Explanation

(b)
Let $f(x) = \sin{x\cos{x = \frac{1}{2}\sin{2x}}}$
We know that$,\ - 1 \leq \sin{2x \leq 1}$
$\Rightarrow \ - \frac{1}{2} \leq \frac{1}{2}\sin{2x \leq \frac{1}{2}}$
Thus, the greatest and least value of $f(x)$ are $\frac{1}{2}$ and $\frac{1}{2}$ respectively