A) decreasing on [-1/2,1]
B) decreasing on R
C) increasing on [-1/2,1]
D) increasing on R
Correct Answer: C
Solution :
\[f(x)=x{{e}^{x\left( 1-x \right)}},x\in R\] \[f'(x)={{e}^{x\left( 1-x \right)}}.\left[ 1+x-2{{x}^{2}} \right]\] \[=-{{e}^{x\left( 1-x \right)}}.\left[ 2{{x}^{2}}-x-1 \right]\] \[=-2{{e}^{x\left( 1-x \right)}}.\left[ \left( x+\frac{1}{2} \right)\left( x-1 \right) \right]\] \[f'\left( x \right)=-2{{e}^{x\left( 1-x \right)}}.A\] where\[A=\left( x+\frac{1}{2} \right)\left( x-1 \right)\] Now, exponential function is always +ve and f'(x) will be opposite to the sign of A which is-ve in \[\left[ -\frac{1}{2},1 \right]\] Hence, f '(x) is +ve in\[\left[ -\frac{1}{2},1 \right]\] \[\therefore \]f(x) is increasing on \[\left[ -\frac{1}{2},1 \right]\]
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