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

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