ASSIGNMENT

Name:………....................................................................... Batch: ……………....………..

TOPIC: QUADRATIC EQUATIONS AND EXPRESSIONS

SINGLE CORRECT TYPE

1. Let a,b,c,a1,b1,c1,∈ R and ax 2 + bx + c > 0 ∀x ∈ R and a1x 2 + b1 x + c1 > 0 ∀x ∈ R . Then

(A) aa1x 2 + bb1 x + cc1 > 0 ∀x ∈ R
(B) aa1x 2 + bb1 x + cc1 < 0 ∀x ∈ R
(C) aa1x 2 + bb1 x + cc1 = 0, will have real roots.
(D) Nothing can be said in general about the nature of roots of, aa1x 2 + bb1 x + cc1 = 0,
2. ( ) (
Roots of the quadratic equation x 2 − 4x + 3 + λ x 2 − 6x + 8 = 0, λ ∈ R will be )
(A) always real (B) real only when λis positive
(C) real only when λ is negative (D) always imaginary
3. If all the real solutions of the equation 4 x − ( a − 3 ) 2 x + ( a − 4 ) = 0 are non positive, then
(A) 4 < a ≤ 5 (B) 0 < a< 4 (C) a> 4 (D) a< 3
4. If the equation 5x 2 − 10x + log1/ 5 a = 0 has real roots then the minimum value of a is
1 1 1
(A) (B) (C) (D) none of these
55 1010 510
5. The value of ‘a’ for which the equation (a 2
) ( )
+ 4a + 3 x 2 + a2 − a − 2 x + ( a + 1) a = 0 has
more than two roots is
(A) 1 (B) 2 (C) -2 (D) –1
6. If the equation ax 2 + bx + 6 = 0 does not have two distinct real roots, then the least value of
3a + b
(A) 3 (B) –3 (C) 2 (D) –2
7. If p and q are odd integers, then the equation x2 + 2px +2 q = 0
(A) has no integral roots (B) has no rational roots
(B) has no irrational roots (D) has no imaginary roots
8. If 4ac > b2 and a + c > b for real numbers a, b and c, then which of the following is true?
(A) a > 0 (B) c > 0 (C) a + b + c > 0 (D) 4a + c > 2b
2 2
9. If ax + bx + c = 0 and cx + bx + a = 0 (a, b, c ∈ R) have a common non–real root, then
(A) –2|a| < b < 2 |a| (B) – 2|c| < |b| < 2|c| (C) a = ± c (D) a = c
2
10. If the equation ax + bx + c = 0 (a < 0) has two roots α and β such that α < –3 and β > 3, then
(A) 9a + 3|b| + c > 0 (B) c > 0 (C) 4a + 2|b| + c > 0 (D) none of these
10 7 4
11. The interval(s) which satisfy x – x + x – x+1 > 0 is (are)
(A) –1 ≤ x ≤ 0 (B) 0 < x < 1 (C) x ≥ 1 (D) –∞ < x < ∞
2 100
12. (aαx + α y + 1) is a polynomial in x and y. If the sum of the coefficients vanishes for some
real α , then possible values of a are
(A) –2 (B) 2 (C) 1 (D) none

COMPREHENSION - I

Consider the quadratic equation ax 2 − bx + c = 0 , a,b,c ∈ N , which has two distinct real roots
belonging to the interval (1, 2).

13. The least value of a is

(A) 4 (B) 6 (C) 7 (D) 5
14. The least value of b is
(A) 10 (B) 11 (C) 13 (D) 15
15. The least value of c is
(A) 4 (B) 6 (C) 7 (D) 5

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COMPREHENSION - II

Consider the inequality 9 x − a3 x − a + 3 ≤ 0, where ' a ' is a real parameter.

16. The given inequality has atleast one negative solution for a ∈
(A) ( -∞, 2) (B) (3, ∞) (C) (-2, ∞) (D) (2, 3)
17. The given inequality has atleast one positive solution for a ∈
(A) (-∞, -2) (B) [3, ∞) (C) (2, ∞) (D) [-2, ∞)
18. The given inequality has atleast one real solution for a ∈
(A) (-∞, 3) (B) [2, ∞) (C) (3, ∞) (D) [-2, ∞)

ASSERTION & REASON TYPE

(A) Statement -1 is true, Statement -2 is false.

