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1988 AHSME Problems/Problem 30

Problem

Let $f(x) = 4x - x^{2}$. Give $x_{0}$, consider the sequence defined by $x_{n} = f(x_{n-1})$ for all $n \ge 1$. For how many real numbers $x_{0}$ will the sequence $x_{0}, x_{1}, x_{2}, \ldots$ take on only a finite number of different values?

$\textbf{(A)}\ \text{0}\qquad \textbf{(B)}\ \text{1 or 2}\qquad \textbf{(C)}\ \text{3, 4, 5 or 6}\qquad \textbf{(D)}\ \text{more than 6 but finitely many}\qquad \textbf{(E) }\infty$

Solution

Note that $x_{0} = 0$ gives the constant sequence $0, 0, ...$, since $f(0) = 4 \cdot 0 - 0^2 = 0$. Because $f(4)=0, x_{0} = 4$ gives the sequence $4, 0, 0, ...$ with two different values. Similarly, $f(2) = 4$, so $x_{0} = 2$ gives the sequence $2, 4, 0, 0, ...$ with three values. In general, if $x_{0} = a_{n}$ gives the sequence $a_{n}, a_{n-1}, ... , a_{2}, a_{1}, ...$ with $n$ different values, and $f(a_{n+1}) = a_{n}$, then $x_{0} = a_{n+1}$ gives a sequence with $n+1$ different values. (It is easy to see that we could not have $a_{n+1} = a_{i}$ for some $i < n + 1$.) Thus, it follows by induction that there is a sequence with $n$ distinct values for every positive integer $n$, as long as we can verify that there is always a real number $a_{n+1}$ such that $f(a_{n+1}) = a_{n}$. This makes the answer $\boxed{\text{E}}$. The verification alluded to above, which completes the proof, follows from the quadratic formula: the solutions to $f(a_{n+1}) = 4a_{n+1} - a_{n+1}^{2} = a_{n}$ are $a_{n+1} = 2 \pm \sqrt{4 - a_{n}}$. Hence if $0 \leq a_{n} \leq 4$, then $a_{n+1}$ is real, since the part under the square root is non-negative, and in fact $0 \leq a_{n+1} \leq 4$, since $4-a_{n}$ will be between $0$ and $4$, so the square root will be between $0$ and $2$, and $2 \pm$ something between $0$ and $2$ gives something between $0$ and $4$. Finally, since $a_{1} = 0 \leq 4$, it follows by induction that all terms satisfy $0 \leq a_{n} \leq 4$; in particular, they are all real.

See also

1988 AHSME (Problems • Answer Key • Resources)
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