the tens place of all the 120 numbers
is 24(1 + 2 + 3 + 4 + 5) tens = 24 × 15 tens = 360 tens Sums for the remaining
places are obtained similarly. Hence, the sum of all the 120 numbers is given
by 360 (1+ 10 + 100 + 1000 + 10000) = 360 × 11111 = 3999960. Ans.(4)
IC : PTpnrhm01
∞
n−1
n=1
a
1
[(1+ b + b 2 + ... ∞)] −
[(1+ ab + (ab )2 +... ∞)]
1− a
1− a
=
1
1
a
1
.
−
=
. Ans.(3)
1− a 1− b (1− a)(1− ab ) (1− ab)(1− b )
110. The required sum to n terms is
+
1
3+ 5
1
+
5+ 7
e
1
[( 3 − 1) + ( 5 − 3 ) +
2
π/4
a
∑ bn−1 − 1− a ∑ (ab )
n=1
=
1
tan 3 θ sec 2 θ dθ − tan θ (sec 2 θ − 1) dθ
+ log |sec θ||0π/4 =
1
a n bn −1
=
1− a
1− a
n =1
n=1
n
∞
∞
bn−1
F 1− a I
∞
∑ GH 1− a JK b n−1
n=1
∑ 1− a − ∑
1+ 3
π/4
π /4
b n−1 =
n =1
2 a cos θ dθ
tan5 θ dθ
0
0
=
z
π
l n 2. Ans.(2)
2
π /4
5
2
l n sin θ dθ = −
0
100. Let x = 2 a sin θ, so dx =
z
12 C
∴ P (E) =
Let x = sin θ ⇒ dx = cos θ dθ
∴ I=
= 10 + 15 = 25
108. n(E) = 5C 1 = 5
0
= − [ − π / 2 log 2 ] − [ − π / 2 log 2 ] = π log 2. Ans.(1)
0
1
∞
(sin θ + cos θ)
dθ
sin θ cos θ
log
π/2
z
15 C
1 + (1 + a) b + (1 + a + a 2) b 2 + (1 + a + a 2 + a 3) b 3 + ... to ∞
(1+ tan 2 θ)
0
∴I=
= 12
109. We have,
2
sec θ dθ
log (tan θ + cot θ) dθ =
π /2
1
107. n(E) =
n(S) =
IJ
K
0
99.
n(S) =
∴ P(E) =
x log x
1+ x
0
π/2
z
ze
x log x
1+ x
0
1
dx
⋅
. Let x = tan θ
x 1+ x 2
π/2
=−
dx −
2 2
(1+ x )
0
=
j
dy = −
2
1
x log x
π/2
∴ I =
2
ze
1
y log y
1+ y
0
0
FG
H
∞
j
1
x log x
0
2
ze
dy = −
2
1
1
⇒ dx = − 2 dy
y
y
dx, put x =
1
y log (1/ y)
1
∞
2 2
1 (1+ x )
ze
0
∴ I= −
x log x
106. n(E) = 7C 1 = 7
=
+ .....+
1
2n − 1 + 2n + 1
j
7 − 5 + ( 2n + 1 − 2n − 1) ]
1
( 2n + 1 − 1) . Ans.(4)
2
111. We have,
Length of a side of S n = Length of a diagonal of S n+1
⇒ Length of a side of S n = 2 (Length of a side of S n+1
⇒
Length of a side of S n+1
1
=
for all n ≥ 1.
Length of side of Sn
2
⇒ Sides of S 1..S2...,S n form a G.P. with common ratio
∴Sn
F 1I
= 10 G
H 2 JK
n−1
=
10
n−1
2 2
Now, area of S n < 1 ⇒
⇒ Area of S n = (side) 2
100
2n−1
1
2
and first term 10.
F I
G 10 JJ
= G
GH 2 n 2− 1 JK
2
=
100
2n −1
< 1⇒ 2 n−1 > 100 ⇒ n − 1≥ 7 ⇒ n ≥ 8 . Ans.(2)
(35) of (48)
112. We have, 1 + | cos x | + cos 2 x + | cos 3x | + ... =
1
= α (say)
1−|cos x|
∴ exp {α.log e 4} = exp {log e 4 α } = 4 α
2x 1 + 12 – 1 = 0 ⇒ x 1 = –
11
2
5
.
3
and –4 + 3y 1 – 1 = 0 ⇒ y 1 =
Now, y 2 – 20y + 64 = 0 ⇒ (y – 4) (y – 16) = 0
⇒y = 4 or y = 16
Since the given expression satisfies the given equation.
(Thus, x 1, y 1 ) satisfies 2x + 6y + 1 = 0 . Ans.(2)
118. Since the origin remains same. So, lengths of the perpendicular from the
Therefore, 4 α = 4 or 4 α = 16
⇒ α = 1 or α = 2 ⇒
1
1
= 1 or
=2
1− |cos x |
1− |cos x |
⇒ | cos x | = 0 or | cos x | = 1/2 ⇒ x = π/2 or x = π/3, 2π/3. Ans.(4)
x y
+ = 1 are equal. Therefore,
p q
and
113. AB ⊥ AD and CD ⊥ CB & mid point of AC = mid point of BD
(γ, δ)
D
(γ, δ)
C
A
(α, β)
1
1
a
Hence ABCD is a square. Ans.(2)
114. From triangle OQ 1Q 2 , by applying cosine formula
Q 1Q 22 = OQ12 + OQ 22 − 2OQ1.OQ 2 cos Q1OQ 2
2
)2
2
2
2
+
1
b
2
1
=
1
p
2
+
1
q
⇒
Thus, the equation of AA’ is
or (x 1 – x 2 + (y 1 – y 2 = x 1 + y 1 + x 2 + y 2 – 2OQ 1.OQ 2 cos θ
115. Let the origin be transferred to the point (h, k) then replacing x, y by x + h,
y + k respectively in the given equation we get
given by
i.e. 2h + k – 7 = 0 and h + 4 k – 5 = 0.
⇒ x = 2 − 2 2 ,y = 1− 2 2
23
3
Solving these we get h =
and k =
7
7
FG 23 , 3 IJ . Ans.(1)
H 7 7K
=
1=
y2
1
a
1
ar
2 2
ar
b
1
bs
1=
bs
2
1
1
1
ab r
2
2
r
1
1
s
1
2
1
s
1
1
1
1
ab r − 1 s − 1 0 Applying R 2 – R 1, R 3 – R 1
2
2
2
r −1 s −1 0
1
=
ab ( r– 1) (s–1)
2
1
1
1
1
1
0
+
1
q
2
. Ans.(2)
x−2
y−1
=
cos π / 4 sin π / 4
y'
121. The equation of tangent to x 2 + y 2 = 25 at (3, 4), is 3x + 4y – 25 = 0 which
meets with the axes at A(25/3, 0) and B(0, 25/4). Therefore the required
area =1/2(OA)(OB) = 1/2(25/3) (25/4) = 625/24 sq. units Ans.(3)
122. If a circle bisects the circumference of another circle, then their radical axis
must pass through the centre of the second circle. Equation of radical axis
is 2 (g – g 1) x + 2 (f – f 1) y + c – c 1 = 0. If it is passes through centre ( –g 1, –f1 )
of second circle then; 2g 1(g – g 1) + 2f 1 (f – f 1) = c – c 1 . Ans.(4)
123. Let the three circle be x 2 + y 2 = a 2 , x 2 + y 2 = b 2, x 2 + y 2 = c 2 such that a 2, b 2,
c 2 are in AP. Let P(x 1, y 1 )be any point from which the length of the tangents
to these circles be PA, PB, PC respectively.
Since a 2, b 2, c 2 are in AP, therefore –a 2, –b 2,–c 2 are in AP
⇒ x 1 2 + y 12 – a 1 2, x 12 + y 12 – b 12, x 12 + y 12 – c 2 are also in AP
r +1 s +1 0
⇒ PA 2, PB 2, PC 2 are in AP. Ans.(2)
124. The given equation is x 3 + x 2 + x + 1 = 0 or (x 2 + 1) (x + 1) = 0 or x = ± i,
x = –1. Ans.(2)
117. The equation of the line joining the points (2, –1) and (5, –3) is given by
−1+ 3
(x – 2) i.e. 2x + 3y – 1 = 0
2−5
2
Then PA 2 = x 12 + y 12 – a 2, PB 2 = x 12 + y 12 – b 2, PC 2 = x 12 + y 12 – c 2.
1
=
ab (r –1) (s–1) (s – r). Ans.(3)
2
y+1=
1
p
120. Let OA = a, OB = b. Since tangents at A and B meet at right angles in P (h, k),
OPAB is a rectangle ⇒ OP 2 = OB 2 + BP 2 = h 2 + k 2 = a 2 + b 2 . Hence locus of p
is x 2 + y 2 = a 2 + b 2 which is concentric circle with given circles. Ans.(3)
The area of the triangle is given by
1
=
Hence the coordinates of A’ are (2–2 2 , 1 – 2 2 ) . Ans.(3)
116. We have, x 1 , = ar, x 2 = ar 2 , y 1 = bs, y 2= bs 2
b
2
π
π
, y = 1 – 4 sin
4
4
⇒ x = 2 – 4 cos
y1
1
b
x −2
y−1
=
= −4
cos π / 4 sin π / 4
⇒ If first degree terms are to be removed, then the coefficient of
X and Y should Vanish.
a
1
x1
2
x2
+
Since AA’ = 4 and A’ lies in third quadrant therefore the coordinates A’ are
(x + h) 2 + (x + h) (y + k) + 2 ( y + k) 2 – 7 (x + h) – 5 (y + k) + 12 = 0
∆=
2
2
2
or x 1x 2 + y 1y 2 = OQ 1.OQ 2 cos Q 1OQ 2. Ans.(3)
The required point is
1
a
119. Since the point A (2, 1) is translated parallel to x – y = 3. Therefore AA’ has
the same slope as that of x – y = 3. Therefore, AA’ passes through (2, 1) and
has the slope = 1. Here tan θ = 1
y
1)
1
1
2,
,sin θ =
⇒ cos θ =
A(
2
2
x'
x
3
A'
y=
x–
B
(γ, β)
)2
x y
+ =1
a b
origin on the line in its positions
125. Ans.(4)
126. The given equation can be written as |x 2 – x + 1|
..... (1)
Since (x 1, 4) and (–2, y 1) line on 2x + 3y –1 = 0, therefore
= |x 2 – 2x + 3| or |(x – 1/2) 2 + 3/4| = |(x – 1) 2 + 2|.
∴ x 2 – x + 1 = x 2 – 2x + 3 or x = 2. Ans.(2)
127. The required equation is (x – 2) (x – 3) (x – 4) = 0
i.e. x 3 – 9x 2 + 26x – 24 = 0. Ans.(3)
(36) of (48)
IC : PTpnrhm01
128. We have 2x 2 + 3|x| + 1 = 2 |x| 2 + 3|x| + 1 > 0.
Thus, the equation 2x 2 + 3|x| + 1 = 0 has no real roots. Ans.(4)
129. For f(θ) = a cos θ + b sin θ, the maximum value is
a 2 + b 2 . Hence a = 5 and b = 12. ∴
5 2 + 12 2 = 13 . Ans.(3)
(n + 1)! (n + 2)!
n!
