# Class 12 Maths MCQ – Calculus Application – Maxima and Minima – 1

This set of Class 12 Maths Chapter 14 Multiple Choice Questions & Answers (MCQs) focuses on “Calculus Application – Maxima and Minima – 1”.

1. For which value of x will (x – 1)(3 – x) have its maximum?
a) 0
b) 1
c) 2
d) -2

Explanation: Let, y = (x – 1)(3 – x) = 4x – x2 – 3
Then, dy/dx = 0
Or 4 – 2x = 0
Or 2x = 4
Or x = 2
Now, [d2y/dx2] = -2 which is negative.
Therefore, (x – 1)(3 – x) will have its maximum at x = 2.

2. What will be the values of x for which the value of cosx is minimum?
a) (2m + 1)π
b) (2m)π
c) (2m + 1)π/2
d) (2m – 1)π

Explanation: Let, f(x) = cosx
Then, f’(x) = -sinx and f”(x) = -cosx
At an extreme point of f(x), we must have,
f’(x) = 0
Or -sinx = 0
Or x = nπ where, n is any integer.
If n is an odd integer i.e., n = 2m + 1 where m is any integer, then at,
x = (2m + 1)π we have, f”(x) = [(2m + 1)π] = -cos(2mπ + π) = -cosπ = -1(-1) = 1
So, f”(x) is positive at x = (2m + 1)π
Hence, f(x) = cosx is minimum at x = (2m + 1)π.

3. What will be the value of x for which the value of cosx is minimum?
a) 0
b) -1
c) 1
d) Cannot be determined

Explanation: Let, f(x) = cosx
Then, f’(x) = -sinx and f”(x) = -cosx
At an extreme point of f(x) we must have,
f’(x) = 0
Or -sinx = 0
Or x = nπ where, n – any integer.
If n is an odd integer i.e., n = 2m + 1 where m is any integer, then at,
x = (2m + 1)π we have, f”(x) = [(2m + 1)π] = -cos(2mπ + π) = -cosπ = -1(-1) = 1
So, f”(x) is positive at x = (2m + 1)π
Hence, f(x) = cosx is minimum at x = (2m + 1)π.
So, the minimum value of cosx is cos(2mπ + π) = cosπ = -1.

4. What will be the point of maximum of the function 2x3 + 3x2 – 36x + 10?
a) -1
b) -2
c) -3
d) -4

Explanation: Let y = 2x3 + 3x2 – 36x + 10 ……….(1)
Differentiating both sides of (1) with respect to x we get,
dy/dx = 6x2 + 6x – 36
And d2y/dx2 = 12x + 6
For maxima or minima value of y, we have,
dy/dx = 0
Or 6x2 + 6x – 36 = 0
Or x2 + x – 6 = 0
Or (x + 3)(x – 2) = 0
Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2
Now, d2y/dx2 = 12x + 6 = 12(-3) + 6 = -30, which is < 0.

5. What will be the point of minimum of the function 2x3 + 3x2 – 36x + 10?
a) 1
b) 2
c) 3
d) 4

Explanation: Let y = 2x3 + 3x2 – 36x + 10 ……….(1)
Differentiating both sides of (1) with respect to x we get,
dy/dx = 6x2 + 6x – 36
And d2y/dx2 = 12x + 6
For maxima or minima value of y, we have,
dy/dx = 0
Or 6x2 + 6x – 36 = 0
Or x2 + x – 6 = 0
Or (x + 3)(x – 2) = 0
Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2
Now, d2y/dx2 = 12x + 6 = 12(2) + 6 = 30, which is > 0.
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6. What will be the maximum value of the function 2x3 + 3x2 – 36x + 10?
a) 71
b) 81
c) 91
d) 0

