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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Applications of Quadratic Equations”.

1. If, (a + 1)x2 + 2(a+1)x + (a – 2) = 0, then, for what parameter of ‘a’ the given equation have real and distinct roots?
a) (-∞, ∞)
b) (-1, ∞)
c) [-1, ∞)
d) (-1, 1)

Explanation: For, real and distinct roots, D > 0
Where, D = b2 – 4ac
In the equation, (a + 1)x2 + 2(a+1)x + (a – 2) = 0
D = [2(a+1)]2 – 4 (a + 1)(a – 2)
= 4a2 + 4 + 8a – 4{a2 – 2a + a – 2}
= 4a2 + 4 + 8a – 4a2 + 4 a + 8 > 0
=> 12a + 12 > 0
=> 12a > -12
=> a > -1
Therefore, a € (-1, ∞)

2. If, (a + 1)x2 + 2(a+1)x + (a – 2) = 0, then, for what parameter of ‘a’ the given equation have equal roots?
a) (-∞, -1)
b) [-1, ∞)
c) (0, 1)
d) Not possible

Explanation: For, equal roots, D = 0
Where, D = b2 – 4ac
In the equation, (a + 1)x2 + 2(a+1)x + (a – 2) = 0
D = [2(a+1)]2 – 4 (a + 1)(a – 2)
= 4a2 + 4 + 8a – 4{a2 – 2a + a – 2}
= 4a2 + 4 + 8a – 4a2 + 4 a + 8 > 0
=> 12a + 12 = 0
=> 12a = -12
=> a = -1
So, from here it is clear that a = -1 is not possible because the equation is becoming linear.

3. If, (a + 1)x2 + 2(a+1)x + (a – 2) = 0, then, for what parameter of ‘a’ the given equation have imaginary roots?
a) (-∞, -1)
b) (-1, ∞)
c) (-1, 1)
d) (-∞, ∞)

Explanation: For, imaginary roots, D > 0
Where, D = b2 – 4ac
In the equation, (a + 1)x2 + 2(a+1)x + (a – 2) = 0
D = [2(a+1)]2 – 4 (a + 1)(a – 2)
= 4a2 + 4 + 8a – 4{a2 – 2a + a – 2}
= 4a2 + 4 + 8a – 4a2 + 4 a + 8 < 0
=> 12a + 12 < 0
=> 12a < -12
=> a < -1
Therefore, a € (-∞, -1)
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4. If x1, x2 are real roots of ax2 – x + a = 0. Then, find the set of all values of parameter ‘a’ for which |x1 – x1| < 1?
a) (1 – 5a)/ a2 < 0
b) (1 – 5a)/ a2 = 0
c) (1 – 5a)/ a2 > 0
d) (1 – 5a)/ a < 0

Explanation: |x1 – x2| < 1
= (x1 – x2)2 < 1
= (x1 + x2)2 – 4 x1 x2 -1 < 0
= (1/a)2 – 4 – 1 < 0
= (1 – 5a)/a2 < 0.

5. What is the value of x if (a + 2b – 3c)x2 + (b + 2c – 3a)x + (c + 2a – 3b) = 0 where a, b, c are in A.P?
a) 1/2
b) 1/4
c) 2/3
d) 3/4

Explanation: If the coefficients of (x2 + x + c) = 0, then
x will always be = 1
Therefore, here, (a + 2b – 3c) + (b + 2c – 3a) + (c + 2a – 3b) = 0
So, x = 1.
As, one of its root is 1 so, we will calculate the other one.
As, a, b, c are in A.P so,
b = (a + c)/2
Thus, product of the roots αβ = (c + 2a – 3b)/(a + 2b – 3c)
As, a root say α = 1, then,
β = (c + 2a – 3(a + c)/2) / (a + 2(a + c)/2 – 3c)
We get the value of β = 1/4

6. What will be the value of f(x) if, 2A, A + B, C are integers and f(x) = Ax2 + Bx + C = 0?
a) Natural Number
b) Unpredictable
c) Integer
d) Complex Number

Explanation: f(x) = Ax2 + Bx + C = 0
So, f(x) = Ax2 + (A + B)x – Ax + C
= Ax2 – Ax + (A + B)x + C
= 2Ax(x – 1)/2 + (A + B)x + C
Therefore, f(x) is an integer.

7. What will be the sum of the real roots of the equation x2 + 5|x| + 6 = 0?
a) Equal to 5
b) Equal to 10
c) Equal to -5
d) Does not exist

Explanation: Since, x2, 5|x| and 6 are positive,
So, x2 + 5|x| + 6 = 0 does not have any real root
Therefore, sum does not exist.