(B) Statement -1 is true, Statement -2 is true; Statement -2 is not correct explanation of
Statement-1.
(C) Statement -1 is true, Statement -2 is true; Statement -2 is the correct explanation of
Statement-1.
(D) Statement -1 is false, Statement -2 is true.

19. Statement – 1: If equations ax 2 + bx + c = 0 and x 2 − 3x + 4 = 0 have exactly one root

common, then atleast one of a,b,c is imaginary.
Statement – 2: If a,b,c are not all real, then equation ax 2 + bx + c = 0 can have one root
real and one root imaginary.
20. Statement – 1: The equation ( x − p )( x − r ) + λ ( x − q )( x − s ) = 0 ,where p < q < r < s , has
non-real roots.
Statement – 2: The equation px 2 + qx + r = 0 ( p, q,r ∈ R ) has non-real roots if q2 − 4pr < 0 .

21. { ( ) }
If 3x 2 − 2 ( a − d) x + a2 + 2 b2 + c 2 + d2 = 2 ( ab + bc + cd) then
(A) a,b,c, d are in G.P (B) a,b,c, d are in A.P
(C) a,b,c, d are in H.P (D) a,b,c, d are in A.G.P
2π 2π
22. Let a = cos + i sin , α = a + a2 + a 4 , β = a3 + a5 + a6 , then the equation whose roots are
7 7
α, β is
(A) x 2 − x + 2 = 0 (B) x 2 + x − 2 = 0 (C) x 2 − x − 2 = 0 (D) x 2 + x + 2 = 0
23. Let S be the set of values of ' a ' for which all the real solutions of
4 x − ( a − 3 ) 2x + ( a − 4 ) = 0 are non-positive, then
(A) a > 4 (B) 0 < a < 4 (C) 4 < a ≤ 5 (D) a < 3
2
24. The equation x + nx + m = 0 , n,m ∈ I cannot have
(A) integral roots (B) non-integral rational roots
(C) irrational roots (D) complex roots
25. The set of values of ' a ' for which each one of the roots of x 2 − 4ax + 2a2 − 3a + 5 = 0 are
greater than 2 is
9 
(A) a ∈ (1, ∞ ) (B) a = 1 (C) a ∈ ( −∞,1) (D) a ∈  , ∞ 
2 
26. A quadratic equation f ( x ) = ax 2 + bx + c (a ≠ 0) has positive distinct roots reciprocal of each
of other. Which one is correct.
(A) f ' (1) = 0 (B) a f ' (1) < 0 (C) a f ' (1) > 0 (D) None of these

27. The equation x 2 + b2 = 1 − 2bx and x 2 + a2 = 1 − 2ax have one and only one common root,
then
(A) a − b = 2 (B) a − b = −2 (C) a + b = 2 (D) a + b = −2

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28. The equation x3 − 6x 2 + 9x + λ = 0 has exactly one root in (1,3 ) then [ λ + 3] is

(A) 0 (B) 1 (C) 2 (D) 3
29. If 0 < a < b < c and the roots of α, β of the equation ax 2 + bx + c = 0 are imaginary, then
(A) α = β (B) α > 1 (C) β < 1 (D) α + β > 1
2
30. If the equation ax + bx + c = 0 (a < 0) has two roots α and β such that α < −3 , β > 3 then
(A) 9a + 3 b + c > 0 (B) 4a + 2 b + c > 0 (C) c > 0 (D) 9a − 3 b + c < 0
2 2
31. If α, β are roots of ax + bx + c = 0 and α + h, β + h are roots of px + qx + r = 0 then
a b c 1b q 1b q b2 − 4ac q2 − 4pr
(A) = = (B) h =  −  (C) h =  +  (D) =
p q r 2a p 2a p a2 p2
32. The graph of the quadratic polynomial y = ax 2 + bx + c is as Y
shown in the figure then

(A) b2 − 4ac > 0 (B) b < c (C) a > 0 (D) c < o

n

 (x − k)
2
33. The number of real roots of the quadratic equation = 0 ( n > 1) is
k =1

34. The least integral value of k such that ( k − 2 ) x 2 + 8x + k + 4 is positive for all real values of x
is.
35. The number of solutions of log4 ( x − 1) = log2 ( x − 3 ) is
36. The number of solutions of the equation x3 + 2x 2 + 5x + 2 cos x = 0 in [0, 2π ] is
37. Let x 2 + x + 1 is divisible by 3. If x is divided by 3, the remainder will be
38. a, b, c are real numbers and are in G.P. One of the roots of the equation ax 2 + bx + c = 0 is
real, then value of b is
39. Match the following for the equation x 2 + a x + 1 = 0 , where a is a parameter.

