Question

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

• A. $2p+1$
• B. $2p-1$
• C. $2p$
• D. $2p-2$

Answer 1

Answer: (B)

$2p-1$

$f\left(x\right)=[n+p\sin x]=n+\left[p\sin x\right]$ ( $n$ is an integer )

$\left[x\right]$ is not differentiable at those points where $x$ is an integer.

So, $f\left(x\right)$ is not differentiable at those points where value of $p\sin x$ is an integer, on the interval $(0,\pi )$

That is,

$\sin x=\frac{1}{p},\frac{2}{p},\frac{3}{p}............\frac{p-1}{p},\frac{p}{p}$

$\sin x=\frac{1}{p},\frac{2}{p},\frac{3}{p}............\frac{p-1}{p},1$

From the above values, $\sin x$ attains the values $\frac{1}{p},\frac{2}{p},\frac{3}{p}............\frac{p-1}{p}$ twice on the interval $(0,\pi )$

But $\sin x=1$ at only one point, $x=\frac{\pi }{2}$

So, total no. of points $=2(p-1)+1=2p-1$