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

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Solution

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

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

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The number of point(s) of discontinuity of f(x)=[5sinx],x∈[0,π] is

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