Column I Column II

(A) No real roots p. a < −2

(B) Two real roots q. φ

(C) Three real roots r. a = −2

(D) Four distinct real roots s. a≥0

40. Match the following question.

Column I (Number of positive integers for which) Column II
(A) one root is positive and the other is negative for the
p. 0
equation ( m − 2 ) x 2 − ( 8 − 2m ) x − ( 8 − 3m ) = 0
(B) exactly one root of equation q. infinite
x 2 − m ( 2x − 8 ) − 15 = 0 lies in interval (0, 1)
(C) the equation x 2 + 2 ( m + 1) x + 9m − 5 = 0 has both
r. 1
roots negative
(D) the equation x 2 + 2 (m − 1) x + m + 5 = 0 has both
s. 2
roots lying on either sides of 1

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2 2
41. Let α, β are roots of the equation sin x + a sin x + b = 0 and also of cos x + c cos x + d = 0. If
2 2 2 2
a0 a + a1b + a2 c + a3 d −a0 + a1 + a2 + a3
cos (α + β) = 2 2 2 2
, then equals
b0 a + b1b + b2 c + b3 d b0 + b1 + b2 + b3
−a0 a1a2 a3
(A) 0 (B) –1 (C) 1 (D)
b0 b1b2 b3
42. f(x) and g(x) are quadratic polynomials and |f(x)| ≥ |g(x)|, ∀ x ∈ R. Also f(x) = 0 have real
roots. Then number of distinct roots of equation h(x) h′′(x) + h(x)2 = 0 are (where h(x) = f(x)
g(x))
(A) 0 (B) 2 (C) 3 (D) 4
2 2
43. Conditions so that equation (a1x + b1x + c1)(a2x + b2x + c2) = 0, (a1 a2 > 0) have four real
roots of
(i) c1c2 > 0 (ii) a1c2 < 0 (iii) a2c1 < 0
(A) only (i) is sufficient
(B) any one of the three statement is sufficient
(C) either of two statements are sufficient
(D) either of two statements are necessary and sufficient

44. If r1, r2, r3 are the radii of the escribed circles of a triangle ABC and r is the radius of its
incircle, then the root(s) of the equation x2 – r(r1r2 + r2r3 + r3r1)x + r1r2r3 – 1 = 0
(A) 1 (B) r1 = r2 + r3 (C) r (D) r1r2r3 – 1
3 2
45. If 1 ≥ a ≥ b ≥ c ≥ 0 and ‘λ’ is a real root of the equation x + ax + bx + c = 0, then maximum
value of |λ| is equal to
3 1
(A) 1 (B) (C) 2 (D)
2 2
46. If (b2 – 4ac)2 (1 + 4a2) < 64a2, a < 0, then maximum value of quadratic expression ax2 + bx + c
is always less than
(A) 0 (B) 2 (C) –1 (D) –2
47. The number of quadratic equation which have real roots and which remain unchanged, if the
roots are squared, is
(A) 2 (B) 4 (C) 6 (D) 3
48. If α and β are such that α + β = 1 and 25(α3 + β3) = 7, then the equation whose roots are α
and β is
2 2
(A) 25x – 5x + 6 = 0 (B) 25x – 25x + 6 = 0
2 2
(C) x + 5x + 6 = 0 (D) x – 35x + 216 = 0
x x
49. If both the roots of (2a – 4)9 – (2a – 3)3 + 1 = 0 are non-negative, then
5 5
(A) 0 < a < 2 (B) 2 < a < (C) a < (D) a > 3
2 4
2
50. If α, β are acute angles such that α + β and α – β are the roots of tan θ – 4 tan θ + 1 = 0, then
2α – 3β is equal to
5π π 3π
(A) (B) (C) 0 (D)
12 8 2
51. The equation 5x2 + ax + 1 = 0, 4x2 + bx + 1 = 0 have a common root, then the sum of the
reciprocals of the other two roots is
(A) b – a (B) 2(b – a) (C) 3(b – a) (D) 9(b – a)
n
52. The number of real roots of quadratic equation  (x − k)
k =1
2
= 0 (n > 1) is

