Question

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

• A. decreasing on $[-1/2,1]$
• B. decreasing on $R$
• C. increasing on $[-1/2,1]$
• D. increasing on $R$

Answer 1

Answer: (C)

$f\left(x\right)=x{e}^{x\left(1-x\right)},x\in R$
$f^{\prime }\left(x\right)=e^{x\left(1-x\right)}.\left[1+x-2x^2\right]$
$=-e^{x\left(1-x\right)}\left[2x^2-x-1\right]$
$=-2x^{x\left(1-x\right)}\left[\left(x+\frac{1}{2}\right)\left(x-1\right)\right]$
$f^{\prime }\left(x\right)=-2e^{x\left(1-x\right)}.A$
where A $=\left(x+\frac{1}{2}\right)\left(x-1\right)$
Now, exponential function is always +ve and $f^{\prime }\left(x\right)$ will be opposite to the sign of A
which is -ve in $\left[-\frac{1}{2},1\right]$
Hence, $f^{\prime }\left(x\right)$ is +ve in $\left[-\frac{1}{2},1\right]$
$\therefore f\left(x\right)$ is increasing on $\left[-\frac{1}{2},1\right]$