question_answer

If \[f(x)=x{{e}^{x(1-x)}}\], then \[f(x)\] is                                             [IIT Screening 2001]

A)            Increasing on \[\left[ -\frac{1}{2},\,1 \right]\]

B)            Decreasing on R

C)            Increasing on R

D)            Decreasing on \[\left[ -\frac{1}{2},1 \right]\]

Correct Answer: A

Solution :

           \[{f}'(x)={{e}^{x(1-x)}}+x.{{e}^{x(1-x)}}.(1-2x)\]                    \[=\,\,{{e}^{x(1-x)}}\{1+x(1-2x)\}={{e}^{x(1-x)}}.(-2{{x}^{2}}+x+1)\]            Now by the sign-scheme for \[-2{{x}^{2}}+x+1\]            \[{f}'(x)\ge 0,\] if \[x\,\in \,\left[ -\frac{1}{2},\,1 \right],\] because \[{{e}^{x}}(1-x)\] is always positive. So, \[f(x)\] is increasing on \[\left[ -\frac{1}{2},\,1 \right]\].