A) \[e\]
B) \[\frac{1}{e}\]
C) \[-\frac{1}{e}\]
D) \[\frac{2}{e}\]
E) \[-e\]
Correct Answer: C
Solution :
Let \[y=x\,{{\log }_{e}}x\] On differentiating w. r. t.\[x,\]we get \[\frac{dy}{dx}=x.\frac{1}{x}+\log x=(1+\log x)\] Again differentiating, we get \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{1}{x}\] Put\[\frac{dy}{dx}=0\]for maxima or minima \[\Rightarrow \] \[1+\log x=0\] \[\Rightarrow \] \[x=\frac{1}{e}\] \[\therefore \] \[{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}_{(x=1/e)}}=e\] \[\therefore \] y is minimum at\[x=1/e\] \[\therefore \] \[{{y}_{\min }}=\frac{1}{e}{{\log }_{e}}\left( \frac{1}{e} \right)=-\frac{1}{e}\]
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