AMC12 2014 A

AMC12 2014 A · Q24

AMC12 2014 A · Q24. It mainly tests Absolute value, Graphs (coordinate plane).

Let $f_0(x)=x+|x-100|-|x+100|$, and for $n\geq 1$, let $f_n(x)=|f_{n-1}(x)|-1$. For how many values of $x$ is $f_{100}(x)=0$?

令 $f_0(x)=x+|x-100|-|x+100|$，对于 $n\geq 1$，令 $f_n(x)=|f_{n-1}(x)|-1$。有几个 $x$ 使得 $f_{100}(x)=0$？

(A) 299 299

(B) 300 300

(D) 302 302

(E) 303\qquad 303\qquad

Answer

Correct choice: (C)

正确答案：（C）

Solution

1. Draw the graph of $f_0(x)$ by dividing the domain into three parts. 2. Apply the recursive rule a few times to find the pattern. Note: $f_n(x) = |f_{n-1}(x)| - 10$ is used to enlarge the difference, but the reasoning is the same. 3. Extrapolate to $f_{100}$. Notice that the summits start $100$ away from $0$ and get $1$ closer each iteration, so they reach $0$ exactly at $f_{100}$. $f_{100}(x)$ reaches $0$ at $x = -300$, then zigzags between $0$ and $-1$, hitting $0$ at every even $x$, before leaving $0$ at $x = 300$. This means that $f_{100}(x) = 0$ at all even $x$ where $-300 \le x \le 300$. This is a $601$-integer odd-size range with even numbers at the endpoints, so just over half of the integers are even, or $\frac{601+1}{2} = \boxed{\textbf{(C) }301}$.

1. 将定义域分为三部分，绘制 $f_0(x)$ 的图像。 2. 应用递归规则几次，发现模式。注意：使用 $f_n(x) = |f_{n-1}(x)| - 10$ 来放大差异，但推理相同。 3. 外推到 $f_{100}$。注意到峰值起始距离 $0$ 为 $100$，每次迭代靠近 $1$，故恰好在 $f_{100}$ 处到达 $0$。 $f_{100}(x)$ 在 $x = -300$ 处达到 $0$，然后在 $0$ 和 $-1$ 间锯齿状变化，每当 $x$ 为偶数时击中 $0$，直到 $x = 300$ 离开 $0$。这意味着 $f_{100}(x) = 0$ 当且仅当 $-300 \le x \le 300$ 且 $x$ 为偶数。这是包含 $601$ 个整数的奇数大小范围，端点为偶数，故偶数个数略多于一半，即 $\frac{601+1}{2} = \boxed{\textbf{(C) }301}$。

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