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  • Need solution for RD Sharma maths class 12 chapter 9 Differentiability exercise Very short answer type question 7

Need solution for RD Sharma maths class 12 chapter 9 Differentiability exercise Very short answer type question  7

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Answer: 1

Hint: if f(x) is not differentiable then L H D \neq R H D

Given:f(x)=\left|\log _{e} x\right| is not differentiable.

Explanation: f(x)=\left|\log _{e} x\right|=\left\{\begin{array}{cc} \log _{e} x & x \geq 1 \\ -\log _{e} x & 0<x<1 \end{array}\right\}

as logarithmic function is differentiable in its domain. So we only have to check at x=1

        \begin{aligned} &L H D=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\\\ &=\lim _{x \rightarrow 1} \frac{-\log _{e} x-0}{x-1} \\\\ &=\lim _{x \rightarrow 1} \frac{-\log x}{x-1} \end{aligned}

applying  L' Hospital rule

        \begin{aligned} &=\lim _{x \rightarrow 1} \frac{-1}{x} \\ &=-1 \end{aligned}

        \begin{aligned} &R H D \text { atx }=1 \\\\ &\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1} \\\\ &\lim _{x \rightarrow 1^{+}} \frac{\log _{e} x-\log 1}{x-1} \end{aligned}

        \begin{aligned} &=\lim _{x \rightarrow 1} \frac{\log _{e} x-0}{x-1} \\\\ &=\lim _{x \rightarrow 1} \frac{\log x}{x-1} \end{aligned}

applying  L' Hospital rule

        \begin{aligned} &=\lim _{x \rightarrow 1} \frac{1}{x} \\\\ &=1 \\\\ &\text { As } L H D \neq R H D_{\text {at }} \mathrm{x}=1 \end{aligned}

\left|\log _{e} x\right| is not differentiable at x=1

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