Difference between revisions of "2014 AMC 12A Problems/Problem 24"
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Let <math>f_0(x)=x+|x-100|-|x+100|</math>, and for <math>n\geq 1</math>, let <math>f_n(x)=|f_{n-1}(x)|-1</math>. For how many values of <math>x</math> is <math>f_{100}(x)=0</math>? | Let <math>f_0(x)=x+|x-100|-|x+100|</math>, and for <math>n\geq 1</math>, let <math>f_n(x)=|f_{n-1}(x)|-1</math>. For how many values of <math>x</math> is <math>f_{100}(x)=0</math>? | ||
− | <math>\ | + | <math>\textf{(A) }299\qquad |
\textbf{(B) }300\qquad | \textbf{(B) }300\qquad | ||
\textbf{(C) }301\qquad | \textbf{(C) }301\qquad | ||
\textbf{(D) }302\qquad | \textbf{(D) }302\qquad | ||
\textbf{(E) }303\qquad</math> | \textbf{(E) }303\qquad</math> | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
Revision as of 14:30, 12 October 2018
Contents
Problem
Let , and for , let . For how many values of is ?
$\textf{(A) }299\qquad \textbf{(B) }300\qquad \textbf{(C) }301\qquad \textbf{(D) }302\qquad \textbf{(E) }303\qquad$ (Error compiling LaTeX. ! Undefined control sequence.)
Solution 1
1. Draw the graph of 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 solutions,for x interval:[1,99], the function gain only one extra solution after because there is no summit on the graph any more, and the answer is thus . (Revised by Flamedragon & Jason,C)
Solution 2
First, notice that the recursion and the definition of require that for all such that , if , then is even. Now, we can do case work on to find which values of (such that ) make even. The answer comes out to be all the even values of in the range , in the domain . So, the answer is or .
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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