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If $\mathrm{f}(x)=x \mathrm{e}^{x(1-x)}, x \in \mathrm{R}$, then $\mathrm{f}(x)$ is
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Verified Answer
The correct answer is:
increasing in $\left[-\frac{1}{2}, 1\right]$
$$
\begin{aligned}
\mathrm{f}(x) & =x \mathrm{e}^{x(1-x)} \\
\therefore \quad \mathrm{f}^{\prime}(x) & =x \mathrm{e}^{x(1-x)}[x(-1)+(1-x)]+\mathrm{e}^{x(1-x)} \\
& =\mathrm{e}^{x(1-x)}\left(x-2 x^2+1\right)
\end{aligned}
$$
For $\mathrm{f}(x)$ to be increasing, $\mathrm{f}^{\prime}(x) \geq 0$
$$
\begin{aligned}
& \Rightarrow \mathrm{e}^{x(1-x)}\left(x-2 x^2+1\right) \geq 0 \\
& \Rightarrow x-2 x^2+1 \geq 0 \\
& \Rightarrow 2 x^2-x-1 \leq 0 \\
& \Rightarrow(2 x+1)(x-1) \leq 0 \\
& \Rightarrow x \in\left[-\frac{1}{2}, 1\right]
\end{aligned}
$$
For $\mathrm{f}(x)$ to be decreasing, $\mathrm{f}^{\prime}(x) \leq 0$
$$
\begin{aligned}
& \Rightarrow(2 x+1)(x-1) \geq 0 \\
& \Rightarrow x \in\left(-\infty,-\frac{1}{2}\right] \cup[1, \infty)
\end{aligned}
$$
\begin{aligned}
\mathrm{f}(x) & =x \mathrm{e}^{x(1-x)} \\
\therefore \quad \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)
\end{aligned}
$$
For $\mathrm{f}(x)$ to be increasing, $\mathrm{f}^{\prime}(x) \geq 0$
$$
\begin{aligned}
& \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]
\end{aligned}
$$
For $\mathrm{f}(x)$ to be decreasing, $\mathrm{f}^{\prime}(x) \leq 0$
$$
\begin{aligned}
& \Rightarrow(2 x+1)(x-1) \geq 0 \\
& \Rightarrow x \in\left(-\infty,-\frac{1}{2}\right] \cup[1, \infty)
\end{aligned}
$$
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