7.19 The range of the function $f(x)=\operatorname{arcsincos} x+\operatorname{arccossin} x$ is
(A) $\left[-\frac{\pi}{2}, \frac{3 \pi}{2}\right]$.
(B) $\left[0, \frac{3 \pi}{2}\right]$.
(C) $\left[-\frac{\pi}{2}, \pi\right]$.
(D) $[0, \pi]$.
(2nd "Hope Cup" National Mathematics Invitational Competition, 1991)
[Solution] (1) When $\frac{\pi}{2}<x<\pi$:
We have $-\frac{\pi}{2}<\frac{\pi}{2}-x<0$, and $0<x-\frac{\pi}{2}<\frac{\pi}{2}$. Thus,
$$
\begin{array}{c}
\operatorname{arcsincos} x=\operatorname{arcsinsin}\left(\frac{\pi}{2}-x\right)=\frac{\pi}{2}-x, \\
\operatorname{arccossin} x=\operatorname{arccoscos}\left(x-\frac{\pi}{2}\right)=x-\frac{\pi}{2}, \\
\therefore \quad f(x)=\left(\frac{\pi}{2}-x\right)+\left(x-\frac{\pi}{2}\right)=0 .
\end{array}
$$
(2) When $\frac{3 \pi}{2}<x \leqslant 2 \pi$:

We have $\frac{\pi}{2}<\frac{5}{2} \pi-x \leqslant \pi$, and $0<x-\frac{3}{2} \pi \leqslant \frac{\pi}{2}$, at this time
$$
\begin{array}{c}
\operatorname{arcsincos} x=\operatorname{arcsinsin}\left(x-\frac{3}{2} \pi\right)=x-\frac{3}{2} \pi, \\
\operatorname{arccossin} x=\operatorname{arccoscos}\left(\frac{5 \pi}{2}-x\right)=\frac{5 \pi}{2}-x, \\
\therefore \quad f(x)=\left(x-\frac{3}{2} \pi\right)+\left(\frac{5 \pi}{2}-x\right)=\pi .
\end{array}
$$

When $0 \leqslant x \leqslant \frac{\pi}{2}$, $\cos x$ is a decreasing function, and $\arcsin x$ is an increasing function on $[-1,1]$, thus $\operatorname{arcsincos} x$ is a decreasing function on $\left[0, \frac{\pi}{2}\right]$.
Similarly, $\operatorname{arccossin} x$ is also a decreasing function on $\left[0, \frac{\pi}{2}\right]$.
Therefore, $\quad 0=f\left(\frac{\pi}{2}\right) \leqslant f(x) \leqslant f(0)=\pi$.
When $\pi<x \leqslant \frac{3}{2} \pi$, $\cos x$ is an increasing function, and $\arcsin x$ is an increasing function on $[-1,1]$, hence $\operatorname{arcsincos} x$ is an increasing function on $\left(\pi, \frac{3}{2} \pi\right]$.
Similarly, $\operatorname{arccossin} x$ is also an increasing function on $\left(\pi, \frac{3}{2} \pi\right]$.
Therefore, $\quad 0=f(\pi)<f(x) \leqslant f\left(\frac{3}{2} \pi\right)=\pi$.
In summary, the range of $f(x)$ on $[0,2 \pi]$ is $[0, \pi]$.
Since $f(x)$ is a periodic function with a period of $2 \pi$, the range of $f(x)$ is $[0, \pi]$. Therefore, the answer is $(D)$.