136. D = (n + 1)! (n + 2)! (n + 3)! . Taking common factors out from 1 st , 2 nd , 3 rd
(n + 2 )! (n + 3)! (n + 4)!
rows, we have
130. First angle = 60°
Second angle = 60 g , = 60 ×
90
degrees = 72°
100
5π
5 × 180
Third angle =
radian =
= 150° .
6
6
1 (n + 1)
(n + 2 )(n + 1)
D = [n!(n + 1)!(n + 2 )!] 1 (n + 2) (n + 3 )(n + 2)
1 (n + 3) (n + 4 )(n + 3)
Operating R 2 → R 2 – R 1 , R 3 → R 3 – R 1
∴ Fourth angle = 360° – (60° + 72° + 150°) = 78°. Ans.(1)
131. Let t 1, t 2 , t 3 denote the three expressions on the left.
t 1 = 3 {(sin x – cos x) 2} 2
= 3 {sin 2 x + cos 2 x – 2 sin x cos x} 2
= 3 (1 – 2 sin x cos x) 2
= 3 (1 + 4 sin 2 x cos 2 x – 4 sin x cos x)
t 2 = 4(sin 6 x + cos 6 x)
= 4 (sin 2 x + cos 2 x) (sin 4 x + cos 4 x – sin 2 x cos 2 x)
= 4{(sin 2 x + cos 2 x) 2 – 2 sin 2 x cos 2 x – sin 2 x cos 2x}
= 4{1 – 3 sin 2 x cos 2 x}
t 3 = 6 {sin 2 x + cos 2 x + 2 sin x cos x}
= 6 (1 + 2 sin x cos x)
∴ t 1 + t 2 + t 3 = 3 + 4 + 6 = 13. Ans.(3)
132. L.H.S = sin 6 θ + cos 6 θ + 3 sin 2 θ cos 2 θ (sin 2 θ + cos 2 θ)
= (sin 2 θ) 3 + (cos 2 θ) 3 + 3 sin 2 θ cos 2 θ (sin 2 θ + cos 2 θ)
= (sin 2 θ + cos 2 θ) 3 = 1. Ans.(2)
133. y =
2x 4
–
1
1
x = − , 0,
2
2
F d yI
GH dx JK
F d yI
GH dx JK
F d yI
> 0, G
H dx JK
< 0,
2
x =− 1/ 2
x=0
R.H.S. = a 0 + a 1x + a 2x 2 + a 3x 3 + ... .
Comparing the coefficient. of x, x 2, x 3
a1 = C 1, a 2 = C 2 + 2, a 3 = C 3 + 2 C 1 ...(1)
But 2a 2 = a 1 + a 3 [ Q a1, a 2 , a 3 are in A.P.]
∴ 2 ( nC2 + 2) = nC 1 + ( nC 3 + 2. n C 1)
⇒ n 3 – 9n 2 + 26n – 24 = 0
⇒ (n – 2) (n – 3) (n – 4) = 0 ⇒ n = 2, 3, 4. Ans.(4)
This passes through (4, 3), therefore
9=4+C⇒C=5
z
4
( 2 x − x ) dx
134. T 5 =
4
T6 =
nc
135.
LM2
MM0
N1
=
OP LM− x
PP MM 0
QNx
LM5x
=M0
MN 0
0
1
10 x − 2
OP LM1
0 P = M0
5 x PQ MN0
5
OP
PP
Q
–b) 5.
⇒
tan2 x(1+ tan2 x ) dx
OP
0 P (given)
1PQ
0 0
1
0
tan 2 x sec 2 x dx =
0
z
LM tan x OP
MN 3 PQ
3
π /4
=
0
1
3
π/4
a 3 + a 5 = (tan3 x + tan 5 x ) dx
0
=
∴ 5x = 1, 10x – 2 = 0 ∴ x =
IC : PTpnrhm01
z
z
π/4
a n–5 (
14 x 7 x
1
0
−4 x −2 x
0
0
(tan2 x + tan4 x ) dx
0
a n c 5 n − 4 . Thus, a n − 4 . Ans.(2)
=
=
=
5
b
5
b n c4
0 7
1 0
−2 1
0
tann x dx ⇒ a 2 + a 4 =
π /4
T 5 + T 6 = 0 ⇒ nc 4 a n–4 b 4 – nc 5a n–5 b5 = 0
⇒
z
π/4
2
7
sq. units. Ans.(1)
120
–b) 4 .
z
π /4
140. a n =
−1/ 2
a n–4 (
dy 1
= ⇒ 2ydy = dx .
dx 2y
Hence circle x 2 + y 2 = 4a 2 as centroid divides median in ratio of 2 : 1. Ans.(4)
x =1/2
∴ Required area =
nc
2n(n − 1)
n(n − 1)(n − 2)
+4 =n+
+ 2n
2
12.3
.
139. Centre (0 , 0), radius = 3a × 2/3 = 2a.
1/2
=
which is divisible by n. Ans.(1)
137. L.H.S. = (1 + 2x 2 + x 4)(1 + C 1x + C 2 x 2 + C 3x 3 +... )
Hence the equation of the curve is y 2 = x + 5. Ans.(3)
>0
2
− 4 = 2(n3 + 5n + 4n 2 + 2) − 4 = 2(n3 + 4n2 + 5n)
Integrating, we get y 2 = x + C.
2
2
2
D
(n!)3
138. Accordingly
2
then
∴
⇒
x2
dy
dy
3
= 8 x − 2 x for max.or min.
=0
dx
dx
∴
1 n +1 (n + 1)(n + 2)
D = [n ! (n+ 1)! (n + 2)! ] 0 1
2(n + 2 )
0 2
4n + 10
=
z
z
π /4
π/4
0
0
tan3 x (1+ tan2 x ) dx =
LM tan x OP
MN 4 PQ
4
π/4
=
0
tan 3 x sec 2 x dx
1
1
. Similarly a 4 + a 6 =
5
4
∴ a 2 + a 4, a 3 + a 5, a 4 + a 6 are in H.P. Ans.(3)
1
. Ans.(4)
5
(37) of (48)
141. Put tan x = t,
z
I = (1+ t 2 )
=−
(1+ t 2 )2
dt
t4
1+ t2
=
z
1+ t4 + 2 t 2
t4
dt =
z z z
dt
t4
+ dt +
2
t2
dt
π
2 = 1 + i = −1 . Ans.(2)
3 π −1− i
cos π + cos
2
cos 0 + cos
146.
z 2 + z1 =
147.
z1 − z 3 1 − i 3
=
z 2 − z3
2
1
(tanx) –3 x + tan x – 2 (tan x) – 1 + C
3
1
= − cot 3 x + tan x – 2 cot x + c
3
FG π IJ + i sin FG − π IJ = e
H 3K H 3K
= cos −
−i
π
3
z3
Therefore, K = – 1/3, L = 1, M = – 2. Ans.(3)
iπ
∴
Then applying condition for tangent on this line w.r.t. circle.
−5 − 4m
1+ m
−
z1 − z 3
= e 3 =1
z2 − z3
π
.
3
and angle between z 1 – z 3 and z 2 – z 3 is
= 5 ⇒ 25 + 16m 2 + 40m = 25 + 25m 2
2
π/3
142. Let equation of line be y = mx or y – mx = 0
z1
∴ triangle is equilateral. Ans.(3)
⇒ 9m 2 – 40m = 0
⇒ m = 0 or m = 40/9. Ans.(2)
F x + y I , z = tan u = x + y
GH x − y JK
x−y
3
148. u = tan –1
143. Direction ratios of line joining (2, 1, – 3) and (–3, 1, 7) are (– 5, 0, 10)
x −1 y z +3
are (3, 4, 5).
= =
3
4
5
and D.R’s of line
3
3
z2
3
∂z
∂z
+ y = 2 z.
∂x
∂y
⇒x
Angle between lines is given by
a12 + b 21 + c 12
25 + 0 + 100
or θ = cos −1
25 10
149. d = H cot α d=(H–h) cot (α – β)
⇒ H cot α = (H–h) cot (α – β)
F 7 I . Ans.(1)
GH 5 10 JK
FG1 + cos π IJ FG 1 + cos 3π IJ FG1 + cos 5 π IJ FG1 + cos 7 π IJ
H
8 K
8 KH
8 KH
8K H
α–β
or H =
hcot( α − β )
Ans.(2)
cot(α − β ) − cot α
FG
H
3π
5π
7π
π
. 2 cos 2
. 2 cos 2
. 2 cos 2
16
16
16
16
= 16 cos
FG
H
= 2 cos
LM
N
= cos
=
=
π
3π
5π
7π
cos
cos
cos
16
16
16
16
7π
π
cos
16
16
π
3π
+ cos
2
8
LM
N
2
OP LMcos π + cos π OP
Q N 2 8Q
2
LM
N
π
1
cos
4
4
OP
Q
=
OP
Q
α
d
z
2π
θ =0
πy 2 dx
Y
2
IJ FG 2 cos 5π cos 3π IJ
K H 16 16 K
3π
π
1
2 cos
× cos
8
8
4
2
IJ
K
h
H
d
150. For the arc OPA, θ varies from 0 to 2π. Also the base is x-axis.
∴ the required volume =
= 2 cos 2
H–h
35
∂u
∂u
+ y = 2.sinu.cos u = sin 2u . Ans.(3)
∂x
∂y
⇒ x.
9 + 16 + 25
∂u
∂u
+ y.sec 2 u = 2.tan u
∂x
∂y
β
144.
a 22 + b 22 + c 22
−15 + 0 + 50
∴ cos θ =
cos θ =
⇒ x.sec 2 u
a1 a 2 + b 1 b 2 + c 1c 2
cos θ =
2a
=π
2
145. T 1 = 1 2 , T 3 = 3 2 , T 5 = 5 2
Hence when n is odd, the last term will be n 2 and the sum of first (n – 1) i.e.
even number of terms is obtained by replacing n by n – 1 in the given formula.
Hence the sum when n is even will be
z
2π
0
= 2 πa 3
2
1
. Ans.(3)
8
x = a (θ – sin θ), y = a ( θ – cos θ)
O
2
c
h
2
a 2 1− cos θ × a (1− cos θ) dθ as x = a (θ − sin θ), y = a (1− cos θ)
z
π
0
= 32 π a 3
z
π/ 2
0
sin6
z FGH
IJ dθ
K
1
L 311O
φ d φ [ where θ = φ ] = 32 πa M(5 / 6) ⋅ ⋅ ⋅ π P = 5π a
2
N 422 Q
(1− cos θ)3 dθ = 2 πa 3
π
0
2 sin 2
1
θ
2
3
3
2 3
.
Ans.(3)
151.
LM x OP + LM x OP + LM x OP = 11x = x + x + x
N 10 Q N 20 Q N 30 Q 60 10 20 30
⇒
x x x
, ,
are all integers.
