Mathematics Questions and Answers – Applications of Quadratic Equations

«
»

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)
View Answer

Answer: b
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, ∞)
advertisement

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
View Answer

Answer: d
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) (-∞, ∞)
View Answer

Answer: a
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)
advertisement
advertisement

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
View Answer

Answer: a
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
View Answer

Answer: b
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
advertisement

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
View Answer

Answer: c
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
View Answer

Answer: d
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.
advertisement

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

Answer: a
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
View Answer

Answer: d
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.
advertisement

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

Answer: d
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, ∞)
View Answer

Answer: b
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
View Answer

Answer: a
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, ∞)
View Answer

Answer: c
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]
View Answer

Answer: b
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
View Answer

Answer: c
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.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter