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
Answer & explanation
Correct answer: option 2
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 $x=\frac{π}{2}$