8. What is the number of solution(s) of the equation |√x – 2| + √x(√x – 4) + 2 = 0?
a) 2
b) 4
c) No solution
d) Infinitely many solutions

Explanation: We have |√x – 2| + √x(√x – 4) + 2 = 0,
|√x – 2| + (√x)2 – 4√x + 2 = 0
|√x – 2| + |√x -2|2 – 2 = 0
|√x – 2| = -2, 1
Thus, √x – 2 = +1, -1 or x = 1, 9

9. Which is the largest negative integer which satisfies (x2 – 1)/(x – 2)(x – 3)?
a) -4
b) -3
c) -1
d) -2

Explanation: By wavy curve method
(x2 – 1)/(x – 2)(x – 3) > 0
So, x = -1, 1, 2, 3
Thus, x € (-∞, -1) ∪ (1, 2) ∪ (3, ∞)
Therefore, the largest negative integer is -2.

10. Which one is the complete set of values of x satisfying log x2 (x + 1) > 0?
a) (1, ∞)
b) (-1, 0) – {0}
c) (-1, 1) – {0}
d) (1, 0) ∪ (1, ∞)

Explanation: If, x2 > 1, then x + 1 > 0
So, x > 0
x € (1, ∞)
If, 0 < x < 1, the 0 < x + 1 < 1
x € (-1, 0)
Thus, x € (1, ∞) ∪ (1, ∞)

11. What is the set of values of p for which the roots of the equation 3x2 + 2x + p(p – 1) = 0 are of opposite sign?
a) (-∞, 0)
b) (0, 1)
c) (1, ∞)
d) (0, ∞)

Explanation: Since the roots of the given equation are of opposite sing,
So, products f the roots < 0
p(p – 1) / 3 < 0
p(p – 1) < 0
0 < p < 1
For real and distinct roots ½ – √21/ 6 < p < ½ + √21/6

12. For what value of θ, 1 lies between the roots of the quadratic equation 3x2 – 3sinθ x – 2cos2θ = 0?
a) 2nπ + π/6 < θ < 2nπ + 5π/6
b) 2nπ + π/3 < θ < 2nπ + 5π/3
c) 2nπ + π/6 ≤ θ ≤ 2nπ + 5π/6
d) 2nπ + π/3 ≤ θ ≤ 2nπ + 5π/3

Explanation: Let, f(x) = 3x2 – 3sinθ x – 2cos2θ
The coefficient of x2 > 0
f(1) < 0
So, 3 – 3sinθ – 2cos2θ < 0
=> 2sin2θ – 3sinθ + 1 < 0
=> (2sinθ – 1)(sinθ – 1) < 0
=> ½ 2nπ + π/6 < θ < 2nπ + 5π/6

13. A real number ‘a’ is called a good number if the inequality (2x2 – 2x – 3) / (x2 + x + 1) ≤ a is satisfied for all real x. What is the set of all real numbers?
a) (-∞, 10/3]
b) (10/3, ∞)
c) [10/3, ∞)
d) [-10/3, ∞)

Explanation: We have, (2x2 – 2x – 3) / (x2 + x + 1) ≤ a ᵾ x € R
=> 2x2 – 2x – 3 ≤ a(x2 + x + 1) ᵾ x € R
=> (2 – a)x2 – (2 – a)x – (3 – a) ᵾ x € R
2 – a < 0 and (2 – a)x2 – 4(2 – a)(3 – a) ≤ 0 ᵾ x € R
So, a > 2 and a ≤ 2 or a ≥ 10/3
=> a ≥ 10/3
Therefore, a € [10/3, ∞)

14. Let S denotes the set of all real values of the parameter ‘a’ for which every solution of the inequality log1/2 x2 ≥ log1/2 (x + 2) is the solution of the inequality 49x2 – 4a4 ≤ 0. What is the value of S?
a) (-∞, -√7) ∪ (√7, ∞)
b) (-∞, -√7] ∪ [√7, ∞)
c) (-√7, √7)
d) [-√7, √7]

Explanation: We have, log1/2 x2 ≥ log1/2 (x + 2)
=> x2 ≤ x + 2
=> -1 ≤ x ≤ 2
And, 49x2 – 4a4 ≤ 0 i.e. x2 ≤ 4a4 / 49
=> -2a2/7 ≤ a ≤ 2a2/7
From the above equations,
-2a2/7 ≤ -1 and 2 ≤ 2a2/7
i.e. a2 € 7/2 and a2 ≥ 7
=> a € (-∞, -√7] ∪ [√7, ∞)
So, S = (-∞, -√7] ∪ [√7, ∞)

15. If, x4 + 4x3 + 6ax2 + 6bx + c is divisible by x3 + 3x2 + 9x + 3. Then, what is the value of a + b + c?
a) 4
b) 6
c) 7
d) 10

Explanation: Here, f(x) = x4 + 4x3 + 6ax2 + 6bx + c so, let its roots be, α, β, γ, δ and
g(x) = x3 + 3x2 + 9x + 3 so, let its roots be, α, β, γ
So, from here we can conclude,
α + β + γ + δ = -4 and α + β + γ = -3
Thus, δ = -1
This means (x + 1)(x3 + 3x2 + 9x + 3)
On solving this equation in simpler form we get,
x4 + 4x3 + 12x2 + 12x + 3
=> 6a = 12 => a = 2
=> 6b = 12 => b = 2
=> c = 3
=> a + b + c = 7

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

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