Question

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

• A. increasing on $\left[-\frac{1}{2},1\right]$.
• B. decreasing on $\mathrm{ℝ}$.
• C. increasing on $\mathrm{ℝ}$.
• D. decreasing on $\left[-\frac{1}{2},1\right]$.

Answer 1

The correct option is A

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

The explanation for the correct option:

Step-1 First derivative of $f$

The given function, $f\left(x\right)=x{e}^{x\left(1-x\right)}$.

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left[f\left(x\right)\right]=\frac{d}{dx}\left[x{e}^{x\left(1-x\right)}\right] \\ \Rightarrow f\text{'}\left(x\right)=\frac{dx}{dx}·e^{x\left(1-x\right)}+x·\frac{d}{dx}\left[e^{x\left(1-x\right)}\right] \\ \Rightarrow f\text{'}\left(x\right)=e^{x\left(1-x\right)}+x{e}^{x\left(1-x\right)}\frac{d}{dx}\left[x\left(1-x\right)\right] \\ \Rightarrow f\text{'}\left(x\right)=e^{x\left(1-x\right)}+x{e}^{x\left(1-x\right)}\frac{d}{dx}\left[x-x^2\right] \\ \Rightarrow f\text{'}\left(x\right)=e^{x\left(1-x\right)}+x{e}^{x\left(1-x\right)}\left(1-2x\right) \\ \Rightarrow f\text{'}\left(x\right)=e^{x\left(1-x\right)}\left(1+x-2x^2\right) \end{gathered}$$

Step-2 Applying condition of increasing function:

For increasing function, $f\text{'}\left(x\right)\ge 0$.

$$\begin{gathered} \Rightarrow e^{x\left(1-x\right)}\left(1+x-2x^2\right)\ge 0 \\ \Rightarrow \left(1+x-2x^2\right)\ge 0\left[\because e^{x\left(1-x\right)}>0\right] \\ \Rightarrow 2x^2-x-1\le 0 \\ \Rightarrow 2x^2-2x+x-1\le 0 \\ \Rightarrow 2x\left(x-1\right)+1\left(x-1\right)\le 0 \\ \Rightarrow \left(x-1\right)\left(2x+1\right)\le 0 \\ \Rightarrow \left(x-1\right)\left(x+\frac{1}{2}\right)\le 0 \\ \Rightarrow x\in \left[-\frac{1}{2},1\right] \end{gathered}$$

Therefore, $f\left(x\right)$ is increasing on $\left[-\frac{1}{2},1\right]$.

Hence, (A) is the correct option.