2.QUADRATIC EQUATIONS AND EXPRESSIONS_Bit_Bank

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QUADRATIC EQUATIONS
OBJECTIVE PROBLEMS
1.
(a) 2, –1
(c)
2.
3.
4.
5.
6.
7.
x+
The solution of the equation
− 1,−
(b) 0, –1,
1
5
−
1
=2
x
will be
1
5
(d) None of these
The roots of the given equation ( p − q ) x 2 + (q − r ) x + (r − p) = 0 are
(a)
p−q
,1
r−p
(c)
r−p
,1
p−q
If
(b)
q −r
,1
p−q
(d)
x 2 + y 2 = 25 , xy = 12
, then
q −r
p −q
1,
x=
(a) {3, 4}
(b) {3, –3}
(c) {3, 4, –3, –4}
(d) {–3, –3}
a(x 2 + 1) − (a 2 + 1)x = 0
The roots of the equation
(a)
a,
1
a
(b) a, 2a
(c)
a,
1
2a
(d) None of these
The value of 2 +
1
1
2+
2 + ...........∞
are
is
(a)
1− 2
(b) 1 +
(c)
1± 2
(d) None of these
2
The number of real solutions of the equation |
(a) 1
(b) 2
(c) 3
(d) 4
The roots of the equation
x4 − 8x 2 − 9 = 0
(a)
±3, ± 1
(b)
(c)
±2, ± i
(d) None of these
x | 2 – 3 | x | +2 = 0
are
±3, ± i
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are
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8.
Let one root of
3− 5
(a)
(c)
9.
where
ax 2 + bx + c = 0
a, b , c
are integers be
3+ 5
, then the other root is
(b) 3
(d) None of these
5
3x +1 +1 = x
The roots of the equation
(a) 0
(b) 1
(c) 0, 1
(d) None
are
10. The value of x = 2 + 2 + 2 + ..... is
(a) –1
(b) 1
(c) 2
(d) 3
11. If
P(x ) = ax 2 + bx + c
and
Q(x ) = −ax 2 + dx + c
ac ≠ 0 ,
where
(a) Four real roots
(b) Two real roots
(c) Four imaginary roots
(d)None of these
12. The real roots of the equation
(b) 1, 4
(c) – 4, 4
(d) None of these
13. If the roots of the equation
P ( x ).Q ( x ) = 0 has
at least
are
x 2 + 5| x| + 4 = 0
(a) – 1, 4
then
( p 2 + q 2 )x 2 − 2 q ( p + r ) x
+
(q 2 + r 2 ) = 0
be real and equal, then
p, q, r will
be
in
(a) A.P.
(b) G.P.
(c) H.P.
(d) None of these
α
14. Let
and
β
be the roots of the equation
(a)
x2 − x −1 = 0
(b)
(c)
x2 + x −1 = 0
(d) x 2 + x + 1 = 0
x2 + x +1 = 0
The equation whose roots are
2x 2 + 6 x + α2 + 1 = 0
be
(a) –1
(b) 1
(c) 2
(d) –2
16. If x 2/3 − 7 x1/3 + 10 = 0, then
(c)
φ
is
x2 − x +1 = 0
15. If the product of the roots of the equation
(a) {125}
α 19 , β 7
x=
(b) {8}
(d) {125, 8}
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is
−α
, then the value of
α
will
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17. The number of roots of the equation | x |2 −7 | x | +12 = 0 is
(a) 1
(b) 2
(c) 3
(d) 4
18. The equation
(x + 1) − (x − 1) = (4 x − 1)
(a) No solution
has
(b) One solution (c)Two solutions
(d) More than two solutions
19. The number of solutions of log 4 ( x − 1) = log 2 ( x − 3)
(a) 3
(b) 1
(c) 2
(d) 0
20. If the roots of the given equation 2 x 2 + 3(λ − 2) x + λ + 4 = 0 be equal in magnitude but
opposite in sign, then λ =
(a) 1
(b) 2
(c) 3
(d) 2/3
21. If a root of the equation x 2 + px + 12 = 0 is 4, while the roots of the equation
x 2 + px + q = 0
same, then the value of q will be
(a) 4
(b) 4/49
(c) 49/4
(d) None of these
22. The equation e x − x − 1 = 0 has
(a) Only one real root
x =0
(b) At least two real roots
(c) Exactly two real roots
(d) Infinitely many real roots
23. The number of solutions for the equation x 2 − 5 | x | + 6 = 0 is
(a) 4
(b) 3
(c) 2
(d) 1
24. The values of 'a' and 'b' for which equation x 4 − 4 x3 + ax 2 + bx + 1 = 0 have four real roots
(a) – 6, – 4
(b) – 6, 5
(c) – 6, 4
(d) 6, – 4
25. If a + b + c = 0, a ≠ 0, a, b, c ∈ Q , then both the roots of the equation ax 2 + bx + c = 0 are
(a) Rational
(b) Non-real
(c) Irrational
(d) Zero
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are
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26. If
a+b +c = 0
, then the roots of the equation
(a) Equal
(b) Imaginary
(c) Real
(d) None of these
27. If the roots of the given equation
(a)
p ∈ (−π ,0 )
(b)
4 ax 2 + 3 bx + 2 c = 0
(cos p − 1)x 2 + (cos p )x + sin p = 0
 π π
p ∈ − , 
 2 2
(c) p ∈ (0, π )
(d)
are
are real, then
p ∈ (0 , 2π )
28. The roots of the equation
(a 2 + b 2 )t 2 − 2(ac + bd )t + (c 2 + d 2 ) = 0 are equal, then
(a)
ab = dc
(b) ac = bd
(c)
ad + bc = 0
(d)
29. The expression
a c
=
b d
x 2 + 2 bx + c
has the positive value if
(a)
b 2 − 4c > 0
(b) b 2 − 4 c < 0
(c)
c2 < b
(d) b 2 < c
30. If the roots of the equations
px 2 + 2 qx + r = 0
and qx 2 − 2 pr x + q = 0 be real, then
(a)
p =q
(b) q 2 = pr
(c)
p 2 = qr
(d) r 2 = pq
31. If
l, m , n
are real and
l≠m,
then the roots of the equation
(a) Complex
(b) Real and distinct
(c) Real and equal
(d) None of these
(l − m )x 2 − 5(l + m )x − 2(l − m ) = 0
32. The least integer k which makes the roots of the equation
(a) 4
(b) 5
(c) 6
(d) 7
33. The condition for the roots of the equation,
(c 2 − ab)x 2 − 2(a 2 − bc )x + (b 2 − ac) = 0
to be equal is
(a)
a=0
(b) b = 0
(c)
c=0
(d) None of these
34. Roots of
ax 2 + b = 0
are real and distinct if
(a)
ab > 0
(b) ab < 0
(c)
