If [x] denotes the integral part of x and $f(x) = [n + p\sin x],\, 0 < x < π, n ∈ I$ and p is a prime number, then the number of points where f(x) is not differentiable is

$2p − 1$

[x] is not differentiable at integral points.

Also $[n + p \sin x] = n + [p \sin x]$

$∴ [p \sin x]$ is not differentiable, where $p \sin x$ is an integer.

But p is prime and $0 < \sin x ≤ 1 [∵ 0 < x < π]$

$∴ p \sin x$ is an integer only when $\sin x =\frac{r}{p}$, where $0 < r ≤ p$ and $r ∈ N$

For $r = p,\, \sin x = 1 ⇒ x = \frac{π}{2}$ in $(0, π)$

For $0 < r < p,\, \sin x =\frac{r}{p}$

$∴x=\sin^{-1}\frac{r}{p}$ or $π-\sin^{-1}\frac{r}{p}$

Number of such values of x = p − 1 + p − 1 = 2p − 2.

∴ Total number of points where f(x) is not differentiable = 1 + 2p − 2 = 2p − 1.