(A) 1 (B) 2 (C) n (D) 0

53. If a, b, c are sides of ∆ABC, then a + b − c is
(A) always positive (B) always negative
(C) positive only when c is smallest (D) none of these
54. If p, q, r, s, t are numbers such that p + q < r + s, q + r < s + t, r + s < t + p, s + t < p + q, then
the smallest numbers are
(A) p and q respectively (B) r and t respectively
(C) r and p respectively (D) q and p respectively

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55. The value of 140 2 − 57 − 40 2 + 57 is equal to

(A) 10 (B) 4 (C) –4 (D) –10
56. PQRS is a common diameter of three circles. The area of the middle circle is the average of
the other two. If PQ = 2 and RS = 1, then the length of QR is
(A) 6 + 1 (B) 6 − 1 (C) 5 (D) 4
3 2
57. Let x1, x2, x3 be the roots of the equation x – x + βx + γ = 0, which are in A.P, then βx + γy =
1 passes through the fixed point
 9 27  9 7  5 27  9 5
(A)  ,  (B)  ,  (C)  ,  (D)  , 
2 2  2 2 2 2  2 2
2 3 19 20 2
58. If the polynomial f(x) = 1 – x + x – x ...... – x + x is expressed as g(y) = a0 + a2y + a2y +
20
...... +a20y where y = x – 4, then the value of a0 + a1 + a2 + ...... + a20 is
521 − 1 520 1 + 520 1 + 521
(A) (B) (C) (D)
6 6 6 6
59. Suppose a and b are rational and α, β be the roots of x2 + 2ax + b = 0, then the equation with
rational coefficients one of whose roots is α + β + α 2 + β2 is
(A) x2 + 4ax + 2b = 0 (B) x2 + 4ax – 2b = 0
(C) x2 – 4ax + 2b = 0 (D) x2 – 4ax – 2b = 0
2
60. If ℓ, m, n are real, ℓ ≠ m, then the roots of the equation (ℓ – m)x – 5(ℓ + m)x – 2(ℓ – m) = 0 are
(A) real and equal (B) complex (C) real and unequal (D) none of these

61. If unity is double repeated root of px2 + q(x2 + x) + r = 0, then

(A) p.r < 0 (B) p.q < 0 (C) p.q.r > 0 (D) none of these
62. The quadratic x2 + ax + b + 1 = 0 has roots which are positive integers, then (a2 + b2) can be
equal to
(A) 50 (B) 37 (C) 61 (D) 19
63. The equation x2 + ax + 1 = 0 and x2 + bx + c = 0 have common real root and the equations
2 2
x + x + a = 0 and x + cx + b = 0 have a common real root. Then a + b + c is equal to
(A) –1 (B) –2 (C) –3 (D) none of these
2
64. If p is a prime and both roots of x + px – 444p = 0 are integers, then p is equal to
(A) 2 (B) 3 (C) 37 (D) none of these
65. If a, b, c be distinct positive numbers, then the natural of roots of the equation
1 1 1 1
+ + = is
x −a x −b x −c x
(A) all real and distinct (B) all real and atleast two are distinct
(C) atleast two real (D) all non-real
66. The equation a8x8 + a7x7 + a6x6 + ...... + a0 = 0 has all its roots positive and real (where a8 = 1,
1
a7 = –4, a0 = 8 ), then
2
1 1 7 1
(A) a1 = 8 (B) a1 = − 4 (C) a2 = 4 (D) a2 = 2
2 2 2 2
2 2 2
67. If a, b, c ∈ R, such that a + b + c < 2(ab + bc + ca), then
(A) either a, b, c are all positive or all negative (B) atleast two of a, b, c are equal
(C) none of a, b, c can be zero (D) a, b, c are all distinct
68. If the equation ax2 + bx + c = 0 and 5x2 + 12x + 13 = 0 have a common root, where a, b and c
are the sides of a ∆ABC, then
(A) ∆ABC is acute angled (B) ∆ABC is right angled
(C) ∆ABC is isosceles (D) ∆ABC is right angled isosceles
5 5
69. The number of the distinct real roots of the equation (x + 1) = 2(x + 1) is
(A) 0 (B) 1 (C) 2 (D) 3
2
70. The set of exhaustive of a for which the equation |ax – 2| = 2x + ax + 4 has atleast one
positive roots, is
(A) (–∞, 0] (B) (–∞, –2] (C) (–∞, –2] ∪ [2, ∞) (D) [2, ∞)

