If f(x)=| _ex|, then:

MCQ

If $\text{f(x)}=|\log_\text{e}\text{x}|,$ then:

✓
$\text{f}\ '(1^+)=1$
B
$\text{f}\ '(1^-)=-1$
C
$\text{f}\ '(1)=1$
D
$\text{f}\ '(1)=-1$

✓

Answer

Correct option: A.

$\text{f}\ '(1^+)=1$

$\text{f(x)}=|\log_\text{e}\text{x}|,=\begin{cases}-\log_\text{e}\text{x}, \text{for}0<\text{x}<1\log_\text{e}\text{x}, \text{for x}\geq1\end{cases}\}$
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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