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for those who are preparing for jmet and xat ...this is a set of questions which can help u out a lot.....
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(1) of (57)
Higher Maths Questions - JMET, XATExercise - 01
Directions: For the following questions choose thecorrect option.
1. A die is thrown twice. The sum is 9. Theprobability that at least 1 number is six, is :
(1)14
(2)12
(3)34
(4)23
2. The condition when ax2 + bx + c = 0 & px2 +qx + r = 0 has reciprocal roots is :
(1)bc
qp
ac
rp
= =, (2)ac
qp
bc
rp
=−
=,
(3)ab
pq
cb
qr
= =, (4) none of these.
3. If y2 = 4ax, ay2 = 4x3, then the angle betweenthese curves is :
(1) θ =FHG
IKJ
−tan 1 13
(2) θπ
=4
(3) θ =FHG
IKJ
−tan 1 14
(4) θ =FHG
IKJ
−tan 1 12
4. The value of log(tan ) :/2
x is0
πz(1) 0 (2)
π4
(3) 2 (4) none of these.
5. If I xnn= z (log ) , which of the relation is true :
(1) In – n In–1 + x log x = 0
(2) In + n In–1 – x (log x)n = 0
(3) In – n In–1 – x (log x)n = 0(4) In + n In–1 – x (log x) = 0
6. If the variance of x1, x2, x3 ....... x4 are σ2 thenthe variance of px1, px2, px3 ... px4 will be :
(1) pσ2 (2) p3σ2
(3) p2σ2 (4) pσ
7. If 1 1 3
a c b c a b c++
+=
+ +, then value of ∠C is :
(1) 60° (2) 45°
(3) 30° (4) 15°
8. The mean deviation from the median for thefollowing data
36, 42, 56, 44, 61, 63, 70, 46, is :
(1) 12.25 (2) 10.25(3) 10.05 (4) none of these.
9. The area enclosed by the region R = {y ≥ x2
and y ≥ |x|} is :
(1)12
(2)13
(3) 1 (4) 0
10. If a, b, c and a', b', c' are the direction ratiosof two vectors then the value aa’ + bb’ + cc’ is:
(1) cosθ (2) sinθ
(3)r rA B⋅ (4)
r rA B×
11. a2, b2, c2 are in A.P. then a2 + b2, b2 + c2, c2 +a2 are in :(1) A.P. (2) H.P.
(3) G.P. (4) none of these.
12. Diagona ls o f a Rhombus are
5 2 6 2 3 5i j k i j k− + − −$ $ & $ $ , then the a rea o frhombus is :
(1) 9 (2) 10
(3)3542
(4)3652
13. A particle is thrown vertically upward from atop of tower from a height 25m. With velocityof 15m/s, then find the time in which particlehits the ground.
(1) 4.25 sec (2) 4.35 sec
(3) 4.8 sec (4) none of these.
(2) of (57)
14. A point on the curve such that (2,0) is at
shortest distance from curve y x= is :
(1)32
32
,FHG
IKJ (2)
32
32
,FHG
IKJ
(3)32
32
,FHG
IKJ (4) 0
32
,FHG
IKJ
15. The distance between the parallel lines 2x2 –4xy + 2y2 + 4x + 6y – 10 = 0 is :
(1) 2√3 (2) √3(3) 3√3 (4) 2√6
16. Area of triangle formed by the lines, x + y =1, y = 3(x – 1), y + 3x = –7 is :
(1)997
(2)334
(3)1006
(4)1013
17. The equation of l ine passing through theintersection of lines 2x + 3y + 1 = 0 & 5x – 3y +4 = 0 and perpendicular to x – 3y + 4 = 0 is :(1) 21y – 63 x + 40 = 0
(2) 21y + 63 x – 40 = 0
(3) 21y + 63 x + 40 = 0
(4) both (1) and (2)
18. Angle of intersection of two circles is given by:
(1) cos θ =+ −F
HGGIKJJ
r r d
r r12
22 2
1 22(2) sinθ =
FHG
IKJ
rd1
(3) tanθ =−
+
FHG
IKJ
m mm m
1 2
1 21(4) none of these.
19. The eccentric angle of the point where the line5x – 3y = 8√2 is a normal to the ell ipse
x yis
2 2
25 91+ = :
(1) 0 (2) π/4
(3) π/3 (4) π/2
20. The intercepts of c irc le on x and y axisrespectively, where the equation of circle isx2 + y2 + 2gx + 2fy + c = 0, are :
(1) 2 22 2g c f c− −,
(2) 2 22 2f c g c− −,
(3) 2 22 2 2 2g f c g f c+ − − +,
(4) none of these.
21. The Walli formula given for sin/
n xdx0
2πz , where
n is even, is given by :
(1)n
nnn
nn
−FHG
IKJ
−−
FHG
IKJ
−−
FHG
IKJ
1 32
54
21 2
... .π
(2)n
nnn
nn
−FHG
IKJ
−−
FHG
IKJ
−−
FHG
IKJ
1 32
54
12 2
... .π
(3)n
nnn
nn
−FHG
IKJ
−−
FHG
IKJ
−−
FHG
IKJ
1 32
54
23
...
(4) none of these.
22. The equation of tangent in point slope form ofhyperbola is given by :
(1) y mx a m b= ± +2 2 2
(2) y mx a m= ± +1 2
(3) y mx a m b= ± −2 2 2
(4) none of these.
23. The value of Ie
dxx
=+−z 1
1:
(1) log |e–x + 1| + c(2) log |ex – 1| + c
(3) log |e–x – 1| + c
(4) log |ex + 1| + c
24. Which of the following functions is an evenfunction ?
(1) f xa a
a a
x x
x xc h =+
−
−
−(2) f x
a
a
x
xc h =+
−
1
1
(3) f x xa
a
x
xc h =−
+
1
1(4) f x x xc h = + +F
HGIKJlog2
2 1
(3) of (57)
25. If 3 4
2 42
3x
x xdx x k x C
+
− −= − + +z log log , fc h then
which of the following is not correct ?
(1) k = −12
(2) f(x) = x2 + 2x + 2(3) f(x) = |x2 + 2x + 2|
(4) k =14
26. If sin θ and cos θ are the roots of the equationax2 –bx + c = 0, then a, b and c satisfy therelation:
(1) a2 + b2 + 2ac = 0(2) a2– b2 + 2ac = 0
(3) a2 + c2 + 2ab = 0(4) a2 – b2 – 2ac = 0
27. The solution of the equation dydx
x yx y
=− −− −
3 4 23 4 3
is
:
(1) (x – y)2 + C = log (3x – 4y + 1)(2) x – y + c = log (3x – 4y + 4)
(3) x – y + C = log (3x – 4y + 3)(4) x – y + c = log (3x – 4y + 1)
28. The decimal conversion of (D6C1)16 is :
(1) (54248)10 (2) (53248)10(3) (54977)10 (4) none of these.
29. A man in a boat rowing away from a cliff 150m high takes 2 minutes to change the angle ofelevation of the top of the cliff from 60° to45°. The speed of the boat is :(1) 1.9 km /hr (2) 0.9 km /hr(3) 2.9 km /hr (4) none of these.
30. A determinant is chosen at random from theset of all determinants of order 2 with elements0 or 1 on ly. The probab i l i ty that thedeterminant chosen is non-zero is :
(1) 3/16 (2) 3/8(3) 1/4 (4) none of these.
31. It is given that
X : 2 3 5 8 9
Y : 4 6 10 16 18
The coefficient of correlation between X and Yis :(1) 0 (2) 0.5(3) 1 (4) 0.25
32. The ratio of the coefficient of x15 to the termindependent of x in [x2 + (2/x)]15 is:(1) 1 : 16 (2) 16 : 1(3) 1 : 2 (4) 1 : 32
33. A variable chord is drawn through the originto the circle x2 + y2 – 2ax = 0. The locus ofthe centre of the circle drawn on this chord asdiameter, is :
(1) x2 + y2 + ax =0(2) x2 + y2 + ay =0(3) x2 + y2 – ax = 0(4) x2 + y2 – ay = 0
34. The complete solution of the equation 7 cos2 x+ sin x cos x – 3 = 0 is given by :
(1) n π + π/2 (n ∈ I)(2) n π – π/4 (n ∈ I)(3) n π + tan–1(4/3) (n ∈ I)
(4) nππ
π+ +FHG
IKJ
−34
43
1, tan k
35. The number 15 is divided into two parts suchthat square of one multiplied with cube of otheris maximum. Its one part is :
(1) 2 (2) 10(3) 6 (4) 8
36. If f(x) = log2 x2 and g(x) = logx4, then :
(1) f(x) = g(x) (2) f(x) = 4/g(x)(3) f(x) = 1/g(x) (4) f(x) = 2/g(x)
37. Number of diagonals of a polygon with n sides,is :(1) n! (2) n! – n(3) nC2 – n (4) nC2
38. Jane has as many brothers as sisters and herbrother Todd has two more sisters than broth-ers. If Todd is the youngest of the four broth-ers, how many offsprings are there in the fam-ily ?
(1) 11 (2) 9(3) 7 (4) 10
39. The 1st terms of two GPs are equal. The 3rdterm of the 1st GP is the same as the 5th termof the 2nd, then the ratio of the 67th term of1st GP to the 133rd term of the 2nd.
(1) 67/135 (2) 1(3) 2 (4) 11/17
(4) of (57)
40. If the shortest and the longest sides of a rightangled triangle are 7 and 25, then the radiusof its incircle is :(1) 3 (2) 12.5(3) 17 (4) 6
41. If f(x) = (x + 3)/(4x – 5), and f(t) = x, then tcould be :
(1)4 35 5xx
−−
(2) 3 15 4
xx
−−
(3)4 15 3
xx
−+
(4)3 54 1
+−
xx
42. Let a = 2i + j – 2k and b = i + j. If c is a vectorsuch that a. c = | c |, |c – a| = 2√2 and theangle between (a × b) and c is 30°, then |(a ×b) × c| =
(1) 23
(2) 32
(3) 2 (4) 3
43. The value of the determinant 1 3 92 4 123 5 15
is :
(1) 0 (2) – 50(3) 24 (4) none of these.
44. If p times the pth term of an A.P. is equal to qtimes the qth term of an A.P., then (p + q)th
term is :
(1) 0 (2) 1(3) 2 (4) 3
45. If ∆ = a2 – (b – c)2, where ∆ is the area oftriangle ABC, then tan A is :
(1)1516
(2)815
(3)817
(4)12
46. The value of sin cos3 2
1
1
x x dx−z is :
(1) 0 (2) 1(3) 1/2 (4) 2
47. The solutions of the trigonometry equation
sin2θ – cosθ=14
, 0 ≤ θ ≤ 2π are :
(1)23 3π π
, (2)π π3
53
,
(3) −π π3
23
, (4)23
53
π π,
48.x x
xdx
2 2
4
1 1 2+ + −z log log xe jis equal to :
(1)13
11
11 2
32
1 2
2+
FHG
IKJ +
FHG
IKJ +
LNM
OQP +
x xClog
(2) − +FHG
IKJ +
FHG
IKJ −
LNM
OQP +
13
11
11 2
32
3 2
2x xClog
(3) 23
11
11 2
32
3 2
2+
FHG
IKJ +
FHG
IKJ +
LNM
OQP +
x xClog
(4) none of these.
49. Probability that a student will succeed inAIMCET is 0.2 and that he will succeed in Puneentrance test is 0.5. If the probability that hewill be successful at both the places is 0.3,then the probability that he does not succeedat both the places is :
(1) 0.4 (2) 0.3(3) 0.2 (4) 0.6
50. The differential equation of family of curveswhose tangent form an angle of π/4 with thehyperbola xy = c2 is :
(1)dydx
x c
x c=
+
−
2 2
2 2 (2)dydx
x c
x c=
−
+
2 2
2 2
(3)dydx
c
x= −
2
2 (4) none of these.
51. The value of cos nx dx0
2 πz is :
(1) n (2)1n
(3) 1 (4) 0
(5) of (57)
52.tan−
+z 1
20
1
1
x
xdx is equal to :
(1)π2
16(2)
π2
8
(3)π2
32(4) 0
53. If α, β are the roots of the equation x2 + px +q = 0, then α2 + β2 will be equal to :
(1) p2 + 2q (2) p2 - 2q(3) q2 + 2p (4) q2 - 2p
54. If the foci of the ellipse x y
b
2 2
2161+ = and the
hyperbola x y2 2
144 81125
− = coincide, then the
value of b2 is :(1) 1 (2) 5(3) 7 (4) 9
55.7
16 6 7 16 6 7+ − −e j e jis a :
(1) irrational number (2) complex number(3) prime integer (4) rational number
56. If 25 tan A = 7 sec A then sin A (1 + sin A) +cos A (1 + cos A) will be equal to :
(1)5625
(2)2556
(3)2825
(4)2528
57. If a normal chord subtends a right angle atthe vertex of the parabola y2 = 4ax, then it isinclined to the axis at an angle :(1) π/2 (2) π/4(3) tan–1 √3 (4) tan–1 √2
58. The distribution of a random variable X is givenin the following table
X : 1 2 3 4P(X = x) : C 2C 3C 4C
Then the value of C is :
(1) 0.1 (2) 0.01(3) 0.0001 (4) 0.001
59. The value of log log log log :3 2 1 44 3 2 1× × × is
(1) 0 (2) 1(3) 2 (4) none of these.
60. If A is a square matrix and A + AT is symmet-ric matrix, then A – AT is :
(1) Unit matrix(2) Symmetric matrix
(3) Skew-symmetric matrix(4) Zero matrix
61. The hexadecimal equivalent of (41819)10 is :
(1) (A53B)16 (2) (B53A)16(3) (A35B)16 (4) (B35A)16
62. lim| |
,x
x x→
−LNM
OQP
FHGG
IKJJ2
3 3
3 3 where [x] is the greatest
integer less than or equal to x is :(1) 0 (2) 8/3(3) 64/27 (4) none of these.
63.ddx
(log (ax)x), where a is a constant, is equal
to :
(1) 1 (2) log ax(3) 1/a (4) log (ax) + 1
64. The angle of elevation of a tower CD at a placeA due south of it is 600; and at a place B duewest of A, the elevation is 300. If AB = 3 km,then height of the tower is :(1) 2√3 km. (2) 2√6 km.(3) (3√3)/2 km. (4) 3√6/4 km.
65. 'a' and 'c' are two distinct positive real numberssuch that a, b, c are in G.P. If b – c, c – a, a – bare in H.P., then a + 4b + c is :
(1) a constant independent of a, b, c(2) independent of a(3) independent of b(4) independent of c
66. If 3 6
5 3 32 50 2 12
+
− + − = x, then the value of
x is :
(1) √3 + √2 (2) √2(3) √6 (4) √3
67. If log107 = 0.8451, how many digits would theexpression 720 contain ?
(1) 16 (2) 17(3) 10 (4) 20
(6) of (57)
68. If A1 is the area of the parabola y2 = 4ax lyingbetween vertex and latus rectum and A2 thearea between latus rectum and double ordinate
x = 2a, then AA
1
2 =
(1) 2√2 – 1 (2) (2√2 – 1)/7(3) (2√2 + 1)/7 (4) none of these.
69. The area of the parabola y2 = 4ax lyingbetween the vertex and the latus rectum isrevolved about x-axis. Then the volumegenerated is :(1) 2πa3 (2) πa3
(3) 4πa3 (4) none of these.
70. limsinx
x x x
x→
− − +0 2
6 2 3 1 =
(1) loge 6 loge 3 loge 2 (2) loge 3 loge 2(3) loge 6 (4) none of these.
71. lim
sin
x
x
t dt
x x→
z+0
3
04 55
=
(1) 0.5 (2) 0.25(3) 0.2 (4) does not exist
72. Consider the statements (i) to (iv) below :A bounded real valued continuous functionattains its infimum on an interval I if(i) I is open and bounded below(ii) I is closed(iii) I is closed and bounded(iv) I is bounded below.Which among the following is correct ?(1) only (iii) is true(2) only (i) is true(3) only (ii) and (iv) are true(4) (ii), (iii) and (iv) all are true
73. Let f(x) = 3 – |sin x|. Then f(x) is :(1) continuous and differentiable every where.(2) continuous no where.(3) differentiable no where.(4) cont inuous everywhere but notdifferentiable at nπ, n ∈ Z
74. If y = (xx)x, then dy/dx =(1) x(xx)x (1 + log x2) (2) x(xx)x (1 – log x2)(3) x(xx)x (1 + log x) (4) none of these
75. The area of the greatest isosceles triangle thatcan be inscribed in a given ellipse having itsvertex coincident with one extremity of themajor axis.
(1) 3 3
4
FHG
IKJ ab (2) −
FHG
IKJ
3 34
ab
(3)3 3
2
FHG
IKJ ab (4) none of these.
76. The sma l les t poss ib le per imeter o f arectangular plot of land of area 36 squaremeters is :(1) 15 m (2) 20 m(3) 24 m (4) 26 m
77. The tangent to the curve y = x2 + 3x will passthrough the point (0, – 9) if it is drawn at thepoint :(1) (0, 0) (2) (1, 4)(3) (– 4, 4) (4) (– 3, 0)
78. The value of | |
,xx
dx a ba
b
<z is :
(1) b – a (2) a – b(3) a + b (4) |b| – |a|
79.dx
e ex x+ +=
−z 3 2
(1) loge
eC
x
x−
++
1
2(2) log
e
e
x
x−
−
1
2 + C
(3) loge
e
x
x+
+
1
2 + C (4) none of these.
80. e dxx =z0
1
(1) 2e – 2 (2) 2e(3) 4e (4) 4e – 2
81. An n → ∞ , then express ion
10
222
32 2n n n n
nn
sin sin sin sin ... sin+ + + + +LNM
OQP
π π π π,
tends to :
(1) 1 (2)π2
(3)2π
(4)π2
6
82. x x x
x xdx
sin cos
sin cos
/
4 40
2
+=zπ
(1)π2
8(2)
π2
16
(3)π2
32(4)
π2
4
83. Assuming the rate of increase is proportionalto the number of inhabitants, the number ofyears in which the population of a city willtreble, if it doubles in 50 years is :(1) 69 (2) 76(3) 79 (4) 81
(7) of (57)
84. If dydx
yx
ex+ = and y(0) = 1, then y =
(1) xex (2) (x – 1)ex
(3) (x + 1)ex (4) none of these.
85. If the equations x2 + 2x + 3λ = 0 and 2x2 + 3x+ 5λ = 0 have a non-zero common root, thenλ is equal to :(1) 1 (2) –1(3) 3 (4) none of these.
86. A and B solve an equation. A commits a mistakein constant term and finds the roots as 8 and2, B commits a mistake in the coefficient of xand finds the roots as – 9, – 1. Then the correctroots are :(1) 8, – 9 (2) 1, 9(3) 2, – 1 (4) none of these.
87. If a b
a b
n n
n n
+ ++
+
1 1 is the H.M. between a and b then
n =(1) 1 (2) 0(3) – 1 (4) none of these.
88. The no. of rectangles that you can find on achess board is :(1) 1296 (2) 64(3) 32 (4) none of these.
89. Let A =− − −
F
HGGG
I
KJJJ
1 1 35 2 62 2 3
then A is :
(1) Nilpotent (2) Idempotent(3) Scalar (4) none of these.
90. If Axx x
and A=FHG
IKJ =
−FHG
IKJ
−2 0 1 01 2
1 then x is equal
to :(1) 1 (2) 2(3) 1/2 (4) none of these.
91.
a b c a ab b c a bc c c a b
− −− −
− −
2 22 22 2
=
(1) (a + b + c)3 (2) (a + b + c)2
(3) ab + bc + ca (4) none of these.
92. A coin whose faces are marked 3 and 5 istossed 4 times, the odds against the sum ofthe numbers throwing less than 15 is :(1) 5 : 11 (2) 11 : 5(3) 6 : 5 (4) 5 : 6
93. If p is the chance that an odd number of acesturn up when n ordinary dice are thrown then1 – 2p =(1) (2/3)n (2) (1/3)n
(3) (1/3)n–1 (4) none of these.
94. From a bag containing 3 rupee coins and 4 fiftypaise coins, a person is asked to draw two coinsan random, the expected value is :(1) Rs. 2 (2) Rs. 3(3) Rs. 2.04 (4) none of these.
95. The probabil ity distribution function of acontinuous distribution is f(x) = ax2, 2 ≤ x ≤ 3f(x) = 0, otherwise, the value of "a" is :(1) 3/5 (2) 3/8(3) 3/13 (4) 3/19
96. Minimum Z = x + y s.t. 3x + 2y ≥ 12, x + 3y ≥11 and x ≥ 0, y ≥ 0 is :(1) 5 (2) 6(3) 4 (4) none of these.
97.n
rn
rn
rr
n C
C C+ −=∑
10
is equal to :
(1)n2
(2)n + 2
2
(3)n n( )+ 1
2(4)
n nn( )( )
−+
12 1
98. [ ]k x dxn
0z , where [x] represents greatest
integer less than or equal to x and n, k ∈ I isequal to :
(1)n n
k( )−12
(2)n n k( )− 1
2
(3) n n kk
( )+ 12
(4)n n k
k( )+ 1
99. The interval for which sin x may be representedby its Maclaurin's series is :
(1) −LNM
OQP
π π2 2
, (2) 02
,πL
NMOQP
(3) [0, π] (4) (– ∞, ∞)
100. The system x + y + z = 0, 2x + y – z = 0, 3x +2y = 0 has :(1) no solution(2) a unique solution(3) an infinite number of solutions(4) none of these.
(8) of (57)
Directions: For the following questions choose thecorrect option.
1. The value of a for which the quadratic equation3x2 + 2(a2 + 1)x + (a2 – 3a + 2) = 0 possessesroots of opposite sign lies in :(1) (– ∞, 1) (2) (– ∞, 0)
(3) (1, 2) (4)32
2,FHG
IKJ
2. If the sum of n terms of an A.P. is cn(n – 1),where c ≠ 0. The sum of the squares of theseterms is :(1) c2n2(n + 1)2
(2)23
1 2 12c n n n( ) ( )− −
(3)23
1 2 12c
n n n( ) ( )+ +
(4) none of these.
3. If the first and the (2n + 1)th terms of an A.P.,a G.P. and an H.P. of positive terms are equaland their (n + 1)th term are a, b and crespectively, then :(1) a = b = c (2) a ≥ b ≥ c(3) a + c = 2b (4) none of these.
4. If the letters of the word RACHIT are arrangedin all possible ways and these words are writtenout as in a dictionary, then the rank of the wordRACHIT is :(1) 365 (2) 481(3) 720 (4) none of these.
5. The coefficient of xk(0 ≤ k ≤ n) in the expansionof E = 1 + (1 + x) + (1 + x)2 + ... + (1 + x)n
is :(1) n+1Ck+1 (2) nCk(3) n+1Cn–k+1 (4) nCn–k–1
6. The remainder when 22003 is divided by 17 is :(1) 1 (2) 2(3) 8 (4) none of these.
7. If
x x xy y yz z z
k k k
k k k
k k k
+ +
+ +
+ +
2 3
2 3
2 3 = (x – y)(y – z)(z – x) (1/
x + 1/y + 1/z) then :(1) k = – 3 (2) k = – 1(3) k = 1 (4) k = 3
8. The solution set of the inequality ||x| – 1| < 1– x (x ∈ R) is :(1) (1, 1) (2) (0, ∞)(3) (– 1, ∞) (4) none of these.
Exercise - 02
9. There are two balls in an urn whose coloursare not known (each ball can be either whiteor black). A white ball is put into the urn. Aball is drawn from the urn. The probability thatit is white is :(1) 1/4 (2) 1/3(3) 2/3 (4) 1/6
10. A speaks the truth in 70 percent cases and Bin 80 percent cases. The probability that theywill contradict each other in describing a singleevent is :(1) 0.36 (2 0.38(3) 0.4 (4) 0.42
11. A pole 50 m high stands on a building 250 mhigh. To an observer at a height of 300 m, thebuilding and the pole subtend equal angles.The distance of the observer from the top ofthe pole is :(1) 25 m (2) 50 m(3) 24√3 m (4) 25√6 m
12. The value of tan–1 (1) + cos–1(– 1/2) + sin–1
(– 1/2) is equal to :
(1)π4
(2)512
π
(3)34π
(4)1312
π
13. The ratio of the greatest value of 2 – cos x +sin2 x to its least value is :
(1)14
(2)94
(3)134
(4) none of these
14. The value of 2(sin6θ + cos6θ) – 3 (sin4θ + cos4θ)+ 1 is equal to :(1) 0 (2) 1(3) 2 (4) 3
15. I f πα π
2< < and
32
2π
β π< < ,
sin , tanα β= = −1517
125
, then sin (β – α) is equal
to :
(1)141221
(2)21221
(3)291121
(4)2185
(9) of (57)
16. The perimeter of a certain sector of a circle is equalto half of the circle of which, it is a part. The circularmeasure of the angle of the sector is :
(1) 2 (2)π2
(3) π – 2 (4) π + 2
17. lim ( ) lim[ ( )],x a x a
f x f x→ →
= where [.] denotes greatest
integer function, then :
(1) lim ( )x a
f x→
is an integer
(2) lim ( )x a
f x→
is non-integer
(3) lim ( )x a
f x→
can not be determined
(4) lim ( )x a
f x→
does not exists
18. Let f x x x xx
n In
( ) sin( / ),,
, ( ).= ≠=
∈RS|T|
1 00 0
2 Then :
(1) lim ( )x
f x→0
exists for n > 1
(2) lim ( )x
f x→0
exists for n < 0
(3) lim ( )x
f x→0
does not exists for any value of n
(4) lim ( )x
f x→0
can not be determined
19. The number of digits in the numeral for 264 is:(1) 16 (2) 18(3) 19 (4) 20
20. If xn > xn–1 > .... > x2 > x1 > 1, then the value
of log log log ....log...
x x x x nx
nn
x
x1 2 3
11
− is equal to :
(1) 0 (2) 1(3) 2 (4) none of these
21. If log45 = a and log56 = b, then log32 is equalto :
(1)1
2 1a +(2)
12 1b +
(3) 2ab + 1 (4)1
2 1ab −
22. The domain of the derivative of function
tan | |
(| | ) | |
− ≤
− >
RS|T|
1 112
1 1
x if x
x if xis :
(1) R – {0} (2) R – {1}(3) R – {–1} (4) R – {–1, 1}
23. The function f(x) = ln( ) ln( )1 1+ − −ax bx
x is not
defined at x = 0. The value which should beassigned to f at x = 0, so that it becomescontinuous at x = 0, is :(1) a – b (2) a + b(3) ln a + ln b (4) none of these
24. Let f(x) be polynomial of degree n, an oddpositive integer, and has monotonic behaviourthen number of real roots of the equation f(x)
+ f(2x) + f(3x) +....+f(nx) = 12
n (n + 1) is
equal to :(1) at least one (2) exactly one(3) almost one (4) none of these.
25. The function f(x) = tan–1 x – x decreases inthe interval :(1) (1, ∞) (2) (–1, ∞)(3) (– ∞, ∞) (4) none of these.
26. The function f(x) = x2 e–2x, x > 0. Then themaximum value of f(x) is :(1) 1/e (2) 1/2e(3) 1/e2 (4) none of these.
27. The function f whose graph passes through the
point (0, 7/3) and whose derivative is x x1 2−
is given by :(1) f(x) = (– 1/3) [(1 – x2)3/2 – 8](2) f(x) = (1/3) [(1 – x2)3/2 + 8](3) f(x) = (– 1/3) [sin–1 x + 7](3) none of these.
28. The value of the integral e e
edx
x x
x
−
+z 1
30
5log
is :
(1) 3 + 2π (2) 4 – π(3) 2 + π (4) none of these.
29. limn
r
n
nr
n r→ ∞= +∑1
2 21
2 equals :
(1) 1 5+ (2) − +1 5
(3) − +1 2 (4) 1 2+
30. The value of dx
x1 30
2
+z tan
/π
is :
(1) 0 (2) 1(3) π/2 (4) π/4
31. The area bounded by the curve y = f(x) = x4 –2x 3 + x 2 + 3 , x-ax is and ord inatescorresponding to minimum of the function f(x)is :(1) 1 (2) 91/30(3) 30/9 (4) 4
(10) of (57)
32. The solution of the equation dy/dx = cos (x –y) is :
(1) yx y
C+−F
HGIKJ =cot
2
(2) xx y
C+−F
HGIKJ =cot
2
(3) xx y
C+−F
HGIKJ =tan
2
(4) none of these.
33. The vectors a b and c→ → →
, are of the same lengthand, taken pairwise, they form equal angles.
If a i j and b j k→ →
= + = +$ $ $ $ , the coordinates of c→
are :(1) (1, 0, 1)(2) (1, 2, 3)(3) (– 1/3, 4/3, – 1/3)(4) both (1) and (3)
34. The value of [ ]a b b c c a→ → → → → →
− − − is :(1) 0 (2) 1(3) 2 (4) none of these.
35. If a i j k b i j k and c→ → →
= + − = − +$ $ $, $ $ $, is a unit
vector perpendicular to the vector a→
and
coplanar with a→
and b→
, then a unit vector d→
perpendicular to both a→
and c→
is :
(1)1
62( $ $ $)i j k− + (2)
1
2( $ $)j k+
(3) 1
62($ $ $)i j k− + (4)
1
2( $ $)j k−
36. The value of x for which the matrix product
2 0 70 1 01 2 1
14 70 1 0
4 2−
L
NMMM
O
QPPP
−
− −
L
NMMM
O
QPPP
x x x
x x x equals an identityy
matrix is :
(1)12
(2)13
(3)14
(4)15
37. If A3 = 0, then I + A + A2 equals :(1) I – A (2) (I – A)–1
(3) (I + A)–1 (4) none of these.
38. The number of values of k for which the systemof equations(k + 1)x + 8y = 4k, kx + (k + 3)y = 3k – 1 hasno solution is :(1) 0 (2) 1(3) 2 (4) infinite.
39. The coordinates of three points O, A and B are(0, 0), (0, 4) and (6, 0), respectively. If a pointP moves so that the area of ∆POA is alwaystwice the area of ∆POB, then P lies on :(1) x – 3y = 0 (2) x + 3y = 0(3) 3x + 4y = 0 (4) both (1) and (2)
40. The medians AD and BE of a triangle withvertices A(0, b), B(0, 0) and C(a, 0) areperpendicular to each other if :
(1) b a= 2 (2) a b= 2
(3) b a= − 2 (4) none of these.
