This set of Class 11 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Applications of Quadratic Equations”.

1. If, (a + 1)x^{2} + 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)

View Answer

Explanation: For, real and distinct roots, D > 0

Where, D = b

^{2}– 4ac

In the equation, (a + 1)x

^{2}+ 2(a+1)x + (a – 2) = 0

D = [2(a+1)]

^{2}– 4 (a + 1)(a – 2)

= 4a

^{2}+ 4 + 8a – 4{a

^{2}– 2a + a – 2}

= 4a

^{2}+ 4 + 8a – 4a

^{2}+ 4 a + 8 > 0

=> 12a + 12 > 0

=> 12a > -12

=> a > -1

Therefore, a € (-1, ∞)

2. If, (a + 1)x^{2} + 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

View Answer

Explanation: For, equal roots, D = 0

Where, D = b

^{2}– 4ac

In the equation, (a + 1)x

^{2}+ 2(a+1)x + (a – 2) = 0

D = [2(a+1)]

^{2}– 4 (a + 1)(a – 2)

= 4a

^{2}+ 4 + 8a – 4{a

^{2}– 2a + a – 2}

= 4a

^{2}+ 4 + 8a – 4a

^{2}+ 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)x^{2} + 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) (-∞, ∞)

View Answer

Explanation: For, imaginary roots, D > 0

Where, D = b

^{2}– 4ac

In the equation, (a + 1)x

^{2}+ 2(a+1)x + (a – 2) = 0

D = [2(a+1)]

^{2}– 4 (a + 1)(a – 2)

= 4a

^{2}+ 4 + 8a – 4{a

^{2}– 2a + a – 2}

= 4a

^{2}+ 4 + 8a – 4a

^{2}+ 4 a + 8 < 0

=> 12a + 12 < 0

=> 12a < -12

=> a < -1

Therefore, a € (-∞, -1)

4. If x_{1}, x_{2} are real roots of ax^{2} – x + a = 0. Then, find the set of all values of parameter ‘a’ for which |x_{1} – x_{1}| < 1?

a) (1 – 5a)/ a^{2} < 0

b) (1 – 5a)/ a^{2} = 0

c) (1 – 5a)/ a^{2} > 0

d) (1 – 5a)/ a < 0

View Answer

Explanation: |x

_{1}– x

_{2}| < 1

= (x

_{1}– x

_{2})

^{2}< 1

= (x

_{1}+ x

_{2})

^{2}– 4 x

_{1}x

_{2}-1 < 0

= (1/a)

^{2}– 4 – 1 < 0

= (1 – 5a)/a

^{2}< 0.

5. What is the value of x if (a + 2b – 3c)x^{2} + (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

View Answer

Explanation: If the coefficients of (x

^{2}+ 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) = Ax^{2} + Bx + C = 0?

a) Natural Number

b) Unpredictable

c) Integer

d) Complex Number

View Answer

Explanation: f(x) = Ax

^{2}+ Bx + C = 0

So, f(x) = Ax

^{2}+ (A + B)x – Ax + C

= Ax

^{2}– 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 x^{2} + 5|x| + 6 = 0?

a) Equal to 5

b) Equal to 10

c) Equal to -5

d) Does not exist

View Answer

Explanation: Since, x

^{2}, 5|x| and 6 are positive,

So, x

^{2}+ 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

View Answer

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 (x^{2} – 1)/(x – 2)(x – 3)?

a) -4

b) -3

c) -1

d) -2

View Answer

Explanation: By wavy curve method

(x

^{2}– 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, ∞)

View Answer

Explanation: If, x

^{2}> 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 3x^{2} + 2x + p(p – 1) = 0 are of opposite sign?

a) (-∞, 0)

b) (0, 1)

c) (1, ∞)

d) (0, ∞)

View Answer

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 3x^{2} – 3sinθ x – 2cos^{2}θ = 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

View Answer

Explanation: Let, f(x) = 3x

^{2}– 3sinθ x – 2cos

^{2}θ

The coefficient of x

^{2}> 0

f(1) < 0

So, 3 – 3sinθ – 2cos

^{2}θ < 0

=> 2sin

^{2}θ – 3sinθ + 1 < 0

=> (2sinθ – 1)(sinθ – 1) < 0

=> ½

13. A real number ‘a’ is called a good number if the inequality (2x^{2} – 2x – 3) / (x^{2} + 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, ∞)

View Answer

Explanation: We have, (2x

^{2}– 2x – 3) / (x

^{2}+ x + 1) ≤ a ᵾ x € R

=> 2x

^{2}– 2x – 3 ≤ a(x

^{2}+ x + 1) ᵾ x € R

=> (2 – a)x

^{2}– (2 – a)x – (3 – a) ᵾ x € R

2 – a < 0 and (2 – a)x

^{2}– 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 log_{1/2} x^{2} ≥ log_{1/2} (x + 2) is the solution of the inequality 49x^{2} – 4a^{4} ≤ 0. What is the value of S?

a) (-∞, -√7) ∪ (√7, ∞)

b) (-∞, -√7] ∪ [√7, ∞)

c) (-√7, √7)

d) [-√7, √7]

View Answer

Explanation: We have, log

_{1/2}x

^{2}≥ log

_{1/2}(x + 2)

=> x

^{2}≤ x + 2

=> -1 ≤ x ≤ 2

And, 49x

^{2}– 4a

^{4}≤ 0 i.e. x

^{2}≤ 4a

^{4}/ 49

=> -2a

^{2}/7 ≤ a ≤ 2a

^{2}/7

From the above equations,

-2a

^{2}/7 ≤ -1 and 2 ≤ 2a

^{2}/7

i.e. a

^{2}€ 7/2 and a

^{2}≥ 7

=> a € (-∞, -√7] ∪ [√7, ∞)

So, S = (-∞, -√7] ∪ [√7, ∞)

15. If, x^{4} + 4x^{3} + 6ax^{2} + 6bx + c is divisible by x^{3} + 3x^{2} + 9x + 3. Then, what is the value of a + b + c?

a) 4

b) 6

c) 7

d) 10

View Answer

Explanation: Here, f(x) = x

^{4}+ 4x

^{3}+ 6ax

^{2}+ 6bx + c so, let its roots be, α, β, γ, δ and

g(x) = x

^{3}+ 3x

^{2}+ 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)(x

^{3}+ 3x

^{2}+ 9x + 3)

On solving this equation in simpler form we get,

x

^{4}+ 4x

^{3}+ 12x

^{2}+ 12x + 3

=> 6a = 12 => a = 2

=> 6b = 12 => b = 2

=> c = 3

=> a + b + c = 7

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