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Question
The minimum value of x loge x is equal to ____________ .
Options
e
`1/e`
`-1/e`
`2/e`
`-e`
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Solution
\[\frac{- 1}{e}\]
\[\text { Here }, \]
\[f\left( x \right) = x \log_e x\]
\[ \Rightarrow f'\left( x \right) = \log_e x + 1\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow \log_e x + 1 = 0\]
\[ \Rightarrow \log_e x = - 1\]
\[ \Rightarrow x = e^{- 1} \]
\[\text { Now,} \]
\[f''\left( x \right) = \frac{1}{x}\]
\[ \Rightarrow f''\left( e^{- 1} \right) = e > 0\]
\[\text { So,} x = e^{- 1}\text { is a local minima }. \]
\[\text { Hence, the minimum value of } f\left( x \right) = f\left( e^{- 1} \right) . \]
\[ \Rightarrow e^{- 1} \log_e \left( e^{- 1} \right) = - e^{- 1} = \frac{- 1}{e}\]
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