1977 AMC 12 Problem 22

Attempt Problem 22 of the 1977 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 1977 AMC 12 solutions, or check the answer key.

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22.

If f(x)f(x) is a real-valued function of the real variable x,x, and f(x)f(x) is not identically zero, and for all aa and bb f(a+b)+f(a−b)=2f(a)+2f(b), \begin{gathered} f(a+b)+f(a-b)\\ =2f(a)+2f(b), \end{gathered} then for all xx and yy
f(0)=1f(0)=1
f(−x)=−f(x)f(-x)=-f(x)
f(−x)=f(x)f(-x)=f(x)
f(x+y)=f(x)+f(y)f(x+y)=f(x)+f(y)
there is a positive number TT such that f(x+T)=f(x)f(x+T)=f(x)
Answer: C
Concepts:functional equationfunctionsubstitution
Small hint:
First substitute a=b=0a=b=0
Big hint:
Next set a=0a=0 and b=xb=x
Solution:
Setting a=b=0a=b=0 gives 2f(0)=4f(0),2f(0)=4f(0), so f(0)=0.f(0)=0. Now set a=0a=0 and b=x.b=x. The equation becomes f(x)+f(−x)=2f(0)+2f(x)=2f(x), \begin{aligned} f(x)+f(-x) &=2f(0)+2f(x)\\ &=2f(x), \end{aligned} and hence f(−x)=f(x)f(-x)=f(x) for every real x.x.
Therefore, the correct answer is C.
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