#### Provide solution for RD Sharma maths class 12 chapter 9 Differentiability exercise Fill in the blanks question  10

Answer: $R-\left\{\frac{1}{2}\right\}$

Hint:

Given: $f(x)=|2 x-1|$

Solution:

$f(x)=|2 x-1|$

$f(x)= \begin{cases}(2 x-1) & x \geq 0 \\ -(2 x-1) & x<0\end{cases}$

\begin{aligned} &L_{h \rightarrow 0^{-}} f^{\prime}\left(\frac{1}{2}\right)=\lim _{h} \frac{f\left(\left(\frac{1}{2}\right)+h\right)-f\left(\frac{1}{2}\right)}{h} \\\\ &=\lim _{h \rightarrow 0^{-}} \frac{-\left[2\left(\frac{1}{2}\right)+h-0\right]}{h} \\\\ &=-1 \end{aligned}

\begin{aligned} &\operatorname{Rf}\left(\frac{1}{2}\right)=\lim _{h \rightarrow 0^{+}} \frac{f\left(\frac{1}{2}-h\right)-f\left(\frac{1}{2}\right)}{h} \\\\ &=\lim _{h \rightarrow 0^{+}} \frac{\left[\left[2\left(\frac{1}{2}\right)-1\right]-0\right]}{h} \\\\ &=1 \end{aligned}

$L\; f^{\prime}\left(\frac{1}{2}\right) \neq R\; f^{\prime}\left(\frac{1}{2}\right)$

Hence $f(x)$ is differentiable $\forall x \in R-\left\{\frac{1}{2}\right\}$