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71. If the quadratic equation ax2 + bx + c = 0 (a, b, c ∈ R, a ≠ 0) and x2 + 4x + 5 = 0 have
common root then a, b, c must satisfy the relation
(A) a > b > c (B) a < b < c
(C) a = k; b = 4k; c = 5k (k ∈ R, k ≠ 0) (D) b2 – 4ac is negative
(2x − 1)
72. If S is the set of all real x such that is
(
2x + 3x 2 + x
3
)
 3  3 1  1 1 1 
(A)  −∞, −  (B)  − , −  (C)  − ,  (D)  ,3 
 2  2 4  4 2 2 

(x − a)(x − b)
73. For real x, the function will assume all real values provided
x−c
(A) a > b > c (B) b ≤ c ≤ a (C) a > c > b (D) a ≤ c b

74. If x satisfies |x – 1| + |x – 2| + |x – 3| > 6, then

(A) 0 ≤ x ≤ 4 (B) x ≤ –2 or x ≤ 4 (C) x < 0 or x > 4 (D) none of these

2 2
 x   x 
75. The equation   +  = a (a – 1) has
 x + 1  x − 1
(A) four real roots if a > 2 (B) two real roots if 1 < a < 2
(C) no real root if a < –1 (D) four real roots if a < –1

76. If α, β are the roots of ax2 + bx + c = 0 and α + k, β + k are the roots of px2 + qx + r = 0, then
1b q b2 − 4ac q2 − 4pr a b c
(A) k =  −  (B) = (C) = = (D) none of these
2a p a2 p2 p q r
77. In a ∆ABC, tan A and tan B satisfy the inequation 3x 2 − 4x + 3 < 0. then
(A) a2 + b2 – ab < c2 (B) a2 + b2 > c2 (C) a2 + b2 + ab > c2 (D) all of the above
2
4   π 
78. The value of ‘x’ satisfying the equation x – 2  x sin  x   + 1 = 0
  2 
(A) 1 (B) –1 (C) 0 (D) no value of ‘x’

Integer type
3 1 1 f(a)
79. If f(x) = 27x + 3
and α, β are the roots of 3x + = 2, then − is equal to _____
x x 10

2  1
80. The equation 2(log3x) – |log3 x| + a = 0 has exactly four real solutions if a ∈  0,  , k must be
 k

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KEY & SOLUTIONS

1. D 2. A 3. A 4. A
5. D 6. D 7. A, B 8. A, B, C, D
9. A, B, D 10. A, B, C 11. A, B, C 12. A, B
13. D 14. B 15. B 16. D
17. C 18. B 19. A 20. D

1. D
We have ax 2 + bx + c > 0 ∀x ∈ R
2
a1x b1x + c1 > 0 ∀x ∈ R
2
So, b − 4ac < 0 , a > 0, c > 0
and b12 − 4a1c1 < 0 , a1 > 0, c1 > 0
2
b < 4ac and b12 < 4a1c1
b2 b12 < 16a1c1ac
2
( bb1 ) − 4a1acc1 < 12a1ac ⋅ c1 (positive)
b2 b12 − 4aa1cc1 is discriminate of aa1x 2 + bb1x + cc1 = 0
If may be negative, zero, or positive that way we can’t say nature of the root.

