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Q.
The interval in which $y=x^2 e^{-x}$ is increasing, is
Application of Derivatives
Solution:
Given,
$y =x^2 e^{-x} \Rightarrow \frac{d y}{d x}=x^2 e^{-x}(-1)+e^{-x}(2 x) $
$ =x e^{-x}(-x+2)=x(2-x) e^{-x}$
For increasing function, $\frac{d y}{d x}>0 \Rightarrow x e^{-x}(2-x)>0$
Case I
$\Rightarrow x>0 \text { and } 2-x>0$
$\Rightarrow x>0 \text { and } x < 2 $
$\Rightarrow 0 < x < 2$
Case II
$\Rightarrow x < 0 \text { and } 2-x < 0$
$\Rightarrow x < 0$ and $x>2$
Hence, there is no value of $x$ exist.
Clearly, it is increasing in $(0,2)$. So, correct answer is $( d )$.