10 20 30
∴ x = multiple of L.C.M. of 10, 20, 30 but
0 < x < 1000
∴ x = 1 × 60, 2 × 60, 3 × 60, .... 16 × 60
(n − 1)n2
(n + 1) n2
. Ans.(1)
+ n2 =
2
2
(38) of (48)
∴ No. of possible values of x = 16. Ans.(2)
IC : PTpnrhm01
152. Let (h, k) be the co-ordinates of the centre of circle C 2. Then its equation is
n
n(n − 1) / 12
.
n(n − 1)(n − 2) / 3.2.1
1
= +2
+3
+..........+n.
1
n
n(n − 1) / 12
.
n
(x – h) 2 + (y – k) 2 = 5 2 = 25. The equation of C 1 is x 2 + y 2 = 4 2 and
∴ the equation of the common chord of C 1 and C 2 is
2hx + 2ky = h 2 + k 2 – 9
= n + (n – 1) + (n – 2) + ..... + 1 =
....(1)
Let p be the length of the ⊥ from the centre (0, 0) of C 1 to (1).
Then p =
2
2
h +k −9
will be of maximum length if p = 0 ⇒ h 2 + k 2 – 9 = 0
....(2)
⇒
9
12
∴ k=m
. Hence the centre of circle C 2 are
5
5
FG 9 , − 12 IJ
H5 5 K
10
and
157.
11
∆=
12
11
∆=
12
2θ
10
a
C5
C7
C9
10
11
12
C8
11
C4
C6
11
=
O
a2
= a sin 2θ + sin 4θ .
2
12
C4
C6
A
+ sin2θ cos2θ)
11
C8
10
1
∴ area of ∆ ABC, ∆ = (BC) ( AD)
2
= 1/2 (2a sin2 θ) a (1 + cos2 θ)
10
C4
C6
11
12
13
d∆
= 2 a 2 cos2 θ + 2a 2 cos4 θ.
dθ
For Max. or min. of ∆ ,
Cm+4
12
12
13
C8
C4 + 10 C5
C6 + 11 C7
D
C
11
12
C8 + 12 C9
C5
C7
C9
13
Cm
Cm+2 = 0
Cm+4
11
12
13
2
B
Cm
Cm+2 = 0
Applying C 2 → C 2 + C 1, we get
153. We have AD = a + a cos2 θ, BC = 2 BD = 2a sin2 θ
∴
2 1 1
= +
b a c
⇒
∴ a, b, c, are in H.P. Ans.(3)
FG − 9 , 12 IJ . Ans.(1)
H 5 5K
=
a
3bc – 3ab – 4ac + 4a 2 – 2bc + 2ca + 4ab – 4a 2 = 0
⇒ bc + ab – 2ac = 0
Putting the value of k in (2), we get
a 2(sin2θ
4a
1 3 b b = 0 ⇒ 0 3b − 4a b − a = 0
0 2c − 4a c − a
1 2c c
h 3
4h
− =
∴k=−
.
k 4
3
h=±
1
1 4a a
Since slope of (1) = 3/4 (given)
∴
n(n + 1)
. Ans.(3)
2
156. For non-trivial solution
. The length of the common chord is 2 h2 − p 2 which
4h2 + 4k 2
∑n =
Cm
Cm+2 = 0
Cm+4
Clearly m = 5 satisfies the above result. ( Q C 2 , C 3 will be identical)]. Ans.(3)
158. |z| – 2 = |z – i| – |z + 5i| = 0
|z| – 2 = 0 ⇒ |z| = 2
⇒ x2 + y2 = 4
d∆
=0
dθ
...(1)
Again |z – i| = |z + 5i|
⇒ cos2 θ + cos4 θ = 0 ⇒ 2θ = π – 4θ
⇒ |x + iy – i| = |x + iy + 5i| ⇒ x 2 + (y – 1) 2 = x 2 + (y + 5) 2
⇒ 6θ = π ⇒ θ = π/6,
⇒ x 2 + y 2 – 2y + 1 = x 2 + y 2 + 10y + 25
d2∆
dθ
2
⇒ 12y = – 24 ⇒ y = – 2
= −4a 2 sin2 θ – 8a 2 sin4 θ < 0,
(1) and (2) meet if
⇒ The area of the triangle is maximum for θ = π/6. Ans.(1)
1 54. Coefficient of p th , (p + 1) th and (p + 2) th terms in expansion (1 + x) n are
nC
n
n
p – 1, Cp, Cp + 1.
Then 2nCp = nCp – 1 + nCp + 1
Trick: Let p = 1, hence nC 0, nC 1 and nC 2 are in A.P.
⇒ 2. nC 1 = nC 0 + nC 2
n(n − 1)
⇒ 4n = 2+ n 2 – n
2
S2 =
2
C1
2
C0
+2
2
C2
2
C1
=
159. Since |z – 1| = 1
∴ z – 1 = cos θ + i sin θ
= 2 cos 2
= 2 cos
θ
θ
θ
+ 2i sin cos
2
2
2
FG
H
θ
θ
θ
cos + i sin
2
2
2
1
1
C1
= 1,
C0
⇒ arg z =
2
1
+ 2. = 2 + 1 = 3.
2
1
(1) and (2) does not hold condition, but (3)
put = 1, 2 .... , S 1 = 1, S 2 = 3 which is correct.
Aliter: C1 + 2. C2 + 3. C3 +...........+n. Cn
C0
C1
C2
Cn −1
IJ
K
θ
θ
∴ i tan (arg z) = i tan
2
2
Also z – 2 = – 1 + cos θ + i sin θ
= − 2 sin 2
By option, (put n = 1, 2 ...)
IC : PTpnrhm01
+ 4 = 4 i.e., x = 0
∴ required locus is a single point (0, – 2). Ans.(3)
⇒ n 2 – 5n + 2 = 0. Ans.(2)
155. Trick : Put n = 1, 2, 3 ...., then S1 =
...(2)
i.e. z = 1 + cos θ + i sin θ
⇒ n2 – n(4p + 1) + 4p 2 – 2 = 0.
⇒ 2n = 1+
x2
n(n + 1)
,
2
∴
FG
H
θ
θ
θ
θ
θ
θ
+ 2 i sin cos
= 2i sin
cos + i sin
2
2
2
2
2
2
IJ
K
θ
z−2
= i tan = i tan (arg z) .
z
2
Thus,
z−2
= i tan (arg z ) . Ans.(2)
z
(39) of (48)
165. Let O be the foot of the tower OA = 3a, AB = 4a,
160. α + β = 2a, αβ = b 2
γ + δ = 2b, γδ =
Clearly
∴ OB = 5a
a2
α +β
=a = γδ
2
P
∴ A.M. of α, β = G.M. of γ, δ. Ans.(1)
h
161. Q ∠APB = ∠BAP = α
P
∴ AB = BP = a
In ∆PBQ, sin 2α =
α
H
a
α
a
A
∴ H = a sin 2α. Ans.(3)
a
α
3α
2α
C
B
90°
α
H
Q
18
72
120
( x − 1)3 + ( x − 1) 4 +
( x − 1)5
3!
4!
5!
163. Here u is not a homogeneous function . Therefore write
x + 2y + 3 z
1+ 2(y / x ) + 3( z / x)
. Thus ω is a homogeneous
ω = sinu = 8 8 8 = x −7 .
x +y +z
1 + ( y / x )8 + ( z / x )8
a
x a−x
0
a
LM
MM
N1
LM
MM
N
OP
PP
Q
0 a + b + cO
OP LM
P
1
b P ( by R → R + R
PP MM
1 −2 : c Q N 1 1 −2
c PQ
a+b+c O
0
PP (by R → 1 (R − R ) .
b
1
3
−2 1/ 3(c − b)PQ
−2 1
[A : B] = 1 −2
Y
dx .
OP
PP
Q
169. The given equations are –2x + y + z = a, x – 2y + z = b, x + y – 2z = c.
Augmented matrix
∴ required area of the loop A = 2 x Area OAPO
z
OPLM
PPMM
QN
168. A is invertible if and only if |A| ≠ 0. Ans.(4)
164. The given curve is ay 2 = x 2(a – x).
ydx = 2
h
5a
∴ h = 5a tan β. Ans.(2)
∴ F(– α) = [F(α)] –1 . Ans.(1)
∂u
∂u
∂u
x + y + z = −7 tan u . Ans.(2)
∂x
∂y
∂z
a
∴ h = 3a tan α & in OPB tan β =
LM
MM
N
∂u
∂u
∂u
∴ (i) becomes, x cos u + y cos u + z cos u = −7 sin u or
∂x
∂y
∂z
0
h
3a
cos α − sin α 0 cos α sin α 0
1 0 0
F (α)F( −α) = sin α cos α 0 − sin α cos α 0 = 0 1 0 = 1 .
0
0
1
0
0
1
0 0 1
∂ω
∂u ∂ω
∂u ∂ω
∂u
.
But,
= cos u ,
= cos u ,
= cos u
∂x
∂x ∂y
∂y ∂z
∂z
z
B
167. We have
∂ω
∂ω
∂ω
+y
+z
= ( −7)ω ..... (i).
∂x
∂y
∂z
= 2
5a
166. |A| = –1, |B| = 3 ⇒ |AB| = –3 ⇒ |3AB| = (3) 3 × (–3) = –81. Ans.(2)
function of degree –7 in x, y, z. Hence by Euler’s Theorem
x
β
4a
∴ In ∆AOP tan α =
= 3(x – 1) 3 + 3(x – 1) 4 + (x – 1) 5 Ans.(3)
O
90°
A
162. Let f(x) = x 5 – 2x 4 + x 3 – x 2 + 2x – 1, then f(1) = 0, f i (1) = 0, f ii (1) = 0 f iii (1)
= 18, f iv (1) = 72 and f v (1) = 120. Substituting these values in the Taylor's
formula, we get f ( x ) =
3a
LM0
MM
N1
0
= 1 −2
P
X
O
A(a,o)
1
1
1
0 0
: a
: b = 1 −2
3
1
3
1
2
+ R3 )
2
Here ρ(A : B) = 3 if a + b + c ≠ 0 and ρ(A) = 2. Hence ρ(A : B) ≠ ρ(A).
∴ the given equations are inconsistent if a + b + c ≠ 0.
Putting x = a sin 2 θ ⇒ dx = 2a sin θ cos θ dθ.
∴ A=
= 4a
=
2
2
a
z
π /2
0
2
a sin2 θ a − a sin 2 θ 2a sin θ cos θ dθ = 4a
3 +1 2 +1
.
