Question

The length of the longest interval, in which $f\left(x\right)=3\sin \left(x\right)-4{\sin }^{3}x$ is increasing, is

• A. $\frac{\mathrm{\pi }}{3}$
• B. $\frac{\mathrm{\pi }}{2}$
• C. $\frac{3\mathrm{\pi }}{2}$
• D. $\mathrm{\pi }$

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{3}$

The explanation for the correct option

Given function, $f\left(x\right)=3\sin \left(x\right)-4{\sin }^{3}x$.

$\Rightarrow f\left(x\right)=\sin \left(3x\right)$

Differentiate the given function with respect to $x$.

$$\begin{gathered} \frac{\mathrm{d}}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left[\sin \left(3x\right)\right] \\ \Rightarrow f\text{'}\left(x\right)=3\cos \left(3x\right) \end{gathered}$$

Now, for increasing intervals $\mathit{f}\mathbf{\text{'}}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{>}\mathbf{0}$.

$$\begin{gathered} \Rightarrow 3\cos \left(3x\right)>0 \\ \Rightarrow \cos \left(3x\right)>0 \\ \Rightarrow 3x\in ......\cup \left(-\frac{\mathrm{\pi }}{2},\frac{\mathrm{\pi }}{2}\right)\cup \left(\frac{3\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{2}\right)\cup \left(\frac{7\mathrm{\pi }}{2},\frac{9\mathrm{\pi }}{2}\right)\cup ...... \\ \Rightarrow \mathrm{x}\in ......\cup \left(-\frac{\mathrm{\pi }}{6},\frac{\mathrm{\pi }}{6}\right)\cup \left(\frac{\mathrm{\pi }}{2},\frac{5\mathrm{\pi }}{6}\right)\cup \left(\frac{7\mathrm{\pi }}{6},\frac{3\mathrm{\pi }}{2}\right)\cup ...... \end{gathered}$$

Thus, the longest interval, in which $f\left(x\right)=3\sin \left(x\right)-4{\sin }^{3}x$ is increasing can be given by $\left[\frac{\mathrm{\pi }}{6}-\left(-\frac{\mathrm{\pi }}{6}\right)\right]=\left(\frac{\mathrm{\pi }}{6}+\frac{\mathrm{\pi }}{6}\right)=\frac{\mathrm{\pi }}{3}$.

Therefore, the length of the longest interval, in which $f\left(x\right)=3\sin \left(x\right)-4{\sin }^{3}x$ is increasing, is $\frac{\mathrm{\pi }}{3}$.