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If the function $\mathrm{f}(\mathrm{x})=\mathrm{xe}^{-\mathrm{x}}, \mathrm{x} \in \mathbb{R}$ attains its maximum value $\beta$ at $x=\alpha$ then $(\alpha, \beta)=$
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Verified Answer
The correct answer is:
$\left(1, \frac{1}{\mathrm{e}}\right)$
$f(x)=x \cdot e^{-x}$
$$
\begin{aligned}
& f^{\prime}(x)=-x \cdot e^{-x}+e^{-x} \\
& =e^{-x}(1-x) \\
& f^{\prime}(x)=0 \Rightarrow x=1 \\
& \text { and } f^{\prime \prime}(x)=x \cdot e^{-x}-e^{-x}-e^{-x}=x e^{-x}-2 e^{-x} \\
& \because f^{\prime \prime}(1)=\frac{1}{e}-\frac{2}{e} < 0
\end{aligned}
$$
$\therefore$ maxima is attained at $x=1$
$$
\begin{gathered}
f(x)_{\max }=1 \cdot e^{-1}=\frac{1}{e} \\
\therefore(\alpha, \beta) \equiv\left(1, \frac{1}{e}\right)
\end{gathered}
$$
$$
\begin{aligned}
& f^{\prime}(x)=-x \cdot e^{-x}+e^{-x} \\
& =e^{-x}(1-x) \\
& f^{\prime}(x)=0 \Rightarrow x=1 \\
& \text { and } f^{\prime \prime}(x)=x \cdot e^{-x}-e^{-x}-e^{-x}=x e^{-x}-2 e^{-x} \\
& \because f^{\prime \prime}(1)=\frac{1}{e}-\frac{2}{e} < 0
\end{aligned}
$$
$\therefore$ maxima is attained at $x=1$
$$
\begin{gathered}
f(x)_{\max }=1 \cdot e^{-1}=\frac{1}{e} \\
\therefore(\alpha, \beta) \equiv\left(1, \frac{1}{e}\right)
\end{gathered}
$$
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