Question

The number of point(s) of discontinuity of $f\left(x\right)=\left[5\sin x\right],x\in [0,\pi ]$ is

Answer 1

Piece-wise definition of $f\left(x\right)=\left[5\sin x\right]$ in $x\in [0,\pi ]$ is :

$x\in [0,{\sin }^{-1}\frac{1}{5}),f\left(x\right)=0$
$x\in [{\sin }^{-1}\frac{1}{5},{\sin }^{-1}\frac{2}{5}),f\left(x\right)=1$
$x\in [{\sin }^{-1}\frac{2}{5},{\sin }^{-1}\frac{3}{5}),f\left(x\right)=2$
$x\in [{\sin }^{-1}\frac{3}{5},{\sin }^{-1}\frac{4}{5}),f\left(x\right)=3$
$x\in [{\sin }^{-1}\frac{4}{5},\frac{\pi }{2}),f\left(x\right)=4$
$x\in \left[\frac{\pi }{2}\right],f\left(x\right)=5$
$x\in (\frac{\pi }{2},\pi -{\sin }^{-1}\frac{4}{5}],f\left(x\right)=4$
$\cdot$
$\cdot$
$\cdot$
$x\in (\pi -{\sin }^{-1}\frac{1}{5},0],f\left(x\right)=0$
$\therefore 9$ points of discontinuity.