2
2
2 = 4a 2 × 3 / 2
3+2+2
2 7/2
2
2
4a 2 × 2 × 1/ 2 1/ 2
2 × 5 / 2 × 3 / 2 1/ 2
(40) of (48)
=
4 a 2 × 2 × 4 8a 2
. Ans.(2)
=
4 × 15
15
z
∴ the given equation have no solution, if (a + b + c) ≠ 0 and will have
infinite solution if a + b + c = 0. But this system does not have unique solution
in any case. Ans.(1)
π /2
0
sin3 θ cos 2 θdθ
170. Applying R 3 – (R 2 + 2R 1) we get
LM1
MM
N0
OP
P
0 PQ
1 1 −1
A~ 1 2 3
0 0
4 and
1 1
1 2
≠ 0 ⇒ ρ ( A ) = 2 . Ans.(1)
171. Let P(t, (1/2)t 2) be a point on the curve x 2 = 2y. The distance of P from the
point (0, 5) is a = √[t 2 + (1/2t 2 – 5) 2]. Let u = a 2 = 1/4 t 4 – 4t 2 + 25 ∴ du/dt = t 3–8t
and d 2 u/dt 2 = 3t 2 – 8. For max. or mini. of u, du/dt = 0 ⇒ t = 0, 2√2, – 2√2.
When t = 0, d 2u/dt 2 = –8 (–ve) ∴ u, hence a, is max. when t = 0, i.e., at point
(0,0). When t = ±2√2, d 2u/dt 2 = +16 ∴ u, hence a, is mini. when t = ±2√2. The
function u(t) is continuous and there are 3 extrema, one max. and two min.
Therefore least value of u is the lesser value of u at the two extremes. But
u at ±2√2 are same, hence the points (±2√2, 4) are closest to the point (0, 5).
Ans.(3)
IC : PTpnrhm01
172. We know that m = np = 10 × 0.002 = 0.02,
c0.02 h
e– 0.02 = 1 – .02 +
=
– ...... = 0.9802 approximately.
2 !
The number of packets containing two defective blades
m e
2!
= 10,000 ×
c h
0.02
2
× 0.9802 = 2 approximately. Ans.(2)
2!
OP
P
ω PQ
1
ω2 =
1 −1
S . Ans.(4)
3
Actual
180. If ν and V are actual and relative
velocities of the rain, then resolving
horizontally and vertically, we have
O
30°
3
45
×
0 .0135
100
100
=
=
= 0 .355 . Ans.(1)
5
25
4
30
3
45
0 .038
×
+
×
+
×
100 100 100 100 100 100
174. P(X ≤ 2) = k + 3k + 5k = 9k > 0.3 or k > 1/30
V sin 30° = 20 and V cos 30° = ν.
51 (p
51 )
(q 49 ) =
FG
H
p
100 !
⇒ =
q
50 ! 50 !
100 C
50 (p
50 )
µR
T cos α – µR = 0
...(1)
R + T sin α – W = 0
...(2)
Aα
Taking moment about A, we get
IJ FG 51!49 ! IJ = 51 ⇒ p = 51 . Ans.(4)
K H 100 ! K 50
101
or
T.AB = W.AG sin α
⇒
T = (W/2) sin α.
or
sin α cos α
tan α
(Dividing numerator and denominator by cos 2 α)
177. If α, β, γ, δ be the roots of the biquadratic
+
or µ =
+ rx + s = 0, then
(1 + α 2) (1 + β2 ) (1 + γ 2) (1 + δ2 ) = (1 – q + s) 2 + (p – r) 2
∴ (1 +
⇒ (1 +
α 2)
(1 +
β 2)
(1 +
β 2)(1
(1 +
+
γ 2)
γ 2)
(1 +
= (1 – 2 +
(1 +
δ2 )
b) 2.
= (a –
1) 2
+ (a –
1
2
P(X)
1/4
3/4
∴ 4 ≤ Distance between the centres of two circles ≤ 8
Ans.(3)
2
– (Mean) 2
1
3
From (1) equation of circles are
x 2 + y 2 – 6x + 6y + 9 = 0 and x 2 + y 2 – 30x + 30y + 225 = 0 . Ans.(4)
184. If the tangents at A, B intersect at an angle of 120°, then the chord AB will
subtend an angle of 60° at the centre O.
2
2
LM c h ωc1− ωh ωc1− ωh OP
c h MM cc hh c1−cω1−hcω1+h ωh c1−cω1−hcω1+h ωhPP
N
Q
O
O LM
1 1
1 P
1
1 P
M
P
1 P 1M ω
− 1− ω
ω P
P
= 1
[∴ 1 + ω + ω
M
ω
ω P 3
MM ωω ωω PPP
1
−1− ω P
P MN1 ω ω PQ
ω
ω Q
∴ ∆ OAB is equilateral
∴ length of the chord AB = radius of the circle = 5. Ans.(2)
185. Radius of inner circle
ω 1− ω
1
ω 1− ω
3ω 1− ω
ω 1− ω
IC : PTpnrhm01
2
3
3
2
its passes through (3, – 6)
(c – 15)(c – 3) = 0 ∴ c = 3, 15.
2
2
(x – c) 2 + (y + c) 2 = c 2 ... (1)
⇒ (3 – c) 2 + (c – 6) 2 = c 2 ⇒ c 2 – 18c + 45 = 0
2
=
(h2 + k2 ) ≤ 8
183. Circle possible in IV quadrant. Equation of circle is
FG1× 1 IJ + FG 2 × 3 IJ = 7 and variance = ΣP.X
H 4K H 4K 4
L 1 3 O F 7 I 3 . Ans.(4)
= M ×1 + ×2 P−G J =
N 4 4 Q H 4 K 16
LM1 1 1 OP
LM1 1 1 OP
ω P so that S = M1 ω ω P
S = M1 ω
MN1 ω ω PQ
MN1 ω ω PQ
LM
MM1
MM1
MN1
. Ans.(2)
16 ≤ h 2 + k 2 ≤ 64. Ans.(4)
Mean =
Also S −1 =
2 sec 2 α − tan2 α
But (h, k) lie on x 2 + y 2 = 36 then h 2 + k 2 = 36
b) 2
⇒4≤
X
2
tan α
2 + tan2 α
=
(x – h) 2 + (y – k) 2 = 4.
δ2 )
178. We have
2
2 − sin2 α
182. Let (h, k) be any point in the set then equation of circle is
Here p = a, q = 2, r = b and s = 1.
α 2)
B
(W/2) sin α cos α – µ[W – (W/2) sin 2 α] = 0
= 1 – [ 20 C 0 (0.9) 20 + 20 C 1 (0.1)(0.9) 19 + 20 C 2 (0.1) 2(0.9) 18 ] = 1 – (0.9) 18 × 4.51
= 0.323. Thus the number of samples having at least three defective parts
out of 1000 samples = 1000 × 0.323 = 323. Ans.(2)
+
T
90°
W
C
T cos α – µ(W – T sin α) = 0
or µ =
qx 2
α
G
Eliminating R between (1) and (2), we get
= 1 – (prob. that either none, or one, or two are non–defective parts).
px 3
L
α
T.AB = W.AL
∴ The probability of at least three defective in a sample of 20.
179.
R
Resolving the forces acting on the rod
horizontally and vertically, we get
(q 50 )
176. Mean number of defective = 2 = np = 20p. ∴ The probability of a defective
part is p = 2/20 = 0.1. and the probability of a non–defective part = 0.9.
x4
C
B
181. All forces are shown in figure.
175. Let X be the number of coins showing heads, and let q = 1 – p.
100 C
Opposite
direction
Dividing, tan 30° = 20/ν ⇒ ν = 20√3 km/hr. Ans.(3)
Thus minimum value of k = 1/30. Ans.(3)
Note that X ~ B (100, p), then, since P(X = 51)= P(X = 50), we have
A
V.)
ν
173. Required Probability
20
R.
V(
= 10,000 ×
2 −m
LM
MM
N
1 1
1
1 ω
3
1 ω2
2
= OR – a = a 2 + a 2 − a = a
e
Radius of outer circle = OR + RQ
2
= 0 & ω3 = 1]
=a 2 +a = a
e
j
2 −1
(a,a)
Q
P aR
O
j
2 + 1 . Ans.(3)
(41) of (48)
186. Let DC be parallel to the line y = x + 3
then equation of DC is x – y + k = 0
E a
D
.. (1)
a
Let length of side be 2a then OE = EC = a
we have | 0 – y| ≤ 5 ⇒ |y| ≤ 5 ⇒ – 5 ≤ y ≤ 5 ⇒ 0 ≤ y ≤ 5 [because y ≥ 0]
A
5−3+k
B
When x = 2, we have | 2 – y| ≤ 5 ⇒ – 5 ≤ 2 – y ≤ 5 ⇒ 0 ≤ y ≤ 7
Hence equation of DC is y = x or y = x – 4. Now only (7, 3) satisfies equation
of DC. Ans.(2)
187. Point circles of the co-axial system are
6 + 7 + 8 + 9 + 10 + 11 + 10 + 9 + 8 + 7 + 6 = 91
or x 2 + y 2 – 2x – 4y + 5 = 0 and x 2 + y 2 – 8x – 6y + 25 = 0
∴ Equation of co-axial system is
+
y2
(x 2
– 2x – 4y + 5) + λ
+
y2
gives 8 values similarly we can show that when x equals 3, 4, 5, 6, 7, 8, 9,
10 there are 9, 10, 11, 10, 9, 8, 7, 6, values of y respectively.
Therefore the number of favourable cases is
(x – 1) 2 + (y – 2) 2 = 0 and (x – 4) 2 + (y – 3) 2 = 0
(x 2
But 0 ≤ y ≤ 5, gives 6 values, when x = 1, we have |1 – y| ≤ 5
⇒ – 5 ≤ 1 – y ≤ 5 ⇒ 0 ≤ y ≤ 6, gives 7 values of y.
= 2 ⇒ 2 + k = ± 2 ⇒ k = 0 or – 4
1+ 1
making 11 × 11 = 121 ways to choose both x and y.