2. A
(x 2
) (
− 4x + 3 + λ x 2 − 6x + 8 = 0 )
x 2 (1 + λ ) − 2x ( 2 + 3λ ) + ( 3 + 8λ ) = 0
2
Discriminant. D = 4 ( 2 + 3λ ) − 4 (1 + λ )( 3 + 8λ )

(
D = 4 λ2 + λ + 1 )
if λ∈ R then D > 0
so root of given quadratic always real

3. A
4 x − ( a − 3 ) 2x + ( a − 4 ) = 0
x ≤ 0, Let y = 2x
y2 − ( a − 3 ) y + ( a − 4 ) = 0
The roots of Quadratic must lie b/w (0, 1]
2
(a) ( a − 3 ) − 4 ( a − 4 ) ≥ 0
a2 + 9 − 6a − 4a + 16 ≥ 0
a2 − 10a + 25 ≥ 0
a∈R
a−3
(b) 0 < ≤1
2
0 <a−3 ≤ 2
0<a≤5
(c) f(0) = a – 4 > 0 a > 4
(d) f(1) = 1 – a + 3 + a – 4 ≥ 0 a ∈ (4, 5]

4. A
5x2 − 10x + log1/ 5 a = 0 has real roots.
100 − 20 log1/ 5 a ≥ 0
5
 1
log1/ 5 a ≤ 5 = log1/ 5  
5

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1
a > 0 and a ≥
55
1
minimum value of a =
55

5. D
(a 2
) ( )
+ 4a + 3 x 2 + a2 − a − 2 x + ( a + 1) a = 0
2
( a + 1)( a + 3 ) x + ( a + 1)( a − 2 ) x + ( a + 1) a = 0
( a + 1) ( a + 3 ) x2 + ( a − 2 ) x + a  = 0
if a = -1, then quadratic have more than two roots.

6. D
f(x) = ax 2 + bx + 6 = 0
f(0) = 6 positive
f(x) = ax 2 + bx + 6 ≥ 0 ∀x ∈ R
f(3) = 9a + 3b + 6 ≥ 0
3(3a + b) ≥ -6
3a + b ≥ -2

7. A, B
2 2
x can not be odd integer for if x is odd, x is odd but 2px + 2q is even; so x +2 px + 2q ≠ 0
2
x can not be even integer for if x is even, x +2 px is a multiple of 4 but 2q is not.
2
So x + 2px + 2q ≠ 0
Also (x +p)2 = p2 – 2q
 If x is fraction then (x +p)2 is also a fraction but p2 –2q is an integer. So, roots cannot be
integer or rational numbers.

8. A, B, C, D
Let f(x) = ax2 +bx +c
2
Its discriminate D = b – 4ac < 0
And f(–1) = a– b +c > 0
So, f(x) > 0 for all x ∈ R
So, a > 0
f(0) >0  c > 0
f(1) > 0  a +b +c > 0
f( –2) > 0  4a – 2b + c > 0 .

9. A, B, D
2 2
D1 = b –4ac < 0, D2 = b – 4ac < 0, as the root is non–real
 Both roots will be common.
a b c
 = = =1 a=c
c b a
Now, b2 – 4ac < 0  b2 – 4a2 (or 4c2) < 0
 |b| < 2 |a| ( or 2|c|).

10. A, B, C
2
Let f(x) = ax + bx + c
–3 and 3 are lying between the roots of f(x) = 0
 f(–3) > 0 and f(3) > 0
 9a + 3|b| + c > 0
Also, 0, –2, 2 are lying between the roots
therefore c > 0 and 4a + 2|b| + c > 0

11. A, B, C
For x ≤ 0, All terms are +ve

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For 0 < x ≤ 1, x10 + x4(1 – x3) + 1 –x > 0
 Expression is positive∀ x ∈(–∞,1],
10 4 3
x + x (1 –x ) + 1 –x > 0
7 3 3
For x > 1, x (x – 1) + x(x – 1) + 1 > 0 ∴ Expression is + ve ∀ x ∈ R

12. A, B
Sum of the coefficients is (a( + (2 + 1)100 which is zero for some ( if quadratic equation (2 +
a( + 1 = 0 (in () has real roots i.e. a2 – 4 ( 0 i.e. a (–2 or a (2.

