Q61 / 150 · MHT CET 2024 (02 May Shift 1) · Mathematics
The length of the longest interval, in which the function $3 \sin x-4 \sin ^3 x$ is increasing, is
- Correct answerA$\frac{\pi}{3}$
- B$\frac{\pi}{2}$
- C$\frac{3 \pi}{2}$
- D$\pi$
Solution
$\begin{aligned}
& \mathrm{f}(x)=3 \sin x-4 \sin ^3 x \\
& \Rightarrow \mathrm{f}(x)=\sin 3 x \\
& \Rightarrow \mathrm{f}^{\prime}(x)=3 \cos 3 x
\end{aligned}$
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 $=\frac{\pi}{6}-\left(\frac{-\pi}{6}\right)=\frac{\pi}{3}$
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 $=\frac{\pi}{6}-\left(\frac{-\pi}{6}\right)=\frac{\pi}{3}$
Hence, the correct answer is option: 'A'.