a, b > 0
(d) a, b < 0
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x 2 + 5x + k = 0
are
imaginary is
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35. The expression
36.
y = ax 2 + bx + c
has always the same sign as c if
(a)
4 ac < b 2
(b)
(c)
ac < b 2
(d) ac > b 2
x 2 + x + 1 + 2 k ( x 2 − x − 1) = 0 is
4 ac > b 2
a perfect square for how many values of k
(a) 2
(b) 0
(c) 1
(d) 3
p, q ∈ {1, 2, 3, 4 } .
37. Let
The number of equations of the form
(a) 15
(b) 9
(c) 7
(d) 8
38. If the roots of equation
x 2 + a 2 = 8 x + 6a
a ∈ [2, 8 ]
(b) a ∈ [−2, 8]
(c) a ∈ (2, 8 )
(d) a ∈ (−2, 8)
(a)
39. If a root of the equation
px 2 + qx + 1 = 0
having real roots is
are real, then
ax 2 + bx + c = 0
be reciprocal of a root of the equation
then a ′x 2 + b ′x + c ′ = 0 , then
(a)
(cc ′ − aa ′) 2 = (b a ′ − cb ′)(ab ′ − b c ′)
(b) (b b ′ − aa′)2 = (ca′ − b c ′)(ab ′ − b c ′)
(c)
(cc ′ − aa′)2 = (b a ′ + cb ′)(ab ′ + b c ′)
(d) None of these
40. If α and
β
are the roots of the equation
(a)
−
3
7
(b)
3
7
(c)
−
3
5
(d)
3
5
41. If the roots of the equation
Ax 2 + Bx + C = 0
are
α 2, β 2 ,
(a)
B 2 − 2 AC
A2
(b)
(c)
B 2 − 4 AC
A2
(d) None of these
42. If α and
(a) b
(c)
1−b
β
4 x 2 + 3x + 7 = 0
are
α, β
, then
1
α
+
1
β
=
and the roots of the equation
then value of p will be
2 AC − B 2
A2
are the roots of the equation
x 2 − a( x + 1) − b = 0
then
(b) – b
(d) b − 1
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(α + 1)(β + 1) =
x 2 + px + q = 0
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43. If the sum of the roots of the quadratic equation ax 2 + bx + c = 0 is equal to the sum of the
squares of their reciprocals, then a / c, b / a, c / b are in
(a) A.P.
(b) G.P.
44. If the roots of the equation
(a) 3
(c) H.P.
(d) None of these
are same, then the value of m will be
x 2 + 2 mx + m 2 − 2 m + 6 = 0
(b) 0
(c) 2
45. If the sum of the roots of the equation
(d)–1
be equal to the sum of their squares,
ax 2 + bx + c = 0
then
(a)
a(a + b ) = 2 bc
(b) c(a + c) = 2 ab
(c)
b (a + b ) = 2 ac
(d) b(a + b ) = ac
46. If
α, β
(a)
are the roots of
x 2 + px + 1 = 0
and
γ ,δ
are the roots of
x 2 + qx + 1 = 0 ,then q 2 − p 2 =
(α − γ )( β − γ )(α + δ )( β + δ )
(b) (α + γ )(β + γ )(α − δ )(β + δ )
(c)
(α + γ )( β + γ )(α + δ )( β + δ )
(d) None of these
47. If
(a)
2+i 3
is a root of the equation
48. If the roots of the equation
(c)
( p, q) =
(d) (−4, − 7)
(c) (4, 7)
(a)
where p and q are real, then
(b) (4, − 7)
(−4 , 7 )
cx 2 + bx + a = 0
x 2 + px + q = 0 ,
ax 2 + bx + c = 0
be
α
and
β
, then the roots of the equation
are
−α , − β
(b) α, 1
1 1
,
(d) None of these
α β
β
49. The quadratic in b, such that A.M. of its roots is
(a) t 2 − 2 At + G 2 = 0
(b) t 2 − 2 At − G 2 = 0
(c) t 2 + 2 At + G 2 = 0
(d) None of these
50. If the sum of the roots of the equation
A
and G.M. is G, is
x 2 + px + q = 0
is three times their difference, then
which one of the following is true
(a)
9 p 2 = 2q
(b)
2q 2 = 9 p
(c)
2 p 2 = 9q
(d) 9 q 2
= 2p
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51. A two digit number is four times the sum and three times the product of its digits. The
number is
(a) 42
(b) 24
(c) 12
(d) 21
52. If the product of roots of the equation, mx 2 + 6 x + (2m − 1) = 0 is –1, then the value of m will
be
(a) 1
(b) – 1
(c)
1
1
1
+
=
x + p x +q r
53. If the roots of the equation
1
3
(d)
−
1
3
are equal in magnitude but opposite in sign, then
the product of the roots will be
(a)
p2 + q2
2
(c)
p2 − q2
2
54. If
α, β
(b) – (p
2
(d) – (p
+ q2)
2
2
− q2)
2
are the roots of the equation
(a)
2
a
(b)
2
b
(c)
2
c
(d)
−
ax 2 + bx + c = 0
, then
α
aβ + b
+
β
aα + b
=
2
a
55. The equation formed by decreasing each root of
(a) a = – b
(b) b = – c
(c) c = – a
(d) b = a + c
56. If the ratio of the roots of
x 2 + bx + c = 0
(a)
r 2c = b 2q
(b) r 2 b = c 2 q
(c)
rb 2 = cq 2
(d) rc 2 = bq 2
57. If the ratio of the roots of
and
ax 2 + 2 bx + c = 0
ax 2 + bx + c = 0
x 2 + qx + r = 0
is same as the ratio of the roots of
b2
q2
=
ac
pr
b
q
=
ac pr
(b)
(c)
2b q 2
=
ac
pr
(d) None of these
2 x 2 + 8 x + 2 = 0,
then
be the same, then
then
(a)
by 1 is
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px 2 + 2 qx + r = 0 ,
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58. If the sum of the roots of the equation
x 2 + px + q = 0
is equal to the sum of their squares,
then
(a)
p2 − q2 = 0
(b)
(c)
p 2 + p = 2q
(d) None of these
59. Let
α, β
p 2 + q 2 = 2q
G.P., then integral values of
(a) – 2, – 32
60. If the roots of
and γ ,δ be the roots of
x2 − x + p = 0
be the roots of
p, q
are
. If
α, β ,γ ,δ
are in
are respectively
(b) – 2, 3
ax 2 + bx + c = 0
x2 −4x +q = 0
(c)– 6, 3
α, β
(d)– 6, – 32
and the roots of
Ax 2 + Bx + C = 0
are
α − k , β − k , then
B 2 − 4 AC
b 2 − 4 ac
is equal to
(a) 0
(c)
(b) 1
A
 
