2005 AMC 12A Problems/Problem 20

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Problem

For each $x$ in $[0,1]$, define \[ f(x)=2x, if 0x12;f(x)=22x, if 12<x1. \] 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 19
Followed by
Problem 21
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All AMC 12 Problems and Solutions