The interval in which $y=x^2 e^{-x}$ is strictly increasing is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → $(0,2)$
$y=x^2 e^{-x}$
so $\frac{dy}{dx}=2xe^{-x}-x^2e^{-x}=0$
so $x(2-x)e^{-x}=0$
$x=0,2$
Using wavy curve method
y is increasing strictly in (0, 2)