Q. The function $f(x) = x^{2}e^{- x}$ increases in the interval
A) (0, 2)
B) (2, 3)
C) (3, 4)
D) (4, 5)
Explanation
(a)
Given, $f(x) = x^{2}e^{- x}$
$\Rightarrow f^{'}(x) = 2xe^{- x} - x^{2}e^{- x}$
For $f(x)$ To be increasing, $f^{'}(x) > 0$
$\Rightarrow 2xe^{- x} - x^{2}e^{- x} > 0$
$\Rightarrow e^{- x}(2 - x) > 0$
$\Rightarrow x \in (0,2)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.