1971 AMC 12 Problem 30

Attempt Problem 30 of the 1971 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1971 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

30.

Given the linear fractional transformation f1(x)=2x−1x+1, f_1(x)=\frac{2x-1}{x+1}, define fn+1(x)=f1(fn(x))f_{n+1}(x)=f_1(f_n(x)) for n=1,n=1, 2,2, 3,3, ….\ldots. Assuming f35(x)=f5(x),f_{35}(x)=f_5(x), it follows that f28(x)f_{28}(x) is equal to:
xx
1x\frac{1}{x}
x−1x\frac{x-1}{x}
11−x\frac{1}{1-x}
None of these
Answer: D
Concepts:functionrecursiontransformation
Small hint:
Cancel five iterates by composing with the inverse transformation
Big hint:
If g=f1−1g=f_1^{-1}, then f28=g2f_{28}=g^2 once f30f_{30} is the identity
Solution:
The transformation is invertible. From f35=f5,f_{35}=f_5, composing with f5−1f_5^{-1} shows that f30f_{30} is the identity. Solving y=2x−1x+1y=\frac{2x-1}{x+1} for xx gives g(y)=f1−1(y)=y+12−y. g(y)=f_1^{-1}(y)=\frac{y+1}{2-y}. Since iterates have period 30,30, f28=f−2=g2.f_{28}=f_{-2}=g^2. Direct composition gives g(g(x))=11−x. g(g(x))=\frac{1}{1-x}.
Therefore, the correct answer is D.
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.