If $f(x)=x e^{x(1-x)}$, then $f(x)$ is

• A

  increasing in $R$
• B

  decreasing in $R$
• C

  decreasing in $\left[-\frac{1}{2}, 1\right]$
• Correct answer
  D

  increasing in $\left[-\frac{1}{2}, 1\right]$

Solution

Given, $f(x)=x e^{x(1-x)}$ $f^{\prime}(x)=e^{x(1-x)}+x e^{x(1-x)}(1-2 x)$ $=e^{x(1-x)}(1+x(1-2 x))$ $=-e^{x(1-x)}\left(2 x^2-x-1\right)$ $=-e^{x(1-x)}\left(2 x^2-2 x+x-1\right)$ $=-e^{x(1-x)}(x-1)(2 x+1)$
So, $f(x)$ is increasing when $f^{\prime}(x) \geq 0$ and decreasing when $f^{\prime}(x) \leq 0$. Hence, $f$ is increasing in $\left[\frac{-1}{2}, 1\right]$.

Hence, the correct answer is option: 'D'.