Question

The function $f(x) = \sin^4 x + \cos^4 x$ increases if}

• A. $0<x<\frac{\pi}{8}$
• B. $\frac{\pi}{4}<x<\frac{\pi}{2}$
• C. $\frac{3\pi}{8}<x<\frac{5\pi}{8}$
• D. $\frac{5\pi}{8}<x<\frac{3\pi}{4}$

Hint:

$\sin^4 x + \cos^4 x = 1 - \frac{1}{2} \sin^2 2x$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A function increases where its derivative $f'(x)>0$.
We simplify the function first to differentiate it easily.
Step 2: Key Formula or Approach:
$\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2 2x$.
Derivative of $\sin^2 ax$ is $2a \sin ax \cos ax = a \sin 2ax$.
Step 3: Detailed Explanation:
$f(x) = 1 - \frac{1}{2}\sin^2 2x$.
$f'(x) = -\frac{1}{2}(2 \sin 2x \cdot \cos 2x \cdot 2) = - \sin 4x$.
Function increases if $f'(x)>0 \implies -\sin 4x>0 \implies \sin 4x<0$.
This happens for $\pi<4x<2\pi \implies \pi/4<x<\pi/2$.
Step 4: Final Answer:
The function increases for $\frac{\pi}{4}<x<\frac{\pi}{2}$.