This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Derivatives Application – Increasing and Decreasing Functions”.

1. What is the nature of function f(x) = 7x-4 on R?

a) Increasing

b) Decreasing

c) Strictly Increasing

d) Increasing and Decreasing

View Answer

Explanation: Let x1 and x2 be any two numbers in R.

Then x1<x2 => 7×1 < 7×2.

=> 7×1 – 4 < 7×2 – 4.

As f(x1) < f(x2), thus the function f is strictly increasing on R.

2. What is the nature of function f(x) = x^{3} – 3x^{2} + 4x on R?

a) Increasing

b) Decreasing

c) Constant

d) Increasing and Decreasing

View Answer

Explanation: f(x) = x

^{3}– 3x

^{2}+ 4x

f’(x) = 3x

^{2}– 6x + 4.

f’(x) = 3(x

^{2}– 2x + 1) + 1.

=> 3(x-1)

^{2}+ 1>0, in every interval of R. Therefore the function f is increasing on R.

3. Find the interval in which function f(x) = x^{2} – 4x + 5 is increasing.

a) (2, ∞)

b) (-∞, 2)

c) (3, ∞)

d) (-∞, ∞)

View Answer

Explanation: f(x) = x

^{2}– 4x + 5.

f’(x) = 2x – 4. Therefore f’(x) = 0 gives x = 2.

Now this point x=2 divides the line into two disjoint intervals and the interval namely (2, ∞) is increasing on f(x).

4. Find the interval in which function f(x) = x^{2} – 4x + 5 is decreasing.

a) (2, ∞)

b) (-∞, 2)

c) (3, ∞)

d) (-∞, ∞)

View Answer

Explanation: f(x) = x

^{2}– 4x + 5.

f’(x) = 2x – 4. Therefore f’(x) = 0 gives x = 2.

Now this point x=2 divides the line into two disjoint intervals and the interval namely (-∞, 2) is decreasing on f(x).

5. Find the interval in which function f(x) = sinx+cosx, 0 ≤ x ≤ 2π is decreasing.

a) (π/4, 5π/4)

b) (-π/4, 5π/4)

c) (π/4, -5π/4)

d) (-π/4, π/4)

View Answer

Explanation: f(x) = sinx+cosx.

f’(x) = cosx – sinx. Now f’(x) = 0 gives sinx = cosx which gives that x=π/4, 5π/4 as 0 ≤ x ≤ 2π.

Therefore on checking the values we get f is decreasing in (π/4, 5π/4).

6. Find the interval in which function f(x) = sinx+cosx is increasing.

a) (5π/4, 2π)

b) [0, π/4) and (5π/4, 2π]

c) (π/4, -5π/4)

d) (-π/4, π/4)

View Answer

Explanation: f(x) = sinx+cosx.

f’(x) = cosx – sinx. Now f’(x) = 0 gives sinx = cosx which gives that x= π/4, 5π/4 as 0 ≤ x ≤ 2π.

The points x = π/4 and x = 5π/4 divide the interval [0, 2π] into three disjoint intervals which are

[0, π/4), (π/4, 5π/4) and (5π/4, 2π].

Therefore on checking the values we get f is increasing in [0, π/4) and (5π/4, 2π].

7. Is the function f(x) = 3x+10 is increasing on R?

a) True

b) False

View Answer

Explanation: f(x) = 3x+10.

f’(x) = 3, which shows 3 > 0 for all x ∈ R.

Thus function f(x) is increasing.

8. Find the intervals in which f(x) = 2x^{2} – 3x is increasing.

a) (-1/4, ∞)

b) (-3/4, ∞)

c) (1/4, ∞)

d) (3/4, ∞)

View Answer

Explanation: f(x) = 2x

^{2}-3x.

f’(x) = 4x – 3.

As we know f’(x) = 0, x=-3/4. This shows that function f is increasing in interval (-3/4, ∞) for all x ∈ R.

9. Find the intervals in which f(x) = x^{2} + 2x – 5 is strictly increasing.

a) x>1

b) x<-1

c) x>-1

d) x>2

View Answer

Explanation: f(x) = x

^{2}+2x -5.

f’(x) = 4x – 3.

As we know f’(x) = 0, x=-3/4, which shows that function f is increasing in interval (-3/4, ∞) for all x ∈ R.

10. Nature of the function f(x) = e^{2x} is _______

a) increasing

b) decreasing

c) constant

d) increasing and decreasing

View Answer

Explanation: f(x) = e

^{2x}.

f’(x) = 2e

^{2x}.

As we know 2e

^{2x}> 0, so it always has a value greater than zero.

Which shows that function f is increasing for all x ∈ R.

**Sanfoundry Global Education & Learning Series – Mathematics – Class 12**.

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