Explanation: Let y = 2x3 + 3x2 – 36x + 10 ……….(1)
Differentiating both sides of (1) with respect to x we get,
dy/dx = 6x2 + 6x – 36
And d2y/dx2 = 12x + 6
For maxima or minima value of y, we have,
dy/dx = 0
Or 6x2 + 6x – 36 = 0
Or x2 + x – 6 = 0
Or (x + 3)(x – 2) = 0
Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2
Now, d2y/dx2 = 12x + 6 = 12(-3) + 6 = -30 < 0
Putting x = -3 in (1) we get its maximum value as,
2x3 + 3x2 – 36x + 10 = 2(-3)3 + 3(-3)2 – 36(-3) + 10
= 91

7. What will be the minimum value of the function 2x3 + 3x2 – 36x + 10?
a) -31
b) 31
c) -34
d) 34

Explanation: Let y = 2x3 + 3x2 – 36x + 10 ……….(1)
Differentiating both sides of (1) with respect to x we get,
dy/dx = 6x2 + 6x – 36
And d2y/dx2 = 12x + 6
For maxima or minima value of y, we have,
dy/dx = 0
Or 6x2 + 6x – 36 = 0
Or x2 + x – 6 = 0
Or (x + 3)(x – 2) = 0
Therefore, either x + 3 = 0 i.e., x = -3 or x – 2 = 0 i.e., x = 2
Now, d2y/dx2 = 12x + 6 = 12(2) + 6 = 30 > 0
Putting x = 2 in (1) we get its minimum value as,
2x3 + 3x2 – 36x + 10 = 2(2)3 + 3(2)2 – 36(2) + 10
= -34

8. What will be the maxima for the function f(x) = x4 –8x3 + 22x2 –24x + 8?
a) 0
b) 1
c) 2
d) 3

Explanation:We have, x4 –8x3 + 22x2 –24x + 8 ……….(1)
Differentiating both sides of (1) with respect to x, we get,
f’(x) = 4x3 – 24x2 + 44x – 24 and f”(x) = 12x2 – 48x + 44 ……….(2)
At an extremum of f(x), we have f’(x) = 0
Or 4x3 – 24x2 + 44x – 24 = 0
Or x2(x – 1) – 5x(x – 1) + 6(x – 1) = 0
Or (x – 1)(x2 – 5x + 6) = 0
Or (x – 1)(x – 2)(x – 3) = 0
So, x = 1, 2, 3
Now, f”(x) = 12x2 – 48x + 44
f”(1) = 8 > 0
f”(2) = -4 < 0
f”(3) = 8 < 0
So, f(x) has maximum at x = 2.

9. What will be the minima for the function f(x) = x4 – 8x3 + 22x2 – 24x + 8?
a) -1
b) 0
c) 2
d) 3

Explanation: We have, x4 – 8x3 + 22x2 – 24x + 8 ……….(1)
Differentiating both sides of (1) with respect to x, we get,
f’(x) = 4x3 – 24x2 + 44x – 24 and f”(x) = 12x2 – 48x + 44 ……….(2)
At an extremum of f(x), we have f’(x) = 0
Or 4x3 – 24x2 + 44x – 24 = 0
Or x2(x – 1) – 5x(x – 1) + 6(x – 1) = 0
Or (x – 1)(x2 – 5x + 6) = 0
Or (x – 1)(x – 2)(x – 3) = 0
So, x = 1, 2, 3
Now, f”(x) = 12x2 – 48x + 44
f”(1) = 8 > 0
f”(2) = -4 < 0
f”(3) = 8 < 0
So, f(x) has minimum at x = 1 and 3.

10. What is the nature of the function f(x) = 2/3(x3) – 6x2 + 20x – 5?
a) Possess only minimum value
b) Possess only maximum value
c) Does not possess a maximum or minimum value

Explanation: We have, f(x) = 2/3(x3) – 6x2 + 20x – 5 ……….(1)
Differentiating both side of (1) with respect to x, we get,
f’(x) = 2x2 – 12x + 20
Now, for a maximum and minimum value of f(x) we have,
f’(x) = 0
Or 2x2 – 12x + 20 = 0
Or x2 – 6x + 10= 0
So, x = [6 ± √(36 – 4*10)]/2
x = (6 ± √-4)/2, which is imaginary.
Hence, f’(x) does not vanishes at any point of x.
Thus, f(x) does not possess a maximum or minimum value.

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