The derivative of the function f(x)= ^{4} x+ ^{4} x, 0 ≤ x ≤ 2 π is positive for

Correct Answer is : π4<x<π2π/4 < x < π/24π<x<2π

Correct Answer is : π4<x<π2π/4 < x < π/24π<x<2π

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MHTCET 17 Apr 2026 Shift 1 Paper

Question : 124 of 150

Marks: +1, -0

f(x)=\cos ^{4} x+\sin ^{4} x, \quad 0 \leq x \leq 2 \pi

is positive for

Solution:

Concept:

Differentiate the given function and find the interval where

f'(x) > 0

Explanation:

Given

f(x) = \cos^4 x + \sin^4 x

Differentiating:

f'(x) = 4\cos^3 x(-\sin x) + 4\sin^3 x(\cos x)

f'(x) = 4\sin x\cos x(\sin^2 x - \cos^2 x)

Using

\sin^2 x - \cos^2 x = -\cos 2x

and

2\sin x\cos x = \sin 2x

f'(x) = -2\sin 2x \cos 2x = -\sin 4x

We need

f'(x) > 0

, so

-\sin 4x > 0

, i.e.,

\sin 4x < 0

For Option B:

\frac{\pi}{4} < x < \frac{\pi}{2}

Multiplying by 4:

\pi < 4x < 2\pi

(\pi, 2\pi)

\sin 4x < 0

, hence

f'(x) > 0

Answer:

The derivative is positive for

\frac{\pi}{4} < x < \frac{\pi}{2}

Correct option: B.

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