Let [x] denote the greatest integer less than or equal to x and g(x) be given by

$g(x)=\left\{\begin{array}{cl} [f(x)], & x \in(0, \pi / 2) \cup(\pi / 2, \pi) \\ 3 \quad, & x=\frac{\pi}{2} \end{array}\right.$

where, $f(x)=\frac{2\left(\sin x-\sin ^n x\right)+\left|\sin x-\sin ^n x\right|}{2\left(\sin x-\sin ^n x\right)-\left|\sin x-\sin ^n x\right|}, n \in R^{+},$

then at $x=\frac{\pi}{2}, g(x)$ is

continuous and differentiable when n > 1

Clearly,

$0<\sin x<1$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$.

CASE I:  When n > 1

In this case, we have

$\sin x>\sin ^n x$  for all  $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$

$\Rightarrow \sin x-\sin ^n x>0$  for all  $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$

$\Rightarrow \left|\sin x-\sin ^n x\right|=\sin x-\sin ^n x$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$

∴  $f(x)=\frac{2\left(\sin x-\sin ^n x\right)+\left(\sin x-\sin ^n x\right)}{2\left(\sin x-\sin ^n x\right)-\left(\sin x-\sin ^n x\right)}=3 $

$\Rightarrow [f(x)]=3$  for all  $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$

Thus, we have

$g(x)=3$  for all  $x \in(0, \pi)$

Clearly, it is continuous and differentiable at $x=\pi / 2$

CASE II: When 0 < n < 1

In this case, we have

$\sin x<\sin ^n x$  for all  $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$

$\Rightarrow \sin x-\sin ^n x<0$  for all  $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$

$\Rightarrow \left|\sin x-\sin ^n x\right|=-\left(\sin x-\sin ^n x\right)$ for all $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$

∴  $f(x)=\frac{2\left(\sin x-\sin ^n x\right)-\left(\sin x-\sin ^n x\right)}{2\left(\sin x-\sin ^n x\right)+\left(\sin x-\sin ^n x\right)}=\frac{1}{3}$

$\Rightarrow [f(x)]=0$  for all  $x \in(0, \pi / 2) \cup(\pi / 2, \pi)$

Thus, we have

$g(x)= \begin{cases}0, & \text { for all } x \in(0, \pi / 2) \cup(\pi / 2, \pi) \\ 3, & \text { for } x=\pi / 2\end{cases}$

Clearly, it is discontinuous and hence non-differentiable also at $x=\pi / 2$.