The interval in which y=x^2 e^ -x is increasing is : - Vidyadip

MCQ

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

A
$(-\infty,\ \infty)$
B
$( -2, 0)$
C
$(2, \ \infty)$
✓
$(0, 2)$

✓

Answer

Correct option: D.

$(0, 2)$

$\text{f}\text{(x)} = [\text{y} = \text{x}^2 \text{e}^{-\text{x} } ]$
$\Rightarrow \ \frac{\text{dy}}{\text{dx}}=\text{x}^2\frac{\text{d}}{\text{dx}}\text{e}^{-\text{x}}+\text{e}^{-\text{x}}\frac{\text{d}}{\text{dx}}\text{x}^2$
$=\text{x}^2e^{-\text{x}}(-1)+\text{e}^{-\text{x}}(2\text{x})$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=-\text{x}^2\text{e}^{-\text{x}}+2\text{xe}^{-\text{x}}$
$=\text{xe}^{-\text{x}}(-\text{x}+2)$
$\Rightarrow \ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}(2-\text{x})}{\text{e}^\text{x}}$
In option $(D), \frac{\text{dy}}{\text{dx}} > 0$ for all $x$ in the interval $(0, 2).$

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