41. If the equation of one tangent to the circle withcentre at (2, – 1) from the origin is 3x + y =0, then the equation of the other tangentthrough the origin is :(1) 3x – y = 0 (2) x + 3y = 0(3) x – 3y = 0 (4) x + 2y = 0
42. If O is the origin and OP, OQ are distincttangents to the circle x2 + y2 + 2gx + 2fy + c= 0, the circumcentre of the triangle OPQ is :(1) (– g, – f) (2) (g, f)(3) (– f, – g) (4) none of these
43. The vertex of the parabola y2 + 6x – 2y + 13= 0 is :(1) (1, – 1) (2) (– 2, 1)
(3)32
1,FHG
IKJ (4) −
FHG
IKJ
72
1,
44. The product of the length of the perpendicularsfrom the two foci on any tangent to thehyperbola (x2/a2) – (y2/b2) = 1 is :(1) a2 (2) 2a2
(3) b2 (4) 2b2
45. A stra ight l ine touches the rectangularhyperbola 9x2 – 9y2 = 8 and the parabola y2 =32x. An equation of the line is :(1) 9x + 3y – 8 = 0 (2) 9x – 3y + 8 = 0(3) 3x + 9y + 8 = 0 (4) 3x – 9y – 8 = 0
46. Two of the straight lines given by 3x3 + 3x2 y– 3x y2 + d y3 = 0 are at right angles if :(1) d = – 1/3 (2) d = 1/3(3) d = – 3 (4) d = 3
(11) of (57)
47. If 9x2 + 2h xy + 4y2 + 6x + 2 fy – 3 = 0represents two parallel lines, then the distancebetween them is :
(1)2
13(2)
3
13
(3)4
13(4)
5
13
48. π is the period of the function :
(1)( sin )
cos ( cos )1
1+
+x
x ec x
(2) sin x(3) sin 2x + cos 3x(4) cos (sin x) + cos (cos x)
49. Which of the following function is non periodic?(1) f(x) = {x}, the fractional part of thenumber x(2) f(x) = cot (x + 7)
(3) fx) = 1 – sin
cotcos
tan
2 2
1 1x
xxx+
−+
(4) f(x) = x + sin x
50. The range of the function f xx
x( ) =
+
2
4 1 is :
(1) (0, 1/2) (2) [0, 1/2](3) [0, ∞) (4) [0, 2]
51. limsin ( cos )
x
x
x→ 0
2
2π
equals :
(1) – π (2) π(3) π/2 (4) 1
52. The left-hand derivative of f(x) = |x| sin (πx)at x = k, where k is an integer, is :(1) (– 1)k (k – 1)π (2) (– 1)k–1 (k – 1)π(3) (– 1)k kπ (4) (– 1)k–1 kπ
53. Let f : (0, ∞) → R and F(x) = f t dtx
( )0z . If F(x2)
= x2(1 + x), then f(4) equals :
(1)54
(2) 7
(3) 4 (4) 2
54. Let f : R → R be a function defined by f(x) =max. {x, x3}. The set of all points where f(x)is NOT differentiable is :(1) {– 1, 1} (2) {– 1, 0}(3) {0, 1} (4) {– 1, 0, 1}
55. If f(x) = xex(1–x), then f(x) is :
(1) increasing on −LNM
OQP
12
1, .
(2) decreasing on R.(3) increasing on R.
(4) decreasing on −LNM
OQP
12
1, .
56. Let g(x) = 1 + x – [x] and f(x) =
− <=>
RS|T|
1 00 01 0
,,,
xxx
.
Then for all x, f(g(x)) is equal to :(1) x (2) 1(3) f(x) (4) g(x)
57. If f : [1, ∞) → [2, ∞) is given by f(x) = x + 1x
then f–1(x) equals :
(1)x x+ −2 4
2(2)
x
x1 2+
(3)x x− −2 4
2(4) 1 42+ −x
58. The domain of f(x) = log ( )2
2
3
3 2
x
x x
+
+ + is :
(1) R – {– 1, – 2}(2) (– 2, ∞)(3) R – {– 1, – 2, – 3}(4) (– 3, ∞) – {–1, –2}
59. The triangle formed by the tangent to the curvef(x) = x2 + bx – b at the point (1, 1) and thecoordinate axes, lies in the first quadrant. Itis area is 2, then the value of b is :(1) – 1 (2) 3(3) – 3 (4) 1
60. The value of cos
,2
10
x
adx a
x+>
−zπ
π
, is :
(1) π (2) aπ
(3)π2
(4) 2π
61. The equation of the common tangent touchingthe circle (x – 3)2 + y2 = 9 and the parabolay2 = 4x above the x-axis is :
(1) 3 3 1y x= + (2) 3 3y x= − +( )
(3) 3 3y x= + (4) 3 3 1y x= − +( )
(12) of (57)
62. The number of integer values of m, for whichthe x-coordinate of the point of intersection ofthe lines 3x + 4y = 9 and y = mx + 1 is also aninteger, is :(1) 2 (2) 0(3) 4 (4) 1
63. The equation of the directrix of the parabolay2 + 4y + 4x + 2 = 0 is :(1) x = – 1 (2) x = 1
(3) x = −32
(4) x =32
64. Let AB be a chord of the circle x2 + y2 = r2
subtending a right angle at the centre. Thenthe locus of the centroid of the triangle PAB asP moves on the circle is :(1) a parabola (2) a circle(3) an ellipse (4) a pair of straightlines.
65. In the binomial expansion of (a – b)n, n ≥ 5,the sum of the 5th and 6th terms is zero. Then
ab
equals :
(1)n − 5
6(2)
n − 45
(3)5
4n−(4)
65n −
66. Let f(x) = (1 + b2)x2 + 2bx + 1 and let m(b)be the minimum value of f(x). As b varies, therange of m(b) is :
(1) [0, 1] (2) 012
,OQP
FHG
(3)12
1,LNM
OQP (4) (0, 1]
67. The number o f d i s t inc t rea l roots o f
sin cos coscos sin coscos cos sin
x x xx x xx x x
= 0 in the in terva l
− ≤ ≤π π4 4
x is :
(1) 0 (2) 2(3) 1 (4) 3
68. Let α, β be the roots of x2 – x + p = 0 and γ, δbe the roots of x2 – 4x + q = 0. If α, β, γ, δ arein G.P., then the integral values of p and qrespectively, are :(1) – 2, – 32 (2) – 2, 3(3) – 6, 3 (4) – 6, – 32
69. Let Tn denote the number of triangles whichcan be formed using the vertices of a regularpolygon of n sides. If Tn+1 – Tn = 21, then nequals :(1) 5 (2) 7(3) 6 (4) 4
70. Let the positive numbers a, b, c, d be in A.P.Then abc, abd, acd, bcd are :(1) NOT in A.P./G.P./H.P.(2) in A.P.(3) in G.P.(4) in H.P.
71. If the sum of the first 2n terms of the A.P. 2,5, 8, ..., is equal to the sum of the first n termsof the A.P. 57, 59, 61, ..., then n equals :(1) 10 (2) 12(3) 11 (4) 13
72. Which o f the fo l low ing funct ions i sdifferentiable at x = 0 ?(1) cos (|x|) + |x| (2) cos (|x|) – |x|(3) sin (|x|) + |x| (4) sin (|x|) – |x|
73. The number of solutions of log4 (x – 1) = log2(x– 3) is :(1) 3 (2) 1(3) 2 (4) 0
74. Let f(x) = α x
xx
+≠ −
11, . Then, for what value
of α is f(f(x)) = x ?(1) √2 (2) – √2(3) 1 (4) –1
75. Let
a i k b x i j x k and c y i x j x y k→ → → → → → → → → → →
= − = + + − = + + + −, ( ) ( )1 1 .
Then [ ]a b c→ → →
depends on :(1) only x (2) only y(3) neither x nor y (4) both x and y
76. I f a b and c→ → →
, a re un i t vec tors , then
| | | | | |a b b c c a→ → → → → →
− + − + −2 2 2 does not exceed:(1) 4 (2) 9(3) 8 (4) 6
77. Let PQ and RS be tangents at the extremitiesof the diameter PR of a circle of radius r. If PSand RQ in tersec t a t a po in t X on thecircumference of the circle, then 2r equals :
(1) PQ RS. (2)PQ RS+
2
(3)2PQ RSPQ RS
.+ (4)
PQ RS2 2
2+
(13) of (57)
78. Area of the parallelogram formed by the linesy = mx, y = mx + 1, y = nx and y = nx + 1equals:
(1)| |
( )
m n
m n
+
− 2 (2)2
| |m n+
(3)1
| |m n+(4)
1| |m n−
79. If sin .... cos ...− −− + −FHG
IKJ + − + −
FHG
IKJ =1
2 31 2
4 6
2 4 2 4 2x
x xx
x x π for
0 < |x| < √2, then x equals :
(1)12
(2) 1
(3) – 12
(4) – 1
80. The maximum value of (cos α1) . (cos α2) ...(cos αn), under the restrictions 0 1, 2, ..., αn ≤
π2
and (cot α1) . (cot α2) ... (cot αn) = 1 is :
(1)1
2 2n/(2)
1
2n
(3)12n
(4) 1
81. If α βπ
+ =2
and β + γ = α, then tan α equals :
(1) 2(tan β + tan γ) (2) tan β + tan γ(3) tan β + 2 tan γ (4) 2 tan β + tan γ
82. A man from the top of a 100 metres high towersees a car moving towards the tower at anangle of depression of 30°. After some time,the angle of depression becomes 60°. Thedistance (in metres) travelled by the car duringthis time is :
(1) 100 3 (2)200 3
3
(3)100 3
3(4) 200 3 .
83. If correlation coeff ic ient r in a bivariatedistribution is positive, then A.M. of tworegression coefficients byx and bxy is :(1) > r (2) < r(3) ≥ r (4) ≤ r
84. Let (x i, y i), i = 1, 2, ..., n be ranks of nindividuals allotted in two characters A and B.
If xi + yi = n + 1 for i = 1, 2, ..., n, then rankcorrelation between the characters A and B is(1) 1 (2) 0(3) – 1 (4) none of these.
85. If the integers m and n are chosen at randombetween 1 and 100, then the probability thata number of the form 7m + 7n is divisible by 5equals :
(1)14
(2)17
(3)18
(4)149
86. Given two events A and B. If odds against Aare as 2 : 1 and those in favour of A ∪ B are as3 : 1, then :
(1)12
34
≤ ≤P B( ) (2)512
34
≤ ≤P B( )
(3)14
35
≤ ≤P B( ) (4) none of these.
87. Let a, b, c are positive real numbers. Thefollowing system of equation in x, y, z
x
a
y
b
z
c
x
a
y
b
z
c
x
a
y
b
z
c
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
21 1 1+ − = − + = − + + =, ,
has :(1) no solution.(2) unique solution.(3) infinitely many solutions.(4) finitely many solutions.
88. A person goes in for an examination in whichthere are four papers with a maximum of mmarks for each paper. The number of ways inwhich one can get 2m marks is :(1) (2m+3)C3
(2)13
1 2 4 12( ) ( )m m m+ + +
(3)13
1 2 4 32( )( )m m m+ + +
(4) none of these.
89. For x ∈ R, let [x] denote the largest integerhas than or equal to x, then value of
E =LNM
OQP + +
LNM
OQP + +
LNM
OQP + + +
LNM
OQP
13
13
1100
13
2100
13
99100
... is
(1) 30 (2) 31(3) 32 (4) 33
(14) of (57)
90. The tangent at a point P(a cos θ, b sin θ) on
the ellipse x
a
y
b
2
2
2
21+ = meets the auxiliary
circle in two points. The chord joining themsubtends a right angle at the centre. Then theeccentricity of the ellipse is given by :(1) (1 + sin2 θ)–1/2 (2) (1 + cos2 θ)–1/2
(3) (1 + sin2 θ) (4) (1 + cos2 θ)1/2
91. The di f ferent ia l equat ion whose generalsolution is y = cx2 – x is :(1) y' = 2y + x (2) y' = 2x + y(3) y' = 2y + x2 (4) none of these.
92. Solution of (y log x – 1)y dx = x dy is :
(1) xx x
clog1 1
+ + (2)12x
x x clog + +
(3)1 1x
xx
clog + + (4) none of these.
93. A carpenter has 20 and 15 square meters ofplywood and sunmica respectively. He producesproducts A and B. Product A requires 2 and 1square meters and product B requires 1 and 3square meters o f p lywood and sunmicarespectively. The profit on one piece of productA is Rs. 30 and on one piece of product B isRs. 20. For maximum profit, the number ofpieces A and B manufactured are :(1) 2, 9 (2) 1, 8(3) 3, 7 (4) none of these.
94. The remainder when 246 is divided by 47 is :(1) 1 (2) 3(3) 2 (4) 5
95. For a certain frequency distribution A.M. =24.6, Median 26.1. Its Mode is :(1) 1.5 (2) 50.7(3) 25.35 (4) 29.1
96. In a distribution Mean = 65, Median = 70 andcoefficient of Skewness is – 0.6. The coefficientof variation is :(1) 50% (2) 38.46%(3) 48.46% (4) none of these
97. If the coefficient of rank correlation betweenmarks in Mathematics and marks in Physicsobtained by a certain group of students is 0.8.If the sum of the squares of the difference inranks is given to be 33, then the number ofstudents in a group is :(1) 11 (2) 10(3) 30 (4) none of these
98. A random variable X takes the values – 1, 0, 1its Mean is 0.6. If P(x = 0) = 0.2, then P(x =1) is equal to :(1) 0.6 (2) 0.7(3) 7 (4) none of these.
99. If the dif ference between the mean andvariance of Binomial distribution for 5 trials is
59
the distribution is of the form :
(1) 523
13
5
C
r r
r
FHG
IKJ
FHG
IKJ
−
(2) 513
23
5
C
r r
r
FHG
IKJ
FHG
IKJ
−
(3) 513
235
5 0
CFHG
IKJ
FHG
IKJ (4) none of these.
100. In a Poisson distribution, the probability P(X= 0) is twice the probability P(X = 1). Thenthe Mean of the distribution is :(1) 1 (2) 1/2(3) 1/3 (4) 1/4
(15) of (57)
Directions: For the following questions choose thecorrect option.
1. Let α and β be the roots of the equation x2 + x+ 1 = 0. The equation whose roots are α19, β7
is :(1) x2 – x –1 = 0 (2) x2 – x + 1 = 0(3) x2 + x –1 = 0 (4) x2 + x +1 = 0
2. If ln (a + c), ln (a – c), ln (a – 2b + c) are inA.P., then :(1) a, b, c are in A.P.(2) a2, b2, c2 are in A.P.(3) a, b, c are in G.P.(3) a, b, c are in H.P.
3. Let p, q ∈ {1, 2, 3, 4}. The number of equationsof the form px2 + qx + 1 = 0 having real rootsis :(1) 15 (2) 9(3) 7 (4) 8
4. The function f defined by f(x) = (x + 2) e–x is :(1) decreasing for all x.(2) decreasing in (–∞, –1) and increasing in(–1, ∞).(3) increasing for all x.(4) decreasing in (–1, ∞) and increasing in (–∞, –1).
5. Let f(x) = 0 0
02,,
.x
x x<RST ≥
Then, for all x which of
the following is not true ?(1) f' is differentiable.(2) f is differentiable.(3) f' is continuous.(4) f is continuous.
6. Let g(x) =x f(x) where f x xx
x
x( ) sin ,
,= ≠
=
RS|T|
10
0 0 Att
x = 0 :(1) g is differentiable but g' is not continuous.(2) both g and f are not differentiable.(3) both g and f are differentiable.(4) g is differentiable and g' is continuous.
7. The number of points of intersection of the twocurves y = 2 sin x and y = 5x2 + 2x + 3 is :(1) 0 (2) 1(3) 2 (4) ∞
8. The equation 2x2 + 3y2 – 8x + 18y + 35 = krepresents :(1) no locus if k > 0(2) an ellipse if k < 0(3) a point if k = 0(4) a hyperbola if k > 0
9. Let p and q be the position vectors of P and Qrespectively, with respect to O and |p| = p,|q| = q. The point R and S divide PQ internallyand externally in the ratio 2 : 3 respectively.If OR and OS are perpendicular, then :(1) 9p2 = 4q2 (2) 4p2 = 9q2
(3) 9p = 4q (4) 4p = 9q
10. Let 0 < x < π4
. Then (sec 2x – tan 2x) equals
:
(1) tan –xπ4
FHG
IKJ (2) tan –
π4
xFHG
IKJ
(3) tan x +FHG
IKJ
π4
(4) tan2
4x +FHG
IKJ
π
11. Let A, B, C be three mutually independentevents. Consider the two statements S1 andS2
S1 : A and B ∪ C are independentS2 : A and B ∩ C are independent
Then :(1) Both S1 and S2 are true(2) Only S1 is true(3) Only S2 is true(4) Neither S1 nor S2 is true.
12. For the vector 13
2 2( $ – $ $)i j k+ , which of the
following is not ture ?(1) It is a unit vector.
(2) It makes an angle π3
with the vector
( $ – $ $)2 4 3i j k+ .
(3) It is parallel to the vector –$ $ – $i j k+FHG
IKJ
12
.
(4) I t i s perpend icu la r to the vec tor
( $ $ – $)3 2 2i j k+ .
13. A box contains 24 identical balls of which 12are white and 12 are black. The balls are drawnat random from the box one at a time withreplacement. The probability that a white ballis drawn for the 4th time on the 7th draw is :
(1)564
(2)2732
(3)532
(4)12
14. If y = 4x – 5 is tangent to the curve y2 = px3 +q at (2, 3), then :(1) p = 2, q = –7 (2) p = –2, q = 7(3) p = –2, q = – 7 (4) p = 2, q = 7
Exercise - 03
(16) of (57)
15. Let n be a pos i t i ve in teger such that
sinπ π2 2n n
+ cos =n2
. Then :
(1) 6 ≤ n ≤ 8 (2) 4 < n ≤ 8(3) 6 ≤ n < 8 (4) 4 < n < 8
16. If the lengths of the sides of triangle are 3, 5,7, then the largest angle of the triangle is :
(1)π2
(2)56π
(3)23π
(4)34π
17. An unbiased die is tossed until a numbergreater than 4 appears. The probability thateven number of tosses are needed is :(1) 1/2 (2) 2/5(3) 1/5 (4) 2/3
18. If we consider only the principal values of theinverse trigonometric functions, then the value
of tan cos – sin–1 –11
5 2
4
17
FHG
IKJ is :
(1)293
(2)293
(3)3
29(4)
329
19. Let 2 sin2 x + 3 sin x – 2 > 0 and x2 – x –2 <0 (x is measured in radians). Then x lies in theinterval :
(1)π π6
56
,FHG
IKJ (2) –1,
56πF
HGIKJ
(3) (–1, 2) (4)π6
2,FHG
IKJ
20. The locus of a variable point whose distancefrom (–2, 0) is 2/3 times its distance from the
line x = –92
is :
(1) ellipse (2) parabola(3) hyperbola (4) none of these
21. The equations to a pair of opposite sides ofparallelogram arex2 – 5x + 6 = 0 and y2 – 6y + 5 = 0. Theequation to its diagonals are :(1) x + 4y = 13 and y = 4x – 7(2) 4x + y = 13 and 4y = x – 7(3) 4x + y = 13 and y = 4x – 7(4) y – 4x = 13 and y + 4x = 7
22. I f ω (≠ 1 ) i s a cube root o f 1 , then
1 11 1
1
2 2
2+ +
+=
ii
i
ω ωω
ω– –1 –
–1 – – –1
(1) 0 (2) 1(3) i (4) ω
23. Let n (> 1) be a positive integer. Then thelargest integer m such that (nm + 1) divides(1 + n + n2 + ..... + n127) is :(1) 127 (2) 63(3) 64 (4) 32
24. The radius of the circle passing through the
foci of the ellipse x y2 2
16 91+ = , and having its
centre at (0, 3) is :(1) 4 (2) 3
(3) 12 (4)72
25. The minimum value of the expression sin α +sin β + sin γ, where α, β, γ are real numberssatisfying α + β + γ = π is :(1) positive (2) zero(3) negative (4) – 3
26. The function f(x) = | px – q | + r | x |, x ∈ (–∞, ∞), where p > 0, q > 0, r > 0, assumes itsminimum value only at one point if :(1) p ≠ q (2) r ≠ q(3) r ≠ p (4) p = q = r
27. The value of [ sin ]22
xπ
πz dx where [.] represents
the greatest integer function, is :
(1) –53π
(2) – π
(3)53π
(4) – 2π
28. I f f(x) = A s inπx
B f2
12
2FHG
IKJ +
FHG
IKJ =, ' and
f x dxA
( ) ,=z 2
0
1
π then the constants A and B are
respectively :
(1) π π2 2
and (2) 2 3π π
and
(3) 0 and –4π
(4)4π
and 0
(17) of (57)
29. The variance of the population of observations100, 101, 102, 103, 99 is :(1) 101 (2) 100(3) 1 (4) 2
30. Difference of the two quartiles is 8 and sum ofthe two quartiles is 22 and Mode 11, Mean is8, then the coefficient of skewness is :(1) 0.5 (2) – 0.5(3) 1 (4) none of these
31. The coefficient of correlation between x and yis 0.6. Their covariance is 4.8, var (x) = 9.Then σy is :
(1)83
(2)38
(3)89
(4) none of these
32. A bag contains 6 tickets numbered from 1 to6. A person draws two tickets at random. Ifthe sum of the numbers on the tickets drawnis even he gets Rs. 10/- otherwise he lossesRs. 5/- then the expectation of his gain is :(1) 7 (2) 1(3) 2 (4) 3
33. If the sum of mean and variance of binomialdistribution is 4.8 for five trials, the distributionis :
(1)15
45
5
+FHG
IKJ (2)
13
23
5
+FHG
IKJ
(3)25
35
5
+FHG
IKJ (4) None of these
34. When will current not flow from A to B :
A B
p
q
r
(1) p, q, r are open(2) p is close q, r are open(3) p, r are open q is closed(4) p, q are closed, r is open
35. If ax = by = cz and abc = 1, then xy + yz + zxis equal to :(1) 0 (2) 1(3) xyz (4) none of these.
36. The no. of zeros that are between the decimalpoint and the first significant figure in (0.5)100
if log (0.5) = 1 6990. is :(1) 30 (2) 20(3) 10 (4) 40
37. If a2 + 4b2 = 12ab then log (a + 2b) =
(1)12
(log a + log b – log 2)
(2) log a2
FHG
IKJ + log
b2
+ log 2
(3)12
(log a + log b + 4 log 2)
(4)12
(log a – log b + 4 log 2)
38. The value of
log log log ..... log4 4 4 4114
115
116
11
255+
FHG
IKJ + +
FHG
IKJ + +
FHG
IKJ + + +
FHG
IKJ =
(1) 0 (2) 1(3) 2 (4) 3
39. If x = 3 + 8 , then the value of xx
44
1+ is :
(1) 1154 (2) 1514(3) 5141 (4) 4115
40. If the division of 2x3 + 3x2 + ax + b by :(i) x – 2 gives the remainder 2(ii) x + 2 gives the remainder –2, then thevalues of a and b are respectively(1) –7, –12 (2) 7, 12(3) 12, 7 (4) None of these
41. If x is very small as compared with 1, then
( – ) ( )–
1 7 1 213
34x x+ is equal to :
(1) 1236
+ x (2) 1256
– x
(3) 1236
– x (4) 1256
+ x
42. The co-efficient of x17 in the expansion of (x –1) (x – 2) ..... (x – 18) is :(1) 342 (2) – 171
(3)1712
(4) 684
43. If the absolute term in the expansion of
xa
x–
2FHG
IKJ is 405 term then the value of a is :
(1) ± 3 (2) ± 2(3) ± 1 (4) ± 4
44. If mx2 + 7xy – 3y2 + 4x + 7y + 2 can beexpressed as the product of two linear factorsthen m=(1) 1 (2) 0(3) 2 (4) – 2
(18) of (57)
45. If α, β be the roots of x2 – 3x + a = 0 and γ, δthe roots of x2 – 12x + b = 0 and numbers α,β, γ, δ (in order) form an increasing G.P. then :(1) a = 3; b = 12 (2) a = 12; b = 3(3) a = 2; b = 32 (4) a = 4; b = 16
46. If the quadratic equations ax2 + bx + c = 0and x2 + x + 1 = 0 have a common root, then:(1) a = b = c(2) a ≠ b = c(3) a = b ≠ c(4) a, b, c can taken any values
47. I f θ φπ
– =2
then
cos cos sin
cos sin sin
cos cos sin
cos sin sin
2
2
2
2
θ θ θ
θ θ θ
φ φ φ
φ φ φ
LNMM
OQPP
LNMM
OQPP =
(1) 02 × 2(2) I2
(3)cos ( ) sin( )cos ( ) sin ( – )
θ φ θ φθ φ θ φ
+ ++
LNM
OQP
(4) None of these
48. Let A be an m × n matrix, B and C be n × pmatrices. Then AB = AC ⇒ B = C true :(1) always(2) when A ≠ 0(3) when A is non-singular(4)never
49. The matrix A =L
NMMM
O
QPPP
13
1 2 22 1
2–2
–2 –1 is :
(1) orthogonal (2) involutory(3) idempotent (4) nilpotent
50. If A is a 3-rowed square matrix such that |A|= 2, then |adj [adj (adj A2)]| is equal to :(1) 24 (2) 28
(3) 216 (4) None of these
51. The system of equations x + 4y – 2z = 3, 3x +y + 5z = 7; 2x + 3y + z = 5 has :(1) a unique solution. (2)no solution.(3) an infinite number of solutions.(3) none of these.
52. The sum of infinite number of terms of the
series 145
7
5
10
52 3+ + + +.....
(1)516
(2)336
(3)1736
(4)3516
53. I f a, b, c are in H.P. then the value of
b ab a
b cb c
++
+=
– –(1) 1 (2) 2(3) 3 (4) none of these
54. A man has to pay off a debt of Rs. 40,000 in50 annua l ins ta l lments wh ich form anar i thmet ic progress ion. After paying 40installments he dies leaving one fourth of thedebt unpaid. Then the first instalment amountis :(1) 555 (2) 55(3) 505 (4) none of these
55. A force F i j k→ → → →
= +2 5– λ is applied at the point A(1, 2, 5). If its moment about the point
B = (–1, –2, 3) is 16 6 2i j k→ → →
+FHG
IKJ– λ then λ is
equal to :(1) 2 (2) –1(3) 0 (4) –2
56. A unit vector in xy plane makes an angle 45°
with the vector i j→ →
+ and angle of 60° with
the vector 3 4i j→ →
– . Then the vector is :
(1) i→
(2)i j
→ →+2
(3)i j
→ →–
2(4) none of these
57. Volume of a parallelopiped whose coterminous
edges are 2 3 4 2 2 3i j k i j k i j k→ → → → → → → → →
+ + +– , – , – is
:(1) 5 units (2) 6 units(3) 7 units (4) 8 units
58. The number of solutions of the L.P.P MinimiseZ = x1 + x2 Subject to the constraintsx1 + x2 ≤ 1, – 3x1 + x2 ≥ 3 ; x1, x2 ≥ 0 are :(1) unique (2) infinitely many(3) no solution (4) more than onesolution
59. The maximum value of Z = 4x + 3y subject tothe constraints 3x + 2y ≥ 160, 5x + 2y ≥ 200,x + 2y ≥ 80 x, y ≥ 0 is(1) 320 (2) 230(3) 300 (4) none of these
(19) of (57)
60. Let f (x) be a polynomial of degree three suchthat f (0) = 1, f(1) = 2 and 0 is a critical pointof f(x) such that f(x) does not have a local
extremum at 0. Then f x
xdx
( )2 1+z is equal to :
(1) x – log (x2 + 1) + tan–1 x + C(2) x + (1/2) log (x2 +1) – tan–1 x + C(3) (1/2) x2 + (1/2) log (x2 + 1) – tan–1 x + C(4) (1/2) (x2 – log (x2 + 1)) + tan–1 x + C
61. If f(x) = lim–
,–
–n
n n
n nx x
x x→ ∞ + 0 < x < 1, n ∈ N then
(sin ) ( )–1 x f xz dx is equal to :
(1) – [x sin–1 x + 1 2– x ] + C
(2) x sin–1 x + 1 2– x + C(3) constant(4) None of these
62. Let f be a pos i t i ve funct ion and
I x f x x dx I f x x dxk
k
k
k
1 21 1
1 1= =z z( ( – )) , ( ( – ))– –
where 2k – 1 > 0 then I1/I2 is :(1) 2 (2) k(3) 1/2 (4) 1
63. The limn
nS→ ∞
i f
Sn n n n n
n = + + + ++
12
1
4 1
1
4 4
1
3 2 12 2 2– –.....
