Question

The function $\sin^{4}x + \cos^{4}x$ increase if

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

Answer 1

Answer: (B)

$\frac{\pi}{4} < x < \frac{3\pi}{8}$

Answer 2

$f(x) = \sin^{4}x + \cos^{4}x$= $(\sin^{2}x + \cos^{2}x)^{2} - 2\sin^{2}x\cos^{2}x$

$= 1 - \frac{4\sin^{2}x\cos^{2}x}{2} = 1 - \frac{\sin^{2}2x}{2} = 1 - \frac{1}{4}(2\sin^{2}2x)$

$= 1 - \left( \frac{1 - \cos 4x}{4} \right) = \frac{3}{4} + \frac{1}{4}\cos 4x$

Hence function $f(x)$ is increasing when $f^{'}(x) > 0$

$f^{'}(x) = - \sin 4x > 0$ ⇒ $\sin 4x < 0$

Hence $\pi < 4x < \frac{3\pi}{2}$ or $\frac{\pi}{4} < x < \frac{3\pi}{8}$.