The number of different values of y for a given value of x can be determined
as follows. When x = 0,
2
O
(5, 3)
∴ a 2 + a 2 = 4 ⇒ a = √2
Now OE =
191. There are 11 ways to choose x and 11 ways to choose y,
C
– 8x – 6y + 25) = 0
If it pusses through origin, then 5 + 25λ = 0
∴
..... (i)
λ=–
192. n(S) = 2 4 = 16 (because each of the four places can be filled in 2 ways)
1
5
The zero determinants are
on substituting in (i)
1 1 1 1
−1 −1 −1 1 1 − 1 −1 1
1 −1 −1 −1
,
,
,
,
,
,
,
1 1 −1 −1
1 1 −1 1 1 − 1 1 −1 −1 1 −1 −1
5x 2 + 5y 2 – 10x – 20y + 25 – x 2 – y 2 + 8x + 6y – 25 = 0
⇒ 2x 2 + 2y 2 – x – 7y = 0. Ans.(3)
188. No common tangents exists for concentric circles. Ans.(1)
n(E) = 8, ∴ P(E) =
189. Let no. of red balls are x and no. of blue balls are y.
x
Probability of drawing two red balls =
C2
Now, P(A) = 1 – P ( A) = 1 – [P (all boys) + P (all girls)]
y
= 1−
C2
x +y
C2
x
Probability of drawing 1 blue and 1 red ball =
By the question-
C2
x +y
=5
C2
y
Again ,
x
C1 × y C1
x +y
C2
=6
y
⇒
=
n
2
n
9y 2
∴ required probability =
x
= 3 ⇒ x – 3y + 3 = 0
y−1
– 21y + 12 =
5y 2
– 5y
=
9
1
> ;
16 3
=
n+1
2n
n + 1 2 n−1 − 1
2 n−1 − 1
. n−1 ∴ n = (n + 1)
2n
2
2 n−1
FG 3 IJ
H 4K
3
=
27 1
> ;
64 3
But
FG 3 IJ
H 4K
9 × 10
= 45
2
FG 2 IJ
H 3K
4
=
4 2
=
6 3
16
. Ans.(2)
81
196. We have P(X + Y = 3) = P(X = 0, Y = 3)
+ P(X = 1, Y = 2) + P (X = 2, Y = 1) + P(X = 3, Y = 0)
= P(X = 0) P(Y = 3) + P(X = 1) P(Y = 2) + P(X = 2) P(Y =1 )
+ P(X = 3)P (Y = 0) ( Q X and Y are independent)
4
=
FG 1 IJ C FG 1IJ + C FG 1 IJ C FG 1IJ + C FG 1IJ C FG 1IJ
H 2K H 2K H 2K H 2K
H 2K H 2K
F 1 I F 1I F 1 I
C G J C G J = G J {(1)(35)+(5)(21) + (10)(7) + (10)(1)}
H 2K H 2K H 2K
5
7
= 5 C0
81 1
<
256 3
7
=
5
3
5
5
3
5
+
∴ he must fire 4 times. Ans.(3)
(42) of (48)
n−1
55
11
=
.Ans.(2)
1000 200
∴ Required probability =
2 1
1 3
= , where q = 1 – p = 1− = .
3 3
4 4
2
FG 1 IJ FG 1 IJ
H 2K H 2K
Probability of favourable number in one draw =
1
is obtained
Hence compute successive powers of q until q <
3
FG 3 IJ
H 4K
+n
195. Total number = 6, favourable number = 2, 3, 4, 5, i.e. 4.
n
3 1
> ;
4 3
2
=
n
C2
– 16y 12 = 0
which q n is less than 1−
n
n
Three digits numbers = 9 + 8 + 7 + ...1 =
y = 1 is not possible. So, y = 3; ∴ x = 6. Ans.(1)
=
,
FG 1 IJ
H 2K
One digit number = 1. Two digits numbers = 9
190. The probability of not hitting the target is q n. Thus we seek the smallest n for
1
2n − 1
194. The number of numbers whose sum is 9 is
y 2 – 4y + 3 = 0, (y – 3) (y – 1) = 0 ⇒ y = 3, 1.
FG 3 IJ
H 4K
2 n−1 − 1
⇒ n.2 n–1 = n.2 n–1 – n + 2 n–1 – 1 ⇒ 2 n–1 = n + 1 ⇒ n = 3. Ans.(1)
x(x – 1) = 5y. (y – 1)
(3y – 3) (3y – 4) = 5y (y – 1) ⇒
4y 2
n−1
n
C2
C2
x +y
n
P (AB) = P (one girl) =
x!
y!
y!
×
=6
⇒
( x − 1)! (y − 1)!
2 !( y − 2 )!
⇒
x+ y
FG 1 + 1 IJ = 1− 1
H2 2 K 2
P(B) = P (all boys) + P (one girl) =
C2
x!
y!
=5
⇒
2 !( x − 2 )!
2 !(y − 2)!
⇒
C1 × y C1
C2
x +y
n(E) 8 1
=
= . Ans.(4)
n(S) 16 2
193. The condition for A, B to be independent is P(AB) = P (A). P (B)
C2
x +y
Probability of drawing two blue balls =
x
91
. Ans.(3)
121
so that the required probability is
7
7
7
7
1
2
5
5
2
7
7
1
12
0
220
55
. Ans.(1)
=
2 12 1024
IC : PTpnrhm01
197. P(A ∪ B) = 0.6, P(A ∩ B) = 0.2
⇒ P(A) + P(B) – P(A ∩ B) = 0.6 ⇒ P(A) + P(B) = 0.8
The combined mean salary of all the 85 employees in the three establishments
is given by :
. . Ans.(3)
⇒ 1− P( A) + 1− P( B ) = 0.8 ⇒ P( A) + P( B ) = 12
198.
∂u
3 x 2 − 3yz
=
∂ x x 3 + y 3 + z 3 − 3 xyz
X=
. . . ( 1 ) . S i n c e u i s s y m m e t r i c i n x , y, z s o ,
=
2
∂u
3 (y − xz)
... (2)
=
∂ y x 3 + y 3 + z 3 − 3 xyz
∴
2
d 3 = X 3 − X = 340 − 320 = 20. The variance σ 2 of the combined salaries of all
the workers in the three establishments is given by
σ2 =
2
∂ u ∂ u ∂ u 3( x + y + z − xy − yz − zx)
+
+
=
∂ x ∂y ∂ z
x 3 + y3 + z3 − 3 xyz
F ∂ + ∂ + ∂ IJ u = 3 ... (4).
⇒ G
H ∂ x ∂y ∂ z K x + y + z
L ∂ + ∂ + ∂ OP u = LM ∂ + ∂ + ∂ OP ⋅ LM ∂ + ∂ + ∂ OP u
Now, M
N∂ x ∂y ∂ z Q N∂ x ∂ y ∂ z Q N∂ x ∂y ∂ z Q
L ∂ ∂ + ∂ OP LM 3 OP
=M +
N∂x ∂y ∂z Q Nx + y + z Q
L 1 − 1 − 1 OP = − 9 ⇒ k = −9 .
= 3 M−
MN (x + y + z) (x + y + z) (x + y + z) PQ ( x + y + z)
2
2
2
2
20(2500 + 225) + 25(1600 + 400) + 40(2025 +400)
85
=
20 × 2725 + 25 × 2000 + 40 × 2425
85
54500 + 50000 + 97000 201500
=
= 2370.59 ⇒ σ = 48.69 . Ans.(1)
85
85
204. In the usual notations, we are given
=
n1 = 100,n2 = 150,n3 = 250,X1 = 50, X2 = 55, X 3 = 60, σ 21 = 100,σ 22 = 121,σ 23 = 144
The arithmetic mean of the combined group of 500 items is given by
x=
199. Putting t = 0 on both sides
=
0 −1 −3
1
−3
2 −3 = 1(0 − 9) − 3 (4 + 6) = − 39 . Ans.(2)
4
0
200. Observe that u is a homogeneous function of 2nd degree. If f is homogeneous
function
⇒x
n th d e g r e e
of
then,
x
∂f
∂f
+ y . = n.f . H e r e , n = 2 ,
∂x
∂f
n1x 1 + n 2 x 2 + n3 x 3 100 × 50 + 150 × 55 + 250 × 60
=
n2 + n2 + n3
100 + 150 + 250
5000 + 8250 + 15000 28250
=
= 56.5
500
500
d 1 = x 1 − x = 50 − 56.5 = −6.5 or d 12 = 42.25,
d 2 = x 2 − x = 55 − 56.5 = −15
. or d 22 = 2.25
d 3 = x 3 − x = 60 − 56.5 = 3.5 or d 23 = 12.25 , The variance σ 2 of the combined
series of 150 items is given by
σ2 =
∂f
∂u
+ y = 2u . Ans.(2)
∂x
∂y
n1( σ 12 + d 12 ) + n2 ( σ 2 2 + d 2 2 ) + n3 (σ 3 2 + d 3 2 )
n1 + n2 + n 3
=
2
Ans.(2)
E=
6100 + 7500 + 13600 27200
=
= Rs.320.
85
85
Let d 1 = X1 − X = 305 − 320 = −15, d 2 = X2 − X = 300 − 320 = −20. and
3 ( z 2 − xy)
∂u
=
... (3).
and
∂ z x 3 + y 3 + z 3 − 3 xyz
2
n1X 1 + n2 X 2 + n3 X 3 20 × 305 + 25 × 300 + 40 × 340
=
20 + 25 + 40
n1 + n2 + n3
1
[n1(σ12 + d 12 ) + n 2 (σ 22 + d 22 ) + n3 (σ 32 + d 23 )]
n1 + n2 + n 3
=
1
[100 (100 + 42.25 ) + 150 (121+ 2.25) + 250 (144 + 12.25)]
500
1 2
1 2
2
2
2
Also we know : σ = ΣX − ( X ) = ΣX − ( 4.4 ) or
n
5
=
1
{100 × 142.25 + 150 × 123.25 + 250 × 156.25 }
500
ΣX2 = 5 × 8.24 + 5 ×19.36 = 138 or X 21 + X 22 = 138 − (12 + 2 2 + 6 2 ) = 97 .....(2)
=
From (1), we get X 1 = 13 – X 2 .
Substituting in (2) we have (13 – X 2) 2 + X 2 2 = 97
or 169 + X 22 – 26X 2 + X 2 2 = 97 or 2X 22 – 26X 2 + 72= 0 or X 22 – 13 X 2 + 36 = 0
1
71775.0
(14225 + 18487.50 + 39062.50) =
= 143.55 .
500
500
Hence, σ = 143.55 = 1198
. . Ans.(4)
201. Here we are given : n = 5, X = 4. 4 ∴ ΣX = X 1 + X 2 + X 3 + X 4 + X 5
= 5 × 4.4 = 22 or X 1 + X 2 = 22 – (1 + 2 + 6) = 13 .....(1)
13 ± 169 − 144
∴ X2 =
= 9 or 4 . Hence X 1 = 4 or 9. Ans.(3)
2
202. Let X 1, X 2, .... be the amount found in the pockets of passengers first, second,
third and so on. Then we are given n = 20, Σ X 2 = 2000 and σ = 6.
σ2 =
Now
∑( X − X) 2 =
n
2000
Σ X2
− X 2 or σ 2 =
− X2
20
n
2
∴ X = 100 − 36 = 64 . Hence X = 8 . In other words Rs.8 is the amount that
each of the passengers will have to pay as fine. Ans.(2)
203. Let the number of employees, average monthly salary and standard
deviations of the three establishments be respectively given below with the
usual notations:
n1 = 20
n 2 = 25
n 3 = 40
2
205. In the usual notations we are given n1 + n2 = 100, x = 8, σ = 10.5 or σ = 10.5
n1 = 50, x 1 = 10,σ1 = 2 or σ12 = 4 ,
n2 = 50, σ 1 = 2 or σ12 = 4, n2 = 50, x 2 = ? σ 2 = ?
x=
n1x1 + n2 x 2
or 100 x = 50 x1 + 50 x 2 .
n1 + n2
∴ 50 x 2 = 100 x − 50 x 1 = 100 × 8 − 50 × 10 = 300 or x 2 =
Also (n1 + n 2 ) σ2 = n1(σ12 + d 12 ) + n2 ( σ 22 + d 22 )
300
=6
50
...... (1)
X1 = 305
X2 = 300
X3 = 340
= 4 , d 2 = x 2 − x = 6 − 8 = −2 or d 22 = 4 .
