Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiability5 Marks
Question
Write the points where $f(x) = |log_e x$| is not differentiable.
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Answer
Given: $\text{f(x)}=|\log_\text{e}\text{x}|=\begin{cases}-\log_\text{e}\text{x}, & 0<\text{x}<1\\\log_\text{e}\text{x}, & \text{x}\geq1\end{cases}$
Clearly f(x) is differentiable for all x > 1 and for all x < 1. So, we have to check the differentiability at x = 1.
(LHL at x = 1)
$\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{-}}\frac{-\log\text{x}-\log1}{\text{x}-1}$
$=-\lim_\limits{\text{x}\rightarrow1^{-}}\frac{\log\text{x}}{\text{x}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1-\text{h})}{1-\text{h}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1-\text{h})}{-\text{h}}$
$=-1$
(RHL at x = 1)
$\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\text{f(x)}-\text{f}(1)}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{-\log\text{x}-\log1}{\text{x}-1}$
$=\lim_\limits{\text{x}\rightarrow1^{+}}\frac{\log\text{x}}{\text{x}-1}$
$=-\lim_\limits{\text{h}\rightarrow0^{+}}\frac{\log(1+\text{h})}{1+\text{h}-1}$
$=\lim_\limits{\text{h}\rightarrow0^{-}}\frac{\log(1+\text{h})}{\text{h}}$
$=1$
Thus, (LHL at x = 1) $\neq$ (RHL at x = 1)
So, f(x) is not differentiable at x = 1.
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