#### Explain solution rd sharma class 12 chapter 9 Differentiability exercise Multiple choice question, question 28

(b)

Hint: put $\mathrm{LHD}=\mathrm{RHD} \text { at } \mathrm{x}=1 / 2$

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

Explanation:

$f(x)= \begin{cases}(2 x-1) \sin x & x>1 / 2 \\ -(2 x-1) \sin x & x<1 / 2\end{cases}$

f(x) is differentiable at $x=\frac{1}{2}$

LHD at $x=\frac{1}{2}$

\begin{aligned} &\lim _{x \rightarrow \frac{1^{-}}{2}} \frac{f(x)-f\left(\frac{1}{2}\right)}{x-\frac{1}{2}}=\lim _{h \rightarrow 0} \frac{f\left(\frac{1}{2}-h\right)-f\left(\frac{1}{2}\right)}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{-\left(2\left(\frac{1}{2}-h\right)-1\right) \sin \left(\frac{1}{2}-h\right)-0}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{(-1+2 h+1) \sin \left(\frac{1}{2}-h\right)}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0}-2 \sin \left(\frac{1}{2}-h\right)=-2 \sin \frac{1}{2} \\ \end{aligned}

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

\begin{aligned}&=\lim _{h \rightarrow 0} \frac{2 h \sin \left(\frac{1}{2}+h\right)}{-h} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} 2 \sin \left(\frac{1}{2}+h\right)=2 \sin \frac{1}{2} \\ &L H D=R H D\end{aligned}

Therefore f(x) is not differentiable at $x=\frac{1}{2}$

$\mathrm{R}-{1 / 2}$