Explanation

(c)
Given, $f(x) = x^{2}e^{- 2x}$
$\therefore\ \ \ \ f^{'}(x) = 2xe^{- 2x} - 2x^{2}e^{- 2x}$
$= 2x(1 - x)e^{- 2x}$
For maxima or minima, put $f^{'}(x) = 0$
$2x(1 - x)e^{- 2x} = 0$
$\Rightarrow \ \ \ \ \ \ x = 0,\ 1$
Now, $f^{''}(x) = 2x( - 1)e^{- 2x} + 2(1 - x)^{e^{- 2x}} - 2.2x(1 - x)e^{- 2x}$
$f^{''}(0) = 0 + 2e^{0} = 2 > 0$, minima
And $f^{''}(1) = - 2e^{- 2} + 0 - 0 = - \frac{2}{e^{2}} < 0$, maxima
Thus, maximum value is $f(1) = 1.e^{- 2} = \frac{1}{e^{2}}$