MCQ
If $\text{f(x)}=|\log_\text{e}\text{x}|,$ then:
- ✓$\text{f}'(1^+)=1$
- B$\text{f}'(1^-)=-1$
- CBoth $A$ and $B$
- D$\text{f}'(1)=-1$
$\text{f(x)}=|\log_\text{e}\text{x}|,=\begin{cases}-\log_\text{e}\text{x}, & \text{for}0<\text{x}<1\\\log_\text{e}\text{x}, & \text{for x}\geq1\end{cases}$
Differentiability at $x = 1,$
We have,
$(\text{LHL}$ at $x = 1)$
$=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\text{f(c)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{-\log\text{x}-\log12}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow1^{-}}\frac{\log\text{x}}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1-\text{h})}{1-\text{h}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1-\text{h})}{-\text{h}}$
$=-1$
$(\text{RHL}$ at $x = 1)$
$=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\text{f(c)}-\text{f}(1)}{\text{x}-1}$
$=\lim\limits_{\text{x}\rightarrow1^{+}}\frac{\log\text{x}-\log1}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1+\text{h})}{\text{x}-1}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\log(1+\text{h})}{\text{h}}$
$=1$
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