Answer:
\[f\,(x)=\sin 3x\Rightarrow f'(x)=3\cos 3x\] ?(i) Also, \[0\le x\le \frac{x}{2}\Leftrightarrow 0\le 3x\le \frac{3\pi }{2}\] ?(ii) (a) \[f\,(x)\]is increasing \[\Leftrightarrow f'(x)\ge 0\Leftrightarrow 3\cos 3x\ge 0\Leftrightarrow \cos 3x\ge 0\][from Eq. (i)] \[\Leftrightarrow 0\le 3x\le \frac{\pi }{2}\Leftrightarrow 0\le x\le \frac{\pi }{6}\Leftrightarrow x\in \left[ 0,\,\,\frac{\pi }{6} \right]\] \[f\,(x)\]is increasing on \[\left[ 0,\,\,\frac{\pi }{6} \right]\] (b) \[f\,(x)\]is increasing \[\Leftrightarrow f'(x)\le 0\] \[\Leftrightarrow 3\cos 3x\le 0\Leftrightarrow \cos 3x\le 0\] [from Eq. (i)] \[\Leftrightarrow \frac{\pi }{2}\le 3x\le \frac{3\pi }{2}\Leftrightarrow \frac{\pi }{6}\le x\le \frac{\pi }{2}\Leftrightarrow x\in \left[ \frac{\pi }{6},\,\,\frac{\pi }{2} \right]\] \[f\,(x)\]is decreasing on\[\left[ \frac{\pi }{6},\,\,\frac{\pi }{2} \right]\] Hence, \[f\,(x)\] is increasing on \[\left[ 0,\,\,\frac{\pi }{6} \right]\] and decreasing on\[\left[ \frac{\pi }{6},\,\,\frac{\pi }{2} \right]\].
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