Question

Show that the function given by $f\left(x\right)=sinx$ is
strictly increasing in $(0,\frac{\pi }{2})$

strictly decreasing in $(\frac{\pi }{2},\pi )$

neither increasing nor decreasing in $(0,\pi )$

Answer 1

Since, for each $xϵ(0,\frac{\pi }{2}),cosx>0$; have $f^{\prime }\left(x\right)>0$
($\because$ cos x in 1st quadrant is positive)
Hence, f is strictly increasing on $(0,\frac{\pi }{2})$.

Since, for each $xϵ(\frac{\pi }{2},\pi ),cosx<0$; we have f'(x)<0
($\because$ cos x in IInd quadrant is negative)
Hence, f is strictly decreasing on $(\frac{\pi }{2},\pi )$.

When $xϵ(0,\pi )$. We see that $f^{\prime }\left(x\right)>0$ in $\left(\frac{\pi }{2}\right)$ and $f^{\prime }\left(x\right)<0in(\frac{\pi }{2},\pi )$
So, f'(x) is positive and negative in $(0,\pi )$.
Thus, f(x) is neither increasing nor decreasing in $(0,\pi )$