Example 4. The necessary and sufficient condition for $\arccos (-x)$ to be greater than $\arccos x$ is:
(A) $x \in(0,1)$;
(B) $x \in(-1,0)$;
(C) $x \in[0,1]$;
(D) $x \in\left[0, \frac{\pi}{2}\right]$.
Among the four answers, the domain of $\arccos x$, $|x| \leqslant 1$, directly eliminates (D); then by $\arccos (-x) > \arccos x$, using the monotonicity of the arccosine function, we can deduce that $-x < 0$, thus eliminating (B). When $x=0$, we have $\arccos (-x) = \arccos x$, which eliminates (C); therefore, (A) is the correct answer.