CUET Preparation Today
The interval in which $f(x)=x^2 e^{-x}+7$ is increasing is: |
$(-\infty, \infty)$ $(-2,0)$ $(2, \infty)$ $(0,2)$ |
$(0,2)$ |
The correct answer is Option (4) → $(0,2)$ $y=x^2 e^{-x}+7$ $y'=2xe^{-x}-x^2e^{-x}=0$ $e^{-x}x(2-x)=0$ $⇒x=0,2$
f(x) increasing in (0, 2) |
