250+ TOP MCQs on Derivatives Application – Increasing and Decreasing Functions | Class 12 Maths

Mathematics Multiple Choice Questions 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
Answer: c
Clarification: Let x1 and x2 be any two numbers in R.
Then x1 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) = x³ – 3x² + 4x on R?
a) Increasing
b) Decreasing
c) Constant
d) Increasing and Decreasing
Answer: a
Clarification: f(x) = x³ – 3x² + 4x
f’(x) = 3x² – 6x + 4.
f’(x) = 3(x² – 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) = x² – 4x + 5 is increasing.
a) (2, ∞)
b) (-∞, 2)
c) (3, ∞)
d) (-∞, ∞)
Answer: a
Clarification: f(x) = x² – 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) = x² – 4x + 5 is decreasing.
a) (2, ∞)
b) (-∞, 2)
c) (3, ∞)
d) (-∞, ∞)
Answer: b
Clarification: f(x) = x² – 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)
Answer: a
Clarification: 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)
Answer: b
Clarification: 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
Answer: a
Clarification: 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) = 2x² – 3x is increasing.
a) (-1/4, ∞)
b) (-3/4, ∞)
c) (1/4, ∞)
d) (3/4, ∞)
Answer: d
Clarification: f(x) = 2x²-3x.
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) = x² + 2x – 5 is strictly increasing.
a) x>1
b) x<-1
c) x>-1
d) x>2
Answer: c
Clarification: f(x) = x²+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
Answer: a
Clarification: 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.