Let  $f\left(x\right)=\left\{\frac{x-\sin x}{5}\right\}$  where  $\left\{t\right\}$ denotes fractional part of $\mathrm{t}$ . If the number of points in  $(0,20\pi )$ where $f\left(x\right)$ in non derivable is the number of different values of c of L.M.V.T for the twice differentiable function  $g\left(x\right)$ i.e.,  $g\text{'}\left(c\right)=\frac{g\left(b\right)-g\left(a\right)}{b-a}$ for some  $c\in (a,b)$  and the minimum number of points where  $g\text{'}\text{'}$ (x) vanishes is n then the integral part of  $\frac{n}{2}$ is ______

$f\left(x\right)=\left\{\frac{x-\sin x}{5}\right\}$

Consider  $g\left(x\right)=\frac{x-\sin x}{5};g^{\text{'}}\left(x\right)=\frac{1-\cos x}{5}\ge 0forx\in (0,20\pi )$

Thus, g(x) is an increasing function, also the range of  $g\left(x\right)=\frac{x-\sin x}{5}is(0,4\pi )$
At all integral values, f(x) will not be derivable. Hence there are  $\left[4\pi \right]\text{\hspace{0.17em}}=12$ points, where f(x) is not derivable.
Thus, there are 12 values of c, for which  $g^{\text{'}}\left(c\right)=\frac{g\left(b\right)-g\left(a\right)}{b-a}$ has same values.
Thus, by Rolle’s Theorem for $g^{\text{'}}\left(x\right),g^{\text{'}\text{'}}\left(x\right)$ will vanish at (12 – 1) =11 points (minimum).
Hence n=11
Thus,  $\left[\frac{n}{2}\right]=5$