A) 2
B) 3
C) 1/e
D) 1
Correct Answer: C
Solution :
\[f(x)=y={{x}^{-x}}\] Þ \[\log y=-\,x\log x\] Differentiating w.r.t. x, \[\frac{1}{y}.\frac{dy}{dx}=-\left[ x.\frac{1}{x}+\log x \right]\] Þ \[\frac{1}{y}.\frac{dy}{dx}=-[1+\log x]\] Þ \[\frac{dy}{dx}=-{{x}^{-x}}[1+\log x]\] Þ \[\frac{dy}{dx}={{x}^{-x}}\left[ \log \frac{1}{x}-1 \right]\] Put \[\frac{dy}{dx}=0\] Þ \[{{\log }_{e}}\frac{1}{x}={{\log }_{e}}e\] Þ \[\frac{1}{x}=e\Rightarrow x=\frac{1}{e}\].
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