Problem:
Let f0​(x)=x+∣x−100∣−∣x+100∣, and for n≥1, let fn​(x)=∣fn−1​(x)∣−1. For how many values of x is f100​(x)=0 ?
Answer Choices:
A. 299
B. 300
C. 301
D. 302
E. 303
Solution:
For integers n≥1 and k≥0, if fn−1​(x)=±k, then fn​(x)= k−1. Thus if f0​(x)=±k, then fk​(x)=0. Furthermore, if fn​(x)=0, then fn+1​(x)=−1 and fn+2​(x)=0. It follows that the zeros of f100​ are the solutions of f0​(x)=2k for −50≤k≤50. To count these solutions, note that
f0​(x)=⎩⎪⎪⎨⎪⎪⎧​x+200 if x<−100−x if −100≤x<100, and x−200 if x≥100​
The graph of f0​(x) is piecewise linear with turning points at (−100,100) and (100,−100). The line y=2k crosses the graph three times for −49≤k≤49 and twice for k=±50. Therefore the number of zeros of f100​(x) is 99⋅3+2⋅2=301​.
The problems on this page are the property of the MAA's American Mathematics Competitions