a
2
(d)
61. If α, β are the roots of
a
 
A
2
9 x 2 + 6 x + 1 = 0,
(a)
2 x 2 + 3 x + 18 = 0
(b)
x 2 + 6x − 9 = 0
(c)
x 2 + 6x + 9 = 0
(d)
x 2 − 6x + 9 = 0
62. If p and q are the roots of
(a)
p = 1, q = −2
(c)
p = 1, q = 0
63. If
α, β
(b)
x 2 + px + q = 0,
1 1
,
α β
then
p = −2, q = 0
ax 2 + bx + c = 0
and
α + β,
α 2 + β 2, α 3 + β 3
are in G.P., where
then
(a)
∆≠0
(b) b∆ = 0
(c)
cb ≠ 0
(d) c∆ = 0
64. If
1−i
is a root of the equation
(a) – 2
(b) – 1
(c) 1
(d) 2
x 2 − ax + b = 0
65. If α, β are the roots of the equation
1
2
(b)
1
2
(c) 32
(d)
1
4
(a)
−
is
p = − 2, q = 1
(d)
are the roots of
then the equation with the roots
, then
x 2 + 2 x + 4 = 0,
b=
then
1
α3
+
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1
β3
is equal to
∆ = b 2 − 4 ac ,
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66. If a and b are roots of
x 2 − px + q = 0 ,
(a)
1
p
(b)
1
q
(c)
1
2p
(d)
p
q
then
67. Product of real roots of the equation
(a) Is always positive
α, β
(a)
t 2 x 2 +| x | + 9 = 0
(b)Is always negative
(c) Does not exist
68. If
1 1
+ =
a b
(d) None of these
are the roots of
ax 2 + bx + c = 0
, then the equation whose roots are
2 + α, 2 + β
is
ax 2 + x (4 a − b ) + 4 a − 2b + c = 0
(b) ax 2 + x (4 a − b) + 4 a + 2b + c = 0
(c)
ax 2 + x (b − 4 a) + 4 a + 2b + c = 0
(d) ax 2 + x (b − 4 a) + 4 a − 2b + c = 0
69. If one root of the equation
x 2 + px + q = 0 is
the square of the other, then
(a)
p 3 + q 2 − q(3 p + 1) = 0
(b) p 3 + q 2 + q(1 + 3 p) = 0
(c)
p 3 + q 2 + q(3 p − 1) = 0
(d) p 3 + q 2 + q(1 − 3 p) = 0
70. Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are
the roots of the quadratic equation
(a)
x 2 − 18 x − 16 = 0
(b)
(c)
x 2 + 18 x − 16 = 0
(d) x 2 + 18 x + 16 = 0
71. If
α≠β
but
α 2 = 5α − 3
x 2 − 18 x + 16 = 0
and
β 2 = 5β − 3 ,
then the equation whose roots are
(a)
3 x 2 − 25 x + 3 = 0
(b)
x 2 + 5x − 3 = 0
(c)
x 2 − 5x + 3 = 0
(d)
3 x 2 − 19 x + 3 = 0
72. Difference between the corresponding roots of
a≠b,
x 2 + ax + b = 0
then
(a)
a+b +4 = 0
(b) a + b − 4 = 0
(c)
a −b −4 = 0
(d) a − b + 4 = 0
73. If 3 is a root of
x 2 + kx − 24 = 0,
it is also a root of
(a)
x 2 + 5x + k = 0
(b)
x 2 − 5x + k = 0
(c)
x 2 − kx + 6 = 0
(d)
x 2 + kx + 24 = 0
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and
α/β
and
x 2 + bx + a = 0
β /α
is
is same and
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74. If
x , y, z
are real and distinct, then
u = x 2 + 4 y 2 + 9 z 2 − 6 yz − 3 zx − zxy
is always
(a) Non-negative
(b) Non-positive
(c) Zero
(d) None of these
75. If a root of the equations x 2 + px + q = 0 and
(where
and
q≠β
q−β
α−p
(a)
76. If
p ≠α
p β − αq
q−β
(a) (3, 4)
(b) (4, 5)
(c) (4, 3)
(d) (5, 4)
77. If the two equations
x 2 − cx + d = 0
(b) a + c
(c)
(d)
and
x +2
2x 2 + 3x + 6
 1 1
 , 
 13 3 
(b)
(c)
 1 1 
− , 
 3 13 
(d) None of these
is real, the function
(x − a)(x − b)
(x − c)
a>b >c
(b)
a<b <c
(c)
a>c<b
(d)
a<c<b
equation
x 2 − ax + b = 0
have one common root and the second
will assume all real values, provided
a n x n + a n −1 x n −1 + .... + a1 x = 0 , a1 ≠ 0 ,
na n x n −1 + (n − 1)a n −1 x n − 2 + .... + a1 = 0
n ≥ 2,
has a positive root
(b)Equal to
α
α
(d) Smaller than
81. If α and β (α < β) are the roots of the equation
(a)
0 <α < β
(b) α < 0 < β <| α |
(c)
α <β <0
(d) α < 0 <| α | < β
x =α,
has a positive root, which is
(a) Greater than or equal to α
(c) Greater than
(d) None of these
 1 1
− 13 , 3 


(a)
80. If the equation
p β − αq
q−β
takes all value in the interval
(a)
x
or
−ac
78. If x is real, the expression
79. If
q−β
α−p
2(b + d ) =
(a) 0
ac
(c)
x 4 − px 2 + q, then ( p , q ) =
be a factor of
has equal roots, then
is common, then its value will be
)
(b)
x 2 − 3x + 2
x 2 + αx + β = 0
x 2 + bx + c = 0,
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α
where
c < 0 < b,
then
then the
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82. If x is real, then the maximum and minimum values of expression
(a) 4, – 5
(b) 5, – 4
(c) – 4, 5
(d) – 4, – 5
83. If
x 2 − hx − 21 = 0, x 2 − 3 hx + 35 = 0 (h > 0 )
(a) 1
(b) 2
(c) 3
(d) 4
84. If the roots of the equation
(a)
85. If
a<2
b >a,
(b)
then the equation
(a) Both roots in
[a, b]
(b) Both roots in
(−∞, a)
(c) Both roots in
(b, + ∞)
(d) One root in
86. If S is a set of
(−∞, a)
P (x )
has a common root, then the value of
x 2 − 2 ax + a 2 + a − 3 = 0
2≤a≤3
(c)
(x − a)(x − b) = 1
and the other in
3<a≤4
S =0
(b)
S = ax + (1 − a)x 2 ∀a ∈ (0, ∞)
(c)
S = ax + (1 − a)x 2 ∀a ∈ R
h
is equal to
are real and less than 3, then
(d)
a>4
(b, + ∞)
≤2
such that
P (0 ) = 0 , P (1) = 1 , P' (x ) > 0 ∀x ∈ (0, 1) ,
(d) S = ax + (1 − a)x 2 ∀a ∈ (0, 2)
87. The smallest value of
x 2 − 3x + 3
(a) 3/4
(b) 5
(c) –15
(d) –20
in the interval
(−3, 3 / 2)
is
88. The maximum possible number of real roots of equation
(a) 0
(b) 3
(c) 4
(d) 5
89. The solution set of the equation
pqx 2 − ( p + q )2 x + ( p + q )2 = 0
(a)
p q
 , 
q p 
(b)

p
 pq, 
q


(c)
q

 , pq 
p


(d)
p + q p + q
,


q 
 p
will be
has
is polynomial of degree
(a)
x 2 + 14 x + 9
x 2 + 2x + 3
(e)
x5 − 6x 2 − 4x + 5 = 0
is
p − q p − q
,


