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  4. The function f(x)=x2e-x increases in the interval

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Q. The function $ f(x)={{x}^{2}}{{e}^{-x}} $ increases in the interval

J & K CETJ & K CET 2008Application of Derivatives

Solution:

Given, $ f(x)={{x}^{2}}{{e}^{-x}} $
$ \Rightarrow $ $ f'(x)=2x{{e}^{-x}}-{{x}^{2}}{{e}^{-x}} $
For $ f(x) $ to be increasing, $ f'(x)>0 $
$ \Rightarrow $ $ 2x{{e}^{-x}}-{{x}^{2}}{{e}^{-x}}>0 $
$ \Rightarrow $ $ {{e}^{-x}}(2x-{{x}^{2}})>0 $
$ \Rightarrow $ $ x(2-x)>0 $
$ \Rightarrow $ $ x\in (0,2) $