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Question
If \[f\left( x \right) = \left| \log_e x \right|, \text { then}\]
Options
\[f' \left( 1^+ \right) = 1\]
\[f' \left( 1 \right) = - 1\]
\[f' \left( 1 \right) = 1\]
\[f' \left( 1 \right) = - 1\]
Solution
(a)
`f(x) = |log_e x|, = {(-log_e x ,"for " 0< x<1),(log_e x ,"for "x ge 1):}`
\[\text{ Differentiablity at } x = 1, \]
we have ,
\[ (\text { LHD at x } = 1 ) = {lim}_{x \to 1^-} \frac{f(x) - f(1)}{x - 1}\]
\[ = {lim}_{x \to 1^-} \frac{- \log x - \log 1}{x - 1}\]
\[ = - {lim}_{x \to 1^-} \frac{\log x}{x - 1}\]
\[ \]
\[ = - {lim}_{h \to 0} \frac{\log (1 - h)}{1 - h - 1}\]
\[ = - {lim}_{h \to 0} \frac{\log (1 - h)}{- h} = - 1 \]
\[(\text { RHD at x } = 1 ) = {lim}_{x \to 1^+} \frac{f(x) - f(1)}{x - 1} \]
\[ = {lim}_{x \to 1^+} \frac{\log x - \log (1)}{x - 1}\]
\[ \]
\[ = {lim}_{h \to 0} \frac{\log (1 + h)}{x - 1} = {lim}_{h \to 0} \frac{\log (1 + h)}{h} = 1\]
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