–
is :(1) π/2 (2) 2(3) 1 (4) π/6
64. The area of the figure bounded by y2 = 9x andy = 3x is :(1) 1 (2) 1/4(3) 1/2 (4) 2
65. The so lut ion of dydx
ax bcy d
=++
represents a
parabola if :(1) a = 0, c = 0 (2) a = 1, b = 2(3) a = 0, c ≠ 0 (4) a = 1, c = 1
66. The solution of xdy
x y
y
x y2 2 2 21
+=
+
FHG
IKJ– dx is :
(1) y = x cot (c – x) (2) cos–1 y/x = –x +c(3) y = x tan (c – x) (4) y2/x2 = x tan (c – x)
67. Through any point (x, y) of a curve whichpasses through the origin, lines are drawnparallel to the coordinate axes. The curve,given that it divides the rectangle formed bythe two lines and the axes into two areas, oneof which is twice the other, represents a familyof :(1) circles (2) parabolas(3) hyperbolas (4) straight lines
68. If algebraic sum of distances of a variable linefrom points (2, 0), (0, 2) and (–2, –2) is zero,then the line passes through the fixed point :(1) (–1, –1) (2) (0, 0)(3) (1, 1) (4) (2, 2)
69. The slopes of the lines joining the origin tothe points of trisection of the portion of theline 3x + y = 12 intercepted between the axesare the roots of the equation :(1) 2x2 – 15x + 18 = 0(2) 3x2 – 20x + 12 = 0(3) 12x2 – 20x + 3 = 0(4) 18x2 – 15x + 2 = 0
70. The distance between the parallel lines given
by (x + 7y)2 + 4 2 (x + 7y) – 42 = 0 is :
(1) 4/5 (2) 4 2
(3) 2 (4) 10 2
71. If two circles x2 + y2 + 2gx + 2fy = 0 and x2 +y2 + 2g1x + 2f1y = 0 touch each other, then :(1) f1g = fg1 (2) ff1 = gg1
(3) f2 + g2 = f g12
12+ (4) None of these
72. The equation of a circle with origin as centreand pass ing through the vert ices of anequilateral triangle whose median is of length3a is :(1) x2 + y2 = 9a2 (2) x2 + y2 = 16 a2
(3) x2 + y2 = 4a2 (4) x2 + y2 = a2
73. A line is drawn through the point P(3, 11) tocut the circle x2 + y2 = 9 at A and B. Then PA.PB is equal to :(1) 9 (2) 121(3) 205 (4) 139
(20) of (57)
74. Equation of a common tangent to the curvesy2 = 8x and xy = –1 is :(1) 3y = 9x + 2 (2) y = 2x + 1(3) 2y = x + 8 (4) y = x + 2
75. If F1 = (3, 0), F2 = (–3, 0) and P is any pointon the curve 16x2 + 25y2 = 400, then PF1 +PF2 equals :(1) 8 (2) 6(3) 10 (4) 12
76. If (5, 12) and (24, 7) are the foci of a hyperbolapassing through the origin then the eccentricityof the hyperbola is :
(1)38612
(2)38613
(3)38625
(4)38638
77. If ( – )xii
8 91
18
==∑ and ( – )xi
i
8 452
1
18
==∑ then the
standard deviation of x1, x2 ..... x18 is :(1) 4/9 (2) 9/4(3) 3/2 (4) none of these
78. If the median of 21 observations is 40 and ifthe observations greater than the median areincreased by 6 then the median of the new datawill be :(1) 40 (2) 46(3) 46 + 40/21 (4) 46 – 40/21
79. Let X and Y be two variables with the samevariance and U = X + Y, V = X – Y then r (U, V)is equal to :(1) 1 (2) 0(3) –1 (4) none of these
80. An anti-aircraft gun can take a maximum offour shots at an enemy plane moving awayfrom it. The probabilities of hitting the planein the first, second, third and fourth shot are0.4, 0.3, 0.2 and 0.1 respect ive ly. Theprobability that gun hits the plane is :(1) 0.6976 (2) 0.866(3) 0.922 (4) 0.934
81. Out of 21 tickets marked with numbers 1, 2,....., 21, three are drawn at random withoutreplacement. The probabi l i ty that thesenumbers are in A.P. is :
(1)10133
(2)915
(3)14261
(4)13261
82. The probability of India winning a test matchaga ins t West Ind ies i s 1/2. Assumingindependence from match to match, theprobability that in a 5 match series India'ssecond win occurs at the third test is :
(1)18
(2)14
(3)12
(4)23
83. If sin θ + cosec θ = 2, then sinn θ + cosecn θ =(1) 2n (2) 2–n
(3) 2 (4) 2n
84. L tn
n
n
n
n
n
n
n nn→∞+
++
++ +
+
LNMM
OQPP2 2 2 2 2 2 21 2 1
.....( – )
is :
(1) π (2) π2
(3)π4
(4)π2
85. The area of the region bounded by the x axisand the curves defined by
yx if x
x if x=
≤ ≤
≤ ≤
RS||T||tan
–/
cot /
ππ
ππ
33
63 2
is :
(1)–1
log2
2 (2)12
2log
(3) log 2 (4) none of these
86. A loop of the curve y2 = x2 (1 – x2) is rotatedabout the y axis the volume generated is :
(1)π2
2(2)
π2
4
(3) 2π2 (4) none of these
87. The surface area of the anchor-ring generatedby the revolution of a circle of radius a aboutan axis in its own plane at a distance b fromits centre (b > a) is :(1) 4π2a2b2 (2) 2π2a2b2
(3) 2π2ab2 (4) 4π2ab
88. The value of α belonging to the interval [0, 2π]
and satisfying sin sin/
x dx =z 22
απ
α
are :
(1)π π4 6
,RST
UVW (2)π π2
76
,RST
UVW
(3)π π2
32
,RST
UVW (4)π π π π2
32
76
116
, , ,RST
UVW
(21) of (57)
89. limsin
x
xx→
FHG
IKJ =
0
0
(1)π
180(2)
180π
(3) 1 (4) none of these.
90. If yx x
=FHGG
IKJJ +
FHG
IKJsec
( – )sec
–,–1 –11
1
1
1 22 2 then dy/
dx =
(1)1
1 2( – )x(2)
1
1 2( )+ x
(3)3
1 2( – )x(4) none of these.
91. The maximum possible area that can beenclosed by a wire of length 30 cm by bendingit into the form of a sector in square cm is :(1) 10 (2) 225/2(3) 225/4 (4) none of these.
92. When a stone is thrown upwards on certainplanet, its equation of motion is s = 10t – 3t2
(in metres and seconds). It will fall back (onthe planet) again after :(1) (20/3) sec (2) (10/3) sec(3) (5/3) sec (4) none of these.
93.e x
xedx
x
x( )
cos ( )
12
+=z
(1) – cot (x ex) + C (2) tan (x ex) + C(3) tan ex + C (4) none of these.
94.cos
cos sin
/2
0
2x
x xdx
+=zπ
(1) –1 (2) 0(3) 1 (4) 2
95. The number of integral terms in the expansionof (51/2 + 71/8)1024 is :(1) 128 (2) 129(3) 130 (4) 131
96. If every element of a third order determinantof value ∆ is multiplied by 5 then the value ofthe new determinant is :(1) ∆ (2) 5∆(3) 25∆ (4) 125∆
97. The value of 111
2
2
2
///
a a bcb b cac c ab
=
(1) 1 (2) 0(3) abc (4) none of these.
98.x x x
x x
2 2 2 1 12 1 2 1
3 3 1
+ ++ + =
(1) x3 (2) 1(3) –1 (4) none of these.
99. If the vertices of a triangle are (0, 0), (0, 2)and (2 , 0) , the d i s tance between i t sorthocentre and circumcentre is :(1) 0 (2) 1/√2(3) √2 (4) none of these.
100. The angle of elevation of the sun, when thelength of the shadow is √3 times the height ofa pole is :(1) 30° (2) 60°(3) sin–1 (1/√3) (4) none of these.
(22) of (57)
Directions: For the following questions choose thecorrect option.
1. If one vertex of a equilateral triangle is onorigin and second vertex is (4, 0), then its thirdvertex is :
(1) ( , )2 2± (2) ( , )3 2
(3) ( , )2 2 3 (4) ( , )3 2 2
2. A(a, 0) and B(–a, 0) are two fixed point. A pointC moves such that cot A + cot B = λ, thenlocus of C is :(1) x = 2λa (2) x = aλ(3) y = 2a/λ (4) y = a/λ
3. If the l ine 3x + 4y – 24 = 0 meets thecoordinate axes at A and B, then incentre ofthe ∆OAB is:(1) (1, 2) (2) (2, 2)(3) (12, 12) (4) (2, 12)
4. For what values of a and b, the intercepts cutoff on the coordinate axes by the line ax + by+ 8 = 0 are equal in length but opposite insigns to those cut off by the line 2x – 3y + 6 =on the axes :
(1) a =83
, b = –4 (2) a =–83
, b = –4
(3) a =83
, b = 4 (4) a =–83
, b = 4
5. If u = a1x + b1y + c1 = 0 and v = a2x + b2y +
c2 = 0 and aa
bb
cc
1
2
1
2
1
2= = , then u + kv = 0
represents :(1) u = 0(2) a family of concurrent lines(3) a family of parallel lines(4) none of these.
6. If slopes of the lines given by ax2 + 2hxy +by2 = 0 are in the ratio 3 : 1, then h2 =
(1)43ab
(2)ab
(3)ab3
(4)43ab
7. The lines represented by the equation 9x2 +24xy + 16y2 + 21x + 28y + 6 = 0 are :(1) perpendicular (2) parallel(3) coincident (4) none of these.
8. The length of the chord cut off by y = 2x + 1from the circle x2 + y2 = 2 is :
(1)56
(2) 56
(3)6
5(4) 6
5
9. Area of the circle in which a chord of length √2
makes an angle π2
at the centre is :
(1)π2
(2) π
(3) 2π (4) none of these.
10. The point at which the normal at the point A(2, 3) to the circle x2 + y2 + 4x + 6y – 39 = 0will meet the circle again, is :(1) (6, 9) (2) (–6, 9)(3) (6, –9) (4) (–6, –9)
11. If the equation of one tangent to the circle withcentre at (2, – 1) from the origin is 3x + y =0, then the equation of the other tangentthrough the origin is :(1) 3x + y = 0 (2) x – 3y = 0(3) x + 3y = 0 (4) none of these.
12. A circle is inscribed in an equilateral triangleof side a. The area of square inscribed in thecircle is :
(1)a2
3(2)
23
2a
(3)a2
6(4)
a2
12
13. The coordinates of a point on the parabola y2
= 8x whose focal distance is 4, are :
(1)12
12
, ±FHG
IKJ (2) (2, ± 2)
(3) (4, ± 2) (4) (2, ± 4)
14. The eccentricity of the conic x2 – 4x + 4y2 = 4is :
(1)32
(2)23
(3)3
2(4)
32
15. The angle between the tangents drawn from apoint (–a, 2a) to y2 = 4ax is :(1) π/4 (2) π/3(3) π/6 (4) π/2
Exercise - 04
(23) of (57)
16. If the point p(4, – 2) is the one end of thefocal chord PQ of the parabola y2 = x, then theslope of the tangent at Q is :
(1) –14
(2)14
(3) 4 (4) – 4
17. The eccentricity of the hyperbola with latus-
rectum 12 and semi-conjugate axis 2 3, is :(1) 2 (2) 3
(3)32
(4) 2 3
18. If θ be the angle between the unit vectors a→
and b→
then a b→ →
– 2 will be a unit vector if θ isequal to :
(1)π4
(2) π6
(3) π3
(4)π2
19. The moment of the force F i j k→
= +3 2 4$ $ – $ actingat the point (1, –1, 2), about the point (2, – 1,3) is:
(1) 2 7$ – $ – $i j k (2) 2 7 2$ – $ – $i j k
(3) 2 7 2$ – $ $i j k+ (4) 2 7 2$ $ $i j k+ +
20. If a i j k b i j k→ →
= + = + +2 2$ $ – $, $ $ $, and c i j k→
= +$– $ $,2
then a b c→ → →
× =.( )(1) 6 (2) 12(3) 24 (4) none of these.
21. P is the point of intersection of the diagonalsof the parallelogram ABCD. If O is any point,
then OA OB OC OD→ → → →
+ + + =
(1) OP→
(2) 2OP→
(3) 3OP→
(4) 4OP→
22. lim. . .
. . . . .( ) ( )n n n→ ∞ + + + +
+ +LNM
OQP
113
135
157
12 1 2 3
is
equal to :(1) 0 (2) 2
(3)12
(4)13
23. Domain of the function tan–1 x + cos–1 x is :(1) [–1, 1] (2) (–1, 1)(3) R (4) [–1, 1)
24. Range of the function 1 – |x – 2| is :(1) (0, ∞) (2) (–∞, 0](3) (–∞, 1] (4) [1, ∞]
25. The inverse of the function f(x) = loga (x +
x2 1+ ) (a > 0, a ≠ 1) is :
(1)12
(ax – a–xx) (2)12
(ax + a–xx)
(3) ax – a–x (4) ax + a–x
26. If the function f(x) =
k xx
when x
when x
cos–
,
,
ππ
π2 2
32
≠
=
RS||T||
be
continuous at x =π2
, then k =
(1) 3 (2) 6(3) 12 (4) none of these.
27. Let f(x + y) = f(x) f(y) and f(x) = 1 + xg (x)G(x), where limx → 0 g (x) = a and limx → 0 G(x)= b. Then f'(x) is equal to :
(1) (a + b) f(x) (2)ab
f x. ( )
(3) ab f(x) (4) (a – b) f(x)
28. If y = sin–1 ( – )1 x + cos–1 x then dydx
=
(1)–1
( – )x x1(2)
1
1x x( – )
(3)–1
( )x x1 +(4) none of these.
29. ddx
[tan–1 (cot x) + cot–1 (tan x)] =
(1) 1 (2) –1(3) 2 (4)–2
30. If y ex ex ex
= + + + ∞. . . . .
then dydx
=
(1) 11– y
(2)y
y1–
(3)y
y1 +(4) y
y –1
(24) of (57)
31. 10 feet long rod AB moves with its ends ontwo mutually perpendicular straight line OX andOY. If the end A be moving at the rate of 2 cm/sec then when the distance of A from O is 8cm, the rate at which the end B is moving, is :
(1)43
cm/sec (2)83
cm/sec
(3)13
cm/sec (4) none of these.
32. The maximum and minimum value of thefunction 3x4 – 8x3 + 12x2 – 48x + 25 in theinterval [1, 3] are :(1) 39, 16 (2) –16, 39(3) 16, –39 (4) none of these.
33. The equation of the tangent to the curve y = 1– ex/2 at the point of intersection with the y-axis is :(1) x + 2y = 0 (2) 2x + y = 0(3) x + y = 2 (4) none of these.
34.sin
cos
3
21x
xdx
+=z
(1) cos x + tan x + c(2) sec x + tan x + c(3) cos x + sec x + tan x + c(4) none of these.
35. If g x dx g x( ) ( ),=z then g x f x f x dx( ) { ( ) '( )}+zis equal to :(1) g(x) . f'(x) + C(2) g(x). f(x) + C(3) g(x) . f(x) – g(x). f2 (x) + C(4) none of these.
36. ex
xdxx
( )+=z 2 3
(1)e
xC
x
++
2(2) –
ex
Cx
++
2
(3) ex
Cx .( )
1
2 2++ (4) –
( )
e
xC
x
++
2 2
37.x
xdx
4 1–=z
(1)12
1
1
2
2log
–x
xc
+
LNMM
OQPP
+ (2)12
1
1
2
2log
x
xc
+
−
LNMM
OQPP
+
(3)14
1
1
2
2log
x
xc
−
+
FHG
IKJ + (4) none of these.
38. | |x dx− =z 10
2
(1) 3/2 (2) 3(3) 1/2 (4) 1
39. The value of integral x f x dx(sin )0
πz is :
(1) 0 (2) ππ
f x dx(sin )/
0
2z(3)
ππ
20
f x dx(sin )z (4) none of these.
40. The value of ( )ax bx c dx3
2
2
+ +−z depends on :
(1) a (2) b(3) c (4) none of these.
41. Area bounded by the curve xy – 3x – 2y – 10= 0, x-axis and the lines x = 3, x = 4 is :(1) 16 log 2 – 3 (2) 16 log 2 + 3(3) 16 log 2 (4) none of these
42. The volume of the solid generated by therevolution of the curve y = sin x from x = 0 tox = 2π about x axis is :(1) π2 (2) 2π2
(3) 4π2 (4) 4π3
43. The area of the surface of revolution formedby revolving the loop of the curve 9ay2 = x(3a - x) about x axis is :(1) 2π a2 (2) 3π a2
(3) 4π a2 (4) none of these
44. Expansion of x4 – 5x3 + x2 – 3x + 4 into aseries in powers of (x – 4) is:(1) (x – 4)4 + 11 (x – 4)3 + 37 (x – 4)2 + 21(x – 4) – 56(2) (x – 4)4 – 11 (x – 4)3 + 37 (x – 4)2 + 21(x – 4) – 56(3) (x – 4)4 + 11 (x – 4)3 – 37 (x – 4)2 + 21(x – 4) + 56(4) (x – 4)4 – 11 (x – 4)3 + 37 (x – 4)2 – 21 (x– 4) – 56
45. The solut ion of the dif ferential equation
dydx
yx
++−
=1 21 2
0coscos
is :
(1) tan y + cot x = c (2) tan y . cot x = c(3) tan y – cot x = C (4) none of these.
(25) of (57)
46. The solution of the differential equation (x2 +y2) dx = 2xy dy is :(1) x – c(x2 + y2) = 0(2) x – c(x2 – y2) = 0(3) x + c(x2 + y2) = 0(4) x + c(x2 – y2) = 0
47. The curve passing through the point (0, 1) and
satisfying the equation sin dydx
FHG
IKJ = a is :
(1) cosy
xa
+FHG
IKJ =
1(2) cos
xy
a+
FHG
IKJ =
1
(3) siny
xa
−FHG
IKJ =
1(4) sin
xy
a−
FHG
IKJ =
1
48. The solut ion of the dif ferential equation
dydx
y x x+ − =tan sec 0 is :
(1) y tan x = sec x + c(2) y sec x = tan x + c(3) y sec x = cot x + c(4) none of these.
49. The median of the figures 12.6, 8.3, 5.2, 11.1,16.4, 2.9, 13.3 is :(1) 5.2 (2) 11.1(3) 13.2 (4) 8.3
50. The number of icecream cones bought by men,women, boys, girls and children on a day atthe trade fair was, 40, 42, 46, 48 and 44respectively. Their standard deviation is :(1) 2 (2) 3(3) 2.8 (4) 3.1
51. The mode for the fo l l ow ing f requencydistribution :
Class 0 – 5 5 – 10 10 – 15 15 – 20 20 – 25
Frequency 8 16 25 14 7
is :(1) 11.25 (2) 12.25(3) 10.25 (4) 13.25
52. Median for the following data :
No. of Students 6 4 16 7 8 2
Marks 20 9 25 50 40 80
(1) 22 (2) 26(3) 25 (4) 40
53. A student appears for tests I, II and III. Thestudent is successful it he passes either in testsI and II or tests I and III. The probabilities ofthe student passing in tests I, II and III are p,
q and 12
respectively. If the probability that the
student is successful is 12
, then :
(1) p = q = 1 (2) p = q = 12
(3) p = 1, q = 0 (4) p = 1, q = 12
54. Four dice are thrown. Then the probability thatthe sum of the numbers appearing on the diceis 10, is :
(1)1181
(2)781
(3)481
(4)581
55. Pair of dice is rolled together till a sum of either5 or 7 is obtained. Then the probability that 5comes before 7 is :
(1)19
(2)16
(3)25
(4)536
56. In a purse there are 9 five paisa coins and onerupee coin. In another purse there are in all10 five paisa coins. 9 coins are taken from theformer and put into the second and then 9 coinsare taken from the latter and put into the first.The probability that rupee coin is still in thefirst purse is :
(1)19
(2)1019
(3)519
(4) none of these
57. If x y x yi i x y= = ∑ = = =0 12 2 3, , ,σ σ and n =
10, then the coefficient of correlation is :(1) 0.4 (2) 0.3(3) 0.2 (4) 0.1
58. The value of the Spearman's correlat ioncoefficient for a certain number of pairs wasfound to be 0.8. The sum of the squares ofdifferences between corresponding ranks was136. The number of pairs is :(1) 14 (2) 15(3) 16 (4) 17
(26) of (57)
59. The two lines of regression for a bivariatedistribution (x, y) are 3x + 2y = 7 and x + 4y= 9. The regression coefficients byx and bxy arerespectively :
(1)14
23
, (2) − −14
23
,
(3)23
14
, (4) − −23
14
,
60. Fol lowing data are given, related to theadvertisement expenditure and sales of acompany :
Advertisement expenditure (x)
(Rs. In lacs)
Sales (y)(Rs. In lacs)
Mean 10 90
Standard deviation
3 12
Coefficient of correlation = + 0.8.The equations of regression lines are :(1) y = 3.2x + 58, x = 0.2y + 8(2) y = 3.2x – 58, x = 0.2y – 8(3) y = 3.2x – 58, x = 0.2y + 8(4) y = 3.2x + 58, x = 0.2y – 8
61. Taking n = 4, the value of sin ,x dx0
πz byy
Simpson's rule is :
(1)π6
2 1( )+ (2)π6
2 2 1( )−
(3)π6
2 2 1( )+ (4) none of these.
62. A river is 60 metres wide. The depth d (inmetres) of the river at a distance of x metresfrom one bank is given by the following table :
x 0 10 20 30 40 50 60
d 0 3 7 11 8 6 4
By Trapezium rule, the area of the cross-sectionof the river is :(1) 370 sq. metres (2) 375 sq. metres(3) 380 sq. metres (4) 390 sq. metres
63. The vertices of a feasible region by the linearconstraints x – y ≥ 2, x + 2y ≤ 8, x, y ≥ 0, are:(1) (4, 2), (8, 0), (2, 0)(2) (2, 0), (4, 2), (0, 8)(3) (2, 4), (2, 0), (8, 0)(4) (4, 2), (0, 2), (8, 0)
64. A fruit seller invests Rs. 500 in apples andoranges. He has a space to store at the most12 boxes. A box of oranges costs Rs. 50 and abox of apples costs Rs. 25. He can sell orangesat a profit of Rs. 10 per box and apples at aprofit of Rs. 6 per box. Assuming that he cansell all the item that he buys, the number ofboxes of oranges and apples he should buy tomaximize his profit, is :(1) 8, 4 (2) 0, 12(3) 12, 0 (4) 4, 8
65. Ram and Shyam throw with one dice for a prizeof Rs. 88 which is to be won by the player whothrows 1 first. If Ram starts, then mathematicalexpectation for Shyam is(1) Rs. 35 (2) Rs. 40(3) Rs. 30 (4) Rs. 45
66. If X is a Poisson variate such that P (2) = 9P(4) + 90 P (6), then the Mean of X is :(1) ± 1 (2) ± 2(3) ± 3 (4) None
67. Rama has the probability 23
of winning a game.
If the plays 4 games, the probability that thehe wins at least one game is
(1)181
(2)8081
(3)2981
(4) None of these
68. A frequency distribution gives the followingresults :Coefficient of variation = 5Karl Pearson's co-efficient of skewness = 0.5Standard deviation = 2The mean and mode of the distribution are(1) 40, 39 (2) 30, 29(3) 50, 49 (4) None
69. If the lines of regression be x – y = 0 and 4x –
y – 3 = 0, and σx2 1= , then the coefficient of
correlation is :(1) – 0.5 (2) 0.5(3) 1.0 (4) – 1.0
70. Fit a stra ight l ine to the fo l lowing dataregarding x as the independent variable by themethod of least square.
x 0 1 2 3 4
y 1 1.8 3.3 4.5 6.3
(1) y = 1.33 + 0.72x(2) y = 0.72 + 1.33x(3) 16.9 = 5y + 10x(4) y = x + 0.8
(27) of (57)
71. The value of 2 5 6 3 5 14 6 5+ + =– – –
(1) 0 (2) 1 – √5(3) 2 (4) 3 + 2√5
72. If truth values of p and q both are F, truth valueof p ⇒ (q v ~ q) is :(1) T (2) F(3) T and F both (4) none of these.
73. Let U be the universal set containing 700elements. If A, B are sub-sets of U such thatn(A) = 200, n(B) = 300 and n(A ∩ B) = 100.Then n(A' ∩ B) is equal to :(1) 400 (2) 600(3) 300 (4) none of these.
74. If A is a square matrix of order n × n and k isa scalar, then adj (kA) is equal to :(1) λ–n adj A (2) k adj A(3) k–n+1 adj A (4) kn–1 adj A
75. I f 11
11
1−L
NMOQP −LNM
OQP =
−LNM
OQP
−tan
tantan
tanθ
θθ
θa bb a
,
then :(1) a = 1, b = 1(2) a = cos 2θ, b = sin 2θ(3) a = sin 2θ, b = cos 2θ(4) none of these.
76. The following system of equations 3x – 2y + z= 0, λx – 14y + 15z = 0, x + 2y – 3z = 0 hasa solution other than x = y = z = 0 for λ equalto :(1) 1 (2) 2(3) 3 (4) 5
77. If p + q + r = 0 = a + b + c, then the value of
the determinant pa qb rcqc ra pbrb pc qa
is :
(1) 0 (2) pa + qb + rc(3) 1 (4) none of these
78. All the letters of the word AIMCET are arrangedin all possible ways. The number of sucharrangements in which two vowels are notadjacent to each other is :(1) 360 (2) 144(3) 72 (4) 54
79. There are four balls of different colours andfour boxes of colours same as those of theballs. The number of ways in which the balls,one in each box, could be placed such that aball does not go to box of its own colour is :(1) 8 (2) 7(3) 9 (4) none of these.
80. In the expansion of 21
3 2
9
xx
+FHG
IKJ , the term
independent of x is :
(1) 672 (2)1792
9(3) 1344 (4) none of these.
81. The sum of the coefficients of even powers ofx in the expansion of (1 + x + x2 + x3)5 is :(1) 210 (2) 24
(3) 28 (4) 29
82. The largest term in the expansion of (3 + 2x)50
where x = 1/5, is :(1) 5th and 6th (2) 51st(3) 6th and 7th (4) 6th
83. The maximum value of 5 cos θ + 3 cos θπ
+FHG
IKJ3
+
3 is :(1) 8 (2) 10(3) 11 (4) None of these
84. If sin x + sin y = 3 (cos y – cos x) then the
value of sinsin
33
xy
is :
(1) 1 (2) 0(3) –1 (4) None of these
85. If sin 2θ + sin 2φ = 1 and (cos 2θ + cos 2φ)
=32
, then cos2(θ – φ) =
(1)38
(2)58
(3)34
(4)54
86. If sin–1 x + sin–1 (1 – x) = cos–1 x, then x equals:(1) 1, –1/2 (2) 0, 1/2
(3) – ,12
12
(4) None of these
87. If sin–1 x + sin–1 y + sin–1 z =32π
, then value
of x y zx y z
100 100 100101 101 101
9+ +
+ +– is :
(1) 0 (2) 100(3) –8 (4) 9
(28) of (57)
88. The two straight roads intersect at an angleof 60°. A bus on one road is 2 km away fromthe intersection and a car on the other road is3 km away from the intersection. Then thedirect distance between the two vehicles is :
(1) 1 km (2) 2 km
(3) 4 km (4) 7 km
89. If rr
rr1
2
3= , then :
(1) A = 90° (2) B = 90°(3) C = 90° (4) None of these
90. If the radius of the circumcircle of an isoscelestriangle PQR is equal to PQ (= PR), then theangle P is :
(1)π6
(2)π3
(3)π2
(4)23π
91. The sum of the series
131
2 42
353
464
.!
.! !
.!
. . . . .+ + + + ∞
(1) e (2) 2e(3) 3e (4) 4e
92. If log (1 – x + x2) can be expanded in ascendingpowers of x In the form a1x + a2x
2 + a3x3 +
....., then a3 + a6 + a9 + a12 + ..... ∞ =
(1) log 2 (2)13
2log
(3)23
2log (4) None of these
93. The sum of the series 12
13
16
+ + + . . . . . to 9
terms is :
(1) –56
(2) –12
(3) 1 (4) –32
94. If a1, a2, ....., an are In A.P. with commondifference d, then the sum of the followingseriessin d (cosec a1 . cosec a2 + cosec a2 . cosec a3+ ..... + cosec an–1 cosec an) is :(1) tan a1 – tan an(2) cosec a1 – cosec an(3) sec a1 – sec an(4) cot a1 – cot an
95. If the (m + 1)th, (n + 1)th and (r + 1)th termsof an A.P. are In G.P. and m, n, r are in H.P.,then the value of the ratio of the commondifference to the first term of the A.P. is :
(1) –2n
(2)2n
(3) –n2
(4)n2
96. In a G.P. of positive terms, any term is equalto the sum of the next two terms. Then thecommon ratio of the G.P. is :(1) cos 18° (2) sin 18°(3) 2 cos 18° (4) 2 sin 18°
97. Root of the equation 9x2 – 18|x| + 5 = 0belonging to the domain of definition of thefunction f(x) = log (x2 – x – 2) is :
(1) –53
(2) –13
(3)53
13
, (4) – , –53
13
98. If one root of the equation 8x2 – 6x – a – 3 =0 is the square of the other, then the values ofa are :(1) 4, 24 (2) 4, –24(3) –4, –24 (4) –4, 24
99. I f a + b + c = 0 , then
x x xa b c a b c a b c2 2 2–1 –1 –1 –1 –1 –1=
(1) x (2) x2
(3) x3 (4) 0
100. If A = log2 log2 log4 256 + 2 22
log then A
equals(1) 2 (2) 3(3) 5 (4) 4
(29) of (57)
1. To find probability of given case we get sum 9 whenWe have values 4 & 5, 3 & 65 & 4, 6 & 3 in Ist & IInd throw respectively
Now to find probability that at least 1 no. is 6 is 24
12
= . Ans.(2)
2. Let roots of ax2 + bx + c = 0, be α & β.
∴ roots of px2 + qx + r = 0 is 1 1α β
and .
Now, α + β = – b/a, αβ = c/a &
&1 1α β
α βα β
+ =−
⇒+⋅
=−
⇒−
=−q
pq
p
ba
ca
qp
⇒ =bc
qp ..... (1)
Also 1
α β.=
rp
⇒ =ac
rp ...... (2)
∴ (1) & (2) are required conditions. Ans.(1)3. We have the two curves as
y2 = 4 ax ....... (1)ay2 = 4x3 ....... (2)The angle between two curves is angle between tangents drawn at point ofintersection of two curves. The point of intersection is :⇒ a ⋅ 4ax = 4x3 ⇒ 4x3 – 4a2x = 0⇒ 4x (x2 – a2) = 0 ⇒ x = 0, x = ± aHere x = (–a) is not possible,∴ we try to find the slopes of tangent at x = a, for y2 = 4ax.
2 442
1 12 1ydydx
adydx
ay
mx a x a y a= ⇒FHG
IKJ =
FHG
IKJ = ⇒ == = =. ( ),
For ay2 = 4x3
⇒ 2 126
3 322
2
2aydydx
xdydx
xay
mx a a a
= ⇒ =FHG
IKJ = ⇒ =
= =,
Now tan .θ =−
+
FHG
IKJ =
−+ ×
FHG
IKJ =
FHG
IKJ =
m m
m m1 2
1 213 1
1 3 124
12
⇒ m2 = 3
∴ =FHG
IKJ
−θ tan .1 12
Ans.(4)
4. I x= z log (tan )/
0
2π⇒ = −zI xlog (tan ( / ))
/π
π2
0
2
= z log (cot )/
x dx0
2π= − = −z log tan
/x I
0
2π
∴ I + I = 0, 2I = 0 ⇒ I = 0. Ans.(1)
5. In = x (log x)n – xx
n x dxn. . (log ) .1 1−z
⇒ In = −
−zx x n x dxn n(log ) log .c h 1
⇒ In = x(log x)n – n . In–1 dx⇒ In + n In–1 = x(log x)n
⇒ In + n In–1 – x(log x)n = 0. Ans.(2)6. The variance of given series is σ2.
Now, series whose variance is to be found ispx1, px2, px3 ........ pxn = p2σ2. Ans.(3)
7.1 1 3
a c b c a b c++
+=
+ +⇒
+ + ++ +
=+ +
( ) ( )( )( ) ( )b c a ca c b c a b c
3
⇒ (a + b + 2c) (a + b + c) = 3 (a + c) (b + c)⇒ a2 + ab + 2ac + ab + b2 + 2bc + ca + cb + 2c2 = 3 [ab + ac + bc + c2]⇒ a2 + b2 – c2 + 2ab + 3ac + 3bc = 3ab + 3ac + 3bc
⇒ a2 + b2 – c2 =2
2ab
⇒ a b c
ab
2 2 2
212
+ −=
⇒ cos C = 12
= cos 60° ∴ C = 60°. Ans.(1)
8. Arranging the numbers in increasing order 36, 42, 44, 46, 56, 61, 63, 70no. of observation = 8.
∴ median = A. Mean of n n
th th
2 21
FHG
IKJ +
FHG
IKJ and observation.
