question_answer

Separate \[\left[ 0,\,\,\frac{\pi }{2} \right]\] into subintervals in which \[f(x)=\sin \,3x\] is (a) Increasing                  (b) decreasing

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\[\left[ \frac{\pi }{6},\,\,\frac{\pi }{2} \right]\].