#### Explain solution RD Sharma class 12 chapter 9 Differentiability exercise Very short answer type question  8 maths

Answer:$\pm 1$

Hint: $\log x$  is differentiable in its domain.

Given:$f(x)=|\log | x \|$

Explanation:

$|x|= \begin{cases}-x & , \infty

$\log |x|= \begin{cases}\log (-x),-\infty

$|\log | x \|=\left\{\begin{array}{c} -\log (-x),-0

We have to check differentiability at $\pm 1$

\begin{aligned} &\text { LHD at } x=1\\ &\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\log (1-h)-\log 1}{-h}\\\\ &=\lim _{h \rightarrow 0} \frac{1}{\frac{1-h}{-1}}=-1 \quad \text { [L-Hospital rule] } \end{aligned}                    ......(i)

$=-1$

\begin{aligned} \operatorname{RHD} \text { at } x=1 \\ \lim _{x \rightarrow T^{-}} \frac{f(x)-f(1)}{x-1}=& \lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\log (1+h)-\log 1}{h} \\ &=\lim _{h \rightarrow 0} \frac{1}{1+h}=1 \quad[\mathrm{~L} \text {-Hospital rule }] \end{aligned}    ......(ii)

As $LHD \neq RHD$

So the function is not differentiable at $x=1$

\begin{aligned} &\text { At } x=-1 \\ &\text { LHD at } x=-1 \\ &\begin{aligned} \lim _{x \rightarrow-1^{-}} \frac{f(x)-f(-1)}{x-(-1)} &=\lim _{h \rightarrow 0} \frac{f(-1-h)-f(-1)}{-1-h-(-1)} \\ &=\lim _{h \rightarrow 0} \frac{-\log \{-(-1-h)\}-[-\log \{-(-1)\}]}{-h} \\ &=\lim _{h \rightarrow 0} \frac{\frac{-1}{1+h}-0}{-1}=1 \quad \text { [L-Hospital rule] } \end{aligned} \end{aligned}    .......(iii)

$=1$

\begin{aligned} \mathrm{RHD} \text { at } x=-1 \\ \lim _{x \rightarrow-1^{+}} \frac{f(x)-f(-1)}{x-(-1)} &=\lim _{h \rightarrow 0} \frac{f(-1+h)-f(-1)}{-1+h+1} \\ &=\lim _{h \rightarrow 0} \frac{-\log \{-(-1+h)\}-[-\log \{-(-1)\}]}{h} \\ &=\lim _{h \rightarrow 0} \frac{-1}{1-h}=-1 \quad[\mathrm{~L}-\mathrm{Hospital} \text { rule }] \end{aligned}    ......(iv)

As $LHD \neq RHD$

So the function is not differentiable at $x=-1$