A.M. of 4th & 5 th observation 46 56
2102
251
+= = .
x d = |× – m|36 1542 944 746 556 561 1063 1270 19
Σd = 082
∴ Mean deviation from the median Σdn
= =0828
10.25. Ans.(2)
9. ∴ = zR Shaded area of I quadrantst2 )
y = x y = - x
= − = × =z2 216
13
2
0
1( ) .x x dx sq.units Ans.(2)
10. We haverA = ai + +bi ck
rB = a1i b i c k+ +1 1 s
then r rA B = (ai + bi + ck) (a1⋅ + + = + +i b i c k aa bb cc1 1 1 1 1) . Ans.(3)
11. We have a2, b2, c2 are in A ?Then subtracting a2 + b2 + c2 from all terms the 3 terms we havea2 – (a2 + b2 + c2), b2 – (a2 + b2 + c2), c2 – (a2 + b2 + c2) are in A.P.⇒ – (b2 + c2), – (a2 + c2), – (a2 + b2) in A.P.⇒ b2 + c2, a2 + c2, a2 + b2 are in A.P. Ans.(1)
12. The diagonals of a Rhombus are,d1 = 5 i – 2 j + 6kd2 = 2 i – 3 j + 5k
The area of Rhombus A d d= ×12 1 2| |
r r
r rd d
i j k
1 2 5 2 62 3 5
× = −−
= i [– 10 + 18] – j [25 – 12] + k [–15 + 4]
= 8i – j(13) – 11 k. ∴ × = + + =| |r rd d1 2 64 169 121 354
∴ =Area3542
. Ans.(3)
Exercise - 01 - Solutions
(30) of (57)
13. A particle is thrown vertically upward to point Cwe have AB = 25 m, u = 15 m/sec.Let at point C, v = 0.∴ v2 – u2 = 2as. A↑ u = 15m/sec.
C
B⇒ s = −
− ×15
9.8 2
2
= 11.47 m.
Time required to travel AC distance
–15 = –9.8 × t ⇒ t = 159.8
= 1.533 sec.
Time required to travel BC distance.
⇒ (36.47) = 0 × t + 12
(9.8)t2 ⇒ t2 = 72.94
9.8= 7.4428 ⇒ t = 2.72816 sec.
Total time = 1.53 + 2.728= 4.25 sec. Ans.(1)14. Shortest distance of point (2, 0)
from the curve y = √x ≥ y2 = xLet this point be p(x1, y1) on the curve then y1
2 = x1.Now Let D = (x1 – 2)2 + y1
2.we want D to be minimum.D = (x1
2 – 4x1 + 4 + x1) = D = x12 – 3x1 + 4
dDdx
= 2x1 – 3. For minimum ⇒ dDdx
= 0
⇒ x1 = 3/2 and for shortest distance x y1 132
32
= =,
Required point is 32
32
,FHG
IKJ . Ans.(3)
15. Distance between parallel lines 22g ac
a b a−
+( )
Now x2 – 2xy + y2 + 2x + 3y – 5 = 0, here a = 1, b = 1, h = – 1,g = 1, f = 3/2, c = – 5.
∴ =+
=D 21 5
22 3 . Ans.(1)
16. x + y = 1y = 3 x – 3y + 3 x = – 7.We have a formula for finding area of triangle formed by 3 lines.arx + bry + cr = 0 (r = 1, 2, 3)
as ∆ =1
2 1 2 3
1 1 1
2 2 2
3 3 3
2
c c c
a b ca b ca b c
∴ x + y – 1 = 03x – y – 3 = 03x + y + 7 = 0
∴ Area = ∆ =− +
−− −
14 2 6 2
1 1 13 1 33 1 7
2
[ ] [ ] [ ]
∆ =−
− − + −1
964 30 1 6
2( ) ( ) = − − =
196
401600
962c h
∆ =100
6. Ans.(3)
17. 2x + 3y + 1 ........ (1)5x – 3y + 4 = 0 .........(2)To find point of intersection, adding (1) & (2)
7x + 5 = 0, x =−57
on multiplying (1) by 5 and (2) by 2 and subtracting10x + 15y + 3 = 010x – 6y + 8 = 021y – 5 = 0
y =521
∴ Point is −FHG
IKJ
57
521
,
Now slope of any line ⊥ to x – 3y + 4 = 0is 3y = x + 4 required slope m = – 3.
∴ required line y x−FHG
IKJ = − +
FHG
IKJ
521
357
21y – 5 = – 63x – 4521y + 63 x + 40 = 0. Ans.(3)
18. Standard result. Ans.(1)19. Here a = 5, b = 3
∴ any point on the ellipse (5 cosφ, 3 sinφ). Normal at this point to the givenellipse is 5x sec φ – 3y cosec φ. = 25 – 9 = 16...... (1)Also the equation of the given normal is 5x – 3y = 8√2 ...... (2)comparing the equation of normals (1) & (2).
⇒ = =5
53
316
8 2
sec cosφ φec⇒ sin φ = cos φ = ⇒ =
1
2 4φ
π. Ans.(2)
20. Standard result. Ans.(1)
21. The Wallies formula is sin .............. ./ n x dx
n
n
n
n⋅ =
− −
−z 1 3
212 20
2 c h c hc h
ππ. Ans.(2)
22. Standard result. Ans.(3)
23. Ie
dxx
=+−z 1
1
Ie
edx
x
x=
+z 1I = log|1 + ex| + c. Ans.(4)
24. For (a), f xa a
a af x
x x
x x− =
+−
= −−
−c h c h, so f is an odd function.
(b), f xa
a
a
af x
x
x
x
x− =
+−
=+−
= −−
−c h c h1
1
1
1, so f is an odd function
(c), f x xa
ax
a
af x
x
x
x
x− = −
−
+=
−
+=
−
−c h c h c h1
1
1
1, so f is an even function.
(d),f(– x) = log2 x x2 1+ −FH IK = – f(x) so f is an odd function. Ans.(3)
25. Since x3 – 2x – 4 = (x – 2) (x2 + 2x + 2) and the integrand is a proper rationalfunction, we use the method of partial fractions to get
3 4
2 4 2 2 23 2
x
x x
Ax
Bx C
x x
+− −
=−
++
+ +⇒ 3x + 4 = A(x2 + 2x + 2) + (Bx + C) (x – 2)Putting x = 2, A = 1 and comparing the coefficients of x2, we get0 = A + B, i.e., B = –1.Putting x = 0, we have 4 = 2A – 2C, i.e., C = –1.
Thus 3 4
2 4
12
1
2 23 2
x
x x xdx
x
x xdx
+− −
=−
−+
+ +z zz= − − + + +log log x x x C2
12
2 22 Hence K = –1/2, so that
f(x) = |x2 + 2x + 2| = x2 + 2x + 2, because x2 + 2x + 2 > 0. Ans.(4)26. Since sin θ and cos θ are roots of the given quadratic equation,
we have sin θ + cos θ = b/a and sin θ cos θ = c/a⇒ (sin θ + cos θ )2 = b2/a2
⇒ sin2θ + cos2θ + 2 sin θ cos θ = b2 / a2
⇒ 1 22
2+ =
ca
b
a⇒ a2 + 2ac – b2 = 0. Ans.(2)
27. Put 3x – 4y = X ⇒ 3 – 4dydx
dXdx
dydx
dXdx
= ⇒ = −FHG
IKJ
14
3 .
Therefore the given equation is reduced to
34
14
23
14
4 8 3 3
4 31
4 3− =
−−
⇒ − =− − −
−=
+−
dXdx
XX
dXdx
X X
XXX
c h
c h c h
⇒ −−+
= ⇒ − −+
FHG
IKJ =
XX
dX dxX
dX dx31
14
1
⇒ –X + 4 log (X + 1) = x + constant⇒ 4 log (3x – 4y + 1) = x + 3x – 4y + constant⇒ log (3x – 4y + 1) = x – y + c. Ans.(4)
(31) of (57)
28. (D6C1)16 = D × 163 + 6 × 162 + C × 161 + 1 × 160
= 13 × 163 + 6 × 162 + 12 × 16 + 1 × 160
= 53248 + 1536 + 192 +1= (54977)10. Ans.(3)
29. Let OP be the cliff of height 150 m, A the initial and B the final position of theboat after 2 minutes. Then OB = 150 cot 45° = 150 andOA = 150 cot 60° = 150 (1/√3).
150 m∴ AB = 150 1
1
3
150 3 3
350 3 3−
FHG
IKJ =
−= −
m
e j e j
Hence, the speed of the boat is
50 3 3
10009 3 3
212
.602
−=
−=
e j[9 – 3(1.73)]
12
=(9 – 5.19) = 1.9 km/h. Ans.(1)
30. A determinant of order 2 is of the form ∆ =a bc d
It is equal to ad – bc.
The total number of ways of choosing a, b, c and d is 2 × 2 × 2 × 2 = 16. Now∆ ≠ 0 if and only if either ad = 1, bc = 0 or ad = 0, bc = 1. But ad =1,bc = 0 if and only if a = d = 1 and the least one of b, c is zero.Thus, ad = 1, bc = 0 in three cases. Similarly, ad = 0, bc = 1 in three cases.Thus, the probability of the required event is 6/16 = 3/8. Ans.(2)
31. Since x = 2y, ∀ x and y It is a perfect positive relationship and hencer = 1. Ans.(3)
32. The (r + 1)th term in [x2 + (2/x)]15 is given by
Tr+1 = 15Cr(x2)15–r
2x
rFHG
IKJ =15Crx
30–2r. 2r
rx = 15 30 3 2C xr
r r− ×
Tr+1 will contain x15 if r = 5, and T r+1 will be independent of x if r =10.Thus, the required ratio 15C525 : 15C10210 = 1 : 32 [Q 15C5 = 15C10].Ans.(4)
33. Let (h, k) be the centre of the required circle. Then (h, k) being the midpointof the chord of the given circle, its equation ishx + ky – a(x + h) = h2 + k2 – 2ahSince it passes through the origin, we have–ah = h2 + k2 – 2ah ⇒ h2 + k2 – ah = 0 Hence the locus of (h, k) isx2 + y2 – ax = 0. Ans.(3)
34. The given equation is equivalent to7 cos2x + sin x cos x – 3(sin2 x + cos2 x) = 0or 3 sin2 x – sin x cos x – 4 cos2 x = 0Since cos x ≠ 0 does not satisfy the given equation we divide throughout bycos2x and get3tan2 x – tan x – 4 = 0 ⇒ (3 tan x – 4) (tan x + 1) = 0⇒ tan x = 4/3 or tan x = –1. That is, x = kπ + tan–1(4/3)or x = n π + 3π/4 (n, k ∈ I). It should be noted that the solutions given by (2)or (3) are not complete; the general solution is given by (4) only. Ans.(4)
35. Let two parts be x and y, so that x + y = 15. We have to maximizez = x2 y3 = (15 – y)2 y3 = y5 – 30y4 + 225 y3
dzdy
= 5y4 – 120y3 + 675y2 = 0 or y2(y – 15) (y – 9) = 0
Naturally y ≠ 0, y ≠ 15 and hence y = 9, and x = 6. Ans.(3)36. f(x) = log2x2 = 2log2x and g(x) = logx4 = 2/log2x therefore f(x) = 4/g(x).
Ans.(2)37. The total number of lines that can be drawn joining n points taking two at a
time is nC2 of which n are the sides of the polygon. So the number of diagonalsis nC2 – n. Ans.(3)
38. Let Jane have x sisters then she will have x brothersWe are given total number of brothers = 4So total number of sisters = 3 + 2 = 5Total number of offspring = 4 + 5 = 9. Ans.(2)
39. Let the common ratios of the two GPs be r and R. ∴ aR4 = ar2 ∴ r = R2. Sothe 67th term of the first GP = ar66 = aR132 = 133rd term of the second GP.Ans.(2)
40. The sides of the triangle are 7,24,25. Area = 7 × 24/2 = 84.∴ 84 = r(7 + 24 + 25)/2. ∴ r = 3. Ans.(1)
41. If f(t) = x, then (t + 3)/(4t – 5) = x. ∴ t =(3 + 5x)/4x – 1). Ans.(4)
42. a bi j k
i j k× = − = − +2 1 21 1 0
2 2
∴ × = + + = a b 4 4 1 3
|c – a| = 2√2 ⇒ |c – a|2 = 8⇒ | c |2 – 2a.c + | a |2 = 8 ⇒ | c |2 – 2 | c | + 9 = 8⇒ | c |2 – 2 | c | + 1 = 0 ⇒ | c | = 1
∴ | (a × b) × c| = | a × b | | c | sin 30° = 32
. Ans. (2)
43. The value of the determinant
1 3 92 4 123 5 15
= ⋅ = = ⋅313
1 3 92 4 123 5 15
31 3 32 4 43 5 5
3 0
{ Q Two columns of determinant are equal}. Ans.(1)44. p {a + (p – 1) d} = q {a + (q – 1) d}
⇒ ap + (p2 – p) d = aq + (q2 – q)d⇒ a (p – q) + (p2 – q2) d + (q – p) d = 0⇒ (p – q) {a + (p + q – 1) d} = 0⇒ a + (p + q – 1) d = 0 ⇒ Tp + q = 0 {Q p ≠ q}. Ans. (1)
45. ∆ = a2 – (b – c)2 = a2 – (b2 + c2 – 2bc)
= 2bc – (b2 + c2 – a2) = 2bc 12
2 2 2
−+ −F
HGIKJ
b c abc
= 2bc(1 – cos A) = 2bc. 2sin2 A2
FHG
IKJ .....(1)
Also, ∆ = 12
bc sin A =12
(bc) 2 sin A2
FHG
IKJ cos
A2
FHG
IKJ
= bc sin A2
FHG
IKJ cos
A2
FHG
IKJ ....(2)
from (1) and (2) 2 22 2 2
42 2
2bcA
bcA A A A
. sin sin cos sin cos= ⇒ =
⇒ tan A
t say2
14
FHG
IKJ = = c h . Now tan A =
−=
21
8152
tt
. Ans.( 2)
46. Hint : Given function is an odd function of x. Ans. (1)
47. sin2θ – cosθ =14
⇒ − − =( cos ) cos114
2 θ θ
⇒ 4 – 4 cos2 θ – 4 cos θ = 1⇒ 4 cos2 θ + 4 cos θ – 3 = 0⇒ (2cos θ + 3) (2cos θ – 1) = 0
∴ cos θ = 12
, −32
(2nd value is rejected)
∴ θ = 2nπ ±π/3 ⇒ θ = π/3 or2π – π/3 = 5π/3 in the given interval. Ans.(2)
(32) of (57)
48. I x
x 4=
+ + −LNM OQPz x xdx
2 21 1 2log log.
e jI
xx
x
x
x=
++L
NMMOQPPz
11 12
2
2
4
log
I = + +FHG
IKJz 1
11
1 12 2 3x x x
dxlog . . Putting 112+ =
xt and − =
23x
dx dt ,
we get I tt
tt dt= − = −
FHG
IKJ −
RS|T|UV|W|z z1
212 3 2
23
13 23 2 t dt tlog log
= − −FHG
IKJ
RST|UVW|+
12
23
23
23
3 2 3 2log . . t t t C = − + +13
29
3 2 3 2t t Clog t
= − +FHG
IKJ +
FHG
IKJ −
LNM
OQP+
13
11
11 2
32
3 2
2x xClog . Ans.(2)
49. Let A denote the event that the student is selected in AIMCET entrance testand B denotes the event that he is selected in Pune entrance test. Then P(A) = 0.2, P (B) = 0.5 and P (A ∩ B) = 0.3.
Required probability = P A B∩e j = 1 – P (A ∪ B)
=1 – [ P (A) + P (B) – P (A ∩ B)] = 1 – (0.2 + 0.5 – 0.3) = 0.6 Ans. (4)
50. Let the slope of tangent of required family be dydx
m= 1.
Also ycx
=2
, therefore dydx
cx
m say= − =2
2 2 c h
By the given condition, we have tanπ4 1
1 2
1 2
=−
+m m
m m
⇒ 1 + m1m2 = m1 – m2 ⇒ + = −dydx
cx
cx
dydx
2
2
2
21
⇒ +FHG
IKJ = − ⇒ =
−+
dydx
cx
cx
dydx
x cx c
1 12
2
2
2
2 2
2 2. Ans.(2)
51. The value of 0
2πz ⋅cos nx dx
I nx dx= ⋅z0
2π
cos =LNM
OQP = − = ⋅ =
sinsin sin
nxn n n0
21
2n 01
0 0.π
π Ans.(4)
52. Ix dx
x=
⋅
+z −
0
1 1
21
tanPut tan-1 x = t ⇒
1
1 2+=
xdx dt. at x = 0, t = 0
& ,atx t= =14π
∴ I t dt= ⋅z0
4π /
=LNMM
OQPP =
FHG
IKJ =
t2
0
4 2 2
212 16 32
ππ π
/
. . Ans.(3)
53. α, β are roots of the equation x2 + px + q = 0⇒ α + β = – p and αβ = q.α2 + β2 = (α + β)2 – 2αβ = (– p)2 – 2p = p2 – 2p. Ans.(2)
54. For hyperbola
e2 = 1 181 25
144 251
81144
225144
2
2+ = + = + =
b
a
//
∴ = =e1512
54
i.e., e > 1.
Also a2 14425
= or a =125
, hence the foci are (±ae, 0) i.e. = (± 3, 0)
Now the foci coincide, therefore for ellipse ae = 3 or a2e2 = 9
or aba
22
21 9−FHG
IKJ = or a2 – b2 = 9 or 16 – 9 = b2 ∴ b2 = 7. Ans.(3)
55. We have 7
16 6 7 16 6 7+ − −e j e j
=+ + − + −
7
9 7 2 3 7 9 7 2 3 7. . . .e j e j=
+ − −
7
3 7 3 72 2e j e j
=+ − −
= =7
3 7 3 7
7
2 7
12e j e j
= a rational number Ans. (4)
56.tansec
AA
=725
⇒ × = =sincos
cos sinAA
A A725
Now, sin A (1 + sin A) + cos A (1 + cos A)= sin A + sin2 A + cos A + cos2 A = sin A + cos A + 1
725
1725
1725
125
576 12
+ −FHG
IKJ + = + + = + + =
+ +=
725
2425
17 24 25
255625
.
Ans.(1)
57. y = mx – 2am – am3 or mx y
am am−
+=
213
Make the parabola y2 = 4ax homogeneous and since the lines are at rightangles, sum of the coefficients of x2 and y2 is zero.∴ m2 = 2 or m = √2 or tanθ = √2 or θ = tan–1(√2). Ans.(4)
58. P X x i= =∑ c h 1 ∴ (C) + 2(C) + 3(C) + 4(C) = 1 ⇒ 10 C = 1
⇒ C = 1/10 = 0.1. Ans. (1)
59. log log log loglog
log
log
log
log
log
log
log3 2 1 44 3 2 14
3
3
2
2
1
1
4× × × = × × ×e
e
e
e
e
e
e
e
Using form logabb
ae
e
=RS|T|
UV|W|log
log= 1 × 1 × 1 × 1 = 1. Ans.(2)
60. Ans. (3)
61.
16 41819
16 2613 11
16 163 5
16 10 3
0 10
Decimal reminders Hexadecimal equivalent11 B ----------->Last significant hex-digit5 53 3
10 A ----------->Most significant hex-digit(41819)10 = (A35B)16
. Ans.(3)
62. For 2 < x < 3, x3
LNM
OQP =0. Hence lim
| |x
x x→ +
−LNM
OQP
FHGG
IKJJ = =
2
3 3 3
3 323
83
. Ans.(2)
63.ddx
axddx
x(log( ) ) = (x log ax) = log ax + 1. Ans.(4)
64. Let h be the height of the tower CD.Referring to Fig., we have ∠CAD = 600.∠CBD = 300 and AB = 3 km. Since B isdue west of A and A is due south of the tower CD, we have ∠BAC = 900.
B 3km A
C
60°30°
h
D
Now, from right angled triangle ACD,we get AC = h cot 600 = h.(1/√3), and fromright angled triangle BCD, we get BC = hcot300 = h√3. Therefore, from right-angledtriangle BAC, we have BC2 = AB2 + AC2
⇒ (h√3)2 = (3)2 + (h/√3)2 ⇒ 3h2 = 9 + h2/3⇒ 8/3 h2 = 9 ⇒ h2 = 27/8 ⇒ h = (3√3)/(2√2)km. = (3√6)/4 km.Ans. (4)
(33) of (57)
65. As a, b, c are in G.P., b2 = ac and b – c, c – a, a – b are in H.P.
2 1 1c a b c a b
a cb c a b−
=−
+−
=−
− −( )( )
⇒ 2 (b – c) (a – b) = – (a – c)2
⇒ 2 (ab – ac – b2 + bc) = – (√a – √c)2 ( √a + √c)2
⇒ 2 (ab – 2b2 + bc) = – (√a – √c)2 (√a + √c)2
⇒ 2b (√a – √c)2 = – (√a – √c)2 (√a + √c)2
⇒ 2b = – (a + c + 2√ac) =– (a + c + 2b) [since √a – √c ≠ 0]⇒ a + 4b + c = 0 which is a constant independent of a, b and c.Ans.(1)
66. Denominator of the given expression is 5√3 – 4√2 + 5√2 – 4√3 = √3 + √2.and it's Numerator is √3 (√3 + √2). Hence the value of the given expressionis √3. Ans.(4)
67. Given log107 = 0.8451 ⇒ 100.8451 = 7
⇒ =10 70 84512 20 20.e j ⇒ 720 = 1016.9702
[Since the number of digits in 101 = 2, 102 = 3, 103 = 4....]⇒ The required number of digits = 16 + 1 = 17. Ans.(2)
68. A ax dx a x aa a
10
3 2
0
22 4 423
83
= =LNM
OQP =z /
A ax dx a x a a aa
a
a
a
2
23 2
23 2 3 22 4 4
23
83
2= =LNM
OQP = −z / / /( )
AA
1
2
1
2 2 1
2 2 17
=−
=+ . Ans.(3)
69. V y dx a x dx aa a
= = =z zπ π π2
0
3
0
4 2 . Ans.(1)
70. limsin
lim( ) ( )
sinx
x x x
x
x x
x x xx
x→ →
− − +=
− −⋅
0 2 0
2
2
6 2 3 1 2 1 3 1
= loge 2 loge 3. Ans.(2)
71. lim
sin
x
x
t dt
x x→
z+0
3
04 55
(Form 0/0)
=−
+→lim
(sin ) . sin .x
xx x0
3
3 4
1 0 04 25
Qddx
f t dt f xddx
xddx
u xx
x
( ) ( ) ( ). ( )( )
( )
ν
ν
ν νzLNMMOQPP =
RS|T|
UV|W|
m r
By L-Hospital’s rule.
=+
=+→ →
limcos
limcos
x x
x xx x
xx0
2 3
2 3 0
3312 100
312 100
, when x ≠ 0
= =3 0
1214
.cos. Ans.(2)
72. Ans.(4)73. Obviously the given function is continuous everywhere except possibly at x
= nπ, n ∈ Z.
f(nπ – 0) = lim ( lim [ | sin ( ]h h
f n h) n h)|→ →
− = − −0 0
3π π = − =→lim ( sin
hh)
03 3
f n f n h)h
( ) lim (π π+ = +→
00
= − +→lim [ |sin ( |]
hn h)
03 π = − =
→lim ( sin
hh)
03 3
Q f(nπ – 0) = f(nπ + 0)⇒ f(x) is continuous at x = nπ. Hence f(x) is continuous everywhere.
Also Lf nf n h) f n
hn' ( ) lim
( ( )π
π π=
− −−→0
=− − −
−=
FHG
IKJ =
→ →lim
( sin ( )lim
sinh h
h)h
hh0 0
3 3 01
Rf nf n h) f n
hh)hh h
'( ) lim( ( )
lim( sin ( )
ππ π
=+ −
=− − −
→ →0 0
3 3 0
= −FHG
IKJ = −
→lim
sinh
hh0
1
Q Lf’ (nπ) ≠ Rf’(nπ)⇒ f(x) is not differentiable at x = nπ, V n ∈ Z.⇒ (4) is also correct answer. Ans.(4)
74. y x xx x x= =( )2
Take logarithm and differentiate, we get log y = x2 log x.
12
ydydx
x x x= +log
dydx
x x xx x= +( ) ( log )2 1 . Ans.(1)
75. Let ABC be the required triangle with coordinates of B as(a cos θ, b sin θ) and C as (a cos θ, – b sin θ).
B(a cos b sin , θ)θ
C(a cos b sin , – θ)θ
(– a, 0)
k
∆ = + = +12
2 1( cos ) sin (sin ) ( cos )a a b abθ θ θ θ
Which is maximum at θπ
=3
. Maximum area = 3 3
4ab . Ans.(1)
76. Ans.(3)
77.dydx
x= +2 3
From options if we take point as (– 3, 0)Equation of tangent is y = – 3(x + 3) i.e. y = – 3x – 9It definitely passes through (0, – 9). Ans.(4)
78. Ixx
dx a ba
b
= <z | |,
Case I. If a < b ≤ 0, then |x| = – x in [a, b]
∴ =−
= − = − +zzIx
xdx dx b a
a
b
a
b
1 = – b – (– a) = |b| – |a|.
Case II. If a < 0 < b, then | |xx a xx x b
=− ≤ <
≤ ≤RST
00
∴ = + zzIxx
dxxx
dxb
a
| | | |
0
0
=−
+ = − +z z zzxx
dxxx
dx dx dxa
b b
a
0
0 0
0
= a + b = b – (– a) = |b| – |a|Case III : If, 0 ≤ a < b, then |x| = x in [a, b]
∴ Ixx
dx dx b a b aa
b
a
b
=−
= = − = −z z | | | | .
Hence, I = |b| – |a|. Which is given in (4). Ans.(4)
79.dx
e ee dx
e ex x
x
x x+ +=
+ +−z z2 3 3 22
Putting ex = t, we get Idt
t t t tdt=
+ +=
+−
+LNM
OQPzz 2 3 2
11
12
=++
FHG
IKJ =
++
FHG
IKJ +log log
tt
e
eC
x
x
12
1
2 . Ans.(3)
80. Let x t= , so I te dt t et t= = +z2 2 10
1
0
1( )
= 2[2e – 1] = 4e – 2. Ans.(4)
(34) of (57)
81. Given expression
= =FHG
IKJ ⋅ =
→ ∞=
z∑lim . sin sinn
r
n
nr
x dx1
2n 22
0
1
0
π ππ
. Ans.(3)
82.x x x
x xdx
sin cos
sin cos
/
2 40
2
+zπ
Ix x x
x xdx=
−+z ( / ) sin cos
sin cos
/ ππ 24 4
0
2
22 4 4
0
2
Ix x
x x=
+zπ π sin cos
sin cos
/
I = ⋅ =π π π4 4 16
2
. Ans.(2)
83. Let x denote the population at “t” years and let x = x0 when t = 0.
Then dxdt
kx ordxx
k dt k= = , being a constant
⇒ x = cekt ...(1)At t = 0, x = x0, ∴ from (1) x0 = cThus (1) becomes x = x0 e
kt ...(2)when t = 50, x = 2x0,∴ 2x0 = x0 e
50k i.e. e50k = 2when x = 3x0, we have 3x0 = x0 e
kt
or 3 = ekt ⇒ 350 = e50kt = (e50k) t = 2 t
⇒ t = =50 3
279
loglog
years. Ans.(3)
84. Integrating factor = x
Solution of the equation is xy xe x= zxy = xex + ex + CUsing y(0) = 1, C = – 1∴ Solution is xy = xex + ex – 1. Ans.(4)
85. The common root is given by c a c aa b a b
1 2 2 1
1 2 2 1
−−
= −λ putting this in any one of
the given equations, we getλ2 – 2λ + 3λ = 0 λ(λ + 1) = 0 ⇒ λ = – 1 (Q λ ≠ 0)
∴ the common root is – 1. Ans.(2)86. Let α, β be the correct roots, since A commits a mistake in constant term,
there is no change in the sum of the roots.∴ α + β = 10.Since B commits a mistake in the coefficient of x there is no change in theproduct of the roots for him.∴ αβ = 9 ∴ α – β = 8 ∴ α = 9, β = 1. Ans.(2)
87. We have a ba b
aba b
n n
n n
+ +++
=+
1 1 2
or aan+1 + abn+1 + ban+1 + bbn+1 = 2an+1 b + 2abn+1
⇒ aan+1 + bbn+1 = an+1b + abn+1 ⇒ an+1(a – b) = bn+1 (a – b)⇒ an+1 = bn+1 (a ≠ b) (a/b)n+1 = 1 ⇒ n + 1 = 0 ∴ n = – 1. Ans.(3)
88. There are 9 vertical and 9 horizontal lines. To form a rectangle we requiredtwo horizontal and two vertical lines.∴ the total no. of rectangles = 9C2 × 9C2 = 36 × 36 = 1296. Ans.(1)
89. Here we observe that A and A2 3
0 0 03 3 91 1 3
0 0 00 0 00 0 0
=− − −
F
HGGG
I
KJJJ =
F
HGGG
I
KJJJ .
Hence, A is nilpotent matrix of index 3. Ans.(1)
90. Here C11 = x, C12 = – x, C21 = 0, C22 = 2x
|A| = 2x2 ∴ = =LNM
OQP
−AA
adj Ax
C CC C
12
11 21
12 22
1 12| |
=−LNM
OQP =
−
L
NMMMM
O
QPPPP
12
02
12
10
1 22x
xx x
x
x x
But A x
x xx
− =−
FHG
IKJ ∴
−FHG
IKJ =
−
F
HGGG
I
KJJJ ⇒ =1 1 0
1 21 01 2
12
0
12
11
21
∴ =x12
. Ans.(3)
91. Use R1 → R1 + R2 + R3 and proceed to get ∆ = (a + b + c)3. Ans.(1)92. The total number of sample points = n = 24 = 16.
Let m = the number of favourable cases = sum of the coefficients of power ofx less than 15 in the binomial expansion of (x3 + x5)4.Now (x3 + x5)4 = x12(1 + x2)4 = x12[1 + 4x2 + 6x4 + 4x6 + x8]∴ m = sum of coefficient of x12 and x14 = 1 + 4 = 5.∴ the required probability = m/n = 5/16⇒ odds against are (16 – 5) : 5 = 11 : 5. Ans.(2)
93. Ace on a die means to get 1 on a face.∴ the probability of getting an ace on a die is (1/6) and the probability ofnot getting an ace is 1 – (1/6) = (5/6), so that probability of getting 1, 3, 5 ...aces in a throw of n dice are given by the 2nd, 4th, 6th, ... term in the binomialexpansion of (q + p)n where q = 5/6, p = 1/6∴ the required probability P = nC1 pqn–1 + nC3 p
3 qn–3 + ...
=FHG
IKJFHG
IKJ +
FHG
IKJ
FHG
IKJ +
− −n
nn
n
C C1
1
3
3 116
56
16
56
...
=FHG
IKJ + +− −1
65 51
13
3n
n n n nC C[ ...]
