$7 \cdot 20$ The range of the function $f(x)=\operatorname{arctg} x+\frac{1}{2} \arcsin x$ is
(A) $(-\pi, \pi)$.
(B) $\left[-\frac{3 \pi}{4}, \frac{3 \pi}{4}\right]$.
(C) $\left(-\frac{3 \pi}{4}, \frac{3 \pi}{4}\right)$.
(D) $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.
(China High School Mathematics League, 1989)
[Solution] Since the domain of $f(x)$ is $[-1,1]$, within this interval, we have $-\frac{\pi}{4} \leqslant \operatorname{arctg} x \leqslant \frac{\pi}{4}, -\frac{\pi}{4} \leqslant \frac{1}{2} \arcsin x \leqslant \frac{\pi}{4}$.

Noting that $\operatorname{arctg} x$ and $\frac{1}{2} \arcsin x$ are both increasing on $[-1,1]$, the range of $f(x)$ is $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Therefore, the answer is $(D)$.