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

(b)

HINTS: Understand the definition of continuity and differentiability

GIVEN: $f(x)=\left|\log _{e}\right| x \|$

SOLUTION:

\begin{aligned} &f(x)= \begin{cases}|\log x| & x>0 \\ |\log (-x)| & x \geq 1\end{cases} \\ \end{aligned}

\begin{aligned} &f(x)= \begin{cases}\log (-x) & x<-1 \\ -\log (-x) & -11\end{cases} \end{aligned}

As f(x) is an absolute function .So it is continues for all x.

Now for differentiability

X=1

LHD=

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

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

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

As LHD $\neq$ RHD at x=-1

Now at x=1

LHD=

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

RHD=

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

As LHD $\neq$ RHD

Therefore f(x) is not differentiable at $x\pm 1$