Theory of Equations 33

6. If b > a, then the equation (x − a ) (x − b) − 1 = 0 has Analytical & Descriptive Questions

(a) both roots in (a , b) (2000, 1M)
9. If x2 + (a − b)x + (1 − a − b) = 0 where a , b ∈ R, then find
(b) both roots in ( − ∞ , a )
the values of a for which equation has unequal real
(c) both roots in (b, + ∞ )
roots for all values of b. (2003, 4M)
(d) one root in (−∞ , a ) and the other in (b, ∞ )
7. If the roots of the equation x2 − 2ax + a 2 + a − 3 = 0 are 10. Let a , b, c be real. If ax2 + bx + c = 0 has two real roots α
real and less than 3, then (1999, 2M)
and β, where α < − 1 and β > 1, then show that
c b
(a) a < 2 1 + +  < 0
(b) 2 ≤ a ≤ 3 a a  (1995, 5M)
(c) 3 < a ≤ 4 11. Find all real values of x which satisfy x2 − 3x + 2 > 0 and
(d) a> 4 x2 − 2x − 4 ≤ 0. (1983, 2M)
8. Let f (x) be a quadratic expression which is positive for
all real values of x. If g (x) = f (x) + f ′ (x) + f ′ ′ (x), then for Integer & Numerical Answer Type Question
any real x (1990, 2M)
12. The smallest value of k, for which both the roots of the
(a) g (x) < 0 (b) g (x) > 0 equation x2 − 8kx + 16 (k2 − k + 1) = 0 are real, distinct
(c) g (x) = 0 (d) g (x) ≥ 0 and have values atleast 4, is …… . (2009)

Topic 5 Some Special Forms

Objective Questions I (Only one correct option) 7. The largest interval for which
1. The number of real roots of the equation x12 − x9 + x4 − x + 1 > 0 is (1982, 2M)
5 + |2 − 1| = 2 (2 − 2) is
x x x
(2019 Main, 10 April II) (a) −4 < x ≤ 0
(a) 1 (b) 3 (b) 0 < x < 1
(c) 4 (d) 2 (c) −100 < x < 100
(d) −∞ < x < ∞
2. All the pairs (x, y) that satisfy the inequality
sin 2 x − 2 sin x + 5 1 8. Let a , b, c be non-zero real numbers such that
2 ⋅ ≤ 1 also satisfy the equation
sin 2 y 1
∫0 (1 + cos x)(ax2 + bx + c)dx
4 (2019 Main, 10 April I) 8
(1981, 2M)
(a) 2|sin x| = 3 sin y (b) sin x = |sin y| 2
(c) sin x = 2 sin y (d) 2 sin x = sin y = ∫ (1 + cos 8 x)(ax2 + bx + c)dx
0
3. The sum of the solutions of the equation Then, the quadratic equation ax2 + bx + c = 0 has
| x − 2| + x ( x − 4) + 2 = 0 (x > 0) is equal to (a) no root in (0, 2)
(2019 Main, 8 April I) (b) atleast one root in (1, 2)
(a) 9 (b) 12 (c) a double root in (0, 2)
(c) 4 (d) 10 (d) two imaginary roots
4. The real number k for which the equation,
2x3 + 3x + k = 0 has two distinct real roots in [0, 1] Objective Questions II
(2013 Main) (One or more than one correct option)
(a) lies between 1 and 2 (b) lies between 2 and 3
9. Let S be the set of all non-zero real numbers α such that
(c) lies between − 1and 0 (d) does not exist
the quadratic equation αx2 − x + α = 0 has two distinct
5. Let a , b, c be real numbers, a ≠ 0. If α is a root of real roots x1 and x2 satisfying the inequality x1 – x2 < 1.
a 2x2 + bx + c = 0, β is the root of a 2x2 − bx − c = 0 and Which of the following interval(s) is/are a subset of S?
0 < α < β, then the equation a 2x2 + 2bx + 2c = 0 has a root (2015 Adv.)
(a)  – , –
1 
γ that always satisfies (b)  – , 0
(1989, 2M) 1 1

α+β β  2 5  5 
(a) γ = (b) γ = α +
(c)  0,
1 
(d) 
2 2 1 1
 , 
(c) γ = α (d) α < γ < β  5  5 2
6. If a + b + c = 0, then the quadratic equation 10. Let a ∈ R and f : R → R be given by f (x) = x5 − 5x + a.
3ax2 + 2bx + c = 0 has (1983, 1M)
Then,
(a) at least one root in (0, 1) (a) f (x) has three real roots, if a > 4
(b) one root in (2, 3) and the other in (−2, − 1) (b) f (x) has only one real root, if a > 4
(c) imaginary roots (c) f (x) has three real roots, if a < − 4
(d) None of the above (d) f (x) has three real roots, if −4 < a < 4
34 Theory of Equations

