Question:
The function $f(x) = \sin^4 x + \cos^4 x$ increases if}
The function $f(x) = \sin^4 x + \cos^4 x$ increases if}
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$\sin^4 x + \cos^4 x = 1 - \frac{1}{2} \sin^2 2x$.
Updated On: Apr 26, 2026
- $0<x<\frac{\pi}{8}$
- $\frac{\pi}{4}<x<\frac{\pi}{2}$
- $\frac{3\pi}{8}<x<\frac{5\pi}{8}$
- $\frac{5\pi}{8}<x<\frac{3\pi}{4}$
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The Correct Option is B
Solution and Explanation
Step 1: Simplify Function
$f(x) = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$
$f(x) = 1 - \frac{1}{2} (2 \sin x \cos x)^2 = 1 - \frac{1}{2} \sin^2 2x$.
Step 2: Differentiate
$f'(x) = -\frac{1}{2} (2 \sin 2x \cos 2x \cdot 2) = - \sin 4x$.
Step 3: Condition for Increase
$f'(x)>0 \implies - \sin 4x>0 \implies \sin 4x<0$.
This happens when $\pi<4x<2\pi \implies \frac{\pi}{4}<x<\frac{\pi}{2}$.
Final Answer: (B)
$f(x) = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x$
$f(x) = 1 - \frac{1}{2} (2 \sin x \cos x)^2 = 1 - \frac{1}{2} \sin^2 2x$.
Step 2: Differentiate
$f'(x) = -\frac{1}{2} (2 \sin 2x \cos 2x \cdot 2) = - \sin 4x$.
Step 3: Condition for Increase
$f'(x)>0 \implies - \sin 4x>0 \implies \sin 4x<0$.
This happens when $\pi<4x<2\pi \implies \frac{\pi}{4}<x<\frac{\pi}{2}$.
Final Answer: (B)
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