A) \[\frac{\pi }{3}\]
B) \[\frac{\pi }{2}\]
C) \[\frac{3\pi }{2}\]
D) \[\pi \]
Correct Answer: A
Solution :
\[3\sin x-4{{\sin }^{3}}x=\sin 3x\] It is increasing, when \[-\pi /2\le 3x\le \pi /2\] i.e.,\[-\pi /6\le x\le \pi /6\]. The length of interval = \[\left| \,\frac{\pi }{6}-\left( -\frac{\pi }{6} \right)\, \right|\,=\,\frac{\pi }{3}\].
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