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Difference between revisions of "2014 AMC 12A Problems/Problem 24"

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==Solution 2==
 
==Solution 2==
  
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First, notice that the recursion and the definition of <math>f_0(x)</math> require that for all <math>x</math> such that <math>-100 \le x \le 100</math>, if <math>f_{100}(x)=0</math>, then <math>f_0(x)</math> is even. Now, we can do case work on <math>x</math> to find which values of <math>x</math> (such that <math>-100 \le x \le 100</math>) make <math>f_0(x)</math> even. The answer comes out to be all the even values of <math>x</math> in the range <math>-100 \le x \le 100</math>. So, the answer is <math>2\cdot150+1</math> or <math>\boxed{\textbf{(C)}\ 301}</math>.
+
First, notice that the recursion and the definition of <math>f_0(x)</math> require that for all <math>x</math> such that <math>-100 \le x \le 100</math>, if <math>f_{100}(x)=0</math>, then <math>f_0(x)</math> is even. Now, we can do case work on <math>x</math> to find which values of <math>x</math> (such that <math>-100 \le x \le 100</math>) make <math>f_0(x)</math> even. The answer comes out to be all the even values of <math>x</math> in the range <math>-100 \le x \le 100</math>, in the domain <math>-300 \le x \le 300</math> . So, the answer is <math>2\cdot150+1</math> or <math>\boxed{\textbf{(C)}\ 301}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2014|ab=A|num-b=23|num-a=25}}
 
{{AMC12 box|year=2014|ab=A|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:52, 30 July 2014

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 $\boxed{\textbf{(C)}\ 301}$.

See Also

2014 AMC 12A (Problems • Answer Key • Resources)
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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