1977 AHSME Problems/Problem 22

Problem 22

If $f(x)$ is a real valued function of the real variable $x$ , and $f(x)$ is not identically zero, and for all $a$ and $b$ $f(a+b)+f(a-b)=2f(a)+2f(b)$ , then for all $x$ and $y$

$\textbf{(A) }f(0)=1\qquad \textbf{(B) }f(-x)=-f(x)\qquad \textbf{(C) }f(-x)=f(x)\qquad \\ \textbf{(D) }f(x+y)=f(x)+f(y) \qquad \\ \textbf{(E) }\text{there is a positive real number }T\text{ such that }f(x+T)=f(x)$

Solution

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 $\boxed{C}$

~~JustinLee2017

1977 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 23
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1977 AHSME Problems/Problem 22

Problem 22

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