A) \[\frac{1}{e}\]
B) \[\frac{1}{2e}\]
C) \[\frac{1}{{{e}^{2}}}\]
D) \[\frac{4}{{{e}^{4}}}\]
Correct Answer: C
Solution :
\[\because \] \[f(x)={{x}^{2}}{{e}^{-2x}}\] \[\therefore \] \[f(x)=2x{{e}^{-2x}}-2{{x}^{2}}{{e}^{-2x}}\] \[=2x(1-x){{e}^{-2x}}\] Put \[f(x)=0\]for maxima or minima, we get \[2x(1-x){{e}^{-2x}}=0\] \[x=0,1\] Now, \[f\,(x)=2x(-1){{e}^{-2x}}+2(1-x){{e}^{-2x}}\] \[-2\cdot 2x(1-x){{e}^{-2x}}\] \[f\,(0)=0+2{{e}^{0}}=2\] and \[f\,(1)=-\,2{{e}^{-2}}+0-0=-\frac{2}{{{e}^{2}}}<0\] \[\therefore \]\[f(x)\]is maximum at\[x=1\]. Thus, maximum value of\[f(x)=1\cdot {{e}^{-2}}=\frac{1}{{{e}^{2}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