=FHG
IKJ + − −
LNM
OQP
16
12
5 1 5 1n
n n( ) ( ) =FHG
IKJ −
LNM
OQP
16
12
6 4n
n n( )
=FHG
IKJ −
FHG
IKJ
RS|T|UV|W|
LNMM
OQPP = −
FHG
IKJ
LNMM
OQPP
16
12
6 146
12
123
nn
n n
∴ = −FHG
IKJ ⇒ − =
FHG
IKJ2P 1
23
1 2P23
n n
. Ans.(1)
94. The following cases arise
(i) both are rupee coins, its probability = 3
27
2
17
CC
=
(ii) both of them are 50 paise coins, its probability = 4
27
2
27
CC
=
(iii) one rupee coin and the other is 50 paise coin, the probability
=×
= =3
14
17
2
1221
47
C CC
Also 2 rupee coins = 4 fifty paise coins 1 rupee and one fifty paise coin= 3 fifty paise coins
Required expectation =FHG
IKJ × +
FHG
IKJ × +
FHG
IKJ ×
LNM
OQP
17
427
247
3 fifty paise coins,
=FHG
IKJ × =
207
50 1 42paise Rs. . . Ans.(4)
95. By the definition of P.d.f.,
ax dxa
x a2
2
33
2
31
31
319
= ⇒ = ⇒ =z . Ans.(4)
(35) of (57)
96. Required region is shaded region
(0, 6)
(0, 11/3)(2, 3)
(4, 0) (11, 0)
3x + 2y = 12Z(0, 6) = 6Z(2, 3) = 5Z(11, 0) = 11. Ans.(1)
97.n
rn
rn
rr
n nr
nrr
n
r
nCC C
CC n
n r+
= =+
− +−=
+= =
∑ ∑ ∑10
10 0
11
1
=+
+ + −+L
NMOQP =
+11
1 11
22
2( )( ) ( )
( )n
n nn n n
. Ans.(2)
98. [ ] ... ( )/
/
/
/
/
kx dx dx dx dx n k dxk
k
kn
k
k
nkk
nkk
= + + + −z zz z z−
0 1 2 10
1
1
2
0 2
3
1
= + + −1
1 2 1k
n k... ( )
=−
=−1 1
21
2knk nk n nk( ) ( )
. Ans.(2)
99. Let f(x) = sin x then fn(x) = sin (nπ/2 + x), V n ∈ N, V x ∈ R.Apart from the sign, fn(x) = sin x or cos x and
R xxn
xn
n
( )| |
!| sin |= 0 or
| |!
| cos |xn
xn
0 where x0 (0, x)
As xn
n
!→ 0 as n → ∞, since |sin x0| ≤ 1
and |cos x0| ≤ 1, ∴ =→lim ( )
nnR x
00
Thus, the series representing sin x exists V x ∈ R. Ans.(4)
100. Here the coefficient matrix is A = −F
HGGG
I
KJJJ
1 1 12 1 13 2 0
and |A| = 0
∴ the system has infinitely many solutions. Ans.(3)
(36) of (57)
1. The quadratic equation 3x2 + 2(a2 + 1) x + a2 – 3a + 2 = 0 will have two rootsof opposite sign if it has real roots and the product of the roots is, negative,that is, if
4(a2 + 1)2 – 12(a2 – 3a + 2) ≥ 0 and a a2 3 2
30
− +< .
Both of these conditions are met ifa2 – 3a + 2 < 0 i.e. if (a – 1)(a – 2) < 0 or 1 < a < 2. Ans.(3)
2. t r, the rth term of the A.P. is given bytr = Sr – S r–1 = cr(r – 1) – c(r – 1)(r – 2)
= c(r – 1)(r – r + 2) = 2c(r – 1)
We have t t tn12
22 2+ + +... = 4c2 {02 + 12 + 22 + ... (n – 1)2}
=− −
= − −41 2n 1
623
1 2n 12 2cn n
c n n( ) ( )
( ) ( ) . Ans.(2)
3. Ans.(2)4. The letters of the word RACHIT can be arranged in 6! = 720 ways. The number
of words beginning with A is 5!, those beginning with C is 5!, those beginningwith H is 5! and those beginning with I is 5! RACHIT happens to be the firstword beginning with R. Therefore, the rank of the word RACHIT is 4(5!) + 1 =481. Ans.(2)
5. We have E = 1 + (1 + x) + (1 + x)2 + ... + (1 + x)n
=+ −
+ −=
+ −+ +( ) ( )1 11 1
1 11 1xx
xx
n n[sum of a G.P.]
The coefficient of xk in E = coefficient of xk+1 in [(1 + x)n+1 – 1]⇒ n+1Ck+1 = n+1Cn+1–k–1 = n+1Cn–k. Ans.(1)
6. We have 22000 = (24)500 = (17 – 1)500
= 500C0 17500 – 500C1 17499 + ... – 500C499 17 + (– 1)500
= 17 m + 1, where m is some positive integer.⇒ 22003 = 8(22000) = 8(17 m + 1) = 17(8 m) + 8This show that the required remainder is 8. Ans.(3)
7. We can write the given result as
x y zx xy yz z
x y zk k k
111
12 3
2 3
2 3= (x – y)(y – z)(z – x)(yz + zx + xy)
⇒ xk+1 yk+1 zk + 1 ∆1 = (x – y)(y – z)(z – x)(yz + zx + xy)Note that degree of ∆1 is 5.∴ degree of LHS = 3k + 3 + 5 = 3k + 8Also degree of RHS = 5∴ 3k + 8 = 5 or k = – 1. Ans.(2)
8. When x ≥ 0, the inequality becomes|x – 1| < 1 – x or |1 – x| < 1 – x
This inequality is not satisfied by any value of x since |a| ≥ a ∀ a ∈ R.When – 1 ≤ x < 0, the inequality becomes|– 1 – x| < 1 – x or |x + 1| < 1 – x.or x + 1 < 1 – x or 2x < 0 or x < 0Thus, the inequality is satisfied for all x ∈ [– 1, 0).When x < – 1, the inequality becomes|– 1 – x| < 1 – x or |x + 1| < 1 – xor – (x + 1) < 1 – x or – 2 < 0.∴ The inequality is satisfied for all x ∈ (– ∞, – 1).Thus, the solution set is (– ∞, 0). Ans.(4)
9. Let Ei (0 ≤ i ≤ 2) denote the event that urn contains i white and (2 – i) blackballs. Let A denote the event that a white ball is drawn from the urn.We have P(Ei) = 1/3 for i = 0, 1, 2, andP(A | E1) = 1/3, P(A|E2) = 2/3, P(A|E3) = 1.By the total probability rule,
P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + P(E3) P(A|E3)
= + +LNM
OQP =
13
13
23
123
. Ans.(3)
10. Required probability =
P A P B P A P B( ) . ( ) ( ) . ( ) .+ = ⋅ + ⋅ =70
10020
10030
10080
1000 38 . Ans.(2)
11. Let x be the distance of the observer O from the top P of pole PQ, on thebuilding OR (fig.).
Ox
P50 mQ
250 m
R
θθ
Then ∠POQ = ∠QOR = θ(say),PQ = 50 m and QR = 250 m.From right- angle triangle OPQ, we have tanθ = 50/x, and from right-angled triangle OPR,we have tan 2θ = 300/x.
∴ =−
⇒ =⋅
−tan
tan
tan
( / )
( / )2
2
1
300 2 50
1 502 2θ
θ
θ xx
x
⇒ −FHG
IKJ
LNMM
OQPP =3 1
501
2
x
⇒ =FHG
IKJ ⇒ = ⋅ ⋅ ⇒ =
23
5050 50
32
25 62
2
xx x m . Ans.(4)
12. tan ( ) cos sin− − −+ −FHG
IKJ + −
FHG
IKJ = + − =1 1 11
12
12 4
23 6
34
π π π π. Ans.(3)
13. 2 – cos x + 1 – cos2 x = –{cos2 x – cos x} + 3
= − +FHG
IKJ −
RS|T|
UV|W|
+cos x12
14
32
= − +FHG
IKJ
134
12
2
cos x
Maximum value at cos x = −12
, then 134
Minimum value at cos x = 1, i.e. 1 ∴ ratio 134
. Ans.(3)
14. The given expression
= 2 [(sin2θ + cos2θ)3 – 3 sin2θ cos2θ (sin2θ + cos2θ)]
–3 [(sin2θ + cos2θ)2 – 2 sin2θ cos2θ] + 1
⇒ 2[1 – 3 sin2θ cos2θ] – 3[1 – 2 sin2θ cos2θ] + 1
⇒ 2 – 6 sin2θ cos2θ – 3 + 6 sin2θ cos2θ + 1 = 0. Ans.(1)
15. sin , cosα α= = −1517
817
sin , cosβ β= − =1213
513
sin(β – α) = sin β cos α – cos β sin α
= −FHG
IKJ −FHG
IKJ − ×
FHG
IKJ =
1213
1817
513
1517
141221
. Ans.(1)
16. Let ‘l’ be the length of the arc of the circle of radius ‘r’.
∴ l + r + r = πr ∴ l = r(π – 2)
But θ =lr
∴ θ = π – 2. Ans.(3)
17. Ans.(1)18. Ans.(1)19. Let x = 264
Taking log, log x = log 264 = 64 log 2 = 64 (0.3010) = 19.264
Since the characteristic of log x = 19
∴ the number of digits in the numeral for 264 = 19 + 1 = 20. Ans.(3)
20. Value = log log log ....log ( ) log....
x x x x nx
x nn
nx
x
nx x
1 2 3 1
2
1
1−
−−
log log log ....log....
x x x x nx
n
nx
x
x1 2 3 1
2
1
1−
−− = =.....logx x
1 1 1 . Ans.(2)
Exercise - 02 - Solutions
(37) of (57)
21. Since log45 = a ∴ 4a = 5
Since log56 = b ∴ 5b = 6
∴ (4a)b = 6 ⇒ 4ab = 6 ⇒ (22)ab = 6 ⇒ 22ab = 21(3) ⇒ 22ab–1 = 3
∴ log23 = 2ab – 1 ⇒ = =−
loglog3
2
21
31
2 1ab. Ans.(4)
22.
− + < −
− ≤ ≤
+ >
R
S|||
T|||
−
12
1 1
1 112
1 1
1
( )
tan
( )
x x
x x
x x
xx=1O
f x( ) = −π4
f(x) = 1
f x( ) =π4
f(x)
x = –1
Clearly f(x) is not continuous at 1 and –1∴ not diff. at –1 & 1hence domain of derivative is R – {–1, 1} Ans.(4)
23. f(x) = l ln ax n bx
x( ) ( )1 1+ − −
limx→0
f(x) = limx→0
l ln a xax
an bx
bxb
x
( ). lim
( )1 10
++
−−→
= a + b. Ans.(2)
24. Since f(x) is monotonic so f’(x) > 0
or < 0 ∀ x ∈R. Then ddx
FHG
IKJ f(rx) = r f’(rx) also > 0 or < 0 for all x ∈R and for
al l posi t ive integers r . Therefore, f (x) and f ( rx) have same monotonicbehaviour. Since n is odd,f(x) + f(2x) +.....+ f(nx) is a polynomial of odd order and monotonic so attainsall real values only once. Hence the equation
f(x) + f(2x) +....+ f(nx) = n n( )+ 1
2has exactly one solution. Ans.(2)
25. f xx
xx
xx
'( )( )
=+
− =− +
+=
−+
11
11 1
1 12
2
2
2
2 . Since x 2 and 1 + x2 are non-negative
for all x, it fol lows that f ’x) < 0 for all x ∈ R. Therefore, f(x) decreaseseverywhere, and hence on any interval. Ans.(3)
26. f’(x) = – 2x2 e–2x + 2x e–2x = 2(– x2 + x)e–2x
Since e–2x ≠ 0 for any x, f’(x) = 0 ⇒ x = 0, 1.f”(x) = 2(– 2x + 1) e–2x – 4(– x2 + x)e–2x
f”(0) = 2 and f”(1) = 2[– 1]e–2 < 0Hence f is maximum at x = 1 and the maximum value of f = e–2 = 1/e2.Ans.(3)
27. f x x x dx C x x dx C( ) ( )= − + = − − − +z z112
2 12 2
= −FHG
IKJ − +
13
1 2 3 2( ) /x C . So, 73
13
83
= − + ⇒ =C C .
Therefore, f(x) = −FHG
IKJ − −
13
1 82 3 2( ) /x . Ans.(1)
28. Put ex – 1 = t2. The ex dx = 2t dt. Hence, when x = log 5, we have r2 = 4, i.e.,t = 2, and when x = 0, we have t = 0.
∴−
+=
+= −
+
FHGG
IKJJz z zze e
edx
tt
dt dtdt
t
x x
x
1
32
42 1 4
40
5 2
20
2
20
2
0
2log
= − ⋅FHG
IKJ = − ⋅
FHG
IKJ = −−2 4
12 2
2 2 24
41
0
2
tt
tanπ
π . Ans.(2)
29. lim lim/n
r
n
nr
n
nr
n r nr
r n→∞ = →∞ =+=
+∑ ∑1 1
12 21
2
2 21
2
=+
= + = −z x
xdx x
11 5 1
20
22
0
2
. Ans.(2)
30. Idx
x
dx
x=
+=
+ −z z1 1 230
2
30
2
tan tan ( / )
/ /π π
π
=+
=+
= −+
FHG
IKJ = −z z zdx
x
dx
xdx
xdx
1 11
1
1 213
0
2
30
2
30
2
cot tan tan
/ / /π π π π
So I = π4
. Ans.(4)
31. f’(x) = 4x3 – 6x2 + 2x = 2x (2x2 – 3x + 1) = 2x(2x – 1) (x – 1).Since f is a differentiable function, for extremum points of f(x), we must havef’(x) = 0. Hence x = 0, 1/2, 1.Now f”(x) = 12x2 – 12 x + 2, f”(0) = 2, f”(1) = 2and f”(1/2) = 3 – 6 + 2 = – 1.Thus the function has minimum at x = 0 and x = 1. Therefore, the requiredarea is
= − + + = − + +FHG
IKJ =z ( )x x x dx
x x xx4 3 2
0
1 5 4 3
0
1
2 35 2 3
39130
. Ans.(2)
32. Putting u = x – y, we get du/dx = 1 – dy/dx. The given equation can be writtenas 1 – du/dx = cos u ⇒ (1 – cos u) = du/dx.
⇒ duu
dx ec u du dx1
12
22
−= ⇒ =z zcos
cos ( / )
⇒ – x + cot (u/2) = constant or x + cot x y
C−
=2
. Ans.(2)
33. Let c c c c→
= ( , , )1 2 3 . Then | | | | | |c a b c c c→ → →
= = = = + +2 12
22
32 . I t is given
that the angles between the vectors are identical, and equal to φ(say). Then
cos| | | |
φ =⋅
=+ +
=
→ →
→ →
a b
a b
0 1 0
2 2
12
a c
a c
c c→ →
→ →
⋅=
+=
| | | |
1 2
2 2
12
and b c
b c
c c→ →
→ →
⋅=
+=
| | | |
2 3
212
.
Hence c1 + c2 = 1 and c2 + c3 = 1. That is, c1 = 1 – c2 and c3 = 1 – c2.
⇒ 2 = c c c c c c c c12
22
32
22
22
22
22
21 1 3 2 4+ + = − + + − = + −( ) ( )Therefore, c2 = 0 or c2 = 4/3. If c2 = 0, then c1 = 1 and c3 = 1, and ifc2 = 4/3, then c1 = –1/3, c3 = – 1/3.
Hence the coordinates of c→
are (1, 0, 1) or (– 1/3, 4/3, – 1/3). Ans.(4)
34. Since a b b c c a→ → → → → →
− + − + − = 0 , the vectors a b b c and c a→ → → → → →
− − −, are
coplanar, and so [ , ]a b b c c a→ → → → → →
− − − = 0 . Ans.(1)
(38) of (57)
35. Let c x i y j zk→
= + +$ $ $ . Since a b c→ → →
, , are coplanar, so
x y z1 1 11 1 1
0−−
=
⇒ 0 ⋅ x – 2y – 2z = 0 ⇒ y + z = 0.
Also c→
is perpendicular to a→
, so x + y – z = 0.
Therefore x y z−
= =−2 1 1
and x2 + y2 + z2 = 1
( c→
being a unit vector). We have x y z= − = = −2
6
1
6
1
6, ,
Thus c i j k→
= − + −1
62( $ $ $) and d
a ca c
i j k→
→ →
→ →=
×× −
− −
11 1 12 1 1| |
$ $ $
=×
+ + = +→ →
1
6
10 3 3
1
2| |( $ $ $) ($ $)
a ci j k j k . Ans.(2)
36. Product of the two given matrices is
5 0 00 1 00 10 2 5
1 0 00 1 00 0 1
x
x x−
L
NMMM
O
QPPP
=L
NMMM
O
QPPP
if and only if x =15
. Ans.(4)
37. We have (I – A)(I + A + A2) = I + A + A2 – A – A 2 – A 3 = I – 0 = I⇒ (I – A)–1 = I + A + A 2. Ans.(2)
38. For the system of equations to have no solution, we must have
kk k
kk
+=
+≠
−1 8
34
3 1
⇒ (k + 1) (k + 3) = 8k and 8(3k – 1) ≠ 4k(k + 3)But (k + 1) (k + 3) = 8k ⇒ k2 + 4k + 3 = 8kor k2 – 4k + 3 = 0 or (k – 1) (k – 3) = 0 ⇒ k = 1, 3.For k = 1, 8(3k – 1) = 16 and 4k(k + 3) = 16.∴ 8(3k – 1) = 4k(k + 3) for k = 1.For k = 3, 8(3k – 1) = 64 and 4k(k + 3) = 72i.e., 8(3k – 1) ≠ 4k(k + 3) for k = 3.Thus, there is just one value for which the system of equations has nosolution. Ans.(2)
39. Let P(x, y) be the moving point. Then the area of ∆POA is
12
4 0 2| | | |x y x⋅ − ⋅ = , and that of ∆POB is 12
0 6 3| | | |x y y⋅ − ⋅ = − , because
one vertex of the triangle is at the origin. Therefore, from the given condition,we have |2x| = 2|– 3y| or 2x = ± 2(– 3y)⇒ x – 3y = 0 or x + 3y = 0. Ans.(4)
40. The coordinates of D, the mid-point of BC, are (a/2, 0) and those of E, themid-point of AC, are (a/2, b/2).
∴ slope of AD = b
ab
a−
−= −
00 2
2( / )
and slope of BE = ( / )( / )ba
ba
2 02 0
−−
= .
Now, as AD is perpendicular to BE we must have
− ⋅ = − ⇒ = ⇒ = ±2
1 2 22 2ba
ba
b a a b . Ans.(2)
41. Since 3x + y = 0 touches the given circle, its radius equals the length of theperpendicular from the centre (2, – 1) to the line 3x + y = 0.
That is, r =−
+=
6 1
9 1
5
10.
Let y = mx be the equation of the other tangent to the circle from the origin.
Then 2m 1
1
5
1025 1 10 2m 1
2
2 2+
+= ⇒ + = +
mm( ) ( )
⇒ 3m2 + 8m – 3 = 0 which gives two values of m and hence the slopes oftwo tangents from the origin, with the product of the slopes being – 1. Since
the slope of the given tangent is – 3, that of the required tangent is 13
, and
hence its equation is x – 3y = 0. Ans.(3)
42. Since PQ is the chord of contact of the tangents from the origin O to thecircle x2 + y2 + 2gx + 2fy + c = 0, ...(1)equation of PQ is gx + fy + c = 0 ...(2)An equation of a circle through the intersection of (1) and 2) is given byx2 + y2 + 2gx + 2fy + c + λ(gx + fy + c) = 0 ...(3)If the circle (3) passes through O, the origin, then c + λc = 0, i.e., λ = – 1,and the equation of the circle (3) becomes
x2 + y2 + gx + fy = 0Centre of this circle is (– g/2, – f/2), and hence it is the circumcentre of thetriangle OPQ. Ans.(4)
43. The equation of the parabola can be written as (y – 1)2 = – 6(x + 2), which isof the form Y2 = 4aX, where Y = y – 1, a = – 3/2, X = x + 2. The vertex istherefore X = 0 and Y = 0, i.e. x = – 2 and y = 1. Ans.(2)
44. Equation of a tangent to the given hyperbola is
y mx a m b= + −2 2 2 . The products of the perpendiculars from the foci
(± ae, 0) are aem a m b
m
aem a m b
m
+ −
+×
− + −
+
2 2 2
2
2 2 2
21 1
=− −
+=
− − −+
a m b a e m
m
b m a e
m
2 2 2 2 2 2
2
2 2 2 2
21
1
1
( )=
− −+
=b m b
mb
2 2 2
22
1.
Ans.(3)45. Let the equation be y = mx + c. Since it touches the hyperbola
9x2 – 9y2 = 8 or x y2 2
8 9 8 91
/ /− = , (with a2 = b2 = 8/9), we get
c m2 289
1= −( ) ...(i)
Also, since it touches the parabola y2 = 32x, we get
cm
=8
...(ii)
From (i) and (ii) we get,
89
164
1 7222
2 2( ) ( )mm
m m− = ⇒ − =
⇒ m4 – m2 – 72 = 0 ⇒ (m2 – 9) (m2 + 8) = 0⇒ m = ± 3 [m2 = – 8 is rejected]From (ii), c = 8/3 and – 8/3 when m = 3 and – 3 respectively. Therefore, therequired equations are
y = 3x + 83
and y = – 3x – 83
or 9x – 3y + 8 = 0 or 9x + 3y + 8 = 0. Ans.(2)46. Let the lines represented by the given equations be
y = m1x, y = m2 x and y = m3 x.
Then m1m2m3 = – 3/d.
If two of the lines are perpendicular.
Let m1 m2 = – 1 ⇒ m3 = 3/d.
⇒ y = (3/d)x satisfies the given equation
⇒ d(3/d)3 – 3(3/d)2 + 3(3/d) + 3 = 0 ⇒ d = – 3. Ans.(3)
47. Since the given equation represents a pair of parallel lines, we have
h2 = 9 × 4 ⇒ h = ± 6 and 9 3
43 3
0h
h ff −
= .
⇒ 9 (– 12 – f2) – h(– 3h – 3f) + 3(h f – 12) = 0⇒ 3h2 + 6h f – 9f2 – 144 = 0⇒ 108 ± 36f – 9f2 – 144 = 0 (Q h = ± 6)⇒ 9f2 – 36f + 36 = 0 ⇒ f = 2.When h = ± 6, f = 2, the given lines are
9x2 ± 12xy + 4y2 + 6x + 4y – 3 = 0⇒ (3x ± 2y + 1)2 = 4⇒ 3x ± 2y + 3 = 0 or 3x ± 2y – 1 = 0.
The distance between these parallel lines is 3 1
9 4
4
13
+
+= . Ans.(3)
(39) of (57)
48. The function in (1) can be written in a simplified form assin x/cos x = tan x, so it has period π sin x has period 2π.
sin coscos cos2 2 1 22
1 22
x xx x
+ =−
++
Since the period of sin 2x is π and the period of cos 3x is 2π/3, the period ofsin 2x + cos 3x is L.C.M. (π, 2π/3) = 2π. The period of the function in (4) isclearly 2π. Ans.(1)
49. The function in (1) is periodic with period 1. The function in (2) is periodicwith period π. The function (3) on simplification is equal to (1/2) sin 2x, whichis periodic with period π. The function in (4) is non periodic asg(x) = x is non periodic. Ans.(4)
50. Clearly, f(x) ≥ 0 for any x ∈ R. Moreover, (x2 – 1)2 = x4 – 2x2 + 1 ≥ 0
for any x ∈ R, so x
x
2
4 1
12+
≤ . Hence the range of f is 0,12
LNM
OQP . Ans.(2)
51. limsin ( cos )
limcos ( cos ) . . ( sin )
x x
xx
x xx→ →
=−
0
2
2 0
2 22
π π π
= −FHG
IKJ = − − =
→ →π π π πlim
sinlim cos ( cos ) ( ) . ( )
x x
xx
x0 0
222
1 1 . Ans.(2)
52. Lf kf k h) f k
hh
k h k h)}hh h
'( ) lim( ( )
, lim[ ] sin { (
=− −
−> =
− −−→ →0 0
0π
= −−
−= − −
→ →lim ( )
sin (( ) ( ) lim
sin.
h
k
xk
k h)h
kh
h0 01 1 1
π π ππ
π
= (– 1)k (k – 1) . 1 . π. Ans.(1)
53. F x f t dt F x f x f F f fx
( ) ( ) '( ) ( ) ( ) '( ) ( ) ( )= = = − ⇒ = −z0
0 4 4 0 ...(1)
Now, F(x2) = x2(1 + x) ⇒ F’(x2) 2x = 2x + 3x2
⇒ F’(4) × 4 = 4 + 12 = 16 ...(2)∴ From (1) and (2), f(4) – f(0) = 4∴ f(4) = 4. Ans.(3)
54. f(x) = max. {x, x3}
Lf’(0) = lim( ( )
limh h
f h) fh
hh→ →
− −−
=− −
−=
0 0
30 00
Rf’(0) = lim( ( )
limh h
f h) fh
hh→ →
−=
−=
0 0
0 01
∴ Lf’(0) ≠ Rf’(0)Similarly, Lf’(1) = 1 ≠ Rf’(1) = 3
Lf’(–1) = 1 ≠ Rf’(– 1) = 3. Ans.(4)55. f(x) = xex(1–x) ⇒ f’(x) = ex(1–x) (1 + x – 2x2)
∴ f(x) is increasing if f’(x) ≥ 0i.e. if 1 + x – 2x2 ≥ 0 ⇒ 2x2 – x – 1 ≤ 0
⇒ (2x + 1)(x – 1) ≤ 0 ⇒ − −FHG
IKJ
FHG
IKJ − ≤ ⇒ ∈ −
LNM
OQPx x x
12
1 012
1( ) , . Ans.(1)
56. Since 1 + x – [x] > 0, for all real x∴ g(x) > 0 for all real x∴ f(g(x)) = 1. Ans.(2)
57. y f x xx
x yx= = + ⇒ − + =( )1
1 02
⇒ =± −
⇒ =± −
⇒ ∞ = ∞− −xy y
f xx x
f2
12
14
2
4
22, 1( ) ( ) [ , )
∴ =+ −−f x
x x12 4
2( ) {Q x2 – 4 ≥ 0 iff x ∈ (– ∞, – 2] ∪ [2, ∞)]. Ans.(1)
58. f xx
x x( )
log ( )=
++ +2
2
3
3 2
x2 + 3x + 2 ≠ 0 ⇒ (x + 2) (x + 1) ≠ 0 ⇒ x ≠ – 1, – 2.Also, x + 3 > 0 ⇒ x > – 3.∴ Domain of f is (– 3, ∞) – {– 1, – 2}. Ans.(4)
59. y = f(x) = x2 + bx – b ⇒ f’(x) = 2x + b
∴ = +f x bP
'( )( , )1 1
2 . B
y
x
P
A
∴ Equation of tangent at P isy – 1 = (2 + b)(x – 1).It meets x-axis where y = 0
∴ xb
b=
++
12
.
It meets y-axis where x = 0∴ y = – (1 + b)
∴ Area = 12
12
12
1xyb
bb= −
++
FHG
IKJ +( )
∴ −++
=12
12
22( )b
b (given)
⇒ b2 + 6b + 9 = 0 ⇒ b = – 3. Ans.(3)
60. Ix
adx ax=
+>
−z cos
,2
10
π
π
...(1)
Let x = – t ⇒ = −+
=+
−
−z zI
t a
adt
x a
adx
t
t
x
x
cos . cos .2 2
1 1π
π
π
π
...(2)
Adding (1) and (2),
21 2
22I x dx
xdx= =
+
−
− zz coscos
π
π
π
π
212
22 2
I xx
I= +FHG
IKJ ⇒ =
−
sin
π
ππ
. Ans.(3)
61. (x – 3)2 + y2 = 9 ...(1)y2 = 4x ...(2)
Equation of tangent to (2) is y = mx + 1m
.
It touches (1) with centre (3, 0) and radius 3,
if
3m 01
13
1
32
− +
+= ⇒ = ±
m
mm .
Since the tangent lies above x-axis.
∴ = −m1
3. Therefore, equation of tangent is 3 3y x= − +( ) . Ans.(2)
62. 3x + 4y = 9 ...(1)y = mx + 1 ...(2)
Using (2) in (1), we get
3x + 4(mx + 1) = 9 ⇒ x(3 + 4m) = 5 ⇒ =+
x5
3 4m
which is integer if m = – 1, – 2. Ans.(1)63. y2 + 4y = – 4x – 2 ⇒ (y + 2)2 = – 4x + 2 Y
X⇒ (y – (2))2 = − −
FHG
IKJ4
12
x
or Y2 = – 4X when X = x – 12
,
Y = y + 2, ‘a’ = 1
∴ Equation of directrix is X = 1 or x x− = ⇒ =12
132
. Ans.(4)
(40) of (57)
64. Let P be (r cos θ, r sin θ). If (x, y) be the centroid of ∆PAB, then
xr x x
=+ +cos θ 1 2
3
O
P
B(x , y )1 1 A(x , y )2 2
yr y y
=+ +sin θ 1 2
3
=+r ysin θ 2
31 (clearly y1 = y2)
∴ =− −
cos θ3 1 2x x x
r
and sin θ =−3 2 1y yr
∴− −F
HGIKJ +
−FHG
IKJ =
3 3 211 2
21
2x x x
ry y
r ⇒ (3x – x1 – x2)
2 + (3y – 2y1)2 = r2
∴ Locus is a circle. Ans.(2)65. T5 + T6 = 0 ⇒ T5 = – T6 ⇒ nC4 a
n–4(– b)4 = – nC5 an–5(– b)5
⇒ab
CC
nn
n= =
−5
4
45
. Ans.(2)
66. We know that ax2 + bx + c, (a > 0) has min. value =−D
a4∴ (1 + b2)x + 2bx + 1 = 0 has min. value
m(b) = −− −
+=
+( )
( )4 4 4
4 11
1
2 2
2 2
b bb b
. Also, 01
11
2<
+≤
b
∴ Min. value, m(b) ∈ (0, 1]. Ans.(4)
67.
sin cos coscos sin coscos cos sin
x x xx x xx x x
= 0
⇒−
− −−
=sin cos sincos sin cos cos sincos sin cos
x x xx x x x xx x x
0
00
⇒ −−
− =(sin cos )sincoscos
x xxxx
21 0
1 10 1
0 = −−
(sin cos )sincos
cosx x
xx
x
21 0
2 1 00 1
⇒ (sin x – cos x)2 (sin x + 2 cos x) = 0
⇒ tan x = 1, – 12
⇒ x = −FHG
IKJ ∈ −
LNM
OQP
−π π π4
12 4 4
1, tan , . Ans.(2)
68. Since α, β, γ, δ are in G.P., let α = a, β = ar, γ = ar2, δ = ar3.∴ α + β = 1 ⇒ a(1 + r) = 1 ...(1)
αβ = p ⇒ a2r = p ...(2)γ + δ = 4 ⇒ ar2 (1 + r) = 4 ...(3)
γδ = q ⇒ a2r5 = q ...(4)(3) ÷ (1) gives r = ± 2
∴ When r = 2, a =13
When r = – 2, a = – 1∴ For integral values of p and q, p = – 2, q = – 32. Ans.(1)
69. Tn = nC3, Tn+1 = n+1C3∴ Tn+1 – Tn = n+1C3 – nC3 = 21
⇒( ) ( ) ( ) ( )n n n n n n+ −
−− −
=1 1
61 26
21 ⇒n n
n n( )
( )−
+ − + =1
61 2 21
⇒ n(n – 1) = 42 ⇒ n = 7. Ans.(2)70. a, b, c, d are in A.P.
