Let f(x) = [n + p sin x], x ∈ (0, π), n ∈ Z, p is a prime number and…

We know that the greatest integer function [x] is neither continuous nor differentiable at integer points. Therefore, $f(x)=[n+p \sin x]$ is discontinuous and non-differentiable at those points where $n+p \sin x$ is an integer.

Clearly, $n+p \sin x$ will be an integer when $p \sin x$ is an integer.

Now, $p \sin x$ is an integer, if

$\sin x=1, \frac{1}{p}, \frac{2}{p}, \frac{3}{p}, ..., \frac{p-1}{p} ~~~[∵ x \in(0, \pi) ~∴ \sin x>0]$

$\Rightarrow x=\sin ^{-1}(1), \sin ^{-1}\left(\frac{1}{p}\right), \sin ^{-1}\left(\frac{2}{p}\right), ... \sin ^{-1}\left(\frac{p-1}{p}\right)$

But, $\sin (\pi-\theta)=\sin \theta$.

Therefore, $\sin x$ will assume the values $\frac{1}{p}, \frac{2}{p}, \frac{3}{p}, ... \frac{p-1}{p}$ at points

$\pi-\sin ^{-1} \frac{1}{p}, ..., \pi-\sin ^{-1}\left(\frac{p-1}{p}\right)$ also

Hence, the total number of points of discontinuity of f(x) is (2p - 1).