q 
 p
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is
then
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90. If x is real and satisfies
x+2>
x + 4,
(a)
x < −2
(b)
x >0
(c)
−3 < x < 0
(d)
−3 < x < 4
then
91. If α, β and γ are the roots of equation
x 3 − 3x 2 + x + 5 = 0
then
y = ∑ α 2 + αβγ
satisfies the
equation
(a)
y3 + y + 2 = 0
(b)
y3 − y2 − y − 2 = 0
(c)
y 3 + 3y 2 − y − 3 = 0
(d)
y 3 + 4 y 2 + 5 y + 20 = 0
92. If α, β and γ are the roots of
x3 +8 = 0 ,
(a)
x3 −8 = 0
(b)
x 3 − 16 = 0
(c)
x 3 + 64 = 0
(d)
x 3 − 64 = 0 .
93. If
x 2 + 2 ax + 10 − 3 a > 0
for all
x ∈R ,
(a)
−5 < a < 2
(b) a < −5
(c)
a>5
(d)
94. If
α, β , γ
then the equation whose roots are
then
2<a<5
are the roots of the equation
(a) 2
(b) 3
(c) 4
(d) 5
x 3 + 4 x + 1 = 0,
then (α + β )−1 + (β + γ )−1 + (γ + α )−1 =
QUADRATIC EQUATIONS
HINTS AND SOLUTIONS
1.
2.
(d)
x+
1
1
= 2 ⇒ x + − 2 = 0 (∵ x ≠ 0 )
x
x
⇒
x 2 − 2x + 1 = 0
⇒ (x − 1) 2
(c) Given equation is
x =
⇒
x =
α 2,β 2
=0
⇒
x = 1,1 .
( p − q )x 2 + (q − r)x + (r − p ) = 0
(r − q ) ± (q − r) 2 − 4 (r − p )(p − q)
2( p − q )
(r − q) ± (q + r − 2 p)
r−p
⇒x =
,1
2( p − q)
p−q
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and
γ 2 is
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3.
4.
(c)
x 2 + y 2 = 25
xy = 12
and
2
⇒
 12 
x 2 +   = 25 ⇒ x 4 + 144 − 25 x 2 = 0
 x 
⇒
(x 2 − 16 )(x 2 − 9) = 0
⇒
x = ±4
⇒
x 2 = 16
and
x2 = 9
x = ±3 .
and
a(x 2 + 1) − (a 2 + 1)x = 0
(a) Equation
⇒ ax 2 − (a 2 + 1)x + a = 0
⇒
5.
(ax − 1)( x − a) = 0
x = a,
1
a
.
1
x =2+
(b) Let
⇒
2+
⇒
x = 2+
⇒
x =1± 2
1
2 + ..... ∞
1
x
But the value of the given expression cannot be negative or less than 2, therefore
1+ 2
is
required answer.
6.
(d) Given |
⇒
⇒
7.
x | 2 −3 | x | +2 = 0
(| x | −1)(| x | −2) = 0
| x | = 1 and | x | = 2 ⇒ x = ±1, x = ±2 .
(b) Equation
⇒
x4 − 8x 2 − 9 = 0
x4 − 9x2 + x2 − 9 = 0
⇒
8.
x 2 (x 2 − 9 ) + 1(x 2 − 9 ) = 0
( x 2 + 1)( x 2 − 9 ) = 0 ⇒ x = ± i, ± 3
.
(a) If one root of a quadratic equation with rational coefficients is irrational and of the form
α+ β
9.
⇒
, then the other root must also be irrational and of the form
(d) Given equation is
⇒
3x + 1 =
3x + 1 + 1 =
x
x −1
Squaring on both sides, we get
⇒ 2 x + 2x = 0
Thus
10. (c)
x =
⇒
x ≠0
2+x
(Irrational function)
and
⇒
3x + 1 = x + 1 − 2 x
x ≠1,
since equation is non-quadratic equation.
x2 − x −2 = 0
( x − 2)( x + 1) = 0
⇒
x = 2,−1
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α− β
.
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2 + 2 + ..... ≠ −1 ,
But
so it is equal to 2.
11. (b) Let all four roots are imaginary. Then roots of both equations
P( x ) = 0
and
imaginary.
Thus
b 2 − 4 ac < 0 ; d 2 + 4 ac < 0 ,
So, if
b≠0
d≠0
or
b = 0, d = 0 ,
If
P(x ) = ax 2 + c = 0
x2 = −
Or
12. (d)
So
b2 + d2 < 0
, which is impossible unless
b = 0, d = 0
.
at least two roots must be real.
we have the equations.
and
c 2 c
;x =
a
a
Q(x ) = −ax 2 + c = 0
c
a
as one of
and
−
c
a
must be positive, so two roots must be real.
x 2 + 5 | x | + 4 = 0 ⇒ | x | 2 +5 | x | + 4 = 0
⇒|
x | = −1,−4
, which is not possible. Hence, the given equation has no real root.
13. (b) Given equation is
( p 2 + q 2 )x 2 − 2 q( p + r)x + (q 2 + r 2 ) = 0
Roots are real and equal, then
4 q 2 ( p + r) 2 − 4 ( p 2 + q 2 )(q 2 + r 2 ) = 0
⇒ q 2 ( p 2 + r 2 + 2 pr ) − ( p 2 q 2 + p 2 r 2 + q 4
⇒
+ q 2r 2 ) = 0
q 2 p 2 + q 2 r 2 + 2 pq 2 r − p 2 q 2 − p 2 r 2 − q 4 − q 2 r 2 = 0
⇒
2 pq 2 r − p 2 r 2 − q 4 = 0 ⇒ (q 2 − pr ) 2 = 0
Hence
14. (d) Given
∴
x =
But
q 2 = pr .
Thus p, q, r in G.P.
x2 + x +1 = 0
1
1
1
[−1 ± i 3 ] = (−1 + i 3 ), (−1 − i 3 ) = ω , ω 2
2
2
2
α 19 = ω 19 = ω and β 7 = ω 14 = ω 2 .
Hence the equation will be same.
15. (a) According to condition
⇒
α2 +1
2
= −α
α 2 + 2α + 1 = 0 ⇒ α = −1,−1 .
16. (d) Given that
Let
x 2 / 3 − 7 x 1 / 3 + 10 = 0
a = x1/3 ,
. Given equation can be written as
then it reduces to the equation
a 2 − 7 a + 10 = 0 ⇒ (a − 5 )(a − 2) = 0 ⇒ a = 5, 2
Putting these values, we have
a 3 = x ⇒ x = 125
and 8.
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( x 1 / 3 ) 2 − 7( x 1 / 3 ) + 10 = 0
Q(x ) = 0
are
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(| x | −4 )(| x | −3) = 0
17. (d) The equation
⇒
⇒|
| x | = 4 ⇒ x = ±4
18. (a) Given
x | = 3 ⇒ x = ±3 .
(x + 1) − (x − 1) = (4 x − 1)
Squaring both sides, we get
Squaring again, we get
− 2 ( x 2 − 1) = 2 x − 1
x = 5/4
which does not satisfy the given equation. Hence equation
has no solution.
19. (b) log 4 (x − 1) = log 2 (x − 3)
⇒ x 2 − 7 x + 10 = 0
∴ x = 5, 2
but
⇒ x − 1 = ( x − 3) 2
⇒ ( x − 5 )( x − 2) = 0
x −3<0
when
x =5
∴
Only solution is
∴
Hence number of solution is one.
α
20. (b) Let roots are
α + (−α ) =
21. (c) Put
and
−α
x = 4 in x 2 + px + 12 = 0 ,
ex = x +1 ⇒ 1+
⇒
, then sum of the roots
3(λ − 2)
3
⇒ 0 = (λ − 2) ⇒ λ = 2
2
2
Now second equation
22. (a)
