8.52 Let the function $f(x)$ satisfy $f(1+x)=f(1-x)$ for all real numbers. If the equation $f(x)=0$ has exactly two distinct real roots. Then the absolute value of the difference between these two roots is
(A) 1.
(B) 2.
(C) 4.
(D) Uncertain.
(China Sichuan Province High School Mathematics Competition, 1990)
[Solution 1] Let $f\left(x_{1}\right)=f\left(x_{2}\right)=0$, i.e., $x_{1}, x_{2}$ are two distinct roots of $f(x)=0$.

Let $x_{1}=1+x_{0}$, by the given condition, $f\left(1+x_{0}\right)=f\left(1-x_{0}\right)$, we get
$$
\begin{array}{c}
x_{2}=1-x_{0} . \\
\therefore \quad\left|x_{1}-x_{2}\right|=\left|\left(1+x_{0}\right)-\left(1-x_{0}\right)\right|=2\left|x_{0}\right| .
\end{array}
$$

Therefore, the answer is $(D)$.
[Solution 2] A parabola with the axis of symmetry at $x=1$ and intersecting the $x$-axis at two points is a function that satisfies the given condition (as shown in the figure).

From the graph, $| x_{1}-x_{2} |$ is uncertain (as it depends on the shape of the parabola).
Therefore, the answer is $(D)$.