Question

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

• A. $0$
• B. $2(p-1)$
• C. $2p-1$
• D. None of these

Answer 1

Answer: (C)

$2p-1$

Here,

$f\left(x\right)=[n+p\sin x]$ is not differentiable at those points where $n+p\sin x$ is integer.

As $p$ is a prime number.

$\Rightarrow n+p\sin x$ is an integer if $\sin x=1,-1,\frac{r}{p},$

i.e., $x=\frac{\pi }{2},\frac{-\pi }{2},{\sin }^{-1}\frac{r}{p},\pi -{\sin }^{-1}\frac{r}{p},$ where $0\le r\le p-1$

But $x\ne \frac{-\pi }{2},0.$

$\therefore$ Function is not differentiable at $x=\frac{\pi }{2},{\sin }^{-1}\frac{r}{p},\pi -{\sin }^{-1}\frac{r}{p},$ where $0<r\le p-1$

So, the required number of points are, $=1+2(p-1)=2p-1$