Let [x] denote the greatest integer less than or equal to x. Now g(x) is defined as below:

$g\left(x\right)=\{\begin{array}{c}\left[f\left(x\right)\right],x\in (0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi )\\ { }_{3,x=\frac{\pi }{2}}\end{array}$  where  $f\left(x\right)=\frac{2(\sin x-{\sin }^{n}x)+|\sin x-{\sin }^{n}x|}{2(\sin x-{\sin }^{n}x)-|\sin x-{\sin }^{n}x|},n\in R.$ Then which of the following is/are correct 

If n > 1, sin x >sinnx. If 0 < n < 1, sin x <sinnx.
$\therefore$  if   $n>1,f\left(x\right)=\frac{2(\sin x-{\sin }^{n}x)+(\sin x-{\sin }^{n}x)}{2(\sin x-{\sin }^{n}x)-(\sin x-{\sin }^{n}x)}=3$
if   $\begin{array}{l}0<n<1,f\left(x\right)=\frac{2(\sin x-{\sin }^{n}x)-(\sin x-{\sin }^{n}x)}{2(\sin x-{\sin }^{n}x)+(\sin x-{\sin }^{n}x)}=\frac{1}{3}\\ \end{array}$
$\therefore$ if   $n>1,g\left(x\right)=3,x\in (0,\pi )$
$\therefore$  $g\left(x\right)$ is continuous and differentiable at  $x=\frac{\pi }{2}$

If    $0<n<1,g\left(x\right)=0,x\in (0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi )3,x=\frac{\pi }{2}$
Then  $g(\frac{\pi }{2}+0)=0,g(\frac{\pi }{2}-0)=0,g\left(\frac{\pi }{2}\right)=3$ So, $g\left(x\right)$ is not continuous at   $x=\frac{\pi }{2}$

Hence, $g\left(x\right)$ is also not differentiable at   $x=\frac{\pi }{2}$