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 | ||
| − | + | <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} | ||
| − | + | </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} | ||
| − | + | </cmath> | |
== Solution == | == Solution == | ||
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>,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>. | ||
Revision as of 14:19, 25 December 2007
Problem
For each 
in ![$[0,1]$](http://latex.artofproblemsolving.com/e/8/6/e861e10e1c19918756b9c8b7717684593c63aeb8.png)
, define
![\[\begin{array}{clr} f(x) & = 2x, & \text { if } 0 \leq x \leq \frac {1}{2}; \\ f(x) & = 2 - 2x, & \text { if } \frac {1}{2} < x \leq 1. \end{array}\]](http://latex.artofproblemsolving.com/c/2/5/c25a89c6f31ca5a9ad24ed419fdf61430fad13e6.png)
Let ![$f^{[2]}(x) = f(f(x))$](http://latex.artofproblemsolving.com/1/9/a/19a1e133f69e72ede551e9a9ac13d6fc753631b6.png)
, and ![$f^{[n + 1]}(x) = f^{[n]}(f(x))$](http://latex.artofproblemsolving.com/b/2/8/b28e24ffc0a9d6d078390a1cc78472f25aad30de.png)
for each integer 
. For how many values of in is ![$f^{[2005]}(x) = \frac {1}{2}$](http://latex.artofproblemsolving.com/c/f/8/cf8eae18d34cdc34dd8eb5c5567f33e726ef8bb4.png)
?
![\[(\text {A}) \ 0 \qquad (\text {B}) \ 2005 \qquad (\text {C})\ 4010 \qquad (\text {D}) \ 2005^2 \qquad (\text {E})\ 2^{2005}\]](http://latex.artofproblemsolving.com/9/b/d/9bd6c0aa1515605b08a16a33ca750b38371d9417.png)
Solution
For the two functions 
and 
,we can see that as long as 
is between 
and 
, will be in the right domain. Therefore, we don't need to worry about the domain of . Also, every time we change , 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 
) and altogether we have to choose 
times. Thus, 
.
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 | |