Let a = 1, b = 2, c = 3, d = 4∴ abc, abd, acd, bcd become 6, 8, 12, 24 which are in H.P.
because 18
16
112
18
124
112
124
− = − = − = − . Ans.(4)
71.22
2 2 2n 1 32
2 57 1 2 11n n
n n[ ( ) ] [ ( ) ]× + − = × + − ⇒ = . Ans.(3)
72. f(x) = sin (|x|) – |x|
Lf’(0) = lim( ( )
limsin
h h
f h) fh
h hh→ →
− −−
=− −−0 0
0 0 0
= − +FHG
IKJ =
→lim
sinh
hh0
1 0 .
Rf’(0) = lim( ( )
limsin
h h
f h) fh
h hh→ →
+ −=
−= − =
0 0
0 01 1 0 . Ans.(4)
73. log4(x – 1) = log2 (x – 3) ...(1)
⇒ 4 12 3log ( )x x− = −
⇒ 2 1 3 1223 2log ( ) ( )x x x x− = − ⇒ − = −
⇒ x = 2, 5But x = 2 does not satisfy (1).∴ x = 4 is the only solution. Ans.(2)
74. f xx
xf f x
f xf x
x
xxx
x
( ) ( ( ))( )
( )=
+⇒ =
+⇒ = +
++
α αα
α
α1 11
11
⇒ =+ +
⇒ = −xx
x xα
αα
2
11 . Ans.(4)
75. [ ]a b c x xy x x y
C C xy x x
→ → →=
−−
+ −+ =
+
1 0 11 1
1
1 0 01 1
13 1
= 1 + x – x = 1 which is independent of x and y. Ans.(3)
76. Since ( ) ( ) ( )a b b c c a→ → → → → → →
− + − + − = 0
∴ − − −→ → → → → →a b b c c a, , are the component vectors of an equilateral triangle.
∴ maximum angle possible between a b c→ → →
, , taking two at a time is 120°.
∴ − + − + −→ → → → → →
| | | | | |a b b c c a2 2 2
= + + − + +→ → → → → → → → →
2 22 2 2(| | | | | | ) ( . . . )a b c a b b c c a
= − − − −FHG
IKJ =6 2
12
12
12
9
∴ Maximum possible value is 9. Ans.(2)
77. In ∆PQRPQPR
, tan ( ) cot= ° − =90 α α ...(1)
In ∆PRSRSPR
, tan= α ...(2) P2r
90° – αα R
Q S
X
∴ Multiplying (1) and (2),
PQ RS
PR
.2 1=
⇒ = ⇒ =PR PQ RS PQ RS. .2r .
Ans.(1)
(41) of (57)
78. Solving y = mx and y = nx + 1, we get
Bm n
mm n
1− −
FHG
IKJ,
D y = mx + 1
y = mx
y = n
x
y = n
x + 1
C(0, 1)
A(0, 0) B
∴ Area of llgm = 2 × area of ∆ABC
=
− −
=−
0 0 10 1 11
1
1
m nm
m n
m n| |. Ans.(4)
79. Since sin–1 θ + cos–1 θ = π2
∴ − + − = − + −xx x
xx x2 3
24 6
2 4 2 4... ...
⇒ − + −FHG
IKJ = − + −
FHG
IKJ ⇒
+=
+x
x xx
x x xx
x
x1
2 41
2 4 12 1
2
22
2 4 2
2... ...
⇒+
=+
⇒ + = + ⇒ =2
22
22 2 1
2
22 2x
xx
xx x x x . Ans.(2)
80. For 021 2≤ ≤α α απ
, , ... n , and (cot α1) (cot α2) ... (cot αn) = 1,
cos (α i) is maximum when απ
i =4
.
∴ Maximum value of (cos α1) (cos α2) ... (cos αn)
=FHG
IKJFHG
IKJ
FHG
IKJ =
1
2
1
2
1
2
1
2 2... /n . Ans.(1)
81. α βπ
α βπ α β
α β+ = ⇒ + = ⇒
+−2 2 1
tan ( ) tantan tan
tan tan
⇒ 1 – tan α tan β = 0 ⇒ tan α tan β = 1 ...(1)
Now β γ αβ γ
β γα+ = ⇒
+−
=tan tan
tan tantan
1
⇒ tan β + tan γ = tan α – tan α tan β tan γ⇒ tan β + tan γ = tan α – tan γ (By 1)⇒ tan α = tan β + 2 tan γ. Ans.(3)
82. Required distance = DC = y – x30°
30° 60°
100 m
60°
xy
DCar
BC
A
= 100 cot 30° – 100 cot 60°
= −100 3 1001
3
=−F
HGIKJ =100
3 1
3
200 33
.
Ans.(2)83. Ans.(3)
84. We are given x i + y i = n + 1, i = 1, 2, ..., nLet xi – y i = d i∴ 2x i = n + 1 + d i ⇒ d i = 2x i – (n + 1)
⇒ ∑ = ∑ − + + +d x n x ni i i2 2 24 4 1 1( ( ) ( ) )
= ∑ − + ∑ + +4 4 1 12 2x n x n ni i( ) ( )
=+ +
− ++
+ +41 2n 16
4 11
21 2n n
nn n
n n( )( )
( )( )
( )
(Q x1, x2, ..., xn are permutations of 1, 2, 3, ..., n)
=−n n( )2 1
3
∴ rd
n n
n
n ni= −
∑−
= −
−
−= −1
61
1
6n 13
11
2
2
2
2( )
( )
( ). Ans.(3)
85. Total number of cases = 100 × 100.Now, 7m + 7n is divisible by 5 if one of the term has to end with 9 and otherwith 1(7x cannot be divisible by 9)∴ m can be 2, 6, 10, 14, ..., 98 (25 values)and n can be 4, 8, 12, ..., 100 (25 values)Since m and n can interchange.
∴ Required probability =× ×
×=
2 25 25100 100
18
. Ans.(3)
86. P A P A B( ) , ( )= ∪ =13
34
Now, P(A ∪ B) = P(A) + P(B) – P(A ∩ B) ≤ P(A) + P(B)
⇒ ≤ + ⇒ ≤34
13
512
P B P B( ) ( )
Also, B ⊆ A ∪ B ⇒ P(B) ≤ P(A ∪ B) = 34
∴ ≤ ≤5
1234
P B( ) . Ans.(2)
87. Let xa
Xyb
Yzc
Z2
2
2
2
2
2= = =, , , then the given system of equations is
X + Y – Z = 1, X – Y + Z = 1, – X + Y + Z = 1
∴ Coefficient matrix is A =−
−−
L
NMMM
O
QPPP
1 1 11 1 11 1 1
Since |A| ≠ 0, the given equations have unique solution. Ans.(2)88. No. of ways = coeff. of x2m in (1 + x + ... + xm)4
(1 + x + ... + xm)4 = 1 1
1
1 4( )−
−
LNMM
OQPP
+xx
m
= (1 – xm+1)4 (1 – x)–4
= (1 – 4xm+1) = 6x2m+2 ...) × 1 41 2 3
3+ + +
+ + ++
FHG
IKJx
k k kxk...
( ) ( ) ( )!
...
(Note this step.)
=+ + +
− ×+ +F
HGIKJ
( ) ( ) ( )!
( ) ( )!
2m 1 2m 2 2m 33
41 23
2 2xm m m
xm m + other terms
=+ + +( ) ( )m
x m1 2m 4m 33
22 + other terms
∴ No. of ways = ( ) ( )m + + +1 2m 4m 3
3
2
. Ans.(3)
(42) of (57)
89.23
0 666666 6666
100= =. ....
. ....
mfor m and
mfor m
10023
1 66100
23
67 99< ≤ ≤ > ≤ ≤ .
∴ + < + = ≤ ≤13 100
13
23
1 1 66m
for m
and 13 100
13
23
1 67 99+ > + = ≤ ≤m
for m
∴ +LNM
OQP = ≤ ≤ +
LNM
OQP = ≤ ≤
13 100
0 1 6613 100
1 67 99m
for m andm
for m .
∴ E = (0 + 0 ... 66 times) + (1 + 1 + ... 33 times) = 33. Ans.(4)90. Equation of tangent at P(a cos θ, b sin θ) is
xa
yb
cos sinθ θ+ = 1 ...(1)
and equation of auxiliary circle isx2 + y2 = a2 ...(2)Making (2) homogeneous with the help of (1)
we get, x y axa
yb
2 2 22
+ = +FHG
IKJcos sinθ θ
Lines represented by it are perpendicular if coefficient of
x2 + coefficient of y2 = 0 i.e. 1 1 022
22− + −
FHG
IKJ =cos sinθ θe j a
b
or sin22
21 1θ −FHG
IKJ = −
a
b or sin2
2
211θ
e
e−
FHG
IKJ =
or 1 11
2
22
22−
= = +e
eor
esin sinθ θ or e =
+
1
1 2sin θ. Ans.(1)
91.dydx
cx ory
xc= −
+=2 1
12'
Required differential equation yy
xx x=
+FHG
IKJ −
' 12
2
or 2y = xy’ – x or xy’ = 2y + x. Ans.(4)
92. The given equation can be written as 1 1 1 12y
dydx x y x
x+ =. log .
Putting 1y
t= , the equation reduces to dzdx x
zx
x− = −1 1
log .
I.F. = e ex
xdx x− −z = =
11log
The solution is zx x
x dx. log1 1
2= −zzx x
xx x
dx= − z1 1 1log . (using by parts)
= + +1 1x
xx
clog where zy
=1
. Ans.(3)
93. Let no. of pieces of A be x and that of B beConstraints are
2x + y ≤ 20x + 3y ≤ 15
(0, 5) (9, 2)
(0, 20)
(15, 0)(10, 0)
(0, 0)
Objective function isz = 30x + 20yFigures shows the feasible regionz’(0, 0) = 0, z(0, 5) = 100
z(10, 0) = 300z(9, 2) = 310. Ans.(4)
94. 25 = 32 = (– 15) mod 47⇒ 210 = 225 mod 47 = (– 10) mod 47⇒ 220 = 100 mod 47 = 6 mod 47⇒ 240 = 36 mod 47 = (– 11) mod 47245 = (– 11) × (– 15) mod = 47 = 24 mod 47⇒ 246 = 48 mod 47 = 1 mod 47. Ans.(1)
95. Mode = 3(Median) – 2(A.M.)= 3(26.1) – 2(24.6) = 29.1. Ans.(4)
96. We have S.D. = 25
Coefficient of Variation = S D
Mean. .
.× = × =1002565
100 38 46% . Ans.(2)
97. We have rd
n ni= −
∑−
16
1
2
2( )
It is given that r = 0.8 ∑di2 = 33
0.8 = 1 – 6 33
1
6 33
10 2
2 2
( )
( )
( )
( ).
n n n n−⇒
−=
⇒ n (n2 – 1) = 990 ⇒ n (n2 – 1) = 10 (102 – 1) ⇒ n = 10. Ans.(2)98. Let P(X = 1) = a, P(X = – 1) = b ⇒ a + 0.2 + b = 1
⇒ a + b = 0.8 ...(1)1.P(X = 1) + 0.P(X = 0) – 1(P = – 1) = 0.6⇒ a – b = 0.6 ...(2)(1) + (2) ⇒ 2a = 1.4 ⇒ a = 0.7. Ans.(2)
99. np – npq = 59
5, n = 5p (1 – q) = 59
5 559
2⇒ = =p p p.
p p2 19
13
= =; ∴ = − =q p123
Distribution is given by p(x = r) = ⋅FHG
IKJ ⋅
FHG
IKJ
−
513
23
5
C
r r
r
r = 0, 1, 2, 4, 5
Answer is 513
23
0, 1 2, 3, 4, 55
C
r r
rr
FHG
IKJ
FHG
IKJ =
−
; , . Ans.(2)
100. We know that P(X = r) = f(r) = er
r−λ λ
!'r = 0, 1, 2, ...∞
P(X = 0) = e–λ , P(X = 1) = e–λ, λ
P(X = 0) = 2, P(X = 1) ⇒ e–λ = 2.e–λ . λ ⇒ 1 = 2 λ ⇒ =λ12
.
Since the mean of Poisson variable is λ, for the given distribution mean =
12
. Ans.(2)
(43) of (57)
1. The roots of x2 + x + 1 = 0 are ω, ω2. Let a = ω and b = ω2. Thenα19 = ω19 = (ω3)6 ω = ω and β7 = (ω2)7 = ω14 = (ω3)4 = ω2.
The equation whose roots are ω and ω2 isx2 + x + 1 = 0. Ans.(4)
2. In (a + c), In (c – a); ln (a – 2b + c) are in A.P.⇒ 2 ln (c – a) = ln (a + c) + ln (a – 2b + c)⇒ (c – a)2 = (a + c) (a – 2b + c)⇒ (c – a)2 = (a + c)2 – 2b (a + c)⇒ 2b (a + c) = (a + c)2 – (c – a)2 = 4ac
⇒ bac
a c=
+2
⇒ a, b, c are in H.P. Ans.(4)3. For px2 + qx + 1 = 0 to have real roots, we must have
q2 – 4p ≥ 0 or q2 ≥ 4p.If p = 1, then q2 ≥ 4 ⇒ q = 2, 3, 4.If p = 2, then q2 ≥ 8 ⇒ q = 3, 4If p = 3, then q2 ≥ 12 ⇒ q = 4.If p = 4, then q2 ≥ 16 ⇒ q = 4.Thus, the number of equations of the form px2 + qx + 1 = 0 which have realroots is 7. Ans.(3)
4. We have f’ (x) = 1 . e–x + (x + 2) (– 1) e–x
= e–x (1 – x – 2) = – (x + 1) e–x.Since e–x > 0 V x ∈ R,f’(x) > 0, if – (x + 1) > 0 or x + 1 < 0 or x < – 1;f’(x) < 0, if – (x + 1) < 0 or x + 1 > 0 or x > – 1.Thus, f(x) increases on (–∞, –1) and decreases on (–1, ∞). Ans.(4)
5. We have f’ (x) = 0 0
2 0x
x x<>
RSTFor x = 0, we have
Rff x f
xx
xx
x x x' ( ) lim
( ) – ( )–
lim–
lim ( )00
00
00 0
2
0= = = =
→ + → + → +
and Lff x f
x xx x' ( ) lim
(– ) – ( )– –
lim–
–0
00
0 00
0 0= = =
→ + → +
f’ (0) = 0.
Thus f’ (x) = 0 0
2 0if x
x if x≤>
RSTNote that f’ is not differentiable at x = 0.∴ (2), (3), (4) are true. Ans.(1)
6. f (x) =x
xx
x
sin ,
,
10
0 0
≠
=
RS|T|
, g (x) = x f(x) ⇒ g (x) =x
xx
x
2 10
0 0
sin ,
,
FHG
IKJ ≠
=
RS|T|
For x ≠ 0, g’ (x)
= x2 cos 1 1
2x xFHG
IKJFHG
IKJ– + 2x sin
1x
FHG
IKJ = – cos
1x
FHG
IKJ + 2x sin
1x
g'g x g
x
xx
xx
xx x x( ) lim
( ) – ( )–
limsin –
lim sin00
0
10
10
0 0
2
0= =
FHG
IKJ
=FHG
IKJ =
→ → →
∴ g' x xx x
x
x( ) sin – cos ,
,=
FHG
IKJ
FHG
IKJ ≠
=
RS|T|
21 1
0
0 0
g’ (x) is not continuous at x = 0
Q cos 1x
FHG
IKJ is not continuous at x = 0
Also, f is not differentiable at x = 0. Ans.(1)
7. y = 5x2 + 2x + 3 = 5 x x2 25
35
+ +LNM
OQP = + +
FHG
IKJ +
FHG
IKJ
LNMM
OQPP5
25
15
35
15
22 2
x x –
= 5 [(x + 1/5)2 + 3/5 – 1/25] = 5 (x + 1/5)2 + 14/5 > 2.Since, y = 2 sin x ≤ 2, so there cannot be any point of intersection. Ans.(1)
8. 2x2 + 3y2 – 8x – 18y + 35 = k⇒ 2 (x2 – 4x) + 3 (y2 – 6y) + 35 =k⇒ 2 [(x – 2)2 – 4] + 3 [(y – 3)2 – 9] + 35 = k⇒ 2 (x – 2)2 + 3 (y – 3)2 = kFor k = 0, we get 2(x – 2)2 + 3(y – 3)2 = 0 which represents the point (2, 3).Ans.(3)
9. ORp q
p q=++
= +3 2
3 215
3 2( ) and OSp q
p q= =3 2
3 23 2
––
–
As OR ⊥ OS, OR . OS = 0
⇒15
(3p + 2q) . (3p – 2q) = 0
⇒ 9 | p |2 – 4 | q |2 = 0 or 9p2 = 4q2. Ans.(1)10. Sec 2x – tan 2x
= = =12
22
1 22
2
2 2cos–
sincos
– sincos
(cos – sin )
cos – sinxxx
xx
x x
x x
=+
=+
=cos – sincos sin
– tantan
tan / –x xx x
xx
x11
4πc h . Ans.(2)
11. Only (1) is correct. Ans.(1)
12. Let a i j k→
= +13
2 2( $ – $ $), then | | ( ) | |a a→ →
= + + ⇒ =2 19
4 4 1 1
Let b i j k→
= +2 4 3$ – $ $, then cos.
| | | |
( )
( )θ = =
+ +
+ +=
→ →
→ →
a b
a b
13
4 8 3
1 4 16 9
5
29
⇒ θ ≠ π/3
Now, let c i j k a→ →
= + =–$ $ – $ –12
32
⇒ c→
parallel to a→
Now, let d i j k→
= +3 2 2$ $ – $, then a d→ →
= =. ( – – )13
6 4 2 0 ⇒ a d→ →
⊥ . Ans.(2)
13. Probability of getting a white ball at any draw is p = =1224
12
.
The probability of getting a white ball 4th time in the 7th draw= p (getting 3 white balls in 6 draws) × P (white ball at the 7th draw)
= = =( ) ( / ) . .63
671 2
12
20
2
532
C Ans.(3)
14. As (2, 3) lies on y2 = px3 + q, we get 9 = 8p + q.
Also, 2 33
22
2
ydydx
pxdydx
pxy
= ⇒ = ⇒ dydx
atp
p.( )( )( )
2, 33 2
2 32
2
= =
Since slope of tangent is 4, we have 2p = 4 ⇒ p = 2.Since 8p + q = 9, we get q = – 7. Ans.(1)
15. sin (π/2n) + cos (π/2n) = n / 2 ..... (1)⇒ sin2 (π/2 n) + cos2 (π/2n) + 2 sin (π/2 n) cos (π/2 n) = n/4
⇒ sin(π/n) =n n4
14
4–
–=
As sin(π/n) > 0 V n > 1, we get
nn
– 44
0 4.> ⇒ > ..... (2)
Also, sin (π/2 n) + cos (π/2 n) = 24 2n
sinπ π
+FHG
IKJ
⇒n
2 2 4 2n= +
FHG
IKJsin
π π[By (1)]
Since sin (π/4 + π/2n) < 1, we get
n
2 2< 1 or n < 8 ..... (3)
∴ From (2) and (3), 4 < n < 8. Ans.(4)
Exercise - 03 - Solutions
(44) of (57)
16. Since the largest angle is opposite of the largest side, we have
cos C =+
= =3 5 7
2 3 5 3012
2 2 2–( ) ( )
–15– ⇒ C =
23π
. Ans.(3)
17. We have P(S) = P(5, 6) = 2/6 = 1/3 ⇒ P(F) =23
P (an even number of tosses is needed)= P (FS or FFFS or FFFFFS or ..... )= P(F) P(S) + P(F)3 P(S) + P(F)5 P(S) + .....
(G.P. with T1 = P(F) P(S) and common ratio P(F)2)
= = =P F P S
P F
( ) ( )
– ( )
/– /
.1
2 91 4 9
252 Ans.(2)
18. We have cos–1 1
5 2
71
7FHG
IKJ =
FHG
IKJ =tan tan–1 –1 and sin tan–1 –14
17
41
FHG
IKJ =
FHG
IKJ
Now, tan cos – sin–1 –11
5 2
4
17
LNMM
OQPP = tan [tan–1 7 – tan–1 4]
=+ ×
RSTUVWtan tan
––1 7 41 7 4
=FHG
IKJ
RST|UVW| =tan tan–1 3
29329
. Ans.(4)
19. 2 sin2 x + 3 sin x – 2 > 0⇒ (2 sin x – 1) (sin x + 2) > 0⇒ 2 sin x – 1 > 0 [∴ sin x + 2 > 0 V x ∈ R]⇒ sin x > 1/2 ⇒ x ∈ (π/6, 5 π/6)Also x2 – x – 2 < 0⇒ (x – 1) (x + 1) < 0 ⇒ –1 < x < 2As 2 < π/6, we get that x must lie in (π/6, 2). Ans.(4)
20. By definition, it is an ellipse. Ans.(1)21. The sides of the parallelogram are x = 2, x = 3; y = 1 and y = 5. Therefore
angular points are A(2,1), B(2,5), C(3.5), D(3,1).
Equation of AC is y – 1 =5 13 2
2––
( – )x or y = 4x – 7
Equation of BD is y – 5 =1 53 2
2––
( – )x or 4x + y = 13. Ans.(3)
22. Operation R1 + R3, we get
1 11 1 1
10
2 2
2– –– – –– – – –1
ii
i i
ω ω ωω
ω
+
+=
[∴ ω2 + ω = – 1, then R1 and R2 become identical]. Ans.(1)
23. (1 + n + n2 + ..... + n127) =n
nn n
n
128 64 6411
1 11
––
( – ) ( – )–
=
= (1 + n + n2 ..... + n63) (n64 + 1)Thus the largest integer m such that nm + 1 divides 1 + n + n2 ..... + n127 is64. Ans.(3)
24. We have a2 = 16 and 9 = b2 = a2 (1 – e2) = 16(1 – e2),
so e =7
4. Thus the foci are ( , )± 7 0
The radius of the required circle = + = + =( – )7 0 3 7 9 4.2 2 Ans.(1)
25. Taking α = – π/2, β = – π/2 and γ = 2π, we find thatsin α + sin β + sin γ = – 2. Moreover, sin α + sin β + sin γ > – 3 for any α, β,γ. So the minimum value of sin α + sin β + sin γ is negative. Ans.(3)
26. f xpx q rx xpx q rx x q p
px q rx q p x( )
– – ,– , /
– , /=
+ ≤+ + < ≤
+ <
RS|T|
00
Thus f has two points of minimum at x = 0 and x = q/p if r = p. In case p ≠ r thenx = 0 is point of minimum if r > p and x = q/p is a point of minimum if r < p.
y
qy = f(x)
O q/px
When r = p When r < p
q/px
O
q
y
When r > p
q/px
O
q
y
. Ans.(3)
27. [ sin ] [ sin ]/
2 27 62
x dx x dx= zzπ
π
π
π
+ + zz [ sin ] [ sin ]//
/
2 211 6
2
7 6
11 6
x dx x dxπ
π
π
π
= + +zz z(–1) (– ) (– )/
//
/
dx dx d2 17 6
11 67 6
11 6
2
π
π
π
π
π
π
= =(–1) – – – .π π π π6
86 6
53
Ans.(1)
28. f’(x) = Ax
f Aπ π π2 2
2 1 22
1
2cos ' ( / )
FHG
IKJ ⇒ = = ⇒ A =4/π
Now f x dxA
( ) =z 2
0
1
π ⇒
2 42
0
1
0
1Ax dx B dx
π ππ
=FHG
IKJ +z zsin
⇒2 4 2
2 0
1A x
Bπ π
ππ
= +cos ( / )
/ ⇒
2 40.
AB B
π πππ
+ × + ⇒ = Ans.(4)
29. S.D. of five consecutive numbers is n2 21
125 1
122
−=
−=
Variance = 2. Ans.(4)30. Apply the Bowley’s Method of finding coefficient of skewness
=+Q Q Median
Q Q3 1
3 1
2––
Q3 + Q1 = 22 ; Q3 – Q1 = 8We have to find out the value of MedianMode = 3 Median – 2 Mean11 = 3 Median – 2 × 8Median = 9
Coefficient of Skewness =×
= +22 2 9
80.5
–. Ans.(1)
31. We have r =Cov x y
x y
( , )σ σ
⇒ 0.64.8
34 8
3 0 683
= ⇒ =×
=σy
yσ.
.. Ans.(1)
(45) of (57)
32. The number of ways of selecting 2 tickets out of 6 tickets is 6C2 = 15The favourable cases for sum to be even is(1 + 3), (1 + 5), (2 + 4), (2 + 6), (3 + 5), (4 + 6)there are 6 favourable cases to win Rs. 10/-
Probability of gain = =6
1525
Probability of loss = 125
35
– =
Expected gain = 10 × 25
+ (– 5) × 35
= 4 – 3 = 1. Ans.(2)
33. n = 5 ; np + npq = 4.8np(1 + q) = 4.85p (2 –p) = 4.8p2 – 2p + .96 = 025p2 – 50p + 24 = 0
(5p – 6) (5p – 4) = 0 ⇒ p =65
not possible
p q= = =45
145
15
; –
The distribution is 15
45
5
+FHG
IKJ .
Trick : Work from options. Ans.(1)34. By definition current will not flow if p, q, r are open. Ans.(1)35. ax = by = cz = k ⇒ a = k1/x, b = k1/y, c = k1/z
abc = 1 ⇒ k x y z1 1 1
+ += 1 = k° ⇒
1 1 10
x y z+ + =
⇒ xy + yz + zx = 0. Ans.(1)
36. Consider log (0.5)100 = 100 ( . ) .1 6990 3190= (0.5)100 = Antilog 3190.
Hence, there are 30 zeros between the decimal point and the first significantfigure of the number to be obtained from Antilog table. Ans.(1)
37. a2 + 4b2 = 12abAdding 4ab to both sidesa2 + 4b2 + 4ab = 16ab(a + 2b)2 = 16abtaking logarithms2 log (a + 2b) = log (24 ab) = 4 log 2 + log a + log b
log (a + 2b) =12
[log a + log b + 4 log 2]. Ans.(3)
38. Given expression = log4 54
+ log4 65
+ log4 76
+ ..... + log4 256255
= log4 54
65
76
256255
. . ......... = log464 = log4 43 = 3. Ans.(4)
39. x = +3 8
1 1
3 8
1
3 8
3 8
3 83 8
x=
+=
+× =
–
––
∴ xx
+ = + + =1
3 8 3 8 6–
Squaring both sides, we get
xx
xx
+FHG
IKJ = ⇒ + + =
16
12 36
22 2
2⇒ x
x2
2
134+ =
Squaring again, we get
xx
xx
22
22 4
4
134
11156 2+
FHG
IKJ = ⇒ + =( ) – = 1154. Ans.(1)
40. f(x) = 2x3 + 3x2 + ax + bWhen f(x) is divided by x – 2; the remainder is 2.∴ f(2) = 216 + 12 + 2a + b = 2⇒ 2a + b = – 26 ..... (1)When f(x) is divided by x + 2, the remainder is –2.∴ f(– 2) = – 2⇒ 16 + 12 – 2a + b = – 2⇒ –2a + b = 2 ..... (2)Solving (1) and (2) a = – 7, b = – 12. Ans.(1)
41. ( ) ( )/ /1 7 1 2 173
164
1 3 3 4− + = −FHG
IKJ −FHG
IKJ
−x xx x
[Neglecting higher terms as x is small]
= −FHG
IKJ1
236
x [Neglecting x2 term]. Ans.(3)
42. x17 = – (1 + 2 + 3 + ..... + 18)
= +– [ ]182
1 18 = – 9 × 19 = – 171. Ans.(2)
43. t C xa
xC a x xr r
r r
rr r
rr
+ =FHG
IKJ =1
1010
210 5
2e j – – –2(–1) = (–1) ( )–
rr
rr
C a x1010 5
2
This term is independent of x. Hence 10 5r
20
–= ⇒ r = 2
Term independent of x = t3 = (– 1)2 10 C2 a2 = 405 given.
aC
a210
2
405 40545
9 3= = = ⇒ = ± . Ans.(1)
44. If mx2 + 7xy – 3y2 + 4x + 7y + 2comparing the given expression with standard form
a = m 2f = 7 ⇒ f =72
b = – 3 2g = 4 ⇒ g = 2
c = 2 2h = 7 ⇒ h =72
Since the given expression is factorable into two linear factors thenabc + 2fgh – af2 – bg2 – ch2 = 0
∴ – . . . – –6m 272
272
494
3.4 2494
0.+ + =m ⇒ = ⇒ =–73 –146m
m4 4
2 .
