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

View Answer

Explanation: Let, y = (x – 1)(3 – x) = 4x – x

^{2}– 3

Then, dy/dx = 0

Or 4 – 2x = 0

Or 2x = 4

Or x = 2

Now, [d

^{2}y/dx

^{2}] = -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

View Answer

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

View Answer

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 2x^{3} + 3x^{2} – 36x + 10?

a) -1

b) -2

c) -3

d) -4

View Answer

Explanation: Let y = 2x

^{3}+ 3x

^{2}– 36x + 10 ……….(1)

Differentiating both sides of (1) with respect to x we get,

dy/dx = 6x

^{2}+ 6x – 36

And d

^{2}y/dx

^{2}= 12x + 6

For maxima or minima value of y, we have,

dy/dx = 0

Or 6x

^{2}+ 6x – 36 = 0

Or x

^{2}+ 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, d

^{2}y/dx

^{2}= 12x + 6 = 12(-3) + 6 = -30, which is < 0.

5. What will be the point of minimum of the function 2x^{3} + 3x^{2} – 36x + 10?

a) 1

b) 2

c) 3

d) 4

View Answer

Explanation: Let y = 2x

^{3}+ 3x

^{2}– 36x + 10 ……….(1)

Differentiating both sides of (1) with respect to x we get,

dy/dx = 6x

^{2}+ 6x – 36

And d

^{2}y/dx

^{2}= 12x + 6

For maxima or minima value of y, we have,

dy/dx = 0

Or 6x

^{2}+ 6x – 36 = 0

Or x

^{2}+ 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, d

^{2}y/dx

^{2}= 12x + 6 = 12(2) + 6 = 30, which is > 0.

6. What will be the maximum value of the function 2x^{3} + 3x^{2} – 36x + 10?

a) 71

b) 81

c) 91

d) 0

View Answer

Explanation: Let y = 2x

^{3}+ 3x

^{2}– 36x + 10 ……….(1)

Differentiating both sides of (1) with respect to x we get,

dy/dx = 6x

^{2}+ 6x – 36

And d

^{2}y/dx

^{2}= 12x + 6

For maxima or minima value of y, we have,

dy/dx = 0

Or 6x

^{2}+ 6x – 36 = 0

Or x

^{2}+ 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, d

^{2}y/dx

^{2}= 12x + 6 = 12(-3) + 6 = -30 < 0

Putting x = -3 in (1) we get its maximum value as,

2x

^{3}+ 3x

^{2}– 36x + 10 = 2(-3)

^{3}+ 3(-3)

^{2}– 36(-3) + 10

= 91

7. What will be the minimum value of the function 2x^{3} + 3x^{2} – 36x + 10?

a) -31

b) 31

c) -34

d) 34

View Answer

Explanation: Let y = 2x

^{3}+ 3x

^{2}– 36x + 10 ……….(1)

Differentiating both sides of (1) with respect to x we get,

dy/dx = 6x

^{2}+ 6x – 36

And d

^{2}y/dx

^{2}= 12x + 6

For maxima or minima value of y, we have,

dy/dx = 0

Or 6x

^{2}+ 6x – 36 = 0

Or x

^{2}+ 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, d

^{2}y/dx

^{2}= 12x + 6 = 12(2) + 6 = 30 > 0

Putting x = 2 in (1) we get its minimum value as,

2x

^{3}+ 3x

^{2}– 36x + 10 = 2(2)

^{3}+ 3(2)

^{2}– 36(2) + 10

= -34

8. What will be the maxima for the function f(x) = x^{4} –8x^{3} + 22x^{2} –24x + 8?

a) 0

b) 1

c) 2

d) 3

View Answer

Explanation:We have, x

^{4}–8x

^{3}+ 22x

^{2}–24x + 8 ……….(1)

Differentiating both sides of (1) with respect to x, we get,

f’(x) = 4x

^{3}– 24x

^{2}+ 44x – 24 and f”(x) = 12x

^{2}– 48x + 44 ……….(2)

At an extremum of f(x), we have f’(x) = 0

Or 4x

^{3}– 24x

^{2}+ 44x – 24 = 0

Or x

^{2}(x – 1) – 5x(x – 1) + 6(x – 1) = 0

Or (x – 1)(x

^{2}– 5x + 6) = 0

Or (x – 1)(x – 2)(x – 3) = 0

So, x = 1, 2, 3

Now, f”(x) = 12x

^{2}– 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) = x^{4} – 8x^{3} + 22x^{2} – 24x + 8?

a) -1

b) 0

c) 2

d) 3

View Answer

Explanation: We have, x

^{4}– 8x

^{3}+ 22x

^{2}– 24x + 8 ……….(1)

Differentiating both sides of (1) with respect to x, we get,

f’(x) = 4x

^{3}– 24x

^{2}+ 44x – 24 and f”(x) = 12x

^{2}– 48x + 44 ……….(2)

At an extremum of f(x), we have f’(x) = 0

Or 4x

^{3}– 24x

^{2}+ 44x – 24 = 0

Or x

^{2}(x – 1) – 5x(x – 1) + 6(x – 1) = 0

Or (x – 1)(x

^{2}– 5x + 6) = 0

Or (x – 1)(x – 2)(x – 3) = 0

So, x = 1, 2, 3

Now, f”(x) = 12x

^{2}– 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(x^{3}) – 6x^{2} + 20x – 5?

a) Possess only minimum value

b) Possess only maximum value

c) Does not possess a maximum or minimum value

d) Datainadequate

View Answer

Explanation: We have, f(x) = 2/3(x

^{3}) – 6x

^{2}+ 20x – 5 ……….(1)

Differentiating both side of (1) with respect to x, we get,

f’(x) = 2x

^{2}– 12x + 20

Now, for a maximum and minimum value of f(x) we have,

f’(x) = 0

Or 2x

^{2}– 12x + 20 = 0

Or x

^{2}– 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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