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

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

Solution

$f(x)=x^2 e^{-x}$ null $f(x)$ is strictly increasing in $(0,2)$

Asked in: MHT CET 2022 (10 Aug Shift 1)

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