Q.24. The function f(x) = sin4x + cos4x is increasing in
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}
Answer :- B. \frac{\pi}{4}<x< \frac{\pi}{2}
Explanation :-
\mathrm{f}(x) = \sin^{4}x + \cos^{4}x \mathrm{f}^{\prime}(x) = 4\sin^3x\cos x - 4\cos^3x\sin x = 4\sin x\cos x (\sin^2x - \cos^2x) = -2\sin2x\cos2x = -\sin4x \text{If } f(x) \text{ is increasing, then } f^{\prime}(x) > 0 \text{i.e., } -\sin4x > 0 \Rightarrow \pi < 4x< 2\pi \Rightarrow \frac{\pi }{4}< x< \frac{\pi }{2}