x=2
we get
p = −7
x 2 + px + q = 0
have equal roots. Therefore
x x2
+
+ ...... = x + 1
1! 2!
x2
x3
+
+ ...... = 0
2!
3!
x 2 = 0, x 3 = 0, ...... x n = 0
Hence,
x = 0 only
23. (a) Given equation
i.e.,
x 2 − 5 | x | +6 = 0
x2 − 5x + 6 = 0
x 2 − 3x − 2x + 6 = 0
(x − 3) (x − 2) = 0
x = 3, x = 2
and
one real roots.
and
and
and
x2 + 5x + 6 = 0
x 2 + 3x + 2x + 6 = 0
(x + 3) . (x + 2) = 0
x = −3, x = −2 .
i.e., Four solutions of this equation.
24. (d) Let for real roots are
α, β , γ , δ
then equation is
(x − α ) (x − β )(x − γ ) (x − δ ) = 0
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p 2 = 4q ⇒ q =
49
4
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x 4 − (α + β + γ + δ )x 3 + (αβ + βγ + γδ + αδ + βδ + αγ )x 2 − (αβγ + βγδ +αβδ + αγδ )x + αβγδ = 0
x 4 − ∑ α . x 3 + ∑ αβ . x 2 − ∑ αβγ . x + αβγδ = 0
On comparing with
x 4 − 4 x 3 + ax 2 + bx + 1 = 0
∑ α = 4 , ∑ αβ = a
∑ αβγ = −b, αβγδ = 1
Solving
∴b = −4 ; ∴ a = 6
25. (a)
and
b = −4 .
D ≡ b 2 − 4 ac = (−a − c) 2 − 4 ac
(∵ a + b + c = 0 )
= (a + c) 2 − 4 ac = (a − c) 2 ≥ 0
Hence roots are rational.
26. (c) We have
Let
4 ax 2 + 3 bx + 2 c = 0
Let roots are
α
and
β
D = B 2 − 4 AC = 9 b 2 − 4 (4 a)(2 c) = 9 b 2 − 32 ac
(a + b + c) = 0 ⇒ b = −(a + c)
Given that,
Putting this value, we get
= 9(a + c) 2 − 32 ac = 9(a − c) 2 + 4 ac .
Hence roots are real.
27. (c) Given equation
Its discriminant
D≥0
since roots are real
⇒
cos 2 p − 4 (cos p − 1) sin p ≥ 0
⇒
cos 2 p − 4 cos p sin p + 4 sin p ≥ 0
⇒
(cos p − 2 sin p ) 2 − 4 sin 2 p + 4 sin p ≥ 0
⇒
(cos p − 2 sin p ) 2 + 4 sin p(1 − sin p ) ≥ 0
Now
(1 − sin p ) ≥ 0
0 < p <π
or
for all real p,
…..(i)
sin p > 0 for 0 < p < π .
Therefore
4 sin p(1 − sin p) ≥ 0
when
p ∈ (0, π )
28. (d) Accordingly,
29. (d)
(cos p − 1)x 2 + (cos p )x + sin p = 0
{2(ac + bd )} 2 = 4 (a 2 + b 2 )(c 2 + d 2 )
⇒
4a 2 c 2 + 4b 2 d 2 + 8abcd = 4a 2 c 2 + 4a 2 d 2 + 4b 2 c 2 + 4b 2 d 2
⇒
4a 2 d 2 + 4b 2 c 2 − 8abcd = 0 ⇒ 4(ad − bc)2 = 0
⇒
ad = bc ⇒
a c
=
b d
.
x 2 + 2bx + c = (x + b ) 2 + c − b 2
∵ (x + b ) 2 is
a perfect square, therefore the given expression is positive if
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c − b 2 > 0 or b 2 < c
.
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30. (b) Equations
px 2 + 2 qx + r = 0
qx 2 − 2( pr )x + q = 0
4 q 2 − 4 pr ≥ 0 ⇒ q 2 − pr ≥ 0 ⇒ q 2 ≥ pr
have real roots, then from first
4 ( pr ) − 4 q 2 ≥ 0
And from second
⇒
and
.....(i)
(for real root)
pr ≥ q 2
.....(ii)
From (i) and (ii), we get result
q 2 = pr .
(l − m )x 2 − 5(l + m )x − 2(l − m ) = 0
31. (b) Given equation is
D = 25 (l + m ) 2 + 8 (l − m ) 2
Its discriminant
Which is positive, since
are real and
l, m , n
l≠m.
Hence roots are real and distinct.
32. (d) Roots are non real if discriminant < 0
i.e. if
5 2 − 4 .1 k < 0
i.e. if
4 k > 25
i.e. if
k>
25
4
Hence, the required least integer k is 7.
33. (a) According to question,
4 (a 2 − bc ) 2 − 4 (c 2 − ab )(b 2 − ac) = 0
34. (b)
⇒
a(a 3 + b 3 + c 3 − 3 abc ) = 0
⇒
a = 0 or a 3 + b 3 + c 3 = 3 abc
B 2 − 4 AC > 0
35. (b) Let
⇒ 0 − 4 ab > 0 ⇒ ab < 0 .
f ( x ) = ax 2 + bx + c .
If
c>0
If
c<0,
Then
f (0 ) = c
. Thus the graph of
f (x ) > 0
, then by hypothesis
y = f (x ) meets
This means that the curve
y = f (x )
then by hypothesis f (x ) < 0 , which means that the curve
and so it does not intersect with x-axis. Thus in both cases
x-axis i.e.
f (x ) ≠ 0
for any real x. Hence
f (x ) = 0
i.e.
(1 + 2k )x 2 + (1 − 2k )x + (1 − 2k ) = 0
If equation is a perfect square then roots are equal
i.e.,
(1 − 2 k )2 − 4 (1 + 2 k ) (1 − 2 k ) = 0
i.e.,
k=
1 −3
,
2 10
. Hence total number of values = 2.
37. (c) For real roots, discriminant
≥0
⇒ q2 − 4 p ≥ 0 ⇒ q2 ≥ 4 p
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does not meet x-axis.
y = f (x )
y = f (x )
ax 2 + bx + c = 0
so b 2 < 4 ac .
36. (a) Given equation
y-axis at (0, c).
is always below x-axis
does not intersect with
has imaginary roots and
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For
p = 1, q 2 ≥ 4 ⇒ q = 2,3,4
p = 2, q 2 ≥ 8 ⇒ q = 3, 4
p = 3, q 2 ≥ 12 ⇒ q = 4
p = 4 , q 2 ≥ 16 ⇒ q = 4
Total seven solutions are possible.
38. (b) Since the roots
∴
x 2 − 8 x + a 2 − 6a = 0
are real.
64 − 4 (a 2 − 6 a) ≥ 0 Or a 2 − 6 a − 16 ≤ 0
⇒
a ∈ [−2,8 ]
39. (a) Let α be a root of first equation, and then
α2
Hence
1
aα 2 + b α + c = 0 and a'
Therefore
=
b a ′ − b ′c
α
cc ′ − aa ′
=
α
2
+ b′
1
α
1
α
+ c′ = 0
be a root of second equation.
or
c′α 2 + b′α + a′ = 0
1
′
ab − b c ′
(cc '−aa' ) 2 = (ba '−cb ' )(ab '−bc ' ) .
40. (a) Given equation
α+β =−
3
4
1
Now
α
and
+
1
β
4 x2 + 3x + 7 = 0 ,
7
4
αβ =
α + β −3 / 4 −3 4
3
=
=
× =−
αβ
7/4
4
7
7
=
41. (b) α , β are the roots of
So,
B
A
α+β =−
Again
α 2,β 2
α 2 + β 2 = −p
Now
Ax 2 + Bx + C = 0
αβ =
and
and
.
.
C
A
are the roots of
x 2 + px + q = 0
then
(αβ )2 = q
α 2 + β 2 = (α + β ) 2 − 2αβ

