If \(\mathrm{f}(x)=x \mathrm{e}^{x(1-x)}, x \in \mathrm{R}\), then \(\mathrm{f}(x)\) is

\(\mathrm{f}(x) =x \mathrm{e}^{x(1-x)} \)
\( \therefore \mathrm{f}^{\prime}(x) =x \mathrm{e}^{x(1-x)}[x(-1)+(1-x)]+\mathrm{e}^{x(1-x)} \)
\( =\mathrm{e}^{x(1-x)}\left(x-2 x^2+1\right)\)
For \(\mathrm{f}(x)\) to be increasing, \(\mathrm{f}^{\prime}(x) \geq 0\)
\(\Rightarrow \mathrm{e}^{x(1-x)}\left(x-2 x^2+1\right) \geq 0 \)
\( \Rightarrow x-2 x^2+1 \geq 0 \)
\( \Rightarrow 2 x^2-x-1 \leq 0 \)
\( \Rightarrow(2 x+1)(x-1) \leq 0 \)
\( \Rightarrow x \in\left[-\frac{1}{2}, 1\right] \)
For \(\mathrm{f}(x)\) to be decreasing, \(\mathrm{f}^{\prime}(x) \leq 0\)
\(\Rightarrow(2 x+1)(x-1) \geq 0 \)
\( \Rightarrow x \in\left(-\infty,-\frac{1}{2}\right] \cup[1, \infty)\)