Question

The length of the longest interval, in which the function $3sinx-4si{n}^{3}x$ is increasing, is

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

Answer 1

Answer: (A)

Let $f(x)=3sinx-4si{n}^{3}x=si{n}^{3}x$
Since, $sinx$ is increasing in the interval $\left[-\frac{\pi }{2},\frac{\pi }{2}\right]$.
$\therefore -\frac{\pi }{2}\le 3x\le \frac{\pi }{2}$
$\Rightarrow -\frac{\pi }{6}\le x\le \frac{\pi }{6}$
Thus, length of interval $=\left|\frac{\pi }{6}-\left(-\frac{\pi }{6}\right)\right|$
$=\frac{\pi }{3}$