Given the function $f (x) = x^{2} e^{- 2 x}, x > 0.$ Then $f (x)$ has the maximum value equal to

e^{- 1}

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{(2 e)}^{- 1}

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e^{- 2}

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none of these

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Solution

The correct option is A $e^{- 2}$
$f (x) = x^{2} e^{- 2 x}$
$f^{'} (x) = e^{- 2 x} .2 x - 2 x^{2} . e^{- 2 x}$
$f^{'} (x) = 2 e^{- 2 x} . x (1 - x)$
Hence $f (x)$ will attain its maxima at $x = 1$ which is $e^{- 2}$

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Given the function f(x)=x2e−2x,x>0. Then f(x) has the maximum value equal to

The correct option is A e−2f(x)=x2e−2xf′(x)=e−2x.2x−2x2.e−2xf′(x)=2e−2x.x(1−x)Hence f(x) will attain its maxima at x=1 which is e−2

Given the function $f (x) = x^{2} e^{- 2 x}, x > 0.$ Then $f (x)$ has the maximum value equal to

The correct option is A $e^{- 2}$
$f (x) = x^{2} e^{- 2 x}$
$f^{'} (x) = e^{- 2 x} .2 x - 2 x^{2} . e^{- 2 x}$
$f^{'} (x) = 2 e^{- 2 x} . x (1 - x)$
Hence $f (x)$ will attain its maxima at $x = 1$ which is $e^{- 2}$