Question
Maximum value of $=\left(\frac{1}{x}\right)^x$:
✓
Answer
(A)$y=\left(\frac{1}{x}\right)^x$
$\therefore \quad \log y=\log \left(\frac{1}{x}\right)^x=-x \log x$
$\therefore \quad \frac{1}{y} \frac{d y}{d x}=-x \frac{1}{x}-\log x=-(1+\log x)$
$\Rightarrow \quad \frac{d y}{d x}=-y(1+\log x)$
now $\frac{d y}{d x}=0 \Rightarrow 1+\log x=0$
$\Rightarrow \log x=-1 \Rightarrow x=e^{-1}=\frac{1}{e}$
hence maximum value $=\left(\frac{1}{1 / e}\right)^{1 / e}=e^{1 / e}$
$\therefore \quad \log y=\log \left(\frac{1}{x}\right)^x=-x \log x$
$\therefore \quad \frac{1}{y} \frac{d y}{d x}=-x \frac{1}{x}-\log x=-(1+\log x)$
$\Rightarrow \quad \frac{d y}{d x}=-y(1+\log x)$
now $\frac{d y}{d x}=0 \Rightarrow 1+\log x=0$
$\Rightarrow \log x=-1 \Rightarrow x=e^{-1}=\frac{1}{e}$
hence maximum value $=\left(\frac{1}{1 / e}\right)^{1 / e}=e^{1 / e}$
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