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#### Provide solution for rd sharma maths class 12 chapter 9 Differentiability exercise Multiple choice question, question 11

HINTS: Find LHD and RHD at x=1

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

SOLUTION:

$f(x)=|\log x|= \begin{cases}\log _{e} x & 0

LHD at x=1

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

\begin{aligned} &=\lim _{h \rightarrow 0}-\frac{\log (1-h)-f(1)}{-h} \\ \end{aligned}

\begin{aligned}\lim _{h \rightarrow 0}-\frac{\log (1-h)}{-h} &=\lim _{h \rightarrow 0} \frac{1}{1-h} \times-1 \\ &=-1 \end{aligned}

RHD at x=1

\begin{aligned} \lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} &=\lim _{x \rightarrow 1} \frac{\log x-\log 1}{x-1} \\ \end{aligned}

\begin{aligned} &=\lim _{h \rightarrow 0}-\frac{\log (1+h)}{h} \\ &=\lim _{h \rightarrow 0} \frac{1}{1+h} \\ &=1 \end{aligned}

As LHD $\neq$ RHD at x=1

$f^{\prime}\left(1^{-}\right)=-1 , f^{\prime}\left(1^{+}\right)=1$