The interval in which the function g(x) = x^2e^-x is increasing is:

The correct answer is Option (4) → $(0, 2)$

Given function $g(x)=x^2 e^{-x}$

Differentiate w.r.t. $x$

$\frac{dg}{dx}=2x e^{-x}+x^2(-e^{-x})$

$=e^{-x}(2x-x^2)$

$=e^{-x}x(2-x)$

Since $e^{-x}>0$ for all $x$, sign of $\frac{dg}{dx}$ depends on $x(2-x)$.

$x(2-x)>0$ when $0<\text{ x }<2$.

Increasing in $(0,2)$