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).

Answer 1

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