The length of the longest interval, in which the function $3 \sin x-4 \sin ^3 x$ is increasing, is

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{f}(x)=3 \sin x-4 \sin ^3 x \\ & \Rightarrow \mathrm{f}(x)=\sin 3 x \\ & \Rightarrow \mathrm{f}^{\prime}(x)=3 \cos 3 x \end{aligned}$
For $\mathrm{f}(x)$ to be increasing, $\begin{aligned} & \mathrm{f}^{\prime}(x) \geq 0 \\ & \Rightarrow 3 \cos 3 x \geq 0 \\ & \Rightarrow \cos 3 x \geq 0 \\ & \Rightarrow \frac{-\pi}{2} \leq 3 x \leq \frac{\pi}{2} \\ & \Rightarrow \frac{-\pi}{6} \leq x \leq \frac{\pi}{6} \end{aligned}$ $\therefore \quad$ largest length in which $\mathrm{f}(x)$ is increasing $=\frac{\pi}{6}-\left(\frac{-\pi}{6}\right)=\frac{\pi}{3}$

The length of the longest interval, in which the function $3 \sin x-4 \sin ^3 x$ is increasing, is

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