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Difference between revisions of "2005 AMC 12A Problems/Problem 20"

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== Problem ==
 
== Problem ==
 
For each <math>x</math> in <math>[0,1]</math>, define
 
For each <math>x</math> in <math>[0,1]</math>, define
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\[
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<cmath>
 
\begin{array}{clr} f(x) & = 2x, & \text { if } 0 \leq x \leq \frac {1}{2}; \\
 
\begin{array}{clr} f(x) & = 2x, & \text { if } 0 \leq x \leq \frac {1}{2}; \\
 
f(x) & = 2 - 2x, & \text { if } \frac {1}{2} < x \leq 1. \end{array}
 
f(x) & = 2 - 2x, & \text { if } \frac {1}{2} < x \leq 1. \end{array}
−
\]
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</cmath>
 
Let <math>f^{[2]}(x) = f(f(x))</math>, and <math>f^{[n + 1]}(x) = f^{[n]}(f(x))</math> for each integer <math>n \geq 2</math>. For how many values of <math>x</math> in <math>[0,1]</math> is <math>f^{[2005]}(x) = \frac {1}{2}</math>?
 
Let <math>f^{[2]}(x) = f(f(x))</math>, and <math>f^{[n + 1]}(x) = f^{[n]}(f(x))</math> for each integer <math>n \geq 2</math>. For how many values of <math>x</math> in <math>[0,1]</math> is <math>f^{[2005]}(x) = \frac {1}{2}</math>?
−
\[
+
<cmath>
 
(\text {A}) \ 0 \qquad (\text {B}) \ 2005 \qquad (\text {C})\ 4010 \qquad (\text {D}) \ 2005^2 \qquad (\text {E})\ 2^{2005}
 
(\text {A}) \ 0 \qquad (\text {B}) \ 2005 \qquad (\text {C})\ 4010 \qquad (\text {D}) \ 2005^2 \qquad (\text {E})\ 2^{2005}
−
\]
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</cmath>
−
== Solution ==
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== Solution 1 ==
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For the two functions <math>f(x)=2x,0\le x\le \frac{1}{2}</math> and <math>f(x)=2-2x,\frac{1}{2}\le x\le 1</math>,we can see that as long as <math>f(x)</math> is between <math>0</math> and <math>1</math>, <math>x</math> will be in the right domain.  Therefore, we don't need to worry about the domain of <math>x</math>.  Also, every time we change <math>f(x)</math>, the final equation will be in a different form and thus we will get a different value of x.  Every time we have two choices for <math>f(x</math>) and altogether we have to choose <math>2005</math> times. Thus, <math>2^{2005}\Rightarrow\boxed{E}</math>.
+
For the two functions <math>f(x)=2x,0\le x\le \frac{1}{2}</math> and <math>f(x)=2-2x,\frac{1}{2}\le x\le 1</math>,as long as <math>f(x)</math> is between <math>0</math> and <math>1</math>, <math>x</math> will be in the right domain, so we don't need to worry about the domain of <math>x</math>.   
 +
 
 +
 
 +
Also, every time we change <math>f(x)</math>, the expression for the final answer in terms of <math>x</math> will be in a different form(although they'll all satisfy the final equation), so we get a different starting value of <math>x</math>.  Every time we have two choices for <math>f(x</math>) and altogether we have to choose <math>2005</math> times. Thus, <math>2^{2005}\Rightarrow\boxed{E}</math>.
 +
 
 +
Note: the values of x that satisfy <math>f^{[n]}(x) = \frac {1}{2}</math> are <math>\frac{1}{2^{n+1}}</math>, <math>\frac{3}{2^{n+1}}</math>, <math>\frac{5}{2^{n+1}}</math>, <math>\cdots</math> ,<math>\frac{2^{n+1}-1}{2^{n+1}}</math>.
 +
 
