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

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

Answers (1)

Answer:\pm 1

Hint: \log x  is differentiable in its domain.

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

Explanation:

|x|= \begin{cases}-x & , \infty<x<-1 \\ -x & ,-1<x<0 \\ x & , 0<x<1 \\ x & , 1<x<\infty\end{cases}

\log |x|= \begin{cases}\log (-x),-\infty<x<-1 \\ \log (-x),-1<x<0 \\ \log x, & , 0<x<1 \\ \log x \quad, 1<x<\infty\end{cases}

|\log | x \|=\left\{\begin{array}{c} -\log (-x),-0<x<-1 \\ -\log (-x),-1<x<0 \\ \log x \quad, 0<x<1 \\ \log x, 1<x<\infty \end{array}\right.

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

 

Posted by

infoexpert26

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