⇒
α 2 + β 2 = −
⇒
−p=

2
B
C
 −2
A
A
B 2 − 2 AC
A2
42. (c) Given equation
⇒
therefore
⇒p=
2 AC − B 2
A2
x 2 − a( x + 1) − b = 0
x 2 − ax − a − b = 0 ⇒ α + β = a, αβ = −(a + b )
Now
(α + 1)(β + 1) = αβ + α + β + 1
=
−(a + b ) + a + 1 = 1 − b
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43. (c) As given, if
⇒
−
α, β
be the roots of the quadratic equation, then
2 a b 2 b (ab 2 + bc 2 )
= 2 + =
c
a
c
ac 2
⇒
2 a 2 c = ab 2 + bc 2 ⇒
⇒
c a b
, ,
a b c
α
α
and
β
b
a
αβ =
and
α 2β 2
−
b b 2 − 2 ac
=
a
a2
⇒
ax 2 + bx + c = 0
b2
a
2
−2
c
a
α + β = a2 + β 2
b (a + b ) = 2 ac .
α + β = − p, αβ = 1, γ + δ = −q and γδ = 1
(α − γ )(β − γ )(α + δ )(β + δ )
{αβ − γ (α + β ) + γ 2 }{αβ + δ (α + β ) + δ 2 }
=
(1 + p γ + γ 2 )(1 − p δ + δ 2 ) = ( p γ − q γ )(− p δ − q δ )
γ 2 + q γ + 1 = 0 ⇒ γ 2 + 1 = −q γ
47. (a) Since
2+i 3
are roots of
α+β =−
b
a
and
Let the roots of
α′ + β ′ = −
b
c
and
=7 =q
(Since
and similarly
is a root, therefore
product of roots
but
(α + β ) 2 − 2αβ
are in H.P.
=
⇒
=
c
a
α 2 + β 2 = (α + β ) 2 − 2αβ =
Now,
α, β
β
2
α + α = −2m ⇒ α = −m
be two roots of
α+β =−
46. (a) As given,
48. (c)
a b c
, ,
c a b
and α , then
So under condition
⇒
1
α .α = m 2 − 2m + 6 ⇒ m 2 = m 2 − 2m + 6
Then
⇒
+
m =3.
45. (c) Let
⇒
α
2
2a b c
= +
b
c a
are in A.P. ⇒
44. (a) Let roots are
⇒
1
b (b 2 / a 2 ) − (2c / a) b 2 − 2ac
=
=
a
(c 2 / a 2 )
c2
⇒
and
α+β =
. Hence
2−i 3
γ
is a root of x 2 + qx + 1 = 0 )
δ 2 + 1 = −q δ = −γδ ( p − q )( p + q ) = q 2 − p 2 .
will be other root. Now sum of the roots
( p, q ) = (−4 ,7) .
ax 2 + bx + c = 0
αβ =
c
a
cx 2 + bx + a = 0 be α ′, β ′ ,
α ′β ′ =
α + β −b / a −b
=
=
αβ
c/a
c
then
a
c
⇒1+1
α
β
= α′ + β ′
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= 4 = −p
and
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α′ =
Hence
49. (a) If
α, β
α
and
β′ =
1
β
.
are the roots, then
α+β
A=
1
2
⇒α + β
= 2A
and
G = αβ ⇒ αβ = G 2
t 2 − 2 At + G 2 = 0 .
The required equation is
α, β
50. (c) Let
So
are roots of
α + β = −p
Given that
Now
⇒
and
x 2 + px + q = 0
αβ = q
(α + β ) = 3(α − β ) = − p
⇒ α − β = −p
3
(α − β ) 2 = (α + β ) 2 − 4αβ
p2
= p 2 − 4 q or 2 p 2 = 9 q
9
.
51. (b) It is obviously 24.
52. (c) According to condition
2m − 1
1
= −1 ⇒ 3 m = 1 ⇒ m =
m
3
53. (b) Given equation can be written as
x 2 + x ( p + q − 2r) + pq − pr − qr = 0
Whose roots are
α
and
−α
, then the product of roots
− α 2 = pq − pr − qr = pq − r( p + q )
0 = p + q − 2r ⇒ r =
And sum
.....(i)
.....(ii)
p+q
2
.....(iii)
From (ii) and (iii), we get
− α 2 = pq −
=−
54. (d) α + β
=−
{
p+q
1
( p + q ) = − ( p + q ) 2 − 2 pq
2
2
}
(P 2 + q 2 )
.
2
b
c
, αβ =
a
a
(b 2 − 2 ac)
and
α2 + β2 =
Now
α (aα + b ) + β (aβ + b )
β
α
=
+
(aβ + b )(aα + b )
aβ + b aα + b
a2
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=
=
a(α + β ) + b(α + β )
2
2
αβa 2 + ab(α + β ) + b 2
b 2 − ac − b 2
a c − ab 2 + ab 2
2
− 2 ac
=
a2c
55. (b) α , β be the roots of
(b 2 − 2ac)
 b
+ b − 
 a
=
c
b
  2


 a + ab  −  + b 2
a
 a
a
a2
=−
2
a
.
ax 2 + bx + c = 0
∴α + β = −b / a , αβ = c / a
α − 1, β − 1
Roots are
Sum,
α + β − 2 = (−b / a) − 2 = −8 / 2 = −4
(α − 1)(β − 1) = αβ − (α + β ) + 1
Product,
∵
New equation is
∴b /a = 2
56. (c) Let
α, β
Then
i.e.
b = 2a ,
= c /a + b /a +1 = 1
2x 2 + 8 x + 2 = 0
c + b = 0 ⇒ b = −c
also
be the roots of
x 2 + bx + c = 0
.
and
α', β '
x 2 + qx + r = 0
.
α + β = −b , αβ = c, α '+ β ' = −q , α ' β ' = r
It is given that
⇒
(α + β ) 2
⇒
b 2r = q 2c
(α − β )
2
=
α + β α '+ β '
α α'
=
⇒
=
β β'
α − β α '− β '
(α '+ β ' ) 2
(α '− β ' )
2
⇒
b2
b − 4c
57. (b) If the roots of equation
2
=
q2
q − 4r
2
ax 2 + 2 bx + c = 0
are in the ratio m : n, Then we have
mn (2b ) 2 = (m + n) 2 ac
.....(i)
Also if the roots of the equation
px 2 + 2 qx + r = 0
mn (2 q ) 2 = (m + n) 2 pr
.....(ii)
Dividing (i) and (ii), we get
58. (c) Let the roots be α and β ⇒
Given,
But
be the roots of
b2
q2
=
(ac)
( pr )
or
α + β = −p
,
are also in the same ratio
b2
q2
=
ac
pr
.
αβ = q
α + β =α2 + β2
α + β = (α + β ) 2 − 2αβ ⇒ − p = (− p )2 − 2 q
⇒ p 2 − 2q = − p ⇒ p 2 + p = 2q
.
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m :n,
then
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59. (a) Let r be the common ratio of the G.P. α, β, γ, δ then
⇒ α + α r = 1 ⇒ α (1 + r) = 1
∴α + β = 1
…..(i)
αβ = p ⇒ α (αr) = p ⇒ α 2 r = p
…..(ii)
γ + δ = 4 ⇒ αr 2 + αr 3 = 4 ⇒ α r 2 (1 + r) = 4
…..(iii)
and
γδ = q ⇒ αr 2 .αr 3 = q
β = α r, γ = α r 2
⇒ α 2r5 = q
…..(iv)
⇒ (p, q) = (– 2, – 32).
60. (c)
(α − β ) 2 = (α + β ) 2 − 4αβ = (b 2 − 4 ac) / a 2
......(i)
Also {(α − k ) − (β − k )}2
=
{(α − k ) + (β − k )}2 − 4 (α − k )(β − k )
= (− B / A) 2 − 4 (C / A) = (B 2 − 4 AC ) / A 2
From (i) and (ii),
b 2 − 4 AC
(b 2 − 4 ac) / a 2 = (B 2 − 4 AC ) / A 2
 A
= 
b 2 − 4 ac
a
∴
61. (c) Given equation is
⇒α + β =
−6 −2
=
9
3
2
9x 2 + 6x + 1 = 0
and
∴ α − β = (α + β ) 2 − 4αβ
⇒α =
=
4
1
− 4. = 0
9
9
x 2 − (α + β )x + αβ = 0
⇒ x 2 + 6x + 9 = 0
p + q = −p
and
αβ = 1 / 9
−1
−1
,β =
3
3
∴ Equation
62. (a)
.....(ii)
and
.
pq = q ⇒ p = 1
q = −2 .
63. (d) (α 2 + β 2 ) 2
= (α + β ) (α 3 + β 3 )
2
2
 b 2 − 2 ac 



