The interval in which $y = {x^2}{e^{ - x}}$ is increasing, is

(A) $\left( { - \infty ,\infty } \right)$

(B) $( - 2,0)$

(C) $(2,\infty )$

(D) $(0,2)$

Option D is correct

We have, $y = {x^2}{e^{ - x}}$ …(i)

Differentiating (i) w.r.t. $x$, we get
$\cfrac{{dy}}{{dx}} = {x^2}{e^{ - x}}( - 1) + {e^{ - x}}(2x) = x{e^{ - x}}( - x + 2) = x(2 - x){e^{ - x}}$

For critical points, putting $\cfrac{{dy}}{{dx}} = 0$
$\Rightarrow x(2 - x){e^{ - x}} = 0 \Rightarrow x = 0,x = 2$

The above points divide the real line into following disjoint intervals $( - \infty ,\;0),(0,2),(2,\;\infty )$

Interval
Sign of
$\cfrac{{dy}}{{dx}} = x\left( {2 - x} \right){e^{ - x}}$

Nature of function
$y$
$\left( { - \infty ,0} \right)$
$\left( { - ve} \right)\left( { + ve} \right)\left( { + ve} \right) = - ve$

Decreasing
$\left( {0,2} \right)$
$\left( { + ve} \right)\left( { + ve} \right)\left( { + ve} \right) = + ve$

Increasing
$\left( {2,\infty } \right)$
$\left( { + ve} \right)\left( { - ve} \right)\left( { + ve} \right) = - ve$

Decreasing

Thus, $y$ is increasing function on $(0,2)$