 +
== Solution 2==
 +
 
 +
 
 +
We are given that <math>f^{[2005]}(x) = \frac {1}{2}</math>. Thus,  <math>f(f^{[2004]}(x))=\frac{1}{2}</math>. Let <math>f^{[2004]}(x)</math> be equal to <math>y</math>. Thus <math>f(y)=\frac{1}{2}</math> or <math>y=\frac{1}{4}</math> or <math>\frac{3}{4}</math>. Now we know <math>f^{[2004]}(x)</math> is equal to <math>\frac{1}{4}</math> or <math>\frac{3}{4}</math>. Now we know that <math>f(f^{[2003]}(x))=\frac{1}{4}</math> or <math>\frac{3}{4}</math>. Now we solve for <math>f^{[2003]}(x)</math> and let  <math>f^{[2003]}(x)=z</math>. Thus <math>f(z)</math> is equal to <math>\frac{1}{8}</math>,<math>\frac{7}{8}</math>,<math>\frac{5}{8}</math>,and <math>\frac{3}{8}</math>. As we see, <math>f^{[2005]}(x)</math> has 1 solution, <math>f^{[2004]}(x)</math> has 2 solutions, and <math>f^{[2003]}(x)</math> has 4 solutions. Thus for each iteration we double the number of possible solutions. There are 2005 iterations and thus the number of solutions is <math>2^{2005}</math> <math>\Rightarrow\boxed{E}</math>
 +
(not rigorous)
 +
 
 
==See Also==
 
==See Also==
 
{{AMC12 box|ab=A|num-b=19|num-a=21|year=2005}}
 
{{AMC12 box|ab=A|num-b=19|num-a=21|year=2005}}
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 22:49, 28 September 2022

Problem

For each $x$ in $[0,1]$, define \[\begin{array}{clr} f(x) & = 2x, & \text { if } 0 \leq x \leq \frac {1}{2}; \\ f(x) & = 2 - 2x, & \text { if } \frac {1}{2} < x \leq 1. \end{array}\] Let $f^{[2]}(x) = f(f(x))$, and $f^{[n + 1]}(x) = f^{[n]}(f(x))$ for each integer $n \geq 2$. For how many values of $x$ in $[0,1]$ is $f^{[2005]}(x) = \frac {1}{2}$? \[(\text {A}) \ 0 \qquad (\text {B}) \ 2005 \qquad (\text {C})\ 4010 \qquad (\text {D}) \ 2005^2 \qquad (\text {E})\ 2^{2005}\]

Solution 1

For the two functions $f(x)=2x,0\le x\le \frac{1}{2}$ and $f(x)=2-2x,\frac{1}{2}\le x\le 1$,as long as $f(x)$ is between $0$ and $1$, $x$ will be in the right domain, so we don't need to worry about the domain of $x$.


Also, every time we change $f(x)$, the expression for the final answer in terms of $x$ will be in a different form(although they'll all satisfy the final equation), so we get a different starting value of $x$. Every time we have two choices for $f(x$) and altogether we have to choose $2005$ times. Thus, $2^{2005}\Rightarrow\boxed{E}$.

Note: the values of x that satisfy $f^{[n]}(x) = \frac {1}{2}$ are $\frac{1}{2^{n+1}}$, $\frac{3}{2^{n+1}}$, $\frac{5}{2^{n+1}}$, $\cdots$ ,$\frac{2^{n+1}-1}{2^{n+1}}$.

Solution 2

We are given that $f^{[2005]}(x) = \frac {1}{2}$. Thus, $f(f^{[2004]}(x))=\frac{1}{2}$. Let $f^{[2004]}(x)$ be equal to $y$. Thus $f(y)=\frac{1}{2}$ or $y=\frac{1}{4}$ or $\frac{3}{4}$. Now we know $f^{[2004]}(x)$ is equal to $\frac{1}{4}$ or $\frac{3}{4}$. Now we know that $f(f^{[2003]}(x))=\frac{1}{4}$ or $\frac{3}{4}$. Now we solve for $f^{[2003]}(x)$ and let $f^{[2003]}(x)=z$. Thus $f(z)$ is equal to $\frac{1}{8}$,$\frac{7}{8}$,$\frac{5}{8}$,and $\frac{3}{8}$. As we see, $f^{[2005]}(x)$ has 1 solution, $f^{[2004]}(x)$ has 2 solutions, and $f^{[2003]}(x)$ has 4 solutions. Thus for each iteration we double the number of possible solutions. There are 2005 iterations and thus the number of solutions is $2^{2005}$ $\Rightarrow\boxed{E}$ (not rigorous)

See Also

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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