2014 AMC 12A Problems/Problem 24

Problem

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 $f_{100}(x)=0$ ?

$\textbf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$

Solution 1

1. Draw the graph of $f_0(x)$ by dividing the domain into three parts.

2. Look at the recursive rule. Take absolute of the previous function and down by 1 to get the next function.

3. Count the x intercepts of the each function and find the pattern.

The pattern turns out to be $3n+3$ solutions,for x interval:[1,99], the function gain only one extra solution after $f_{99}(x)$ because there is no summit on the graph any more, and the answer is thus $\textbf{(C) }301\qquad$ . (Revised by Flamedragon & Jason,C)

Solution 2

First, notice that the recursion and the definition of $f_0(x)$ require that for all $x$ such that $-100 \le x \le 100$ , if $f_{100}(x)=0$ , then $f_0(x)$ is even. Now, we can do case work on $x$ to find which values of $x$ (such that $-100 \le x \le 100$ ) make $f_0(x)$ even. The answer comes out to be all the even values of $x$ in the range $-100 \le x \le 100$ , in the domain $-300 \le x \le 300$ . So, the answer is $2\cdot150+1$ or $\boxed{\textbf{(C)}\ 301}$ .

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2014amc12a/383

~ dolphin7

2014 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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2014 AMC 12A Problems/Problem 24

Contents

Problem

Solution 1

Solution 2

Video Solution by Richard Rusczyk

See Also