Question

f(x)=sinx+cos2x is not monotonic on the interval[0,π/2]​

Answer 1

Question :-

• Prove that f(x)=sinx+cos2x is not monotonic on the interval[0,π/2]

Answer

Given :-

• f(x) = sinx + cos2x on [0,π/2]

To prove :-

• f(x) is not monotonic on [0,π/2].

Definition :-

• A function that is completely increasing or completely decreasing on the given interval is called monotonic on the given interval

Method :-

The derivative of a function may be used to determine whether the function is increasing or decreasing on any intervals in its domain.

If f′(x) > 0 at each point in an interval I, then the function is said to be increasing on I.

If f′(x) < 0 at each point in an interval I, then the function is said to be decreasing on I.

Let f be a differentiable function on the interval (a,b) then 

• If f '(x) < 0 for x in (a,b), then f is decreasing there.
• If f '(x) > 0 for x in (a,b), then f is increasing there.
• If f '(x) = 0 for x in (a,b), then f is constant.

Solution :-

Consider, f(x) = sinx + cos2x on [0,π/2]

Differentiate w. r. t. x, we get

⇛ f'(x) = cosx - 2 sin2x

⇛ f'(x) = cosx - 2 × 2sinx cosx

■ By applying x = 0, we get

⇛ f'(0) = cos 0 - 2 × 2 × sin0 cos0

⇛ f'(0) = 1 - 0 = 1

⇛ f'(0) > 0

■ By applying x = π/4, we get

⇛ f'(π/4) = cos π/4 - 2 × 2 × sin(π/4) × cos(π/4)

⇛ f'(π/4) = cos π/4 × (1 - 4 × sin π/4)

⇛ f'(π/4) = cos π/4 × (1 - 2 × √2)

⇛ f'(π/4) < 0

Thus, f' is of different signs at 0 and π/4.

So, the given function is not monotonic function on the interval [0,π/2].