Question

Show that the function given by $f\left(x\right)=\sin x$ is
(a) increasing in $(0,\frac{\pi }{2})$
(b) strictly decreasing $(\frac{\pi }{2},\pi )$
(c)neither increasing nor decreasing in $(0,\pi )$

Answer 1

$f\left(x\right)=\sin x$

Differentiation w.r.t $x,$

$f^{\prime }\left(x\right)=\cos x$

When $xϵ(0,\frac{\pi }{2})$

$\Rightarrow \cos x>0$

$\Rightarrow f^{\prime }\left(x\right)>0$

So, $f\left(x\right)$ is increasing

When $xϵ(\frac{\pi }{2},\pi )$

$\Rightarrow \cos x<0$

$\Rightarrow f^{\prime }\left(x\right)<0$

So, $f\left(x\right)$ is decreasing

When $xϵ(0,\pi )$

As we have found that $f\left(x\right)$ is increasing in

$(0,\frac{\pi }{2})$

and decreasing in

$(\frac{\pi }{2},\pi )$

so, $f\left(x\right)$ is neither increasing nor decreasing in $(0,\pi )$