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

Answer: 5

$\mathrm{f}(\mathrm{x})=\left\{\frac{\mathrm{x}-\sin \mathrm{x}}{5}\right\}$
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, ${\mathrm{c}}_1;{\mathrm{c}}_2;\dots \dots ;{\mathrm{c}}_{12}$ 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 .