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Difference between revisions of "1988 AHSME Problems/Problem 30"

(Created page with "==Problem== Let <math>f(x) = 4x - x^{2}</math>. Give <math>x_{0}</math>, consider the sequence defined by <math>x_{n} = f(x_{n-1})</math> for all <math>n \ge 1</math>. For how ...")
 
(Added the solution for the correct problem)
 
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<math>\textbf{(A)}\ \text{0}\qquad
 
<math>\textbf{(A)}\ \text{0}\qquad
 
\textbf{(B)}\ \text{1 or 2}\qquad
 
\textbf{(B)}\ \text{1 or 2}\qquad
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\textbf{(C)}\ \text{3, 4, 5 or 6}\qquad\\
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\textbf{(C)}\ \text{3, 4, 5 or 6}\qquad
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\textbf{(D)}\ \text{more than 6 but finitely many}\qquad\\
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\textbf{(D)}\ \text{more than 6 but finitely many}\qquad
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\textbf{(E) }\infty</math>  
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\textbf{(E) }\infty</math>
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==Solution==
 
==Solution==
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Note that <math>x_{0} = 0</math> gives the constant sequence <math>0, 0, ...</math>, since <math>f(0) = 4 \cdot 0 - 0^2 = 0</math>. Because <math>f(4)=0, x_{0} = 4</math> gives the sequence <math>4, 0, 0, ...</math> with two different values. Similarly, <math>f(2) = 4</math>, so <math>x_{0} = 2</math> gives the sequence <math>2, 4, 0, 0, ...</math> with three values. In general, if <math>x_{0} = a_{n}</math> gives the sequence <math>a_{n}, a_{n-1}, ... , a_{2}, a_{1}, ...</math> with <math>n</math> different values, and <math>f(a_{n+1}) = a_{n}</math>, then <math>x_{0} = a_{n+1}</math> gives a sequence with <math>n+1</math> different values. (It is easy to see that we could not have <math>a_{n+1} = a_{i}</math> for some <math>i < n + 1</math>.) Thus, it follows by induction that there is a sequence with <math>n</math> distinct values for every positive integer <math>n</math>, as long as we can verify that there is always a real number <math>a_{n+1}</math> such that <math>f(a_{n+1}) = a_{n}</math>. This makes the answer <math>\boxed{\text{E}}</math>. The verification alluded to above, which completes the proof, follows from the quadratic formula: the solutions to <math>f(a_{n+1}) = 4a_{n+1} - a_{n+1}^{2} = a_{n}</math> are <math>a_{n+1} = 2 \pm \sqrt{4 - a_{n}}</math>. Hence if <math>0 \leq a_{n} \leq 4</math>, then <math>a_{n+1}</math> is real, since the part under the square root is non-negative, and in fact <math>0 \leq a_{n+1} \leq 4</math>, since <math>4-a_{n}</math> will be between <math>0</math> and <math>4</math>, so the square root will be between <math>0</math> and <math>2</math>, and <math>2 \pm</math> something between <math>0</math> and <math>2</math> gives something between <math>0</math> and <math>4</math>. Finally, since <math>a_{1} = 0 \leq 4</math>, it follows by induction that all terms satisfy <math>0 \leq a_{n} \leq 4</math>; in particular, they are all real.
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== See also ==
 
== See also ==
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{{AHSME box|year=1988|num-b=28|after=Last Question}}   
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{{AHSME box|year=1988|num-b=29|after=Last Question}}   
  
 
[[Category: Intermediate Algebra Problems]]
 
[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:16, 27 February 2018

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)
Preceded by
Problem 29 		Followed by
Last Question
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