AMC12 2005 A

AMC12 2005 A · Q20

AMC12 2005 A · Q20. It mainly tests Functions basics, Basic counting (rules of product/sum).

For each $x$ in $[0,1]$, define \[f(x) = \begin{cases} 2x, \qquad\qquad \mathrm{if} \quad 0 \leq x \leq \frac{1}{2}\\ 2-2x, \qquad \mathrm{if} \quad \frac{1}{2} < x \leq 1. \end{cases}\] 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 $f^{[2005]}(x) = 1/2$?

对每个 $x \in [0,1]$，定义 \[f(x) = \begin{cases} 2x, \qquad\qquad \mathrm{if} \quad 0 \leq x \leq \frac{1}{2}\\ 2-2x, \qquad \mathrm{if} \quad \frac{1}{2} < x \leq 1. \end{cases}\] 令 $f^{[2]}(x) = f(f(x))$，并且对每个整数 $n \geq 2$，令 $f^{[n + 1]}(x) = f^{[n]}(f(x))$。在 $[0,1]$ 中有多少个 $x$ 满足 $f^{[2005]}(x) = 1/2$？

(A) 0 0

(B) 2005 2005

(D) 2005^2 2005^2

(E) 2^2005 2^{2005}

Answer

Correct choice: (E)

正确答案：（E）

Solution

For the two functions $f(x)=2x,0\le x\le \frac{1}{2}$ and $f(x)=2-2x,\frac{1}{2}\le x\le 1$,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 $\frac{1}{2^{n+1}}$, $\frac{3}{2^{n+1}}$, $\frac{5}{2^{n+1}}$, $\cdots$ ,$\frac{2^{n+1}-1}{2^{n+1}}$.

对于两个函数 $f(x)=2x,0\le x\le \frac{1}{2}$ 和 $f(x)=2-2x,\frac{1}{2}\le x\le 1$，只要 $f(x)$ 在 $0$ 与 $1$ 之间，$x$ 就会落在正确的定义域内，因此我们不必担心 $x$ 的定义域问题。另外，每次改变 $f(x)$ 的取法时，最终答案用 $x$ 表示的形式都会不同（尽管它们都满足最终方程），因此会得到不同的起始值 $x$。每一步对 $f(x$) 都有两种选择，而总共需要选择 $2005$ 次。因此共有 $2^{2005}\Rightarrow\boxed{E}$。 Note: the values of x that satisfy $f^{[n]}(x) = \frac {1}{2}$ are $\frac{1}{2^{n+1}}$, $\frac{3}{2^{n+1}}$, $\frac{5}{2^{n+1}}$, $\cdots$ ,$\frac{2^{n+1}-1}{2^{n+1}}$.

Topics