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

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== Problem ==
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For each <math>x</math> in <math>[0,1]</math>, define
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\[
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\begin{array}{clr} f(x) & = 2x, & \text { if } 0 \leq x \leq \frac {1}{2}; \\
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f(x) & = 2 - 2x, & \text { if } \frac {1}{2} < x \leq 1. \end{array}
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\]
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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>?
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\[
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(\text {A}) \ 0 \qquad (\text {B}) \ 2005 \qquad (\text {C})\ 4010 \qquad (\text {D}) \ 2005^2 \qquad (\text {E})\ 2^{2005}
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\]
 
== Solution ==
 
== Solution ==
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{{solution}}
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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>.
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==See Also==
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{{AMC12 box|ab=A|num-b=19|num-a=21|year=2005}}
  
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For the two functions f(x)=2x (0<=x<=1/2) and f(x)=2-2x (1/2<x<=1),we can see that as long as f(x) is between 0 and 1, x will be in the right domain.  Therefore, we don't need to worry about the domain of x.  Also, every time we change f(x), 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 f(x) and altogether we have to choose 2005 times. 
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[[Category:Introductory Algebra Problems]]
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2*2*2...2=2^2005
 

Revision as of 14:18, 25 December 2007

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

For the two functions $f(x)=2x,0\le x\le \frac{1}{2}$ and $f(x)=2-2x,\frac{1}{2}\le x\le 1$,we can see that as long as $f(x)$ is between $0$ and $1$, $x$ will be in the right domain. Therefore, we don't need to worry about the domain of $x$. Also, every time we change $f(x)$, 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 $f(x$) and altogether we have to choose $2005$ times. Thus, $2^{2005}\Rightarrow\boxed{E}$.

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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