Question

Consider $f ( x )=\left\{\frac{ x -\sin x }{5}\right\}$. If the number of points in $(0,20 \pi)$ where $f ( x )$ is non-derivable the different values of cof LMVT for the twice differentiable function $g(x)$ i.e. $g^{\prime}(c)=\frac{g(b)-g(a)}{b-a}$ for some $c \in( a , b )$ then the minimum number of points where $g ^{\prime \prime}( x )$ vanishes is $n$. Find the value of $\left[\frac{ n }{2}\right]$.
Note : $\{y\}$ and $[y]$ denotes fractional part and greatest integer function of y respectively.]

Answer 1

$\mathrm{f}(\mathrm{x})=\left\{\frac{\mathrm{x}-\sin \mathrm{x}}{5}\right\}$
Now, $(x-\sin x)$ is an increasing function and range of $\left(\frac{x-\sin x}{5}\right)$ in $(0,20 \pi)$ is $(0,4 \pi)$
Hence, number of points where $\mathrm{f}(\mathrm{x})$ is non-derivable are 12
Hence, $\mathrm{c}_{1} ; \mathrm{c}_{2} ; \ldots \ldots ; \mathrm{c}_{12}$ are the values of $\mathrm{c}$ where $\mathrm{g}^{\prime}(\mathrm{c})$ is same.
$\therefore$ Using Rolle's minimum number of points where $\mathrm{g}$ " vanishes is $11=\mathrm{n}$
$\left.\therefore\left[\frac{\mathrm{n}}{2}\right]=5 .