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Let $f$ $(x) = x^{2} e^{- 2 x}, x > 0$ . The maximum value of $f$ (x) is

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$\frac{1}{e^{2}}$

$\frac{1}{4 e^{2}}$

$\frac{1}{2 e}$

answer is B.

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Detailed Solution

${⨍ (x)=x}^{2} e^{- 2 x} ⨍^{1} {(x)=2x e}^{- 2 x} {-2e}^{- 2 x} {.x}^{2} = 2 {xe}^{- 2 x} (1 - x) = - 2 {xe}^{- 2 x} (x - 1)$

$∴ x = 1 ⨍ (x)$ has max

${⨍ (1)=1}^{2} {.e}^{- 2} = \frac{1}{e^{2}}$

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