Ans.(3)45. α, β roots of x2 – 3x + a = 0; α + β = 3, α β = a;
(α – β)2 = (α + β)2 – 4 α β, (α – β)2 = 9 – 4a. γ, δ are the roots ofx2 – 12x + b = 0γ + δ = 12, γδ = b; (γ – δ)2 = (γ + δ)2 – 4 γ δ∴ (γ + δ)2 = 144 – 4bWhen a = 2, b = 32; α, β; γ, δ form a G.P.∴ They are 1, 2, 4, 8. Ans.(3)
46. The given quadratic equations are ax2 + bx + c = 0 and x2 + x + 1 = 0
xb b ac
ax=
±=
±– –;
–1 –2 42
1 42
Both the roots of the equation are imaginary. If this equation has a commonroot with ax2 + bx + c = 0 that common root is also imaginary.∴ The other root will also be common to these as imaginary roots occur inpairs.Hence they can have a common root only when both roots are common andconsequently the equations become identical
∴a b c
a b c1 1 1
= = ⇒ = = . Ans.(1)
47. θ φπ
θπ
φ– = ⇒ = +2 2
cos cos sincos sin sin
cos cos sincos sin sin
2
2
2
2θ θ θ
θ θ θφ φ φ
φ φ φ
LNMM
OQPPLNMM
OQPP
sin – sin cos– sin cos cos
cos cos sincos sin sin
2
2
2
2φ φ φ
φ φ φφ φ φ
φ φ φ
LNMM
OQPPLNMM
OQPP
sin cos – sin cos sin cos – sin cos– sin cos cos sin – sin cos cos sin
2 2 2 2 3 3
3 3 2 2 2 2φ φ φ φ φ φ φ φ
φ φ φ φ φ φ φ φ+ +
LNMM
OQPP
=LNM
OQP = ×
0 00 0
0 2 2[ ] . Ans.(1)
48. AB = AC ⇒ B = C is true if and only if A non-singular in that case AB = AC ⇒A–1 (AB) = A–1 (AC)⇒ (A–1 A) B = (A–1 A) C ⇒ B = C. Ans.(3)
(46) of (57)
49. AAT =L
NMMM
O
QPPPL
NMMM
O
QPPP
19
1 2 22 1
2
1 22 1 22
–2–2 –1
–2
–2 –1=
L
NMMM
O
QPPP
=19
9 0 00 9 00 0 9
3I
Hence, (A) is orthogonal. Ans.(1)50. Let B = adj (adj A2)
Then, B is also a 3 × 3 matrix∴ | adj {adj (adj A2)} | = | adjB | = | B |3 – 1 = | B |2
= | adj (adj A2) | 2 = | |( – )A2 3 122LNM OQP
= | A |16 = 216 [Q | A2 | = | A |2]. Ans.(3)
51. ∆ =1 43 1 52 3 1
–2
= 1(1 – 15) – 3(4 + 6) + 2(20 + 2)
= (–14 – 30 + 44) = 0
∆ x =3 47 1 55 3 1
–2= 3(1 – 15) – 7 (4 + 6) + 5(20 + 2)
= (– 42 – 70 + 110) = –2 ≠ 0Thus, ∆ = 0 and ∆x ≠ 0So, the system has no solution. Ans.(2)
52. S = + + + +145
7
5
10
52 3 .....
15
15
4
5
7
5
10
52 3 4S = + + + +.....
S to115
135
3
5
3
52 3– .....FHG
IKJ = + + + + ∞
S S45
1
35
115
134
74
3516
FHG
IKJ = + = + = ⇒ =
–. Ans.(4)
53. a, b, c are in H.P.
⇒ =+
⇒ =+
bac
a cba
ca c
2 2..... (1)
b ab a
c ac a
+=
+– –
3
From (1) (by componendo and dividendo)
b cb c
a ca c
+=
+– –
3 b ab a
b cb c
c ac a
a ca c
++
+=
++
+=
– – – –3 3
2 . Ans.(2)
54. S50 = 40,000
S40 = 40,000 –14
(40,000) = 30,000
S50 = 40,000
⇒502
{2a + (50 – 1) d} = 40,000 ..... (1)
S40 = 30,000
⇒402
{2a + (40 – 1) d} = 30,000 ..... (2)
From (1) and (2)
2a + 49 d =40,000
25= 1600
2a + 39d =30,000
20= 1500
10d = 100 ⇒ d = 102a = 1600 – 490 = 1110a = 555The value of first installment is Rs. 555. Ans.(1)
55. A = (1, 2, 5) B = (–1, –2, 3)
BA i j k i j k i j k→ → → → →
= + + + = + +( ) – (– – )2 5 2 3 2 4 2
moment m BA F→ → →
= × =
→ → →i j k2 4 22 5–λ
= + +→i j k( ) – ( – ) (–2 – )20 2 10 4 8λ λ = + +
→ →i j k( ) – ( ) (–2 – )20 2 6 8λ λ
But m i j k→ →
= +16 6 2– )λ
Hence 20 + 2λ = 16 ⇒ 2λ = –4λ = – 2. Ans.(4)
56. Let a→
is the required Unit Vector a x i y i x y→ → →
= + + =; 2 2 1
Let 45° is the angle between a→
and i j→ →
+ then,
cos 45° =+
FHG
IKJ
+
→ → →
→ → →
a i j
a i j
.
.=
+FHG
IKJ +
FHG
IKJ
+
→ → → →x i y j i j.
1 1 1
1
2 2=
+x y∴ x + y = 1 ..... (1)
60° is the angle between a→
and 3 4i j→ →
– then
cos
. –
. –
. –
60
3 4
3 4
3 4
1 9 16° =
FHG
IKJ
FHG
IKJ
=+
FHG
IKJ
FHG
IKJ
+
→ → →
→ → →
→ → → →a i j
a i j
x i y j i j
12
3 45
3 452
= ⇒ =x y
x y–
– ..... (2)
3x + 3y = 3; 3x – 4y =52
y x= =1
141314
;
a i j= +1314
114
$ $ . Ans.(4)
57. Volume =2 41 23 1
–3–2
–1
= 2 (2 – 2) + 3 (1 + 6) + 4 (–1 –6)
= | 21 – 28 | = 7 Cubic Units. Ans.(3)58. The desired pair (x1, x2) lie in the first
quadrant only in this case the two halfplanes x1 + x2 ≤ 1 and –3x1 + x3 ≥ 3 donot intersect and hence they have nopoint in common.
As there is no point which can lie in boththe regions, there exists no solution tothe L.P.P. Ans.(3)
(47) of (57)
59. 3x + 2y = 160 ..... (1)5x + 2y = 200 ..... (2)x + 2y = 80 ..... (3)Form (1) and (2)– 2x = –40 ⇒ x = 20 ⇒ y = 50The point of intersection of (1) and (2) is (20, 50)(2) – (3) gives 4x = 120 ⇒ x = 30
2y = 50 ⇒ y = 25The point the intersection of (2) and (3) = (30, 25)(1) – (3) gives 2 x = 80 ⇒ x = 40 ; y = 20The point of intersection of (1) and (3) is (40, 20)
(20, 50)
(40, 20)
C
B
D(0,100)
(40, 0) A (80, 0)
(1) cuts the x-axis at 160
30,
FHG
IKJ
(1) cuts the y-axis at (0, 80)(2) cuts the x-axis at (40, 0)(2) cuts the y-axis at (0, 100)(3) cuts the x-axis at (80, 0)(3) cuts the y-axis at (0, 40)f (x, y) = 4x + 3yf (80, 0) = 320 + 0 = 320f (40, 20) = 160 + 60 = 220f (20, 50) = 80 + 150 = 230f (0, 100) = 0 + (100)3 = 300. Ans.(1)
60. Let f(x) = ax3 + bx2 + cx + d. Since f(0) = 1, so d = 1.Moreover, f(1) = 2 implies that a + b + c = 1 since 0 is a critical point,0 = f’(0) = 3a .0 + 2b.0 + c i.e. c = 0. Also f”(x) = 6ax + 2b and since f(x) doesnot have extremum at 0 so 0 = f”(0) = b. Hence a = 1 and
f xx
dxxx
dx xx
x xdx
( )–2
3
2 2 2111
12
21
11+
=++
=+
++
FHG
IKJz z z
=12
[x2 – log (x2 + 1)] + tan–1 x + C. Ans.(4)
61. f xxxn
n
n( ) lim–
=+
=→ ∞
2
211
–1 (0 < x < 1), so
sin ( ( )) – sin–1 –1x f x dx x dx= zz = – [x sin–1 x + 1 2– x ] + C. Ans.(1)
62. Since f x dx f a b x dxa
b
a
b
( ) ( – ) ,z z= + we have
Ik
k
11
= z–
(k + 1 – k – x) f ((k + 1 – k – x) (1 – (k + 1 – k – x)) dx.
= z1– k
k
(1 – x) f ((1 – x) x) dx
= =z zf x x dx x f x x dx I Ik
k
k
k
(( – ) ) – (( – ) ) –– –
1 11
21
1
So 2I1 = I2 and I1/I2 = 1/2. Ans.(3)
63. lim lim– – / – /
. . . .
––
nn
nS
n n n nn
→ ∞ → ∞= + + + +
FHG
IKJ
L
N
MMMMMM
O
Q
PPPPPP
1 1
4 0
1
4 1
1
4 4
1
41
2 2 2
= = = =→ ∞
=∑ zlim
– ( / –sin
––1
nr
n
n r n)
dx
x
x1 1
4 4 2 620
1
20
1
0
1π
. Ans.(4)
64. Required area
O
PQ
R (1, 0)
(1, 3)
= area OPQRO – area ∆ OQR
= × ×z 912
1 30
1
x dx –
= =323
32
12
3 2
0
1
x / – . Ans.(3)
65. The given equation is with separable variables so
(cy + d) dy = (ax + b) dx. Integrating we have cy
dy Kax
bx2 2
2 2+ + = + , K being
the constant of integration. The last equation represents a parabola i f c =0, a ≠ 0 or a = 0, c ≠ 0. Ans.(3)
66. The given equation can be written as
xdy ydx
x ydx
xdy yd x
x y xdx
––
–
/–2 2 2 2 2
1
1+= ⇒ ×
+=
⇒1
1 2 2+FHG
IKJ =
y x
ddx
yx
dx/
– . Integrating we have tan–1 (y/x) = –x + c
⇒ y = x tan (c – x). Ans.(3)67. Let P (x, y) be the point on the curve passing through the origin O (0, 0), and
let PN and PM be the lines parallel to the x- and y-axes, respectively (fig). Ifthe equation of the curve is y = y(x), the area
N
y
O
P (x, y)
Mx
POM equals yd xx
0z and the area PON equals xy yd x
x
–0z .
Assuming that 2 (POM) = PON, we therefore have
2 30 0 0
yd x xy yd x yd x xyx x x
= ⇒ =z z z–
Differentiating both sides of this gives
3 2 2y xdydx
y y xdydx
dyy
dxx
= + ⇒ = ⇒ =
⇒ log | y | = 2 log | x | + C ⇒ y = Cx2, with C being a constant.This solution represents a parabola. We will get a similar result if we hadstarted instead with 2(PON) = POM. Ans.(2)
68. Let the equation of the variable line be ax + by + c = 0then according to the given condition
2 2 20
2 2 2 2 2 2
a c
a b
b c
a b
a b c
a b
+
++
+
++
+
+=
–2 –
⇒ c = 0which shows that the line passes through (0, 0) for all values of a, b. Ans.(2)
69. The given line meets x-axis at A (4, 0) and y-axis at B (0, 12). Let P divide ABin the ratio 1 : 2 and Q divide AB in the ratio 2 : 1 then coordinates of P andQ are (8/3, 4) and (4/3, 8) respectively. So that equations of OP and OQ arerespectively
y =4
8 3/x and y x=
84 3/
or 2y = 3x and y = 6x
whose slopes 3/2 and 6 are the roots of the equation
x x2 32
632
6 0– +FHG
IKJ + × = ⇒ 2x2 – 15x + 18 = 0. Ans.(1)
(48) of (57)
70. The lines given by the equation are
(x + 7y – z 3 2 ) (x + 7y + 7 2 ) = 0
⇒ x + 7y – 3 2 = 0 and x + 7y + 7 2 = 0.
distance between these lines =+
7 2 3 2
1 72 2
– ( – ) = =10 2
5 22. Ans.(3)
71. Both the circles clearly pass through the origin (0, 0). They will thereforetouch each other if they have a common tangent at the origin. Now the tangentto the first circle at (0, 0) is gx + fy = 0, and that of the second circle is g1x +f1y = 0. If these two equations are to represent the same line, we must haveg/g1 = f/f1, i.e. f1g = fg1. Ans.(1)
72. As the centre of the circumcircle of an equilateral triangle is its centroid andthe distance of the centroid from the vertex of the triangle is 2/3 of its medianthrough that vertex. So the distance of the centre from a vertex of the triangleis (2/3) × 3a = 2a and hence the equation of the circumcircle is x2 + y2 = 4a2.Ans.(3)
73. From geometry we know PA. PB = (PT)2 where PT is the length of the tangentfrom P to the circle.Hence PA . PB = (3)2 + (11)2 – 9 = 112 = 121. Ans.(2)
74. Equation of a tangent at (at2, 2at) to y2 = 8x isty = x + at2 where 4a = 8 i.e. a = 2
⇒ ty = x + 2t2 which intersects the curve
xy = – 1 at the points given by x x t
t
+ 2 2e j= –1 clearly t ≠ 0
or x2 + 2t2x + t = 0 and will be a tangent to the curve if the roots of thisquadratic equation are equal, for which 4t4 – 4t = 0 ⇒ t = 0 or t = 1 and anequation of a common tangent is y = x + 2. Ans.(4)
75. The equation of the ellipse can be written as
x y2 2
25 161+ =
Here a2 = 25, b2 = 16 but b2 = a2 (1 – e2) ⇒ 16 = 25 (1 – e2) ⇒ e = 3/5So that foci of the ellipse are (± ae, 0) i.e. (± 3, 0) or F1 and F2By definition of the ellipse, since P is any point on the ellipse
PF1 + PF2 = 2a = 2 × 5 = 10. Ans.(3)76. Let S (5, 12) and S’ (24, 7) be the two foci and P(0, 0) be a point on the conic
then SP S P= + = = = + = =25 144 169 13; 24 7 625 252 2' ( )
and SS' ( ) ( – )= + + = + =24 5 7 12 19 5 3862 2 2 2
since the conic is a hyperbola, S’P – SP = 2a, the length of transverse axisand SS’ = 2ae, e being the eccentricity.
⇒ eSS
S P SP=
+=
''
38612
. Ans.(1)
77. Let di = x i – 8
but σ σx d i id d2 2 22
118
118
= =FHG
IKJ∑∑ – = ×
FHG
IKJ = =
118
459
1852
14
94
2
– –
Therefore σx = 3/2. Ans.(3)78. If the observations greater than the median are increased by 6, the median
will not be affected. Ans.(1)
79. Cov (UV) =1n
UV u v∑ – = ∑1 2 2 2 2
nX Y X Y( – ) – ( – )
=FHG
IKJ
FHG
IKJ∑ ∑1 12 2 2 2
nX X
nY Y– – – = Var X – Var Y = 0
Therefore, r (U, V) = 0. Ans.(2)80. Let Ei denote the event that with shot hits the plane. Then
P(E1) = 0.4, P(E2) = 0.3,P(E3) = 0.2, P(E4) = 0.1
We haveP(E1 ∩ E2 ∩ E3 ∩ E4)
= 1 – P (E’1 ∩ E’2 ∩ E’3 ∩ E’4)= 1 – P(E’1) P(E’2) P(E’3) P(E’4)
[Q E1, E2, E3, E4 are independent]= 1 – (0.6) (0.5) (0.8) (0.9)= 1 – 0.3024 = 0.6976. Ans.(1)
81. The number of ways of selecting three tickets out of
21 is 21C3 = 21 20 19
3 21330.
× ××
=
Let d be the common difference of the AP. We have the following possiblecases :When d = 1In this case the possible A.P’s are 1, 2, 3; 2, 3, 4; 3, 4, 5; ...; 19, 20, 21Thus, there are 19 such A.P’s.When d = 2In this case the possible A.P’s are 1, 3, 5; 2, 4, 6; 3, 5, 7; ...; 17, 19, 21Thus, there are 17 such A.P’sWhen d = 3In this case the possible A.P’s are 1, 4, 7; 2, 5, 8; 3, 6, 9; ...; 15, 18, 21Thus, there are 15 such AP’s.
....................................................
....................................................When d = 10In this case the possible AP is 1, 11, 21Thus, there is just one AP is this case.∴ numbers of favourable ways is
19 + 17 + 15 + ..... + 1 =102
(19 + 1) = 100
Hence, probability of the required event is = =1001330
10133
General Formula If three tickets are selected at random out of (2n + 1) ticketsnumbered 1, 2, 3, ....., (2n + 1), then the probabil i ty they are in A.P. is
nCn
2
2 13
2
3n4n 1+ =
–. Ans.(1)
82. Required probability= P(winning at exactly one match in the first two matches) × P (winning thirdmarch)
=FHG
IKJFHG
IKJ
LNM
OQPLNM
OQP =2
112
12
12
14
C . Ans.(2)
83. sin θ + cosec θ = 2 ⇒ sin θ + 1
2sin θ
=
⇒ sin2 θ – 2 sin θ + 1 = 0⇒ (sin θ – 1)2 = 0⇒ sin θ = 1⇒ cosec θ = 1so that sinn θ + cosecn θ = (1)n + (1)n = 2. Ans.(3)
84. L tn
n
n
n
n
n
n
n nn→∞ ++
++
++ +
+
LNMM
OQPP
2
2 2
2
2 2
2
2 2
2
2 20 1 2 1.....
( – )
=
+FHG
IKJ
+
+FHG
IKJ
+ +
+FHG
IKJ
L
N
MMMMM
O
Q
PPPPP→∞L t
n
n nn
n
n
1 1
10
1
11
1
11
2 2 2.....
–
=
+FHG
IKJ
→∞L t
n rn
n
1 1
12
=+
= =z 1
1 42 01
0
1
xdx x[tan ]–1 π . Ans.(3)
85. The point of intersection of the curves is given by x =π4
The area enclosed between the curves and the x-axis is bounded by
y = tan x from 0 to π4
and y = cot x from π4
to π2
Required area = tan cot/
/
/
x dx x dx+z z0
4
4
2π
π
π
= = =z2 2 204
0
4
tan [log sec ] log//
x dx x ππ
. Ans.(3)
(49) of (57)
86. The curve is as shown in fig
(–1, 0)
(1, 0)
y = xy = – x
Required volume = z 40
1
π xy dx
= z4 12 2
0
1
π x x dx–
= z4 2 2
0
2
π θ θ θπ
sin cos/
d where x = sin θ
π2
4. Ans.(2)
87. By Theorem of Pappus
b
a
Surface of the ring= 2πa × 2πb = 4π2ab. Ans.(4)
88. sin cos cos sin/
x dx = − =z πα α
π
α
22
2
⇒ sin 2α + cos α = 0 ⇒ cos α (2 sin α + 1) = 0
or cos α = 0 or sin α = −12
⇒ =απ π π π2
32
76
116
, , , . Ans.(4)
89. Ans.(1)90. Put x = sin θ, so y = sec–1(sec θ) + sec–1 (sec 2θ) = 3θ = 3 sin–1 x
∴ =−
dydx x
3
1 2. Ans.(3)
91. Let r be the radius and θ the angle of the sector formed by a wire of length30 cm.
B
O Arθ
r
∴ Perimeter = 2r + rθ = 30A = Area of the sector
= 1/2 r2 θ = 1/2 r (30 – 2r)or A = 15r – r2
∴ dA/dr = 15 – 2r and d2A/dr2 = – 2dA/dr = 0 ⇒ r = 15/2Q d2 A/dr2 < 0.∴ A has maximum when r = 15/2.Since there is only one extrema, which is maxima,∴ area A is greatest when r = 15/2.∴ maximum possible area
= 15.15/2 – (15/2)2 = 225/4 sq. cm.∴ Correct answers is (3). Ans.(3)
92. Velocity at highest point = 0
dsdt
t t= − = ⇒ =1 6 0106
Time taken to fall back = 2106
103
× = sec . Ans.(2)
93.e x
xedx
dtt
x
x
( )cos ( ) cos
+=z z1
2 2
tan t + C = tan (xex) + C. Ans.(2)
94.cos
cos sin(cos sin )
/ /20
0
2
0
2xx x
dx x x dx+
= − =z zπ π
. Ans.(2)
95. (51/2 + 71/8)1024 = [(51/2 + 71/8)8 ]128 = [54 + 8C1 57/2.71/8 + ..... + 7]128
= (54 + 7)128 + (non integral terms)⇒ the numbers of integral terms is 129. Ans.(2)
96. Since the multiplication of each element of A row will multiply the value of ∆by 5, if all elements of all 3 rows are multiplied by 5, then the value of thedeterminant = 5 × 5 × 5∆ = 125 ∆. Ans.(4)
97. ∆ = abca abcb abcc abc
111
3
3
3
= =( ) .abcabc
2
3
3
3
1 11 11 1
0 Ans.(2)
98.x x x
x x x
2
32 2 1 1
2 1 2 13 3 1
1+ ++ + = −( ) . Ans.(4)
99. The vertices of the triangle are A(0, 0), B(0, 2) and C(2, 0). Obviously ABC isa right angled triangle with ∠A = 90°. So A(0, 0) is the orthocentre. Themiddle point D(1, 1) of hypotenuse BC is the circum-centre of the triangle.∴ the distance between the orthocentre and the circum-centre
= AD = √[(1 – 0)2] + (1 – 0)2 = √2. Ans.(3)100. AC = h be the height of the pole
∴ length of the shadow isAB = √3 h (given)
C
h
B Aθ
√3 h
Now tan tan(
θ θ= ⇒ =ACAB
h
h)3
⇒ =FHG
IKJ = °θ tan .–1 1
330 Ans.(1)
(50) of (57)
1. Since ABC is an equilateral triangle,∴ AB = BC = CA.
A (0, 0) B (4, 0)
C (x, y)
⇒ AB2 = BC2 = CA2
⇒ 16 = (x – 4)2 + y2 = x2 + y2
⇒ x = 2 and y = 2 3
Hence third vertex is ( ).2, 2 3 Ans.(3)
2. Let the point C be (h, k)C(h, k)
B(– a, 0) (0, 0)
A(a, 0)
In figure, cot Aa h
k=
– and cot B
a hk
=+
According to the condition,cot A + cot B = λ
⇒a h
ka h
k–
++
= λ ⇒ 2ak
= λ . Hence the locus is ya
=2λ
. Ans.(3)
3. The coordinates of A and B are (8, 0) and (0, 6)
O (0, 0) (8, 0)A8
610
B(0, 6)
∴ AB = +8 62 2
= 10
∴ x =+ ++ +
8.0 10.0 6.86 8 10
= =4824
2
and y =+ ++ +
8.6 10.0 6.08 6 10
= =4824
2.
Hence incentre is (2, 2). Ans.(2)4. The equations of line in intercept form are
xa
yb–8 / –8 /
+ = 1 .....(1)
andx y
– 3 21+ = .....(2)
By hypothesis,
–8a
= – (– 3) ⇒ a = –83
and –8b
= – (– 3) ⇒ b = 4. Ans.(4)
5. Sinceaa
bb
cc
1
2
1
2
1
2= = , therefore u = 0 and v = 0 are the same equations. So
they represent the same equation.Thus, u + kv = 0 is either
(1 + k)u = 0 or (1 + k)v = 0i.e., u = 0 or v = 0. Ans.(1)
6. Let the slopes be 3m and m.
Then 4m = – 2hb
and 3m2 = ab
∴ 3.4
2
2hb
ab
= ⇒ hab2 43
= . Ans.(4)
7. We have ah
= =9
1234
,hb
= =1216
34
and gf
= =2128
34
⇒ ah
hb
gf
= = .
Hence lines are parallel. Ans.(2)
8. Centre of circle is (0, 0) and radius is r = 2 .
∴ The length of ⊥ form (0, 0) on 2x – y + 1 = 0 is p = =1
5
1
5
∴ The length of chord = 2 2 2r p– = 2 215
– = =295
6
5.
Ans.(3)
9. Let AB be the chord of length 2 , O be the
cen t re o f t he c i r c l e and l e t OC be theperpendicular from O on AB. Then
O
A BC
45°
AB = 2√
AC BC= = =2
21
2.
In ∆OBC, OB = BC cosec 45° = =1
22 1.
∴ Area of the circle = π(OB)2 = π. Ans.(2)10. Centre of circle is (–2, –3). Let B (x’, y’) be another point.
Now O is the mid point of AB
O (–2, –3)
B (x’, y’)
A (2, 3)
∴x +
=2
22–
⇒ x = – 6
andy'
–+
=3
23
⇒ y = – 9∴ Point is (–6, –9). Ans.(4)
11. Radius of circle =6 1
10
5
10
–=
Any line through origin is y = mx⇒ mx – y = 0Line is tangent to the circle.
∴m
m
.2 1
1
5
102
+
+= ⇒
( )2m 11
52
2
2++
=m
⇒ 2[4m2 + 4m + 1] = 5m2 + 5 ⇒ 3m2 + 8m – 3 = 0⇒ 3m2 + 9m – m – 3 = 0 ⇒ 3m (m + 3) – 1 (m + 3) = 0
⇒ (3m – 1) (m + 3) = 0 ⇒ m =13
and m = – 3
∴ Other tangent is y x=13
⇒ x – 3y = 0. Ans.(2)
12. If p be the altitude, then p = a sin 60 =a2
3 .A
B CD
QP
R S
Ga xr
rx
Since the t r i ang le i s equ i l a te ra ,there fore the cent ro id , o r thocent re ,circumcentre and incentre all coincide.Hence the radius of the inscribed circle
= = =13 2 3
pa
r
∴ diameter = 2r3
=a
.
Now if x be the side of the square inscribed, then angle in a semicircle beinga right angle, therefore x2 + x2 = d2 = 4r2
⇒ 23
22
xa
=
∴ Area = xa2
2
6= . Ans.(3)
13. x1 + 2 = 4 ⇒ x1 = 2.Putting x = 2 in the equation of the parabola,
y2 = 16 ⇒ y = ± 4Hence the coordinates of required points are (2, ± 4). Ans.(4)
Exercise - 04 - Solutions
(51) of (57)
14. The conic section can be written as
(x – 2)2 + 4(y – 0)2 = 8 or ( – )
( )
( – )
( )
x y2
2 2
0
21
2
2
2
2+ =
This is an ellipse whose semi major and semi minor axes are a = 2 2 and
b = 2 respectively. Therefore, its eccentricity is given by
eb
a= 1
2
2– = =128
32
– . Ans.(1)
15. The given point (–a, 2a) l ies on the directr ix x = – a of the parabola y2 =4ax. Thus, the tangents are at right angle. Ans.(4)
16. The equation of the tangent at (4, –2) to y2 = x is
– 2y =12
(x + 4) or x + 4y + 4 = 0
Its slope is –14
. Therefore the slope of the perpendicular line is 4. Since
the tangents at the end of focal chord of a parabola are at r ight angle,therefore the slope of the tangent at Q is 4. Ans.(3)
17.2
12, 2 3 22b
ab a= = ⇒ =
∴ eb
a= + = + =1 1
124
22
2 . Ans.(1)
18. We have a b→ →
=. cos θ
∴ | – |a b→ →
=2 1 ⇒ | | | | –2 .a b a b→ → → →
+ =2 22 2 1
⇒ 1 + 2 – 2 2 cos θ = 1 ⇒ cos θ = 1 2/ ⇒ θ = π/4. Ans.(1)
19. r i j k i j k→
= + +($ – $ $) – ( $ – $ $)2 2 3 = –$ – $i k
Hence moment of force
= r F→ →
× =
$ $ $
–1 –1–4
i j k0
3 2
= +$( ) – $( ) $ (–2)i j k2 7 = 2 7 2$– $ – $i j k . Ans.(2)
20. a b c→ → →
× =.( )–1
–1
2 11 2 11 2
= 2(5) –1 (1) –1 (–3) = 10 – 1 + 3 = 12. Ans.(2)
21. We know that P will be the mid point of AC and BD
D
O
C
BA
P
∴ OA OC OP→ → →
+ = 2 .....(i)
and OB OD OP→ → →
+ = 2 .....(ii)
Adding (i) and (ii), we get OA OB OC OD OP→ → → → →
+ + + = 4 . Ans.(4)
22. We have lim.
. . . . .( ) ( )n → ∞ + + + +
+ +LNM
OQP
113
13.5
15.7
12n 1 2n 3
=FHG
IKJ +
FHG
IKJ + +
+ +FHG
IKJ
LNM
OQP→ ∞lim – – . . . . . –n
12
113
13
15
12n 1
12n 3
=+
LNM
OQP→ ∞lim –n
12
11
2n 3=
++
LNM
OQP→ ∞limn
12
2n 22n 3
=+
+
L
NMMMM
O
QPPPP
= =→ ∞lim . .nn1
2
11
132n
12
112
Ans.(3)
23. Domain of tan–1 x is R and domain of cos–1 x is [–1, 1].Hence domain of tan–1 x + cos–1 x is = R ∩ [–1, 1] = [–1, 1]. Ans.(1)
24. We have 0 ≤ | x – 2 | ≤ ∞⇒ – ∞ ≤ – | x – 2 | ≤ 0 ⇒ – ∞ ≤ 1 – | x – 2 | ≤ 1Hence range of function is (– ∞, 1]. Ans.(3)
25. Let y = f(x) ⇒ x = f–1 (y)
∴ loga (x + x2 1+ ) = y ⇒ x + x2 1+ = ay
⇒1
12x xa y
+ += – ⇒ – –x x a y+ + =2 1
∴ xa ay y
=FHG
IKJ
– –
2 ⇒ f–1 (y) =
12
(ay – a–y)
Hence the inverse of f(x) is given by
f–1 (x) =12
(ax – a–x). Ans.(1)
26. f(x) is continuous at x =π2
⇒ limcos–/x
k xx
f→FHG
IKJ =
FHG
IKJπ π
π2 2 2
⇒k2
3= ⇒ k = 6. Ans.(2)
27. We have
f xf x h) f x
hh' ( ) lim( – ( )
=+
→ 0 = →lim( ) ( – ( )
hf x f h) f x
h0 [Q f(x + y) = f(x) f(y)]
= →f xf h)
hh( ) lim( –
01
=+
→f xhg h) G h)
hh( ) lim( ( –
01 1
= f(x) limh → 0 g(h) G(h) = f(x) limh → 0 G(h) limh → 0 g(h)= ab f(x). Ans.(3)
28. y x x= +sin – cos–1 –11 = 2 cos–1 x
⇒dydx x x
= –2–
.1
1
1
2= –
( – )
1
1x x. Ans.(1)
29.ddx
[tan–1 (cot x) + cot–1 (tan x)]
=FHG
IKJ +
FHG
IKJ
LNM
OQP
ddx
x xtan tan – cot cot ––1 –1π π2 2
= +LNM
OQP
ddx
x xπ π2 2
– –
= =ddx
x[ – ] –π 2 2. Ans.(4)
30. Given y ex ex ex
= + + + ∞. . . . .
