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The function $f(x)=x^{2} e^{-2 x}, x>0$. Then the maximum value of $\mathrm{f}(\mathrm{x})$ is
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The correct answer is:
$\frac{1}{\mathrm{e}^{2}}$
$\begin{array}{l}
\text { Given : } \mathrm{f}(\mathrm{x})=\mathrm{x}^{2} \mathrm{e}^{-2 \mathrm{x}}, \mathrm{x}>0 \\
\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{2} \cdot \mathrm{e}^{-2 \mathrm{x}}(-2)+\mathrm{e}^{-2 \mathrm{x}} \cdot 2 \mathrm{x} \\
\text { put } \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow 2 \mathrm{e}^{-2 \mathrm{x}} \cdot \mathrm{x}(-\mathrm{x}+1)=0 \\
\Rightarrow \mathrm{x}=1 \text { or } \mathrm{x}=0 \\
\mathrm{f}^{\prime \prime}(\mathrm{x})=\left(-4 \mathrm{x}^{2}-6 \mathrm{x}+1\right) \mathrm{e}^{-2 \mathrm{x}} \\
\mathrm{f}^{\prime \prime}(1)=-9 \mathrm{e}^{-2 \mathrm{x}} < 0 \\
\mathrm{f}^{\prime \prime}(0)=\mathrm{e}^{-2 \mathrm{x}}>0 \\
\therefore \text { value of } \mathrm{f}(\mathrm{x}) \text { is maximum at } \mathrm{x}=1 \\
\because \mathrm{f}(\mathrm{x})=\mathrm{x}^{2} \cdot \mathrm{e}^{-2 \mathrm{x}} \Rightarrow \mathrm{f}(1)=\mathrm{e}^{-2}=\frac{1}{\mathrm{e}^{2}}
\end{array}$
\text { Given : } \mathrm{f}(\mathrm{x})=\mathrm{x}^{2} \mathrm{e}^{-2 \mathrm{x}}, \mathrm{x}>0 \\
\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{x}^{2} \cdot \mathrm{e}^{-2 \mathrm{x}}(-2)+\mathrm{e}^{-2 \mathrm{x}} \cdot 2 \mathrm{x} \\
\text { put } \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow 2 \mathrm{e}^{-2 \mathrm{x}} \cdot \mathrm{x}(-\mathrm{x}+1)=0 \\
\Rightarrow \mathrm{x}=1 \text { or } \mathrm{x}=0 \\
\mathrm{f}^{\prime \prime}(\mathrm{x})=\left(-4 \mathrm{x}^{2}-6 \mathrm{x}+1\right) \mathrm{e}^{-2 \mathrm{x}} \\
\mathrm{f}^{\prime \prime}(1)=-9 \mathrm{e}^{-2 \mathrm{x}} < 0 \\
\mathrm{f}^{\prime \prime}(0)=\mathrm{e}^{-2 \mathrm{x}}>0 \\
\therefore \text { value of } \mathrm{f}(\mathrm{x}) \text { is maximum at } \mathrm{x}=1 \\
\because \mathrm{f}(\mathrm{x})=\mathrm{x}^{2} \cdot \mathrm{e}^{-2 \mathrm{x}} \Rightarrow \mathrm{f}(1)=\mathrm{e}^{-2}=\frac{1}{\mathrm{e}^{2}}
\end{array}$
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