COMPREHENSION - I

13. D
14. B
15. B

Sol:- ax 2 − bx + c = 0
Let f ( x ) = ax 2 − bx + c be the corresponding quadratic
expression and α, β be the roots of f ( x ) = 0 . Then,
1 α β 2
f ( x ) = a ( x − a )( x − β )

b
Now, af (1) > 0. af ( 2 ) > 0,1 < − < 2,b2 − 4ax > 0
2a
 a (1 − α )(1 − β ) > 0, a ( 2 − α )( 2 − β ) > 0, 2a < b < 4a,b2 − 4ac > 0
 a2 (1 − α )(1 − β )( 2 − α )( 2 − β ) > 0
 a2 ( α − 1)( 2 − α )( β − 1)( 2 − β ) > 0
As f (1) and f ( 2) both are integers and f (1) > 0 , and f ( 2 ) > 0 , so
f (1) f ( 2 ) > 0  f (1) f ( 2 ) ≥ 1
2
 1≤ a ( α − 1)( 2 − α )( β − 1)( 2 − β )

( α − 1) + ( 2 − α ) 1
1
Now, ≥ ( ( α − 1)( 2 − α ) ) 2  ( α − 1) + ( 2 − α ) ≤
2 4
1
Similarly, ( β − 1)( 2 − β ) ≤
4
1
 ( α − 1)( 2 − α )( β − 1)( 2 − β ) <
16
As α ≠ β , so a2 > 16  a ≥ 5  b2 > 20c and b > 10  b ≥ 11
Also, b2 > 100  c > 5  c ≥ 6
COMPREHENSION - II

f (t ) F1 ( t ) = t 2 + 3
Sol:- Given that 9x 2 − a3 x − a + 3 ≤ 0
Let t = 3x . Then, t 2 − at − a + 3 ≤ 0
(1, 4 )
or t 2 + 3 ≤ a ( t + 1)
(0 , 3 )
where t ∈ R + , ∀x ∈ R
Let f1 ( t ) = t 2 + 3 and f2 ( t ) = a ( t + 1) . −1 O 1

16. D

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Sol:- For x < 0, t ∈ ( 0,1) . That means (1) should have atleast one solution in t ∈ ( 0,1) . From (1), it is
obvious that a ∈ R + . Now f2 ( t ) a ( t + 1) represents a straight line. It should meet the
curve f1 ( t ) = t 2 + 3 , atleast once in t ∈ ( 0,1) f1 ( 0 ) = 3, f1 (1) = 4, f2 ( 0 ) = a, f2 (1) = 2a

If f1 ( 0 ) = f2 ( 0 ) , then a = 3; if f1 (1) = f2 (1) , then a = 2 . Hence, the required range is a ∈ ( 2,3 ) .

17. C
Sol:- For atleast one positive solution, t ∈ (1, ∞ ) . That means graphs of f1 ( t ) = t 2 + 3 and
f2 ( t ) = a ( t + 1) should meet atleast once in t ∈ (1, ∞ ) . If a = 2 , both the curves touch each
other at (1, 4). Hence, the required range is a ∈ ( 2, ∞ ) .

18. B
Sol:- In this case both graphs should meet atleast once in t ∈ ( 0, ∞ ) . For a = 2 both the curves
touch, hence, the required range is a ∈ [2, ∞ ) .

ASSERTION & REASON TYPE

19. A
Sol:- ax 2 + bx + c = 0 has two complex conjugate roots only if all the coefficients are real. If all the
coefficients are not real then it is not necessary that both the roots are imaginary. Hence,
statement 1 is true.
Now, equation x 2 − 3x + 4 = 0 has two complex conjugate roots. If ax 2 + bx + c = 0 has all
coefficients real, then there will be two common roots. But it is only one root common, then
atleast one of a,b,c must be non-real.
Thus, both the statements are true and statement 2 is correct explanation of statement 1.

20. D
Sol:- Statement 2 is obviously true. Let,
f ( x ) = ( x − p )( x − r ) + λ ( x − q) ( x − s ) = 0
Then, f ( p ) = λ (p − q)(p − s )
f ( r ) = λ ( r − q)(r − s )  f (p) f (r ) < 0

21. B 22. D 23. C 24. B

25. D 26. B 27. A, B 28. A, B, C
29. A, B, D 30. A, B, C 31. B 32. A, D
33. 4 34. 4 35. 1 36. 0
37. 1 38. 0 39. A → s; B → r; C → q; D → p
40. A → r; B → r; C → q; D → p

INTEGER TYPE

33. 4
n
Sol:-  (x − k)
k =1
2
=0

number of real root is zero.

