Problem:
A function f is defined for all real numbers and satisfies
f(2+x)=f(2−x) and f(7+x)=f(7−x)
for all real x. If x=0 is a root of f(x)=0, what is the least number of roots f(x)=0 must have in the interval −1000≤x≤1000?
Solution:
First we use the given equations to find various numbers in the domain to which f assigns the same values. We find that
f(x)=f(2+(x−2))=f(2−(x−2))=f(4−x)(1)
and that
f(4−x)=f(7−(x+3))=f(7+(x+3))=f(x+10)(2)
From Equations (1) and (2) it follows that
f(x+10)=f(x)(3)
Replacing x by x+10 and then by x−10 in Equation (3), we get f(x+10)=f(x+20) and f(x−10)=f(x). Continuing in this way, it follows that
f(x+10n)=f(x), for n=±1,±2,±3,…,(4)
Since f(0)=0, Equation (4) implies that
f(±10)=f(±20)=⋯=f(±1000)=0,
necessitating a total of 201 roots for the equation f(x)=0 in the closed interval [−1000,1000].
Next note that by setting x=0 in Equation (1), f(4)=f(0)=0 follows. Therefore, setting x=4 in Equation (4), we obtain 200 more roots for f(x)=0 at x=−996,−986,…,−6,4,14,…,994. Since the zig-zag function pictured below satisfies the given conditions and has precisely these and no other roots, the answer to the problem is 401​.
The problems on this page are the property of the MAA's American Mathematics Competitions
