Question
$\left(\frac{1}{x}\right)^x$ का उच्चतम मान है-

Answer

(A) $e^{1 / e}$
$y=\left(\frac{1}{x}\right)^x$
$
\begin{aligned}
& \therefore & \log y & =\log \left(\frac{1}{x}\right)^x=-x \log x \\
& \therefore & \frac{1}{y} \frac{d y}{d x} & =-x \frac{1}{x}-\log x=-(1+\log x) \\
& \Rightarrow & \frac{d y}{d x} & =-y(1+\log x)
\end{aligned}
$
अब $\frac{d y}{d x}=0 \Rightarrow 1+\log x=0$$
\Rightarrow \log x=-1 \Rightarrow x=e^{-1}=\frac{1}{e}
$
अत: उच्चतम मान $=\left(\frac{1}{1 / e}\right)^{1 / e}=e^{1 / e}$ 

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