Substituting in (1), we get 100 × 10.5 = 50 (4 + 4) + 50 (σ 22 + 4)
or 1050 = 50 × 8 + 50 σ 22 + 200 = 600 + 50 σ 2 2 , or 50 σ 2 2 = 1050 – 600
σ 1 = 50
σ 2 = 40
σ 3 = 45
= 450. ∴ σ 22 =
IC : PTpnrhm01
where d 1 = x 1 − x =10 – 8 = 2 or
d 21
450
= 9 . Hence σ 2 = 3. Ans.(2)
50
(43) of (48)
206.
I
No. of observation n1 = 50
Group
II
III
n 2 = 60
n3 = 90
n 1 + n 2 + n 3 = 200
x2 = ?
x 3 = 115
σ 2 = 7,
σ3 = ?
x = 116
σ = 7.746
x 1 = 113 ,
Mean
Standard deviation σ 1 = 6
Combined group
The formulae for Mean and S.D. of the combined group are (n 1 + n 2 + n 3)
(n 1 + n 2 + n 3)
σ2
2
2
2
2
∴d1 = 116 – 113 = 3, d 2 = 116 – 120 = – 4, d 3 = 116 – 115 = 1
From (2), we have 200 × (7.746) 2 = {50 × (6) 2 + 60 × (7) 2 + 90 σ 3 2 }
+ {50 × (–3) 2 + 60 × (–4) 2 + 90 ×(1) 2}. or
200 × 60 = 1,800 + 2,940 + 90 σ 32 + 450 + 960 + 90.
∴ σ 32 = 64 ⇒ σ 3 = 8. Ans.(1)
207. We know coefficient of variation =
S.D.
× 100 = 80 or
Mean
S.D.
80 × 90
× 100 = 80 . ∴ S.D. =
= 72
90
100
Karl Pearson's coefficient of skewness is
FG IJ
H K
ΣX 2
ΣX
−
N
N
σx =
3 (Mean − Median) 3(90 − 84)
=
= 0.25 . Ans.(2)
72
S.D.
208. Substituting the given values in the Karl Pearson's first measure of skewness,
of ( X, Y ) =
FG 1 IJ FG 1 IJ
H 2K H 2K
= 1−
FG IJ
H K
(2n) ! 1
n!n! 4
= 1−
(2n) ! 1
(n !) 2 4 n
. Ans.(3)
=
= 580 & Σx i y i = 305.
∑ xi = 60 = 6 and
So, cov( x,y ) =
r=
10
n
∑ xi yi −
=
UR
U F
R|
x
y
S|∑ x − e∑n j |V||S|∑y − e∑n j |V| GH
WT
W
T
2
i
2
i
And =
2
i
2
i
60 × 60
305 −
10
60 × 60
400 −
10
IJ FG
KH
We know that
⇒ 0.3 =
9
c h
ch
4 Var y
_
c h
Cov. x,y
ch
Var x
ch
= Var y =
ch
Var y
⇒ 0.3 =
9
90
=
= 7.5
4 × 0.3 12
IJ
K
60 × 60
580 −
10
x=
=
(490 )2
49
1964 −
( 294)2
49
90
= 0 .3 . T h e o r i g i n a l v a l u e o f r r e m a i n s
300
_
σy
σx
=
2
3
FG IJ
H K
FG IJ
H K
σx 2 4
2
2 12
=
= . Ans.(1)
= 2 and b xy = r
σ y 3 12
9
3 4
IJ
K
Σx 15
Σy 29
=
= 3, y =
=
= 5 .8 .
5
5
n
n
The equation of regression line of y on x is
e j
y − y = b yx x − x , b yx =
c hc h = −20 = −0.4
.
50
− c Σx h
nΣxy − Σx Σy
nΣx
2
2
⇒ y = –0.4x + 7. Ans.(4)
217. i.
We have b yx = 3.2 and b xy = 0.8. b yx . b xy = 3.2 × 0.8 = 2.56 > 1.
This is impossible, because 0 ≤ b yx b xy ≤ 1. ∴ The given statement is
false.
212. Given r (x, y) = 0.3, Cov. (x, y) = 9, Var. (x) = 16.
r x, y =
90
90000
5350 −
∴ The equation is y – 5.8 = –0.4 (x – 3) ⇒ y = –0.4 x + (0.4)3 + 5.8
−55
=
= −0.58 . Ans.(4)
−
40 × 220 10 × 88 93.8
−55
−55
=
( ΣY )2
N
( 490 × 294 )
49
216. Here we have Σx = 15, Σy = 29, Σxy = 83, Σx 2 = 55, Σy 2 = 209, n = 5.
FG
H
e∑ je∑ j
yi
n
450 × 200
ΣY 2 −
3030 −
=
unchanged. Ans.(3)
__
xi
( ΣX )2
N
3030 − 2940
b yx = r
F ∑ x y − x yI = F 305 − 6 × 6I = −5 .5 .
GH n JK GH 10
JK
i i
( ΣX )( ΣY )
N
215. We have x = 6,y = 8,σ X = 4, σ y = 12 and r =
∑ yi = 60 = 6 .
_
y=
10
n
ΣXY
− X .Y
N
ΣX 2 −
211. We may write the data as n = 10, Σx i = 60, Σy i = 60, Σ xi2 =400, Σ y2i
_
Co var iance ( X, Y )
or r. σ x .σ y = Covariance (X, Y) and Covariance
σ x σy
ΣXY −
r=
3 (M − Md )
3 ( 50 − Md)
⇒ Md = 55. Ans.(3)
⇒ −1 =
σ
15
∴ x=
ΣX 2
( X )2 or
N
Corrected value of ΣXY = 3090 – (10 × 6) = 3030.
210. Given M = 50, σ = 15, j P = –1.
jP =
=
ΣXY
ΣXY
− 10 × 6 = 1.8 or
− XY = r. σ x . σ y = .3 × 3 × 2 = 1.8 ,
N
N
= 1 – probability of equal number of heads and tails
= 1− 2 n Cn
2
The corrected value of the coefficient of correlation
209. The required probability
n
FG IJ
H K
ΣX 2
ΣX
−
N
N
ΣXY = 50 (1.8 + 60) = 3090.
29.6 − Mode
or Mode = 29.6 − 0.32 × 6.5 = 27.52. Ans.(4)
6.5
2n−n
2
or σ x =
Corrected ΣX2 = 5450 – (10) 2 = 5350. Similarly ΣY2 = N ( σ 2y + Y 2 )
∴
Mean − Mode
, we get .
S.D.
n
2
490
294
= 10 and Y =
= 6.
49
49
ΣX 2 = N ( σ2x + X 2 ) = 50 (9 + 100) = 5450.
We have r =
jp =
0.32 =
40
2
=
30 × 2 3 . Ans.(1)
= 50 (4 + 36) = 2000. Corrected ΣY 2 = 2000 – (6) 2 = 1964. Now we need to
calculate correct value of ΣXY.
=
viz. SK =
=
50 × 72
= Y × n = 6 × 50 = 300 .Corrected total value x Series = 500 – 10 = 490.
Corrected total value y Series = 300 – 6 = 294.
2)
Using (1), we get 200 × 116 = 50 × 113 + 60. x 2 + 90 × 115 or x 2 = 120
40
=
214. Total value of x Series = X × n = 10 × 50 = 500 . Total value of y Series
= (n 1 σ 1 + n 2 σ 2 + n 3 σ 3 ) + (n 1 d 1 + n 2 d 2 + n 3 d 3 ... (2)
2
var x . var y
Corrected means would be X =
...(1)
x = n1x 1 + n2 x 2 + n3 x 3
cov.( x ,y )
213. By using formula rxy =
9
ii.
We have b yx = 0.4 and b xy = –0.2. This is impossible because the
regression coefficients are either both positive or both negative.
∴ The statement is false.
iii.
We have b yx = 0.8 and b xy = 0.2. ∴ r = ± b yx .b xy = + (0.8)( 0.2) = +0.4 .
ch
16 Var y
∴ The statement is true.
Standard deviation of y series = 7.5. Ans.(1)
(44) of (48)
IC : PTpnrhm01
iv.
The regression line of y on x is 40x – 18y = 5 or 18y = 40 x – 5 or
y=
FG 40 IJ x − FG 5 IJ . ∴ b = FG 40 IJ = FG 20 IJ . Similarly
H 18 K H 9 K
H 18 K H 18 K
yx
10 5
= .
8 4
b xy =
This is impossible because both the regression coefficients cannot
be greater than 1. ∴ The statement as false. Ans.(4)
223. Here mean = 30 and σ = 5.
P(|X – 30| ≤ 5) = P(25 ≤ X ≤ 35) = P(–1 ≤ Z ≤ 1) = 2P(0 ≤ Z ≤ 1)
= 2 × 0.3413 = 0.6826
∴ P(|X – 30| > 5) = 1 – P(|X – 30| ≤ 5) = 1 – 0.6826 = 0.3174. Ans.(3)
224. The probability of getting atleast one head = 1 – probability of getting no head
218. The regression equations are 3x + 12y = 19 ...(1) and
0
n
1
1 − n C0
We don’t know exactly as to which of the above equations is regression
equation of x on y. Let us suppose that (1) is regression equation of x on
y and (2) is the regression equation of y on x.
The probability of getting at least two tails.
= 1 – probability of getting no tails – probability of getting 1 tail
(1) ⇒ x = −4y +
FG 19 IJ . ∴ b
H3K
xy
= –4, and (2) ⇒ y = −3 x +
FG 46 IJ . ∴ b
H3K
yx
∴ (1) is the regression equation of y on x and (2) is the regression equation
of x on y.
FG 1 IJ x + FG 19 IJ . ∴ b
H 4 K H 12 K
F 1 I F 46 I
⇒ x = G − J y + G J. ∴ b
H 3K H 9 K
(1) ⇒ y = −
yx
(2)
xy
r = − b yx b xy = −
FG 1 IJ .
H 4K
F 1I
= −G J .
H 3K
=−
b yx .b xy − 1
b xy + b yx
=
.64 − 1
= 0.18 .
2
n−1
n
n
n
n
n
225. Mean np = 20, S.D. = npq = 4 ⇒ npq = 16
16 4
1
1
= ∴ p = ⇒ n ⋅ = 20 ⇒ n = 100 . Ans.(3)
20 5
5
5
16
= 0 .16 . Let the required
100
area be shown by region A. Let B be the region between the ordinate at the
end of A and z = 0. Then, area A + area B = 0.5 ⇒ 0.16 + area B = 0.5
⇒ area B = 0.5 – 0.16 = 0.34. We know that area 0.34 corresponds to z =
1. Thus, area B = area (0 < z < 1) = area (–1 < z < 0). So, the ordinate at the
end of region A is at z = –1. Now, z = –1
⇒
a n d p 2 = Probability that a new machine needs adjustment = 1/21
∴ q 2 = 1 – p 2 = 20/21
Then P 1(r) = Probability that ‘r’ old machines need adjustment = 3 C r p 1r q 1 3-r
= 3C r (10/11) 3-r (1/11) r
and P 2 (r) = Probability that ‘r’ new machine need adjustment =
= 7 C r (1/21) r (20/21) 7-r
7C
rp 2
r
q2
7-r
The probability that just two old machines and no new machine need
adjustment is given (by the compound probability theorem) by the
expression: P 1(2) . P 2(0) = 3C 2(1/11) 2 . (10/11) . (20/21) 7 = 0.016. Ans.(1)
1
300
= 2 bites/m.
