#### Question

Show that f(x) = sin x is increasing on (0, π/2) and decreasing on (π/2, π) and neither increasing nor decreasing in (0, π) ?

#### Solution

\[\text { Here,} \]

\[f\left( x \right) = \sin x\]

\[\text { Domain of sin x is }\left( 0, \pi \right).\]

\[f'\left( x \right) = \cos x\]

\[\text { For x } \in \left( 0, \frac{\pi}{2} \right), \cos x > 0 \left[ \because \cos x\text { is positive in first quadrant} \right]\]

\[f'\left( x \right) > 0\]

\[\text { So,f(x)is increasing for

}\left( 0, \frac{\pi}{2} \right) . \]

\[\text { For x} \in \left( \frac{\pi}{2}, \pi \right), \cos x < 0 \left[ \because \cos x\text { is negative in second quadrant } \right]\]

\[\text { So,f(x)is decreasing for }\left( \frac{\pi}{2}, \pi \right).\]

\[\text { Since }f(x)\text { is increasing on } \left( 0, \frac{\pi}{2} \right) \text { and decreasing on}\left( \frac{\pi}{2}, \pi \right), f\left( x \right) \text { is neither decreasing nor increasing on }\left( 0, \pi \right).\]