Explanation

(a)
$\because\ \ f(x) = xe^{x(1 - x)}$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = e^{x(1 - x) + x.e^{x(1 - x)}}.(1 - 2x)$
$= e^{x(1 - x)}\{ 1 + x(1 - 2x)\}$
$= e^{x(1 - x)}.\left( - 2x^{2} + x + 1 \right)$
It is clear that $e^{x(1 - x)} > 0$ for all $x$
Now, by sign rule for $- 2x^{2} + x + 1$
$f^{'}(x) > 0$, if $x \in \left\lbrack - \frac{1}{2},1 \right\rbrack$
So, $f(x)$ is increasing on $\left\lbrack - \frac{1}{2},1 \right\rbrack$