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Question
The minimum value of \[\frac{x}{\log_e x}\] is _____________ .
Options
e
1/e
1
none of these
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Solution
e
\[\text { Given }: f\left( x \right) = \frac{x}{\log_e x}\]
\[ \Rightarrow f'\left( x \right) = \frac{\log_e x - 1}{\left( \log_e x \right)^2}\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow \frac{\log_e x - 1}{\left( \log_e x \right)^2} = 0\]
\[ \Rightarrow \log_e x - 1 = 0\]
\[ \Rightarrow \log_e x = 1\]
\[ \Rightarrow x = e\]
\[\text { Now,} \]
\[f''\left( x \right) = \frac{- 1}{x \left( \log_e x \right)^2} + \frac{2}{x \left( \log_e x \right)^3}\]
\[ \Rightarrow f''\left( e \right) = \frac{- 1}{e} + \frac{2}{e} = \frac{1}{e} > 0\]
\[\text { So, x = e is a local minima } . \]
\[ \therefore \text { Minimum value of } f\left( x \right) = \frac{e}{\log_e e} = e\]
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