Passage Based Problems equation f ( x ) = 0 has a root in R. Consider f ( x ) = kex − x for

all real x where k is real constant. (2007, 4M)
Read the following passage and answer the questions.
14. The line y = x meets y = kex for k ≤ 0 at
Passage I
(a) no point
Consider the polynomial f ( x ) = 1 + 2x + 3x 2+ 4x3 . Let s be (b) one point
the sum of all distinct real roots of f ( x ) and let t =| s|. (c) two points
(2010) (d) more than two points
11. The real numbers s lies in the interval 15. The positive value of k for which kex − x = 0 has only one
(a)  − , 0 (b)  − 11, −  (c)  − , −  (d)  0,
1 3 3 1 1 root is

 4   4  4 2  4 1
(a) (b) 1
12. The area bounded by the curve y = f (x) and the lines e
(c) e (d) log e 2
x = 0, y = 0 and x = t , lies in the interval
(a)  , 3
3
(b)  , 
21 11
(c) (9, 10) (d)  0, 
21 16. For k > 0, the set of all values of k for which kex − x = 0
4   64 16   64  has two distinct roots, is
(a)  0,  (b)  , 1
1 1
13. The function f ′ (x) is
 e e 
(a) increasing in  − t, −  and decreasing in  − , t 
1 1
(c)  , ∞ 
1
 4  4  (d) (0, 1)
e 
(b) decreasing in  − t, −  and increasing in  − , t 
1 1
 4  4 
(c) increasing in (−t , t )
True/False
(d) decreasing in (−t , t ) 17. If a < b < c < d, then the roots of the equation (x − a )
(x − c) + 2 (x − b) (x − d ) = 0 are real and distinct.
Passage II
(1984, 1M)
If a continuous function f defined on the real line R,
assumes positive and negative values in R, then the Analytical & Descriptive Question
equation f ( x ) = 0 has a root in R. For example, if it is 18. Let −1 ≤ p < 1. Show that the equation 4x3 − 3x − p = 0
known that a continuous function f on R is positive at some has a unique root in the interval [1/2, 1] and identify it.
point and its minimum values is negative, then the (2001, 4M)

Answers
Topic 1 Topic 2
1. (d) 2. (d) 3. (d) 4. (a) 1. (d) 2. (a) 3. (b) 4. –1
5. (b) 6. (b) 7. (c) 8. (a) 5. False
9. (c) 10. (c) 11. (d) 12. (b)
Topic 3
13. (d) 14. (d) 15. (c) 16. (c)
1. (c) 2. x = α 2β, αβ 2
17. (c) 18. (d) 19. (a) 20. (c)
21. (c) 22. (d) 23. (d) 24. (c) Topic 4
25. (b) 26. (d) 27. (a) 28. (a) 1. (d) 2. (a) 3. (a) 4. (a)
29. (b) 30. (b) 31. (c) 32. (b) 5. (b) 6. (d) 7. (a) 8. (b)
33. (a) 34. (b) 35. (a) 36. (a) 9. a > 1 11. x ∈ [1 − 5, 1 ) ∪ [1 + 5, 2 )
37. (d) 38. (a) 39. (c) 40. (b)
12. k = 2
41. (b) 42. 4 43. k =2 44. ( −4, 7 )
45. –5050 46. True 47. False 48. 1210 Topic 5
 2 1 1. (a) 2. (b) 3. (d) 4. (d)
51. y ∈ { −1 } ∪ [1, ∞ ) 52. x ∈ ( −2, − 1 ) ∪  − , − 
 3 2 5. (d) 6. (a) 7. (d) 8. (b)
53. −4 and ( − 1 − 3 ) 54. x = {a (1 − 2 ), a ( 6 − 1 )} 9. (a,d) 10. (b, d) 11. (c) 12. (a)
55. ±2, ± 2 57. (q − s ) 2 − rqp − rsp + sp 2 + qr 2 13. (b) 14. (b) 15. (a) 16. (a)
58. x = a −1/ 2 − 4 /3
59. q − p
2 2 1 
or a 60. (d) 17. True 18. x = cos cos−1 p
3 
61. (d)
 Hints & Solutions
Topic 1 Quadratic Equations Q The root x =
1
∈ (0, 1) for λ = 3
1. Given quadratic polynomials, x + 20x − 2020 and
2 5
∴ λ ∈ (1 , 3].
x2 − 20x + 2020 having a , b distinct real and c, d distinct
complex roots respectively. Hence, option (d) is correct.
So, a + b = − 20, ab = − 2020 4. Given quadratic equation is
and c + d = 20, cd = 2020  π
x2 + x sin θ − 2 sin θ = 0, θ ∈ 0, 
Now, ac(a − c) + ad (a − d ) + bc(b − c) + bd (b − d )  2
= a 2(c + d ) − a (c2 + d 2) + b2(c + d ) − b(c2 + d 2) and its roots are α and β.
= (c + d ) (a 2 + b2) − (c2 + d 2)(a + b) So, sum of roots = α + β = − sin θ
= (c + d )[(a + b)2 − 2ab] − (a + b) [(c + d )2 − 2cd ] and product of roots = αβ = − 2 sin θ
= 20 [(20)2 + 4040] + 20 [(20)2 − 4040] ⇒ αβ = 2(α + β ) …(i)
= 2 × 20 × (20)2 = 40 × 400 = 16000 α 12 + β12
Now, the given expression is −12
2. It is given that α is a common roots of given quadratic (α + β −12)(α − β)24
equations α 12 + β12 α 12 + β12
x2 – x + 2λ = 0 and 3x2 – 10x + 27λ = 0 = = 12
 1 1 24  β + α 12
∴ 3 α – 10 α + 27λ = 0
2  12 + 12 (α − β)  12 12  (α − β)24
α β   α β 
3 α 2 − 3 α + 6λ = 0
12 12
– + –  αβ   αβ 
= 2
= 
0 – 7α + 21λ = 0 ⇒ α = 3λ  (α + β ) − 4αβ
 (α − β ) 
2

