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 \[\left[ -\frac{1}{2},\,1 \right]\].You need to login to perform this action.
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