Let $f\left(x\right)=\left[np\sin x\right],x\in (0,\pi ),n\in Z$ and $p$ is a prime number, where $[·]$ denotes the greatest integer function. Then, find the number of points, where $f\left(x\right)$ is not differentiable.

 Here, $f\left(x\right)=\left[np\sin x\right]$ is discontinuous and non differentiable at those points where $np\sin x$ is an integer.

As, it is given that  $p$ is a prime number.

$\Rightarrow np\sin x$ is an integer if $\mathrm{sinx}=1,-1,\frac{\mathrm{r}}{\mathrm{p}}$ where $0\le \mathrm{r}\le \left(\mathrm{p}-1\right)$.

$\Rightarrow \mathrm{x}=\frac{\mathrm{\pi }}{2},-\frac{\mathrm{\pi }}{2},{\sin }^{-1}\frac{\mathrm{r}}{\mathrm{p}}$

As we know that $\sin \theta =\sin \left(\mathrm{\pi }-\mathrm{\theta }\right)$,

$$\begin{gathered} \Rightarrow \sin \left(\mathrm{\pi }-x\right)=1,-1,\frac{r}{p} \\ \Rightarrow x=\frac{\mathrm{\pi }}{2},\frac{3\mathrm{\pi }}{2},\mathrm{\pi }-{\sin }^{-1}\left(\frac{\mathrm{r}}{\mathrm{p}}\right) \end{gathered}$$

Since $\frac{3\mathrm{\pi }}{2} , -\frac{\mathrm{\pi }}{2}$ are same so we get the points,

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

But in the question we are given the domain as $\mathrm{x}\in \left(0,\mathrm{\pi }\right)$ , so $\mathrm{x}\ne -\frac{\mathrm{\pi }}{2},0$.

Thus the points of discontinuity are  $\mathrm{x}=\frac{\mathrm{\pi }}{2},\mathrm{\pi }-{\sin }^{-1}\left(\frac{\mathrm{r}}{\mathrm{p}}\right),{\sin }^{-1}\left(\frac{\mathrm{r}}{\mathrm{p}}\right)$ where $0<\mathrm{r}\le \left(\mathrm{p}-1\right)$.

So, the required number of points of non-differentiability are $=1+\left(\mathrm{p}-1\right)+\left(\mathrm{p}-1\right)=2 \mathrm{p}-1$.

Hence option-$1$ is the correct answer.