1977 AHSME Problems/Problem 22

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We can start by finding the value of $f(0)$. Let $a = b = 0$ \[f(0) + f(0) = 2f(0) + 2f(0) \Rrightarrow 2f(0) = 4f(0) \Rrightarrow f(0) = 0\] Thus, $A$ is not true. To check $B$, we let $a = 0$. We have \[f(0+b) + f(0-b) = 2f(0) + 2f(b)\] \[f(b) + f(-b) = 0 + 2f(b)\] \[f(-b) = f(b)\] Thus, $B$ is not true, but $C$ is. Thus, the answer is $\boxed{C}$

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