The function {f}(x)= ^{4} x+ ^{4} x increases if

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

5 π/8 {aligned} & {f}(x)= ^{4} x+ ^{4} x \\ & {f}^{}(x)=4 ³ x x-4 ³ x x \\ & =-4 x x≤ft( ² x- ² x) \\ & =-2 2 x 2 x \\ & =- 4 x \\ & { Function is increasing when } \\ & - 4 x > 0 \\ & 4 x 0

The function {f}(x)= ^{4} x+ ^{4} x increases if

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

Test Index

MHTCET 22 Apr 2025 Shift 2 Paper

Question : 112 of 150

Marks: +1, -0

The function

\mathrm{f}(x)=\sin ^{4} x+\cos ^{4} x

increases if

$0 < x < \frac{\pi}{8}$
$\frac{\pi}{4} < x < \frac{\pi}{2}$
$\frac{3 \pi}{8} < x < \frac{5 \pi}{8}$
$\frac{5 \pi}{8} < x < \frac{3 \pi}{4}$
$\begin{aligned} & \mathrm{f}(x)=\sin ^{4} x+\cos ^{4} x \\ & \Rightarrow \mathrm{f}^{\prime}(x)=4 \sin ^{3} x \cos x-4 \cos ^{3} x \sin x \\ & =-4 \sin x \cos x\left(\cos ^{2} x-\sin ^{2} x\right) \\ & =-2 \sin 2 x \cos 2 x \\ & =-\sin 4 x \\ & \text { Function is increasing when } \\ & \Rightarrow-\sin 4 x > 0 \\ & \Rightarrow \sin 4 x < 0 \\ & \Rightarrow \pi < 4 x < 2 \pi \\ & \Rightarrow \frac{\pi}{4} < x < \frac{\pi}{2} \end{aligned} \mathrm{f}^{\prime}(x) > 0$

Validate

Solution:

Concept:

A function increases where its derivative is positive:

\mathrm{f}'(x) > 0

Explanation:

Given:

\mathrm{f}(x)=\sin^4 x+\cos^4 x

Differentiate:

\mathrm{f}'(x)=4\sin^3 x\cos x-4\cos^3 x\sin x

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

=-2\sin 2x \cos 2x=-\sin 4x

For increasing,

\mathrm{f}'(x)>0

-\sin 4x > 0 \Rightarrow \sin 4x < 0

\Rightarrow \pi < 4x < 2\pi

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

This matches option B.

Answer:

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

Go to Question:

Prev Question

Next Question