 =  − b   − b + 3 abc 
2
3




a
a
 a 



⇒
4 a 2 c 2 = acb 2
⇒
As
a ≠ 0 ⇒ c∆ = 0
ac (b 2 − 4 ac ) = 0
64. (d) 1 − i is a root of the equation so
x =1−i
⇒ ( x − 1) = −i ⇒ ( x − 1)2 = (−i)2 ⇒ x 2 − 2 x + 2 = 0
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and δ = α r 3
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By comparison, a = 2, b = 2.
65. (d) Here,
∴
1
α3
+
α + β = −2
1
β3
=
αβ = 4
and
(α + β ) 3 − 3αβ (α + β )
α3 + β3
=
(αβ ) 3
(αβ ) 3
=
(−2)3 − 3(−2)(4 )
66. (d) Roots of given equation
i.e.,
a+b = p
Then
1 1 a+b
p
+ =
=
a b
ab
q
67. (c) Note that for
68. (d) We have
64
−b
a
.
is
a
ab = q
and
b
……(ii)
.
and hence the given equation cannot have real roots.
t ∈ R, t 2 x 2 + | x | + 9 ≥ 9
α+β =
1
4
=
x 2 − px + q = 0
……(i) and
and
Now sum of the roots
c
a
αβ =
= 2 +α + 2 + β = 4 −
And product of the roots
=4+
= 16
(4 )3
b
a
= (2 + α )(2 + β )
c 2b 4 a + c − 2b
−
=
a
a
a
Hence the required equation is
 b  4a + c - 2b
x2 - x  4 -  +
=0
a
 a
⇒
ax 2 − x (4 a − b ) + 4 a + c − 2b = 0
⇒
ax 2 + x (b − 4 a) + 4 a − 2b + c = 0 .
69. (d) Let root of the given equation
Now,
x 2 + px + q = 0
are
α
and
α2 .
α .α 2 = α 3 = q, α + α 2 = − p
Cubing both sides,
α 3 + (α 2 )3 + 3α .α 2 (α + α 2 ) = − p 3
q + q 2 + 3 q(− p) = − p 3
p 3 + q 2 + q(1 − 3 p ) = 0
.
70. (b) Let the two number is
x1 + x 2
=9
2
and
x1
and
x2
x 1 x 2 = 16
x 1 + x 2 = 18
and
x 1 x 2 = 16
Equation
x 2 − (Sum of roots) x + Product of roots = 0
Required equation
x 2 − 18 x + 16 = 0
.
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71. (d) α 2 − 5α + 3 = 0
…..(i)
β 2 − 5β + 3 = 0
…..(ii)
From (i) – (ii),
⇒
⇒
(α 2 − β 2 ) − 5α + 5 β = 0
α 2 − β 2 = 5(α − β ) ⇒ α + β = 5
From (i) + (ii),
⇒
(α 2 + β 2 ) − 5(α + β ) + 6 = 0
⇒
(α 2 + β 2 ) − 5 . 5 + 6 = 0
Then
⇒
α 2 + β 2 = 19
(α + β ) 2 = α 2 + β 2 + 2αβ
⇒ 25 = 19 + 2αβ ⇒ αβ = 3
Then the equation, whose roots are
α β
x 2 − x  +
β α
⇒
72. (a) Let
⇒
α2 =
⇒
3 x 2 − 19 x + 3 = 0 .
x 2 + ax + b = 0 ⇒ x =
are the roots of the equation
− a ± a2 − 4b
2
x 2 + bx + a = 0
− b + b 2 − 4a
− b − b2 − 4a
, β2 =
2
2
α 1 − β1 = a 2 − 4 b
Given,
, is

