2014 AMC 12A Problems/Problem 24


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


~ dolphin7

See Also

2014 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 23
Followed by
Problem 25
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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