A) \[x={{e}^{-2}}\]
B) \[x=e\]
C) \[x={{e}^{-1}}\]
D) \[x=2{{e}^{-1}}\]
Correct Answer: C
Solution :
[c] Let \[f(x)=x\,\,ln\,\,x\] \[f'(x)=\frac{x}{x}+\] ln \[x=1+ln\,\,x\] Put \[f(x)=0\Rightarrow 1+ln\,\,x=0\] \[\Rightarrow \] ln \[x=-1\Rightarrow x={{e}^{-1}}\] Now, \[f''(x)=\frac{1}{x}\] \[f''(x)\left| _{x=e}-1=\frac{1}{{{e}^{-1}}}=e>0 \right.\] Hence, \[f(x)\] attains minimum value at\[x={{e}^{-1}}\].
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