+1 = 0


are the roots of the eqn
α2, β2
Now
β
α
− a + a 2 − 4b
− a − a 2 − 4b
, β1 =
2
2
α1 =
And
So,
19
+1 = 0
3
α1 , β1
and
 α β
 + . = 0
 β α
α 2 + β 2
x 2 − x 
 αβ
⇒ x 2 − x.
α
β
;
α2 − β2 = b 2 − 4 a
α1 − β1 = α 2 − β 2 ⇒ a 2 − 4b = b 2 − 4 a
⇒ a 2 − b 2 = − 4 (a − b) ⇒ a + b + 4 = 0 .
73. (c) Equation
x 2 + kx − 24 = 0
has one root is 3.
⇒ 3 2 − 3 k − 24 = 0 ⇒ k = 5
Put
x =3
and
k =5
in options, only (c) gives the correct answer.
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74. (a)
x , y, z ∈ R
and distinct.
u = x 2 + 4 y 2 + 9 z 2 − 6 yz − 3 zx − 2 xy
Now,
=
1
(2 x 2 + 8 y 2 + 18 z 2 − 12 yz − 6 zx − 4 xy )
2
=
1 2
x − 4 xy + 4 y 2 ) + (x 2 − 6 zx + 9 z 2 ) + (4 y 2 − 12 yz + 9 z 2 )
2
=
1
( x − 2 y ) 2 + ( x − 3 z ) 2 + (2 y − 3 z ) 2
2
{
}
{
}
Since it is sum of squares. So
75. (c) Let the common root be y. Then
u
is always non- negative
y 2 + py + q = 0
and
y2 +α y + β = 0
On solving by cross multiplication, we have
y2
y
1
=
=
pβ − qα q − β α − p
∴
76. (d)
y=
q−β
α−p
x 2 − 3x + 2
Hence
and
pβ − qα
y2
=y=
y
q−β
be factor of
x 4 − px 2 + q = 0
( x 2 − 3 x + 2) = 0 ⇒ ( x − 2)(x − 1) = 0
⇒
x = 2, 1, Putting
So
4 p − q − 16 = 0
.....(i)
p − q −1 = 0
.....(ii)
And
these values in given equation
Solving (i) and (ii), we get (p, q)=(5, 4)
77. (c) Let roots of
∴
x 2 − cx + d = 0
be
α, β
then roots of
be
α,α
α + β = c, αβ = d , α + α = a, α 2 = b
Hence
2(b + d ) = 2(α 2 + αβ ) = 2α (α + β ) = ac
78. (b) If the given expression be y, then
If
x 2 − ax + b = 0
y ≠ 0 then ∆ ≥ 0 for
Or –
⇒
39 y 2 + 10 y + 1 ≥ 0
or
B 2 − 4 AC ≥ 0
(13 y + 1)(3 y − 1) ≤ 0
−1 / 13 ≤ y ≤ 1 / 3
If
79. (d) Let
real x i.e.
y = 2 x 2 y + (3 y − 1)x + (6 y − 2) = 0
y=
y=0
then
x = −2
which is real and this value of
(x − a)(x − b)
( x − c)
Or
y( x − c) = x 2 − (a + b )x + ab
Or
x 2 − (a + b + y )x + ab + cy = 0
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y
is included in the above range
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∆ = (a + b + y ) 2 − 4 (ab + cy )
= y 2 + 2 y (a + b − 2 c) + (a − b ) 2
Since x is real and y assumes all real values, we must have
∆ ≥ 0 for
all real values of y. The
sign of a quadratic in y is same as of first term provided its discriminant
4 (a + b − 2c) 2 − 4 (a − b ) 2 < 0
This will be so if
Or
4 (a + b − 2 c + a − b )(a + b − 2 c − a + b ) < 0
Or
16 (a − c)(b − c) < 0
16 (c − a)(c − b ) = − ve
or
∴ c lies between a and b i.e.,
a<b
Where
B 2 − 4 AC < 0
b < a then
, but if
a<c<b
.....(i)
the above condition will be
b < c < a or a > c > b
.....(ii)
Hence from (i) and (ii) we observe that (d) is correct answer.
80. (d) Let
f ( x ) = a n x n + a n −1 x n −1 + .... + a1 x
;
f (0 ) = 0 ; f (α ) = 0
⇒
f ′( x ) = 0 ,
has atleast one root between
i.e., equation
na n x n −1 + (n − 1)a n −1 x n − 2 + .... + a1 = 0
Has a positive root smaller than
81. (b) Here
Now,
(0, α )
D = b 2 − 4c > 0
α + β = −b < 0
α
.
because c < 0 < b. So roots are real and unequal.
and
αβ = c < 0
∴ One root is positive and the other negative, the negative root being numerically bigger.
As
82. (a) Let
⇒
α < β ,α
y=
is the negative root while β is the positive root. So, | α | > β and α < 0 < β .
x 2 + 14 x + 9
x 2 + 2x + 3
y(x 2 + 2 x + 3) − x 2 − 14 x − 9 = 0
⇒ (y − 1)x 2 + (2 y − 14 )x + 3 y − 9 = 0
≥0
For real x, its discriminant
i.e.
⇒
4 (y − 7 ) 2 − 4 (y − 1)3(y − 3) ≥ 0
y 2 + y − 20 ≤ 0
or
(y − 4 )(y + 5 ) ≤ 0
Now, the product of two factors is negative if these are of opposite signs. So following
two cases arise:
Case I:
y − 4 ≥ 0 or y ≥ 4
and
y+5 ≤0
or
y ≤ −5
This is not possible.
Case II:
y−4 ≤0
or
y≤4
and
y+5 ≥0
or
y ≥ −5
Both of these are satisfied if
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−5 ≤ y ≤ 4
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2hx = 56 or hx = 28
83. (d) Subtracting, we get
Putting in any,
∴  28 
x 2 = 49
2
= 7 2 ⇒ h = 4 (h > 0 )
 h 
84. (a) Given equation is
x 2 − 2 ax + a 2 + a − 3 = 0
D≥0
If roots are real, then
⇒
4 a 2 − 4 (a 2 + a − 3) ≥ 0 ⇒ − a + 3 ≥ 0
⇒
a−3 ≤0 ⇒ a≤ 3
As roots are less than 3, hence
f (3) > 0
9 − 6a + a 2 + a − 3 > 0 ⇒ a 2 − 5a + 6 > 0
⇒ (a − 2)(a − 3) > 0 ⇒
Hence
a<2
either
or
a>3
satisfy all.
85. (d) The equation is
x 2 − (a + b ) x + ab − 1 = 0
∴
∴
a<2
discriminant
Both roots are real. Let them be
α=
(a + b ) − (b − a) 2 + 4
2
Clearly,
α<
,
β=
= (a + b)2 − 4(ab − 1) = (b − a)2 + 4 > 0
α, β
where
(a + b) + (b − a)2 + 4
2
(a + b ) − (b − a) 2
(a + b ) − (b − a)
=
=a
2
2
(∵ b > a)
And
β>
(a + b) + (b − a)2
a+b +b −a
=
=b
2
2
Hence, one root
α
is less than a and the other root β is greater than b.
P(x ) = bx 2 + ax + c
86. (d) Let
As
P(0 ) = 0 ⇒ c = 0
As
P(1) = 1 ⇒ a + b = 1
P(x ) = ax + (1 − a)x 2
Now
As
P ′( x ) = a + 2(1 − a)x
P ′( x ) > 0
for
x ∈ (0, 1)
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Only option (d) satisfies above condition
2
87. (a)
3
3

x 2 − 3x + 3 =  x −  +
2
4

Therefore, smallest value is
88. (b) Let
3
4
3

 − 3, 
2

, which lie in
f (x ) = x 5 − 6 x 2 − 4 x + 5 = 0
Then the number of change of sign in
f (x )
is 2 therefore
f (x )
can have at most two positive
real roots.
f (− x ) = − x 5 − 6 x 4 + 4 x + 5 = 0
Now,
Then the number of change of sign is 1.
Hence
f (x ) can
have at most one negative real root. So that total possible number of real roots is 3.
89. (d) Given equation
( pq ) x 2 − ( p + q )2 x + ( p + q)2 = 0
Let solution set is
Sum of roots =
p +q p +q
,


q 
 p
( p + q)2
pq
⇒
p + q p + q ( p + q)2
+
=
p
q
pq
Similarly, product of roots =
p + q p + q ( p + q)2
×
=
p
q
pq
⇒
90. (b) Given ,
( p + q)2
pq
.
x + 2 > x + 4 ⇒ ( x + 2)2 > ( x + 4 )
⇒ x 2 + 4 x + 4 > x + 4 ⇒ x 2 + 3x > 0
⇒ x (x + 3) > 0
⇒ x < – 3 or x > 0
91. (b) Given equation
Then
⇒x >0 .
x 3 − 3x 2 + x + 5 = 0 .
α + β + γ = 3 , αβ + βγ + γα = 1 , αβγ = −5
y = Σα 2 + αβγ = (α + β + γ ) 2 − 2 (αβ + βγ + γα ) + αβγ
=
9 −2−5 = 2
∴ y=2
It satisfies the equation
92. (d) Let
y = x2
. Then
y3 − y2 − y − 2 = 0
.
x = y
∴
x 3 + 8 = 0 ⇒ y3/2 + 8 = 0
⇒
y 3 = 64 ⇒ y 3 − 64 = 0
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Thus the equation having roots
α 2,β 2
and
γ 2 is x 3 − 64 = 0 .
93. (a) According to given condition,
4 a 2 − 4 (10 − 3 a) < 0
⇒ a 2 + 3 a − 10 < 0
⇒ (a + 5)(a − 2) < 0 ⇒
−5 < a < 2 .
94. (c) If α, β, γ are the roots of the equation.
x 3 − px 2 + qx − r = 0
∴ (α + β ) −1 + (β + γ ) −1 + (γ + α ) −1 =
Given,
⇒
p2 + q
pq − r
p = 0, q = 4 , r = −1
p2 + q 0 + 4
=
=4.
pq − r
0 +1
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