$10 \cdot 15$ Let $f(x)=4 x-x^{2}$. Given $x_{0}$, consider the sequence defined by $x_{n}=f\left(x_{n-1}\right)$ for all $n \geqslant 1$, how many real numbers $x_{0}$ are there such that the sequence $x_{0}, x_{1}, x_{2}, \cdots$ takes only a finite number of different values?
(A) 0.
(B) 1 or 2.
(C) $3,4,5$ or 6.
(D) More than 6 but finite.
(E) Infinitely many.
(39th American High School Mathematics Examination, 1988)
[Solution] Note that when $x_{0}=0$, we have $f(0)=4 \cdot 0-0^{2}=0$, and since $x_{n}=f\left(x_{n-1}\right)$, we get $\left\{x_{n}\right\}=\{0\}$.
Solving $4 x-x^{2}=0$, we get $x=0$ or $x=4$;
When $x_{0}=4$, we obtain the sequence $4,0,0,0,0, \cdots$
Solving $4 x-x^{2}=4$, we get $x=2$;
When $x_{0}=2$, we obtain the sequence $2,4,0,0,0, \cdots$
Solving $4 x-x^{2}=2$, we get $x=2 \pm \sqrt{2}$;
When $x_{0}=2+\sqrt{2}$, we obtain the sequence $2+\sqrt{2}, 2,4,0,0, \cdots$
When $x_{0}=2-\sqrt{2}$, we obtain the sequence $2-\sqrt{2}, 2,4,0,0, \cdots$
Continuing this process, if we assume $x_{0}=a_{n}$, we get a sequence of $n$ different values:
$$
a_{n}, a_{n-1}, \cdots, a_{3}, a_{2}, a_{1}, \cdots
$$

Then, when $x_{0}$ satisfies $4 x_{0}-x_{0}^{2}=a_{n}$, we get $x_{0}=2 \pm \sqrt{4-a_{n}}$. Note that when $0 \leqslant a_{n} \leqslant 4$, $x_{0}$ is a real number, and $0 \leqslant x_{0} \leqslant 4$. In this case, we can let $a_{n+1}=x_{0}$, and we get a sequence of $n+1$ different values:
$$
a_{n+1}, a_{n}, a_{n-1}, \cdots, a_{2}, a_{1}, \cdots
$$

Clearly, there are infinitely many such $x_{0}$. Therefore, the answer is $(E)$.