2
hours = 150 minutes. Average no. of insect bites =
2
150
The value of m = 2 bites/m. So prob. of zero insect bite is P ( X = n) =
e −m ⋅ mn
n!
with m = 2, n = 0
e −2 ⋅ 2 0
= e –2 = .1353. ∴ Prob. of no. insect bite is 0.1353.
0!
The ecologist was in the forest for 150 minutes.
∴ He was free from insect bites for = 150 × 0.1353 = 20.2 minutes.
∴ For about 20 minute intervals he was free from insect bites. Ans.(3)
222. Here the standard variable X follows normal distribution with mean 12 and sd 4.
X -12
follows N(0, 1).
4
F 4 − 12 ≤ Z ≤ 20 − 12 IJ
P(4 ≤ X ≤ 20) = P G
H 4
4 K
= P(–2 ≤ Z ≤ 2) = P(–2 ≤ Z ≤ 0) + P(0 ≤ Z ≤ 2)
= P(0 ≤ Z ≤ 2) + P(0 ≤ Z ≤ 2) = 2P(0 ≤ Z ≤ 2) = 2Z(2) = 0.9544. Ans.(3)
IC : PTpnrhm01
FG 1 IJ ⋅ 1 = 1− 1 − n ⋅ 1 .
H 2K 2 2 2
FG1− 1 IJ − FG1− 1+ n IJ = 5 ⇒ n = 5 ⇒ n = 5 . Ans.(1)
H 2 K H 2 K 32 2 32
− n Cn−1
Required area 16% of total area = 16% of 1 =
220. Let p 1 = P r o b a b i l i t y t h a t a n o l d m a c h i n e n e e d s a d j u s t m e n t = 1 / 11
∴q 1 = 1 – p 1 = 10/11
Now Z =
.
2n
226. We have x = 50 and σ = 10.
Ans.(2)
P ( X = 0) =
0
From the questions
∴q=
FG − 1 IJ FG − 1 IJ = −0 .2887 . Ans.(4)
H 4K H 3K
219. Given b yx = 1.6, b xy = 0.4. Therefore, tanθ =
= 1− n Cn
= 1−
FG 1 IJ FG 1 IJ
H 2K H 2K
n
= –3.
∴ b xy . b yx = (–4)(–3) = 12 > 1. This is impossible. ∴ Our supposition is
wrong.
221.
FG 1 IJ FG 1IJ
H 2K H 2K
9x + 3y = 46 ...(2).
x−x
x − 50
= −1⇒
= −1⇒ x = 40.
10
σ
∴ 16% of area to the left = P(z < –1) = P(x < 40). Ans.(2)
r r
$
227. a + tb = $i + 2 $j + 3k$ + t − $i + 2 $j + k$ = (1 – t) $i + (2 + 2t) j + (3 + t) k$ .
e
j e
j
r r
r
If a + tb is perpendicular to c ,then
FG a + t b IJ ⋅ c = 0
H K
→
→
→
⇒ [(1 – t) $i + (2 +2t) $j + (3 + t) k$ ]. [3 $i + $j ] = 0 ⇒ 3 (1 – t) + (2 + 2t) = 0
⇒ t = 5. Ans.(3)
r r r
$
228. The resultant of forces F = P + Q = (2 $i – 3 j + k$ ) + ( $i + 5 $j – 3 k$ )
r
= 3 $i + 2 $j – 2 k$ . Displacement d = p.v. of B – p.v. of A = (3 $i + 7 $j + 2 k$ )
$
$
– (–2 $i + 5 $j + 7 k ) = 5 $i + 2 $j – 5 k .
r r
$
∴ Work W = F ⋅ d = (3 $i + 2 j – 2 k$ ) . (5 $i + 2 $j – 5 k$ ) = 15 + 4 + 10 = 29
units. Ans.(2)
r
229. Let r = li$ + mj$ + nk$ be the required unit vector perpendicular to each of the
r
r
r
vectors a and b . Then l² + m² + n² = 1. ( ∴ r is a unit vector)
e
je
j
e
je
j
∴ l $i + mj$ + nk$ ⋅ 3 $i + 2 $j − k$ = 0 and li$ + mj$ + nk$ ⋅ 12 $i + 5 $j − 5k$ = 0 or,
3l + 2m – n = 0 and 12l + 5m – 5n = 0. Solving these equations, we get
l
m
n
= =
=
−5 3 −9
l 2 + m2 + n 2
c−5h + c3 h + c−9h
2
2
2
=
1
115
∴ The required unit vector
→
−5 $i + 3 $j − 9k$
r = l $i + m$j + nk$ or, r =
. Ans.(3)
115
→
(45) of (48)
LM
N
OP LM
Q N
LM a b c OP ⋅ b R|SQ LM a b
N
Q T| N
L
O F I
= Ma b c P ⋅ b⋅G c × a J
N
Q H K
→ →
a× b
Trick : Unit vector perpendicular to both a and b is
.
→ →
a× b
→
→ → →
→ → →
→
e
→
je
1
1 $ $ $
$ $ $
230. The required area. A = 2 a × b = 2 i + 4 j − k × i + j + 2k
j
2
=
^
^
j
k
a× b = 2
1
1 = −5 i + 5 j + 5 k ∴ a × b =
→
231.
c h c h
1 $ $ $ 1 2
2
9 i − 3 j − 3k =
9 + −3 + − 3
2
2
i
→
1
3
99 =
11 . Ans.(3)
2
2
LM a
N
^
→
^
c−5h + 5
→
2
3 4 −1
2
→
→
a × b
→
=
→
→
→
→
U| L
VW| MN
OP
Q
→
OP
Q
→
→
→ → →
→
→ →
e
5 3
→
OP
Q
→
OP LM
QN
→
→ →
b =0 ∴ a× b b× c c× a
OP LM
Q N
→ → →
→
→→ →
2
R|SQ L b
|T MN
→ → →
OP LM
Q N
→ → →
c a = a b c
→→ →
→→→ →
OPU|V
Q|W Ans.(3)
→→→ →
→
→
→
→
→→→ →
→→→ →
→
→
→
→
→→→ →
→→→ →
... (1)
= (b × c )×( r × a ) = [b c a ] r −[ b c r ] a
+ 52 = 5 3
∴ given expression
→→→ →
→→→ →
→→→ →
→→→ →
....(2)
3[ a b c ] r − {[ a b r ] c + [ b c r ] a + [ c a r ] b }
j = − $i + $j + k$ . Ans.(4)
→
→
→
→
→→→ →
→→→ →
Again ( a × b ) × ( r × c ) = [ a r c ] b − [ b r c ] a
3
→→→ →
→→→ →
....(3)
→→ → →
→→→ →
From (1) and (3), [ a b c ] r − [ a b r ] c = [ a r c ] b − [ b r c ] a
232. Let 2 $i – $j + k$ and $i – 2 $j + 2 k$ are the p.v. of O and P.
e
je
→→→ →
j
e
je
j
→→ → →
→→→ →
1→ → 1→ →
| b × c | = | b − c |h , w h e r e h = t h e l e n g t h o f t h e
2
2
2 3 7 . A r e a o f ∆A B C =
perpendicular from A to the line BC. ∴ h =
= −1 −1 1 = $i + $j + 2k$ . Ans.(2)
3 1 −2
→
→→→ →
→→→ →
k$
→
→→→ →
∴ from (2) given expression = 3[ a b c ] r − [ a b c ] r = 2[ a b c ] r . Ans.(1)
$ $ $
$ $ $
∴ Moment about O = − i − j + k × 3 i + j − 2k
$j
→→→ →
∴ [a b c ] r = [a b r ]c +[c a r ]b +[b c r ]a
$ $ $
$ $ $
$ $ $
∴ OP = i − 2 j + 2 k − 2 i − j + k = − i − j + k .
$i
→ → →
= (c × a )×( r × b ) = [ c a b ] r −[ c a r ] b .
5 − $i + $j + k$
a × b
→
236. ( a × b ) × ( r × c ) = ( a × b . c ) r − ( a × b . r ) c = [ a b c ] r − [ a b r ] c
r
r
∴Unit vector perpendicular to both a and b
=
OP
Q
b c ⋅ b c a = a b c
→ → →
^
^
→ → →
→
→ → →
$i $j k$
→ →
∴ a × b = 1 4 −1 = $i (8 + 1) – $j (2 + 1) + k$ (1 – 4) = 9 $i – 3 $j – 3 k$ .
1 1 2
A =
→
= a b c ⋅b− a b b ⋅ c
→
→
→
→
→
|b× c|
. Ans.(2)
|b− c|
238.
→
233. Let a = 2 $i − $j + λk$, b = $i + 2 $j − 3k$, c = 3 $i − 4 $j + 5 k$
→
→
→
→
→
→
→
→
r = 3 i − j + t( i + 2 j + 3 k ) ∴ a vector parallel to the given line is
→
→
b = i + 2 j + 3 k . ∴ unit vector along the line
LM
N
→ → →
OP
Q
2
−1
λ
Since the vectors are coplanar. ∴ a b c = 0 or 1 2 −3 = 0
3 −4 5
→
=
→
→
i +2 j +3 k
1+ 4 + 9
→
=
→
→
i +2 j +3 k
14
or, 2 (10 – 12) + 1(5 + 9) + λ(–4 – 6) = 0 ⇒ 10 – 10λ = 0. ⇒ λ = 1. Ans.(1)
→
F I
234. Since G a × b J ⋅ c
H K
→
→
→ →
→
If
→
→
→
→
→ →
→ →
→
→
→
235. a × b b × c c × a
→
→
→
( i +2 j +3 k)
14
=
1+ 2 + 3
14
=
6
14
. Ans.(3)
→
239. Since b = r × a ∴ we have
→ →
b c , then the parallelopiped is rectangular i.e., the
r r
r r
edges are at right angle to one another. ∴ a . b = 0, b . c = 0
r r
c . a = 0. ∴ The correct answer is (2). Ans.(2)
LM
N
→
is the volume with adjacent sides a , b and c .
FG a × b IJ ⋅ c = a
H K
→
→
→
∴ projection = ( i + j + k ).