34. 4
Sol:- ( k − 2) x 2 + 8x + (k + 4 ) > 0 ∀x ∈ R
D = 64 – 4(k – 2) (k + 4) < 0 and k – 2 > 0
(k + 56) (k – 4) > 0 k>2
k < -6 or k > 4
k>4
k=5

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35. 1
Sol:- log4 ( x − 1) = log2 ( x − 3 )
x – 1 > 0 and x – 3 > 0  x>3
1
log2 ( x − 1) = log2 ( x − 3 )
2
( x − 1) = ( x − 3 )2
x − 1 = x 2 + 9 − 6x
x 2 − 7x + 10 = 0
( x − 2) ( x − 5 ) = 0
x = 2 or x = 5
x = 2 is not accept able ( ∵ x > 3)
so, no. of solution is = 1

36. 0
Sol:- x3 + 2x 2 + 5x + 2 cos x = 0
2
y = x(x + 2x + 5)

-π -π/2 π/2 π

-2
y = -2cos x

x 3 + 2x 2 + 5x = −2 cos x
( )
y = x x 2 + 2x + 5 = −2 cos x

y = x (x 2
+ 2x + 5 )
if x ∈ [0, 2π] then number of solution is zero.

37. 1
Sol:- Case I x = 3m  9m2 + 3m + 1 not diviyible by 3
Case II x = 3m + 1  9m2 + 9m + 3  divisible by 3
Case III x = 3m + 2  9m2 + 15m + 7  not divisible by 3
 If x 2 + x + 1 is divisible by 3, x = 3m + 1
∴ Remainder = 1

38. 0
Sol:- a, b, c are in G.P.
b2 = ac
one of the roots of the equation ax 2 + bx + c = 0 is real then Discriminate
D = b2 − 4ac = b2 − 4b2 = −3b2 ≥ 0
b=0

COLUMN MATCHING
39. A – s ; B– r ; C– q ; D– p ;
Sol:- Obviously when a ≥ 0 , we have no roots as all the terms are followed by + ve sign . Also
for a = −2 , we have x 2 − 2 x + 1 = 0
or

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x − 1 = 0  x = ±1
Hence, the equation has two roots.
Also when a < −2 , for given equation

−a ± a 2 − 4
x = >0
2
Hence, the equation has four roots as −a > a2 − 4 . Obviously, the equation has no three
roots for any value of a .
40. A – r ; B– r ; C– q ; D– p ;
Sol :- A. ( m − 2 ) x 2 − ( 8 − 2m ) x − ( 8 − 3m ) = 0 has roots of opposite signs. The product of roots is
8 − 3m 3m − 8
− <0  <0  2 < m < 8/3
n−2 m−2
B. Exactly one root of equation x 2 − m ( 2x − 8 ) − 15 = 0 lies in interval (0, 1)
f ( 0 ) f (1) < 0  ( 0 − m ( −8 ) − 15 ) (1 − m ( −6 ) − 15 ) < 0
 ( 8m − 15 )( 6m − 15 ) < 0  15 / 8 < m < 15 / 6
2
C. x + 2 ( m + 1) x + 9m − 5 = 0 has both roots negative. Hence, sum of roots is
−2 (m + 1) < 0 or m > −1
Product of roots is 9m − 5 > 0  m > 5 / 9
2
Discriminant, D ≥ 0  4 (m + 1) − 4 ( 9m − 5 ) ≥ 0
 m2 − 7m + 6 ≥ 0  m ≤ 1 or m ≥ 6
5 
Hence, for (1, (2) and (3), we get m ∈  ,1 ∪ [ 6, ∞ )
9 
D. f ( x ) = x 2 + 2 (m − 1) x + m + 5 = 0 has one root less than 1 and the other root greater than
1. Hence, f (1) < 0  1 + 2 ( m − 1) + m + 5 < 0
 m < −4 / 3
41. C 42. D 43. C 44. D
45. A 46. B 47. D 48. B
49. B 50. C 51. D 52. D
53. A 54. A 55. D 56. B
57. A 58. D 59. A 60. C
61. D 62. A 63. C 44. C
65. D 66. B 67. C 68. B
69. D 17. B 71. C, D 22. A, D
73. B, D 74. C 75. A, B, D 76. A, B
77. A, C 78. A, B 79. 1 80. 8

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