Question

Let $f(x)=4x-x^2.$ Given $x_0,$ consider the sequence defined by $x_n=f(x_{n-1})$ for all $n\ge1.$ 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?

$0$

$1$ or $2$

$3,$ $4,$ $5$ or $6$

more than $6$ but finitely many

infinitely many

Answer: E

Concepts:function iteration preimages induction finite orbits

Difficulty rating: 2930

Small Hint:

Start with $0,$ then find numbers mapping successively to $0$

Big Hint:

For every $a\le4,$ solve $4x-x^2=a$ and choose a new real preimage

Solution:

The starting values $0,4,2$ give finite orbits $0$ , $4\to0$ , and $2\to4\to0.$ More generally, if $a_n$ begins a finite chain ending at $0,$ solve $4a_{n+1}-a_{n+1}^2=a_n.$ Its real solutions are $a_{n+1}=2\pm\sqrt{4-a_n}.$ Choosing a preimage not already in the chain extends it by one new value. Repeating produces infinitely many distinct starting values with finite orbits.

Thus the correct answer is E.

1988 AMC 12 Problem 30

Answer: E

Difficulty rating: 2930

Small Hint:

Big Hint:

Solution:

Problem 30 in Other Years