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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
\[\text { Here,} \]
\[f\left( x \right) = \sin x\]
\[\text { Domain of sin x is }\left( 0, \pi \right).\]
\[f'\left( x \right) = \cos x\]
\[\text { For x } \in \left( 0, \frac{\pi}{2} \right), \cos x > 0 \left[ \because \cos x\text { is positive in first quadrant} \right]\]
\[f'\left( x \right) > 0\]
\[\text { So,f(x)is increasing for
}\left( 0, \frac{\pi}{2} \right) . \]
\[\text { For x} \in \left( \frac{\pi}{2}, \pi \right), \cos x < 0 \left[ \because \cos x\text { is negative in second quadrant } \right]\]
\[\text { So,f(x)is decreasing for }\left( \frac{\pi}{2}, \pi \right).\]
\[\text { Since }f(x)\text { is increasing on } \left( 0, \frac{\pi}{2} \right) \text { and decreasing on}\left( \frac{\pi}{2}, \pi \right), f\left( x \right) \text { is neither decreasing nor increasing on }\left( 0, \pi \right).\]
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