OP = R|SFG a × b IJ × FG b × c IJ U|V ⋅ FG c × a IJ
Q |TH K H K |W H K
→
→
→
→
→
→
→
→
→
→
→→ →
→→ →
→→ →
a × b = a ×( r × a ) = (a . a ) r − ( a. r ) a = (a . a ) r
→
∴ r =
→
→→
[Q a . r = 0 ]
→
a× b
→→
. Ans.(2)
a. a
→
because
FG a × b IJ × FG b × c IJ = R|SFG a × b IJ ⋅ c U|V b − R|SFG a × b IJ ⋅ b U|V ⋅ c
H K H K |TH K |W |TH K |W
→
→
(46) of (48)
→
→
→
→
→ →
→
→
→
→
IC : PTpnrhm01
→
→
→
→
→
→
→
→
→
240. Since (l a + m b + n c ) × (l b + m c + n a ).(l c + m a + n b ) = 0 .
l m n
→→→
∴ n l m [a b c ]= 0 .
m n l
→
→
→
n
l
m
m
n
l
→
→
→
→
→
→
→
(18 − x 2 )3 / 2
1
9
x→3
....(1)
→
→
→
Also ( α i + β j + γ k ).( i − 2 j + 3 k ) = 0
∴ α – 2β + 3γ = 0
→
→
....(2)
→
→
→
246.
→
Again ( α i + β j + γ k ).(2 i − j + k ) = −6
∴ 2α – β + γ = –6
....(3)
(1) – (3) gives 4β – 2γ = 6 ⇒ 2β – γ = 3
....(4)
(2) – (3) gives –3β + 5γ = 6
....(5)
→
FG
H
2m
=
–n,
m
=
–
2n.
From
+ 5mn +
2m
=
2n 2
= 0 or, (2m + n)
–
n,
we
1
6
1
,m =
6
, n=−
2
6
1
6
,m =
2
n→∞
.
6
, n=−
2n
1
6
{ f ( x )} 2 − 16
( x − 1)
= Lim
x→1
2 f ( x ) f '( x ) − 0
= 2 f (1) f ' (1)
1
⇒ lim
n→∞
Put x =
x −3 y +1 z−2
=
=
, it follows that the
2
7
−3
required line passing through A(2i – j + k) has the direction of 2i + 7j – 3k.
π
=
2
243. Since the required line is parallel to
Hence, vector equation of the required line is r = 2i – j + k + λ(2i + 7j – 3k)
where λ is a parameter. Ans.(2)
l.0 + m.2 + n(– 1) = 0 i.e. 0.l + 2m – n = 0
(l 1 l 2 + m 1m 2 + n 1n 2) = 0.
IJ
K
.cos
x
23
2
n
2
3
n
x
1
x
x
x
x
cos .cos 2 .cos 3 ........cos n
2
2
2
2
x
2n × x
x
2n
x
2n
=
=
sin x
sin x
sin
1
FG cos x .cos x .....cos x IJ
H 2 2
2 K
2
. Ans.(4)
244. The given lines will be coplanar if we can find a line perpendicular to each
of them. Let l, m, n, be the direction cosines of the line perpendicular to the
first two lines. Then l.1 + m.(– 2) + n.2 = 0 i.e.l – 2m + 2n = 0
FG
H
x
x
x
x
x
= 2 .2 sin 2 .cos 2 .cos
. cos
2
2
2
2
2
2n =
sin x
⇒ lim
m n m+n
l
l 2 + m2 + n2
1
=
=
=
=
=
Again taking m = – 2n, we have
.
2 −1 2 − 1 −1
1+ 4 + 1
6
Hence l = –
x
23
2n sin
have
m n
m + n m + n −l l
l2 + m 2 + n 2
1
=
= =
=
=
=
=
−1 1
−1
1 −2 1− 2
1+ 1+ 4
6
∴l=
x →1
n
From (i), l = – (m + n), substituting this value of l in (ii),
or
= Lim
IJ .cos x .cos x
K 2 2
x F
sin x = 2 . sin
G cos 2x .cos 2x .cos 2x ....cos 2x IJK
2 H
242. The given relations are l + m + n = 0 ... (i) and mn – 2nl – 2lm = 0 ..... (ii)
we get mn + 2n (m + n) + 2(m + n) m = 0 or,
(m + 2n) = 0
( x − 1)
2 2 2 sin
→
∴(2) gives α – 6 + 9 = 0 ∴ α = –3 ∴ x = −3 i + 3 j + 3 k . Ans.(1)
2m 2
4
247. sin x = 2 sin
∴ β = 3 ∴ (4) gives 6 – γ = 3 ∴ γ = 3
→
f(x)
t2
= 2 × 4 × f’(1) = 8f’(1). Ans.(1)
∴ 5(4) + (5) gives 10β – 3β = 15 + 6 or 7β = 21
→
Lim
x→1
(By L’ Hospital rule)
1
. Ans.(4)
9
= f’(3) =
→
= (18 − x 2 ) −1/ 2
f ( x ) − f ( 3)
f '( x) − 0
= Lim
x → 3 1− 0
( x − 3)
∴ Lim
→
241. Let x = α i + β j + γ k . Then (α i + β j + γ k ).(2 i + 3 j − k ) = 0
∴ 2α + 3β – γ = 0
(18 − x )
f'(3 ) =
= 0 ⇒ l + m + n = 0 . Ans.(2)
→
2
x
∴ f ' (x) =
l +m+ n l +m+ n l +m + n
→→→
Since [ a b c ] ≠ 0 .∴
1
245. Since f(x) =
248.
lim
x →0
n
π
both sides
2
1
. Ans.(1)
1 1 1 1 1 1 1 1 1
.
+
.
+
+
2 2 2 2 2 2 2 2 2
f ( x 2 ) − f( x )
2 xf '( x 2 ) − f '( x )
= lim
f ( x ) − f ( 0 ) x →0
f '( x )
= −1+ lim
x→0
2 x f '( x 2 )
= −1 . (Q f’(x) > 0). Ans.(2)
f '( x )
l
m n
= = . This line will be perpendicular to the
−2 l 2
third line if l.1 + m.2 + n.0 = 0 i.e. (– 2).1 + 1.2 + 0 = 0 which is clearly true.
Hence the three given lines, being also concurrent, are coplanar. Ans.(1)
By cross-multiplication,
IC : PTpnrhm01
(47) of (48)
n
249. un =
=
1
np
LMF 1I
MNGH 1+ n JK
n
LM 1
N 2n
1/2
1/2
p
−
1
8n 3 /2
OP LF 1 1
PQ = n MNGH1+ 2n − 8n
−1
2
n
O 1
+ ....P =
Q 2n
p +1/2
−
IJ OP
K Q
p
1
8np +3 / 2
250.
+.... − 1
L.H.S. =
+.... .
= Lt
The first term suggests that the auxiliary series should be taken as
So take v n =
F I = lim
GH JK
u
1
then lim n→∞ n
vn
n p+1 2
quantity. But the auxiliary series
FG
H
when p +
Lim [ x 2 + x + sin x ]
x→0
1
∑ np+1/2
n→∞
1
∑ np+1 2
h→ 0
.
LM 1 − 1 +....OP = 1 =a finite
N 2 8n Q 2
LMh
MN
Lt f ( x) = Lt f (0 − h) = Lt [h2 − h − sin h]
x → 0−
h→0
h→ 0
I OP
JK P
Q
F
GH
h 3 h3
+
− ...
3! 5 !
7.(2)
17.(1)
27.(2)
37.(2)
47.(1)
57.(1)
67.(1)
77.(1)
87.(3)
97.(1)
107.(1)
117.(2)
127.(3)
137.(4)
147.(3)
157.(3)
167.(1)
177.(3)
187.(3)
197.(3)
207.(2)
217.(4)
227.(3)
237.(2)
247.(1)
8.(2)
18.(4)
28.(3)
38.(2)
48.(3)
58.(4)
68.(4)
78.(4)
88.(1)
98.(1)
108.(2)
118.(2)
128.(4)
138.(3)
148.(3)
158.(3)
168.(4)
178.(4)
188.(1)
198.(4)
208.(4)
218.(4)
228.(2)
238.(3)
248.(2)
2
−h− h −
= Lt
h→0
LM−2h + h
MN
2
+
OP
PQ
h3 h5
−
+ ...
3! 5!
= 0.
Ans.(2)
is always divergent except
IJ
K
1
1
> 1 i.e., p > .
2
2
Hence the given series is always divergent except when p >
1
. Ans.(2)
2
Objective key
1.(2)
11.(3)
21.(3)
31.(2)
41.(2)
51.(2)
61.(3)
71.(2)
81.(2)
91.(4)
101.(4)
111.(2)
121.(3)
131.(3)
141.(3)
151.(2)
161.(3)
171.(3)
181.(2)
191.(3)
201.(3)
211.(4)
221.(3)
231.(4)
241.(1)
(48) of (48)
2.(2)
12.(4)
22.(4)
32.(2)
42.(2)
52.(3)
62.(2)
72.(4)
82.(2)
92.(3)
102.(4)
112.(4)
122.(4)
132.(2)
142.(2)
152.(1)
162.(3)
172.(2)
182.(4)
192.(4)
202.(2)
212.(1)
222.(3)
232.(2)
242.(4)
3.(1)
13.(3)
23.(1)
33.(2)
43.(2)
53.(2)
63.(2)
73.(3)
83.(2)
93.(2)
103.(4)
113.(2)
123.(2)
133.(1)
143.(1)
153.(1)
163.(2)
173.(1)
183.(4)
193.(1)
203.(1)
213.(1)
223.(3)
233.(1)
243.(2)
4.(1)
14.(3)
24.(2)
34.(3)
44.(2)
54.(3)
64.(3)
74.(4)
84.(2)
94.(1)
104.(3)
114.(3)
124.(2)
134.(2)
144.(3)
154.(2)
164.(2)
174.(3)
184.(2)
194.(2)
204.(4)
214.(3)
224.(1)
234.(2)
244.(1)
5.(1)
15.(1)
25.(3)
35.(2)
45.(4)
55.(4)
65.(4)
75.(2)
85.(4)
95.(3)
105.(3)
115.(1)
125.(4)
135.(4)
145.(1)
155.(3)
165.(2)
175.(4)
185.(3)
195.(2)
205.(2)
215.(1)
225.(3)
235.(3)
245.(4)
6.(1)
16.(4)
26.(3)
36.(2)
46.(4)
56.(2)
66.(4)
76.(2)
86.(2)
96.(1)
106.(2)
116.(3)
126.(2)
136.(1)
146.(2)
156.(3)
166.(2)
176.(2)
186.(2)
196.(1)
206.(1)
216.(4)
226.(2)
236.(1)
246.(1)
9.(4)
19.(4)
29.(2)
39.(1)
49.(2)
59.(2)
69.(4)
79.(3)
89.(1)
99.(2)
109.(3)
119.(3)
129.(3)
139.(4)
149.(2)
159.(2)
169.(1)
179.(4)
189.(1)
199.(2)
209.(3)
219.(2)
229.(3)
239.(2)
249.(2)
10.(2)
20.(2)
30.(3)
40.(3)
50.(3)
60.(2)
70.(3)
80.(2)
90.(3)
100.(1)
110.(4)
120.(3)
130.(1)
140.(3)
150.(3)
160.(1)
170.(1)
180.(3)
190.(3)
200.(2)
210.(3)
220.(2)
230.(3)
240.(2)
250.(2)
IC : PTpnrhm01