The set of all points, for which $\mathrm{f}(x)=x^2 \mathrm{e}^{-\dot{x}}$ strictly increases, is

The set of all points, for which $\mathrm{f}(x)=x^2 \mathrm{e}^{-\dot{x}}$ strictly increases, is
  1. $(0,2)$
  2. $(2, \infty)$
  3. $(-2,0)$
  4. $(-\infty, \infty)$

Solution

$\mathrm{f}^{\prime}(x)=2 x \mathrm{e}^{-x}-x^2 \mathrm{e}^{-x}=x \mathrm{e}^{-x}(2-x)$
Since f is increasing, $\mathrm{f}^{\prime}(x)\gt0$ $\begin{aligned} & \Rightarrow x \mathrm{e}^{-x}(2-x)\gt0 \\ & \Rightarrow x(2-x)\gt0 \\ & \Rightarrow 0 \lt x \lt 2 \\ & \Rightarrow x \in(0,2) \end{aligned}$

Asked in: MHT CET 2024 (16 May Shift 1)

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