1977 AHSME Problems/Problem 22

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 $\boxed{C}$