If $g(x)=\left\{\begin{matrix}[f(x)],&x∈(0,\frac{π}{2})∪(\frac{π}{2},π)\\3,&x=π/2\end{matrix}\right.$ where [*] denotes the greatest integer function and $f(x)=\frac{2(\sin x-\sin^nx)+|\sin x-\sin^nx|}{2(\sin x-\sin^nx)-|\sin x-\sin^nx|},n∈R$, then

g(x) is continuous and differentiable at $x = π/2$, when $n > 1$

For $0 < n < 1, \sin x < \sin^n x$ 

and for $n > 1, \sin x > \sin^n x$

Now, for $0 < n < 1$,

$f(x)=\frac{2(\sin x-\sin^nx)-(\sin x-\sin^nx)}{2(\sin x-\sin^nx)+(\sin x-\sin^nx)}=\frac{1}{3}$

and for n > 1,

$f(x)=\frac{2(\sin x-\sin^nx)+(\sin x-\sin^nx)}{2(\sin x-\sin^nx)-(\sin x-\sin^nx)}=3$

For $n > 1, g(x) = 3, x ∈ (0, π)$

$∴ g(x)$ is continuous and differentiable at

$x=\frac{π}{2}$, and for 0 < n < 1,

$g(x)=\left\{\begin{matrix}[\frac{1}{3}]=0,&x∈(0,\frac{π}{2})∪(\frac{π}{2},π)\\3,&x=π/2\end{matrix}\right.$

$∴g(x)$ is not continuous at $x=\frac{π}{2}$

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