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