If [x] denotes the integral part of x and in (0, π), we define $f(x)=\left\{\begin{matrix}\left[\frac{2(\sin x-\sin^nx)+|\sin x-\sin^nx|}{2(\sin x-\sin^nx)-|\sin x-\sin^nx|}\right],&x≠\frac{π}{2}\\3,&x=\frac{π}{2}\end{matrix}\right.$
Then for n > 1,
Answer & explanation
Correct answer: option 2
For $0<x<\frac{π}{2}$ or $\frac{π}{2}<x<π$, $0 < \sin x < 1$.
∴ for $n > 1, \sin x > \sin^nx$
$∴ f(x)=\left[\frac{3(\sin x-\sin^nx)}{\sin x-\sin^nx}\right]=3,x≠\frac{π}{2}$ and $f(x) = 3, x=\frac{π}{2}$
Thus in (0, π) , f(x) = 3.
Hence f(x) is continuous and differentiable at $x=\frac{π}{2}$.