The function $f(x) = x^2e^{-2x}$ increases on

$(0,1)$

The correct answer is Option (2) → $(0,1)$

Given: $f(x) = x^{2} e^{-2x}$

Differentiate:

$f'(x) = e^{-2x}(2x - 2x^{2}) = 2x e^{-2x}(1 - x)$

Since $e^{-2x} > 0$ for all $x$, the sign of $f'(x)$ depends on $2x(1 - x)$.

$f'(x) > 0 \Rightarrow 2x(1 - x) > 0$

This holds when $x \in (0, 1)$.

Therefore, the function increases on $(0, 1)$.