So, 9λ2 – 3λ + 2λ = 0
12
1 1  2(α + β) 
⇒ λ= ⇒α= [Qλ ≠ 0] =  [from Eq. (i)]
 (α + β ) − 8 (α + β) 
9 3 2

1
2×  
12
 
12
2λ 9 =2 =
2
=
2
[Q α + β = − sin θ]
As αβ = 2λ ⇒ β = =  
α 1 3  (α + β ) − 8  − sin θ − 8
3 212
1 =
9× (sin θ + 8)12
9λ 9 =3
and αγ = 9λ ⇒ γ = =
α 1 5. Given quadratic equation is x2 + px + q = 0, where
3 p, q ∈R having one root 2 − 3 , then other root is 2 + 3
2 (conjugate of 2 − 3 ) [Q irrational roots of a quadratic
×3
βγ 3 equation always occurs in pairs]
∴ = = 18
λ 1 So, sum of roots = − p = 4 ⇒ p = −4
9 and product of roots = q = 4 − 3 ⇒ q = 1
3. Given quadratic equations Now, from options p2 − 4q − 12 = 16 − 4 − 12 = 0
f (x) = (λ2 + 1)x2 − 4λx + 2 = 0 have exactly one root in 6. Given quadratic equation is
the interval (0, 1).
(m2 + 1)x2 − 3x + (m2 + 1)2 = 0 …(i)
So, D > 0 ⇒ 16λ2 − 4(λ2 + 1)2 > 0
Let the roots of quadratic Eq. (i) are α and β, so
⇒ 8λ − 8 > 0 ⇒ λ2 > 1
2
3
α+β= and αβ = m2 + 1
⇒ λ ∈ (−∞ , − 1) ∪ (1, ∞ ) …(i) m2 + 1
and f (0) f (1) < 0
According to the question, the sum of roots is greatest
⇒ 2(λ2 + 1 − 4λ + 2) < 0 ⇒ λ2 − 4λ + 3 < 0
and it is possible only when ‘‘(m2 + 1) is minimum’’ and
⇒ (λ − 3)(λ − 1) < 0 ⇒ λ ∈ (1, 3) …(ii) ‘‘minimum value of m2 + 1 = 1, when m = 0’’.
From Eqs, (i) and (ii), we get ∴α + β = 3 and αβ = 1, as m = 0
λ ∈ (1, 3) Now, the absolute difference of the cubes of roots
And if λ = 3, then the quadratic equation is = |α 3 − β3|
10x2 − 12x + 2 = 0 = |α − β||α 2 + β 2 + αβ|
⇒ 5 x2 − 6 x + 1 = 0 = (α + β )2 − 4αβ |(α + β )2 − αβ|
1 = 9 − 4 |9 − 1|= 8 5
⇒ x = 1,
5