The function $\mathrm{f}(x)=\sin ^4 x+\cos ^4 x$ is increasing in

The function $\mathrm{f}(x)=\sin ^4 x+\cos ^4 x$ is increasing in
  1. $0 < x < \frac{\pi}{8}$
  2. $\frac{\pi}{4} < x < \frac{\pi}{2}$
  3. $\frac{3 \pi}{8} < x < \frac{5 \pi}{8}$
  4. $\frac{5 \pi}{8} < x < \frac{3 \pi}{4}$

Solution

$\begin{aligned} \therefore \quad \mathrm{f}(x) & =\sin ^4 x+\cos ^4 x \\ \therefore \quad \mathrm{f}^{\prime}(x) & =4 \sin ^3 x \cos x-4 \cos ^3 x \sin x \\ & =4 \sin x \cos x\left(\sin ^2 x-\cos ^2 x\right) \\ & =+2 \sin 2 x \cos 2 x \\ & =-\sin 4 x \end{aligned}$ $\therefore \quad$ If $\mathrm{f}(x)$ is increasing, then $\mathrm{f}^{\prime}(x)>0$ $\text { i.e., } \begin{aligned} -\sin 4 x>0 & \Rightarrow \pi < 4 x < 2 \pi \\ & \Rightarrow \frac{\pi}{4} < x < \frac{\pi}{2} \end{aligned}$

Asked in: MHT CET 2023 (12 May Shift 1)

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