1975 AMC 12 Problem 21

Attempt Problem 21 of the 1975 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 1975 AMC 12 solutions, or check the answer key.

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21.

Suppose f(x)f(x) is defined for all real numbers x;x; f(x)>0f(x)\gt0 for all x;x; and f(a)f(b)=f(a+b)f(a)f(b)=f(a+b) for all aa and b.b. Which of the following statements are true?
I. f(0)=1f(0)=1
II. f(−a)=1f(a)f(-a)=\frac{1}{f(a)} for all aa
III. f(a)=f(3a)3f(a)=\sqrt[3]{f(3a)} for all aa
IV. f(b)>f(a)f(b)\gt f(a) if b>ab\gt a
III\mathrm{III} and IV\mathrm{IV} only
I,\mathrm{I}, III\mathrm{III} and IV\mathrm{IV} only
I,\mathrm{I}, II\mathrm{II} and IV\mathrm{IV} only
I,\mathrm{I}, II\mathrm{II} and III\mathrm{III} only
All are true.
Answer: D
Concepts:functional equationsubstitutioncounterexample
Small hint:
Substitute a=0, b=−a,a=0,\ b=-a, and then add aa three times
Big hint:
Test the monotonicity claim with the constant function f(x)=1f(x)=1
Solution:
Taking a=0a=0 and using positivity gives f(0)=1.f(0)=1. Taking b=−ab=-a gives f(a)f(−a)=1.f(a)f(-a)=1. Also f(3a)=f(a)3,f(3a)=f(a)^3, so positivity permits the positive cube root in III. But f(x)=1f(x)=1 satisfies the functional equation and is not strictly increasing, so IV need not hold.
Therefore, the correct answer is D.
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