Question

Prove that the function given by $f\left(x\right)=\cos x$ is
$\left(a\right)$ decreasing in $(0,\pi )$
$\left(b\right)$ increasing in $(\pi ,2\pi )$, and
$\left(c\right)$ neither increasing nor decreasing in $(0,2\pi )$

Answer 1

Given:

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

Differentiating w.r.t. $x$, we get

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

When $xϵ(0,\pi ),$ then

$f^{\prime }\left(x\right)<0[\because \sin x>0]$

Therefore, $f\left(x\right)$ is decreasing in $(0,\pi )$

When $xϵ(\pi ,2\pi ),$then

$f^{\prime }\left(x\right)>\left)\right[\because \sin x<0]$

Therefore, $f\left(x\right)$ is increasing in $(\pi ,2\pi )$

When $xϵ(0,2\pi )$, then

From above we know that $f\left(x\right)$ is increasing in $(\pi ,2\pi )$ and decreasing in $(0,\pi )$, so
$f\left(x\right)$ is neither increasing nor decreasing in $(0,2\pi )$
Hence all statements are proved.