⇒ y = ex + y⇒ log y = x + y
Differentiating w.r. to x,
11
ydydx
dydx
. = + ⇒ dydx
yy
11
–FHG
IKJ = ⇒
dydx
yy
=1–
. Ans.(2)
(52) of (57)
31. From figure,x2 + y2 = 100 .....(1)
When x = 8 ⇒ y = 6
OA
x
y
B
y
10 cm
x
Differentiating (1) w.r. to x,
2 2 0xdxdt
ydydt
+ =
∴ dydt
xy
dxdt
= – . = ⋅–86
2
= –83
cm/sec. Ans.(2)
32. Let f(x) = 3x4 – 8x3 + 12x2 – 48x + 25∴ f’(x) = 12x3 – 24x2 + 24x – 48Now f’ (x) = 0 ⇒ x3 – 2x2 + 2x – 4 = 0
⇒ x2 (x – 2) + 2(x – 2) = 0 ⇒ (x – 2) (x2 + 2) = 0 ⇒ x = 2Now f(1) = 3 – 8 + 12 – 48 + 25 = –16
f(2) = 48 – 64 + 48 – 96 + 25 = – 39and f(3) = 243 – 216 + 108 – 144 + 25 = 16,Hence, maximum value = 16 and minimum value = – 39. Ans.(3)
33. On y-axis x is zero. ∴ y = 1 – e0 = 1 – 1 = 0
Now the slope of tangent is mdydx
= at (0, 0) = –e0. 12
12
= –
Hence equation of the tangent is y – 0 = –12
(x – 0)
or x + 2y = 0. Ans.(1)
34.sin
cos
sin
cossec
3
2
3
221x
xdx
x
xdx x dx
+= +z z z =
FHG
IKJz 1 2
2
– cos
cos
x
xsin x dx + tan x + c
Let cos x = z ⇒ – sin x dx = dz
∴ I = –1 2
2– zzz dz + tan x + c =
FHG
IKJz 1
12–
z dz + tan x + c
= z +1z
+ tan x + c = cos x + sec x + tan x + c. Ans.(3)
35. g x f x f x dx( ) ( ( ) '( ))+z = +z zg x f x dx g x f x dxII I( ) . ( ) ( ) . '( )
=FHG
IKJ +zzz zf x g x dx
ddx
f x g x dx dx g x f x dx( ) . ( ) – . ( ) ( ) ( ) . '( )
= + +zzf x g x f x g x dx f x g x C( ) . ( ) – ' ( ) . ( ) '( ) . ( )
= f(x) . g(x) + C. Ans.(2)
36. ex
xdx e
x
xdxx x.
( ).
–
( )+=
++zz 2
2 2
23 3 =
+ +
LNMM
OQPPz e
x xdxx .
( )–
( )12
222 3
=+
++
RS|T|UV|W|
LNMM
OQPPz e
x
ddx x
dxx .( ) ( )
1
2
1
22 2 =+
+e
xC
x
( ),
2 2
Q e f x f x dx e f x Cx x: ( ( ) '( )) . ( ) .+ = +z Ans.(3)
37.x
xdx
4 1−z . Put x2 = t ⇒ 2x dx = dt
∴−
=−
=−+
FHG
IKJ + =
−+
FHG
IKJ +z zx
xdx
dt
t
tt
cx
xc4 2
2
21
12 1
14
11
14
1
1log log .
Ans.(3)
38. I x dx x dx x dx= − = − + −z z z| | ( ) ( )1 1 10
2
0
1
1
2
= −FHG
IKJ
LNMM
OQPP + −
LNMM
OQPPx
x xx
2
0
12
1
2
2 2= −
FHG
IKJ −
LNM
OQP + − −
FHG
IKJ
LNM
OQP1
12
0 012
1 = + =12
12
1 .
Ans.(4)
39. Let I x f x dx= z (sin )0
π
...(1)
⇒ = − − ⇒ = −z zI x f x dx I x f x dx( ) (sin ( ) ( ) (sin )π π ππ π
0 0...(2)
Adding (1) and (2), we have
22
0 0
I f x dx I f x dx= ⇒ =z zπππ π
(sin ) (sin ) . Ans.(3)
40. I ax bx c dx ax bx dx c dx= + + = + +− −−z zz( ) ( )3
2
23
2
2
2
2
1
= +−
02
2c x (Q ax3 + bx is an odd function of x)
= c[2 + 2] = 4c.Hence, value of given integral depends on c. Ans.(3)
41. The given curve is xy – 3x – 2y – 10 = 0 ⇒ =+−
yxx
3 102
∴ Required area
= 3 10
23
162
3 16 23
4
3
4
3
4xx
dxx
dx x x+−
= +−
FHG
IKJ = + −z z log ( )
= [12 + 16 log 2 – 9] = 16 log 2 + 3. Ans.(2)
42. V = π π π ππ
ππ ππ
y dx xdx xdx2
0
22 2
0
2
0
222 2 4
12 2
= = = =z zz sin . sin ./
.Ans.(1)
43. S = 20
3
ππ
ydsdx
dxaz ;
18ay dydx
= 1.(3a – x)2 + x.2 (3a – x).3 (a – x)
dydx
= ( )( ) ( )3
6 69
12
a x a xay
a xa
ax
dsdx
dydx
− −=
−∴ = +
FHG
IKJ
= +−RS|T|
UV|W|=
+1
4 2
2( )
( )
a xax
a x
ax
∴ =− − +zS x
a x
a
a x
a
a x
a xdx
a
23
3
3
3 20
3
π . ( )( ) ( ) ( )
= + − = =zπ ππ
33 2
39 32 2 3 2
0
3
aa ax x dx
aa a
a
( ) . . Ans.(2)
44. f(x) = x4 – 5x3 + x2 – 3x + 4 ⇒ f(4) = (4)4 – 5 (4)3 + (4)2 – 3(4) + 4= 256 – 320 + 16 – 12 + 4 =– 56. f'(x) = 4x3 – 15 x2 + 2x – 3⇒ f'(4) = 256 – 240 8 – 3 = 21. f'' (x) = 12x2 – 30x + 2⇒ f' ' (4) = 74. f' ' '(x) = 24 x – 30 ⇒ f' ' ' (4) = 66. fiv(x) = 24
f x ff x f x f x f xiv
( ) ( )'( )( )
!''( )( )
!'''( )( )
!( )( )
!= +
−+
−+
−+
−4
4 41
4 42
4 43
4 44
2 3 4
= − + − + − + − +−
56 21 4742
4663
424( 4
42 3
4
( ) ( )!
( ))
!x x x
x
= (x – 4)4 + 11 (x – 4)3 + 37 (x – 4)2 + 21 (x – 4) – 56. Ans.(1)
45. We have dydx
yx
dyy
dxx
++−
= ⇒+
+−
=1 21 2
01 2 1 2
0coscos cos cos
Integrating, dy
y
dy
x
C
2 2 22 2cos sin+ =zz
⇒ − = ⇒ − =12
12 2
tan cot tan coty xC
y x C . Ans.(3)
(53) of (57)
46. The given equation is dydx
x yxy
=+2 2
2...(1) which is homogeneous.
Putting x = νx and dydx
xdydx
= + ⋅ν , the equation (1) becomes
d dxx
ννν
ν1
2
2+−
= ; Integrating we get 2
1
12
νν
ν−
=z zdx
dx
⇒ – log (1 – ν2) = log x + log c ⇒ − −FHG
IKJ = ⇒
−=log log1
2
2
2
2 2
y
xcx
x
x ycx
⇒ x = c(x2 – y2) ⇒ x – c(x2 – y2) = 0. Ans.(2)
47. We have dydx
a dy a dx= ⇒ =− −sin sin .1 1
Integrating we get y = sin–1 a.x + c .But the curve passes through (0, 1). ∴ c = 1.Hence the curve is y = sin–1 a.x + 1 ⇒ y – 1 = (sin–1 a)x
⇒ siny
xa
−FHG
IKJ =
1. Ans.(3)
48. The given equation is dydx
y x x+ =tan sec , which is linear..
I.F. = ex dxtanz ∴ e log sec x = sec x.
∴ Solution is, sec . sec sec tanx y x dx c y x x c= + ⇒ = +z 2 . Ans.(2)
49. The terms can be put in ascending order of magnitude as2.9, 5.2, 8.3, 11.1, 12.6, 13.3, 16.4
Here number of terms is odd and is equal to 7.
Therefore moedian = 7 1
24 11 1
+= =th term th term . . Ans.(2)
50. Let assumed mean A be 45. Then table for calculation of S.D. is :
x d = x – A = x – 45 d2
40 – 5 25
42 – 3 9
46 1 1
48 3 9
44 – 1 1
Total ∑d = – 5 ∑d2 = 45
Standard deviation
σ =∑
−∑F
HGIKJ = −
−FHG
IKJ = − = =
dn
dn
2 2 2455
55
9 1 2 2 2 8. . Ans.(3)
51. Here the greatest frequency 25 is in 10 – 15 which is therefore the modalclass. Mode is given by
Mode = l11 0
1 0 22+
−− −
×f f
f f fi ...(1)
Here l1 = 10, l2 = 15, f1 = 25, f0 = 16, f2 = 14 and i = 5Using in (1) we get
Mode = 1025 16
2 25 16 145 10
920
5 12 25+−
× − −× = + × = . . Ans.(2)
52. We have the following table after arranging the marks in ascending order.
Marks Frequency Cumulative frequency
9 4 4
20 6 10
25 16 26
40 8 34
50 7 41
80 2 43
Here N is odd and equal to 43.
Hence median =+F
HGIKJ
43 12 th item = value of the 22nd item.
For the table it is obvious that all times from 11 to 26 have their value 25.Hence 22nd item is in the range 11 – 26 and, therefore its value is 25. Ans.(3)
53. Let the events of passing the tests I, II and III be denoted by A, B and Crespectively. It is easy to see that the events A, B, C are indepedent events.Now according to the question,
P A B A C[( ) ( )]∩ ∪ ∩ =12
⇒ ∩ + ∩ − ∩ ∩ =P A B P A C P A B C( ) ( ) ( )12
⇒ + − =P A P B P A P C P A P B P C( ) . ( ) ( ) . ( ) ( ) ( ) ( )12
⇒ + ⋅ − ⋅ =pq p pq12
12
12
⇒ p(q + 1) = 1 ...(1)The values p = 1, q = 0 in (3) satisfy (1) and then the correct answer is (3).Ans.(3)
54. The required probability
=+ + + + +coefficient of x in x x x x x x10 2 3 4 5 6 4
46( )
=+ + + + +coefficient of x in x x x x x6 2 3 4 5 4
41
6( )
The numerator = coefficient of x6 in (1 + x + x2 + x3 + x4 + x5)4
= coefficient of x6 in (1 – x6)4 (1 – x)–4
= coefficient of x6 in (1 – 4x6 + ...) × (1 + 4x + 10x2 + ... + 84x6 + ...)= 84 – 4 = 80.
Therefore the required probability = 80
6
5814 = . Ans.(4)
55. With a pair of dice 5 and 7 can be thrown in 4 and 6 ways respectively. Hencenumber of ways of throwing neither 5 nor 7 = 36 – (4 + 6) = 26.The probability of throwing neither 5 nor 7 in a single throw with a pair of two
dice = 2636
1318
= and the probability of throwing a five = 4
3619
= .
Since five (different from 7) has been thrown already, 5 may be thrown atnext attempt, when attempt to throw 5 or 7 has failed at the first step and the
corresponding probabi l i ty = 1318
19
× or at the th i rd at tempt for which
probability = 1318
19
2FHG
IKJ ⋅
FHG
IKJ and so on. In this way the required probability
= + ⋅ +FHG
IKJ ⋅ + =
−=
19
1318
19
1318
19
19
11318
25
2
... . Ans.(3)
(54) of (57)
56. There are two cases such that one rupee coin is still in the first purse whichare :(1) One rupee coin is not in 9 coins drawn from the first purse.(2) One rupee coin is in the 9 coins drawn from the first purse.
Probability for (1) is 1 1
10109
199
199C
CC
× =
and probability for (2) is 9
81
110
9
188
11
199
81190
C CC
C CC
××
×= .
Therefore the required probability = 1
1081
1901019
+ = . Ans.(2)
57. We have x y x yi i x y= = ∑ = = =0, 0, 12, 2, 3σ σ and n = 10.
∴ = =∑ − −
rx y
x yn
x x y y
y
i i
x y
cov ( , )
var ( ) . var ( )
( ) ( )1
σ
=∑
1n
x yi i
x yσ σQ x y= =0, 0c h
=×
×=
110
12
2 30 2. . Ans.(3)
58. Let the number of pairs be n.Here, R = 0.8 and ∑D2 = 136.
∴ = −∑
−= −
×−
⇒ − =×
RD
n n n nn n1
61
0 8 16 136
11
6 1360 2
2
2 22
( ).
( )( )
.
⇒ (n – 1) n(n + 1) = 30 × 136⇒ (n – 1) n(n + 1) = 15 × 16 × 17 ⇒ n = 16. Ans.(3)
59. Let us assume that the line of regression of y on x be3x + 2y = 7 ...(i)
and the line of regression of x on y bex + 4y = 9 ...(ii)
∴ byx = slope of the line (i) = −32
and 1
b xy = slope of the line (ii) = − ⇒ = −
14
4b xy .
Now, b byx xy⋅ = −FHG
IKJ − = >
32
4 6 1( ) .
∴ Our assumption is wrong.Hence, the line of regression of y on x is x + 4y = 9 and the line of regressionof x on y is 3x + 2y = 7.∴ byx = slope of the line (ii)
= −14
and 1
b xy = slope of the line (i) = − ⇒ = −
32
23
b xy .
Hence, b and byx xy= − = −14
23
. Ans.(2)
60. We have x y rx y= = = = =10, 90, 3, 12, 0 8σ σ . .
Equation of regression line of y on x is
y yr
x y xy
x− = − ⇒ − = × −
σ
σ( ) . ( )10 90 0 8
123
10
⇒ y – 90 = 3.2(x – 10) ⇒ y = 3.2x + 58 ...(1)Equation of regression line of x on y is
x xr
y y x yx
y− = − ⇒ − = × −
σσ
( ) . ( )10 0 83
1290
⇒ x – 10 = 0.2(y – 90) ⇒ x = 0.2y – 8 ...(2)Hence, the equation of regression lines arey = 3.2x + 58 and x = 0.2y – 8. Ans.(4)
Trick : Work from option value of x y, satisfies the regression equations.
61. Here y = sin x, a = 0 and b = π.
∴ =−
=−
=hb a
nπ π0
4 4
∴ = = = = =x x x x x1 1 2 3 50,4 2
34
π π ππ, , , .
The table for the values of y corresponding to these values of x is
x
y x
04 2
34
01
21
1
20
π π ππ
= sin
By Simpson’s rule, we have
= + + + +z hy y y y y
34 20 4 1 3 2
0
[( ) ( ) ]π
= + + +FHG
IKJ + ×
LNMM
OQPP
π2
0 0 41
2
1
22 1( )
= + = +π π12
4 2 26
2 2 1[ ] ( ) . Ans.(3)
62. Here h = 10 and n = 6.
∴ The required area = h
y y y y y y y2
20 6 1 2 3 4 5[( ) ( )]+ + + + + +
= + + + + + +102
0 4 2 3 7 11 8 6[( ) ( )]
= 5(4 + 70) sq. metres.= 370 sq. metres. Ans.(1)
63. The equations corresponding to the given inequations
x – y ≥ 2, x + 2y ≤ 8 are x – y = 2 ⇒ +−
=x y2 2
1
and x + 2y = 8 ⇒ + =x y8 4
1
It is clear that (0, 0) does not lie in the graph of x – y > 2 but lies in the graphof x + 2y < 8.
Y
X1 2 3 4 5 6 7 8
4
32
1
0–1
Y’–2
A(2, 0) C(8, 0)
(4, 2)
x + 2y = 8x –
y = 2
B
Therefore, the feasible region is ABC whose vertices are A(2, 0), B(4, 2) andC(8, 0). Ans.(1)
64. Let the fruit seller buy x boxes of oranges and y boxes of apples. Then,x ≥ 0, y ≥ 0 ...(i)
Since the fruit seller cannot store more than 12 boxes, thereforex + y ≤ 12 ...(ii)
Taking into account the cost of articles and availability of money, we get50x + 25y ≤ 500
or 2x + y ≤ 20 ...(iii)The objective function P is given by
P = 10x + 6y ...(iv)We have to maximize (iv) subject to the constraints (i), (ii) and (iii).The shaded region in the following figure is the feasible region.The corners of the feasible region are the points O(0, 0), C(10, 0), E(8, 4)and A(0, 12).At the point O(0, 0),
P = 10 × 0 + 6 × 0 = 0At the point C(10, 0)
P = 10 × 10 + 6 × 0 = 100At the point E(8, 4),
P = 10 × 8 + 6 × 4 = 104
(55) of (57)
At the point (0, 12),P = 10 × 0 + 6 × 12 = 72.
Y
D(0, 20)20
15
(0,. 12)10
5
0 O
5 10 15 20 X
C B(12, 0)
E(8, 4)
2x + y = 20
Ax+y=12
Thus, the profit will be maximum when the fruit seller buys 8 boxes of orangesand 4 boxes of apples. Ans.(1)
65. Probability of winning of Shyam
= × + × × × +56
16
56
56
56
16
....
= + +FHG
IKJ +
LNMM
OQPP
536
12536
2536
2
. . . . . =536
1
12536
– = × =
536
3611
511
Mathematical expectation for Shyam = Rs.511
× 88 = Rs. 40. Ans.(2)
66. P(2) = 9P(4) + 90 P(6)
⇒ = +− − −e m e m e mm m m2 4 6
29
490
6! ! ! ⇒ = +1912
90360
2 4m m
or m4 + 3m2 – 4 = 0 m ⇒ m = ± 1. Ans.(1)
67. Ram’s probability of winning the game is 23
13
= ⇒ =p q
The p robab i l i t y t ha t he w i l l no t w in a t l eas t one game i s =
40
0 4
4
23
13
11
3
181
CFHG
IKJ
FHG
IKJ =
FHG
IKJ =.
The probability that he will win atleast one game is = =1181
8081
– . Ans.(2)
68. Let M, M0 denote the Mean and Mode of the distr ibution co-eff icient ofvariation = 5
⇒ × =σM
100 5 or M = 20σ = 20 × 2 = 40
Karl Pearson’s co-efficient of Skewness =M M– 0
σ
∴ 0.5 =40
20– M
or 1 = 40 – M0
∴ M0 = 39. Ans.(1)
69. y = x and x y= +14
34
. Hence r = ⋅FHG
IKJ =1
14
0 5. . Ans.(2)
70. Suppose a strianght line to be fitted to the given data is as followsy = a + bx ...(1),then the normal equations are as :∑y = na + b∑x ...(2)and ∑xy = a∑x + b∑x2 ...(3)Now, from the given data, we have
x y xy x2
0 1 0 0
1 1.8 1.8 1
2 3.3 6.6 4
3 4.5 13.5 9
4 6.3 25.2 16
Total ∑x = 10 ∑x = 16.9 ∑xy = 47.1 ∑x2 = 30
Here n = 5. Now substituting these values in the normal equations, we get16.9 = 5a + 10b ...(4)47.1 = 10a + 30b ...(5).Solving (4) and (5), we have a = 0.72 and b = 1.33.Thus, the required equation of the straight line isy = 0.72 + 1.33x [Putting a and b in (1)]. Ans.(2)
71. G.E. = + + ×2 5 6 3 5 9 5 2 9 5– – – –
= + +2 5 6 3 5 3 5– – –e j = +2 5 9 4 5– –
= + + ×2 5 5 4 2 5 4– – = + = =2 5 5 2 4 2– – .e j Ans.(3)
72. By truth table :
p q – q q v ~ q p ⇒ (q v ~ q)
F F T T T
Ans.(1)73. Given n(U) = 700
n(A) = 200, n(B) = 300, n(A ∩ B) = 100n(A ∪ B) = n(A) + n(B) – n(A ∩ B)
= 200 + 300 – 100 = 400.Now n(A’ ∩ B’) = n(A ∪ B)’ = n(U) – n(A ∪ B) = 700 – 400 = 300. Ans.(3)
74. We have (kA)(adj kA) = |kA| = In⇒ k(A adj kA) = k–n |A| In, (Q |kA| = kn |A|)⇒ A(adj kA) = kn–1 |A| In⇒ A adj (kA) = kn–1 A (adj A) (Q A adj A = |A| In)⇒ adj (kA) = kn–1 (adj A). Ans.(4)
75. We have 1
11
1
1−L
NMOQP −LNM
OQP =
−LNM
OQP
−tan
tantan
tanθθ
θa bb a
⇒−L
NMOQP ⋅
+
−LNM
OQP =
−LNM
OQP
11
11
112
tantan ( tan )
tantan
θθ θ
θθ
a bb a
⇒+
− −−
LNMM
OQPP =
−LNM
OQP
1
11 2
2 12
2
2tantan tantan tanθ
θ θθ θ
a bb a
⇒
−+
−+
+−+
L
N
MMMMM
O
Q
PPPPP=
−LNM
OQP ⇒
−LNM
OQP =
−LNM
OQP
1
1
2
12
1
1
1
2 2
2 2
2
2 2
2
2
2
tan
tan
tan
tantan
tan
tan
tan
cos sin
sin cos
θθ
θθ
θθ
θθ
θ θθ θ
a b
b a
a b
b a
⇒ a = cos 2θ, b = sin 2θ. Ans.(2)76. The system of equations has infinitely many (non-trivial) solutions, if
∆ = 0. i.e., if
3 2 114 15
1 2 30
−−
−=λ
⇒ 3(42 – 30) – λ(6 – 2) + 1(– 30 + 14) = 0 ⇒ λ = 5. Ans.(4)
(56) of (57)
77. ∆ =pa qb rcqc ra pbrb pc qa
= pa(qra2 – p2bc) – qb(q2ac – b2pr) + rc(pqc2 – r2ab)= pqr(a3 + b3 + c3) – abc(p3 + q3 + r3) ...(i)
Q a + b + c = 0 and p + q + r = 0∴ a3 + b3 + c3 = 3abc and p3 + q3 + r3 = 3pqr∴ from (1), we have
= pqr.3abc – abc.3pqr.∆ = 0. Ans.(1)
78. Word ‘AIMCET’ has three consonants and three vowels.Three consonants can be arranged in 3! ways.Now we have four places; two between consonants and two are the outerplaces. If we put vowels at these places then two vowels do nt come together.At these four places three vowels can be arranged in 4P3 ways.Hence required number of ways = 3! × 4P3 = 6 × 12 × 2 = 144. Ans.(2)
79. If there are n things of different colours and m boxes of same different coloursthen arrangements when one thing is placed in one box but not of the samecolours
= − + − + + −LNMM
OQPPn
nn1
11
12
13
11
... ( )
Hence in the above sum, the number of arrangements
= − +LNMM
OQPP4
12
13
14 = 12 – 4 + 1 = 9. Ans.(3)
80. T C xx
C xr rr
r
r
r
rr r
+−
−− −= ⋅
FHG
IKJ =1
9 92
99
9 221
3
2
3( ) = ⋅ ⋅
−−9
99 32
3C xr
r
rr
Let (r + 1)th term be independent of x.∴ 9 – 3r = 0 ⇒ r = 3
Hence, independent term is = =93
6
323
17929
C . Ans.(2)
81. (1 + x + x2 + x3)5 = [(1 + x) (1 + x2)]5 = (1 + x)5 . (1 + x2)5
The sum fo coefficients of even powers= 24 . 25 = 24 + 5 = 29
(Q every term of expansion (1 + x2)5 contians even powers). Ans.(4)82. We have Tr + 1 = 50Cr (2x)r 350 – r
∴TT
rr
xr
r
+ =FHG
IKJ1 51 2
3–
Where xTT
rr
r
r= =
FHG
IKJ =+1
551 2
15102 2r
15r1,
– –
Now,TT
rr
r
+ > ⇒ > ⇒ <1 1102 2r
15r1 6.
–
∴ 6th and 7th term are numerically the greatest term. Ans.(3)
83. We have 5 cos θ + 3 cos cos – sin sinθπ
θπ
3 23
LNM
OQP+
= 5 cos θ + 3 .12
3 3[cos – sin ]θ θ + = +132
3 32
3.cos – sinθ θ
The maximum value of given expression
= + +12
169 27 3 = × + = + =12
14 3 7 3 10. Ans.(2)
84. We havesin x + sin y = 3(cos y – cos x)
⇒ sin x + 3 cos x = 3 cos y – sin y .....(1)⇒ r cos (x – α) = r cos (y + α),
where r = =1013
, tan α
⇒ x – χ = ± (y + α) ⇒ x = – y or x + y = 2αClearly, x = –y satisfies (1)
∴sinsin
– sinsin
–133
33
xy
yy
= = . Ans.(3)
85. sin 2θ + sin 2φ =12
⇒ sin2 2θ + sin2 2φ + 2sin 2θ sin 2φ =14
.....(i)
Again, cos 2θ + cos 2φ =32
⇒ cos2 2θ + cos2 2φ + 2cos 2θ cos 2φ =94
.....(ii)
Adding (i) and (ii), we get
1 + 1 + 2 cos (2θ – 2φ) =104
⇒ 2 cos (2θ – 2φ) =52
2–
⇒ cos 2(θ – φ) =14
⇒ 2cos2 (θ – φ)–1 =14
⇒ cos2 (θ – φ) =58
. Ans.(2)
86. We havesin–1 x + sin–1 (1 – x) = cos–1 x
⇒ sin {sin–1 x + sin–1 (1 – x)} = sin (cos–1 x)
⇒ x x x x x1 1 1 1 12 2 2– ( – ) – ( – ) –+ = ⇒ x x x x1 1 12 2– ( – ) –=
Squaring x2 = 0 or 2x – x2 = 1 – x2 ⇒ x = 0, x = 1/2. Ans.(2)
87. We know that | sin–1 x| ≤π2
. Therefore
sin–1 x + sin–1 y + sin–1 z =32π
⇒ sin–1 x = sin–1 y = sin–1 z = π/2⇒ x = y = z = sin π/2 = 1
∴ x y zx y z
100 100 100101 101 101
93
93
0+ ++ +
= =– – . Ans.(1)
88. Let the two roads intersect at A. If the bus and the car are at B and C on thetwo roads respectively, then c = AB = 2km, b = AC = 3 km. The distancebetween the two vehicles = BC = a km.Now, A = 60°⇒ cos A = cos 60°
⇒b c a
bc
2 2 2
212
+=
– ⇒ a km= 7 . Ans.(4)
89. We have
rr
rr1
2
3= ⇒ r . r3 = r1r2 ⇒
∆ ∆ ∆ ∆s s c s a s b
.– –
.–
=
⇒( – ) ( – )
( – )s a s b
s s c= 1 ⇒ tan2
C2
1= ⇒ tanC2
1= ⇒ C2
45= °
⇒ C = 90°. Ans.(3)90. In ∆PQR, the radius of the circumcircle is given by
PQR
QRP
PRQ2 2 2sin sin sin
.= = But it is given that the radius is PQ = PR.
∴ PQ PRPQ
RQR
RPQ
Q= = = =
2 2 2sin sin sin⇒ sin R = sin Q =
12
⇒ ∠ = ∠ =R Qπ6
⇒ ∠ = ∠ ∠ =P R Qππ
– – .23
Ans.(4)
91. The nth term of given series is
Tn n
nn =+( )!
2 =+
=+n
nn
n21
1 31( – ) !
–( – ) !
⇒ Tn nn = =
12
31( – ) ! ( – ) !
∴ The sum of the series,
= e + 3e = 4e. Ans.(4)
(57) of (57)
92. loge (1 – x + x2) = log (1 + x3) – log (1 + x)
– – – – – – . . . . .xx x x x x x x x2 3 4 5 6 7 8 9
2 3 4 5 6 7 8 9+ + + +
LNMM
OQPP
= + +FHG
IKJ +
FHG
IKJ– – – – –x
xx
x xx
23
4 56
21
13 4 5
12
16
– – . . . . .x x
x7 8
9
7 813
19
+ +FHG
IKJ ∞
∴ The value of a3 + a6 + a9 + ..... ∞
=FHG
IKJ
FHG
IKJ +
FHG
IKJ ∞1
13
12
16
13
19
– – – – . . . . .
= + + ∞FHG
IKJ + ∞
FHG
IKJ1
12
13
14
13
112
13
– – . . . . . – . – – . . . . .
=FHG
IKJ + + ∞
FHG
IKJ1
13
112
13
14
– . – – . . . . .
=23
2.loge Ans.(3)
93. S = × +FHG
IKJ
LNM
OQP =
92
212
9 116
32
( – ) – – . Ans.(4)
94. Given a1, a2, a3, ....., an are in A.P. with common difference d.⇒ d = a2 – a1 = a3 – a2 = a4 – a3 = ..... = an – an – 1
Now, sin d. cosec a1 . cosec a2 =sin ( – )sin sin
a aa a
2 1
1 2
=sin cos – cos sin
sin sina a a a
a a2 1 2 1
1 2
= cot a1 – cot a2
Similarly, sin
sin sind
a a2 3= cot a2 – cot a3 .....
and sinsin sin–
da an n1
= cot an – 1 – cot an
Hence, sin d[cosec a1 cosec a2 + cosece a2 cosec a3 + ..... + cosec an–1 .cosec an]= cot a1 – cot a2 + cot a2 – cot a3 + cot a3 – cot a4 + ..... + cot an–1 – cot an= cot a1 – cot an. Ans.(4)
95. Let a and d be the first term and common difference of A.P. ThenTm + 1 = a + mdTn + 1 = a + ndTr + 1 = a + rd
Now, Tm + 1, Tn + 1 and T r + 1 are in G.P.⇒ (a + nd)2 = (a + md) (a + rd)⇒ a(2n – m – r) = d (mr – n2)
or .....(1)
Also, m,n,r in H.P. ⇒ .....(2)
From (1) and (2)
da
m rmr n n
m rm r n
=+
=+
+FHG
IKJ =
2n 2 2n2n
22
– ( )–
– ( )( ) –
– . Ans.(1)
96. Given a = ar + ar2 ⇒ r2 + r – 1 = 0 ⇒ r =± +–1 1 4
2
Since r is not negative r =+
=–1 5
2sin 18°. Ans.(2)
97. We have f(x) = log (x2 – x – 2) which is defined forx2 – x – 2 > 0 ⇒ (x – 2) (x + 1) > 0 ⇒ x < – 1 and x > 2
Now when x < 0 then the equation9x2 – 18 |x| + 5 = 0 becomes 9x2 + 18x + 5 = 0⇒ 9x2 + 15x + 3x + 5 = 0 ⇒ (3x + 1) (3x + 5) = 0
⇒ x = – , –13
53
Again when x > 0 then the equation 9x2 – 18 |x| + 5 = 0 becomes9x2 – 18x + 5 = 0
⇒ (3x – 1) (3x – 5) = 0 ⇒ x =13
53
,
Hence root of equation lying in the domain of f(x) is –53
. Ans.(1)
98. Let the roots of 8x2 – 6x – (a + 3) = 0 be α and α2. Then
α α+ = =2 68
34
⇒ 4αx2 + 4α – 3 = 0 ⇒ 4α2 + 6α – 2α – 3 = 0
⇒ (2α + 3) (2α – 1) = 0 ⇒ α =12
32
and – and
When α =+
=12
38
18
thena
– ⇒ a = – 4
When x thena
=+
=–32
38
278
⇒ a = 24
Hence the values of a are –4, 24. Ans.(4)
99. Given a + b + c = 0 ⇒ a3 + b3 + c3 = 3abc
x x xa b c a b c a b c2 2 2–1 –1 –1 –1 –1 –1. .
= x x xabc
bac
cab
2 2 2. . .
=+ +
xabc
bac
cab
2 2 2
= = =+ +
x x xa b c
abcabc
abc
3 3 3 33 . Ans.(3)
100. A = log2 log2 log4 44 + 2 22
log = log2 log2 4 + 2 ⋅ 2log22
= log2 log2 22 + 4 = log 22 + 4 = 5. Ans.(3)
