# 1975 AHSME Problems/Problem 21

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## Problem

Suppose $f(x)$ is defined for all real numbers $x; f(x) > 0$ for all $x;$ and $f(a)f(b) = f(a + b)$ for all $a$ and $b$. Which of the following statements are true? $I.\ f(0) = 1 \qquad \qquad \ \ \qquad \qquad \qquad II.\ f(-a) = \frac{1}{f(a)}\ \text{for all}\ a \\ III.\ f(a) = \sqrt{f(3a)}\ \text{for all}\ a \qquad IV.\ f(b) > f(a)\ \text{if}\ b > a$ $\textbf{(A)}\ \text{III and IV only} \qquad \textbf{(B)}\ \text{I, III, and IV only} \\ \textbf{(C)}\ \text{I, II, and IV only} \qquad \textbf{(D)}\ \text{I, II, and III only} \qquad \textbf{(E)}\ \text{All are true.}$

## Solution $I: f(0) = 1$

Let $b = 0$. Our equation becomes $f(a)f(0) = f(a)$, so $f(0) = 1$. Therefore $I$ is always true. $II: f(-a) = \frac{1}{f(a)} \text{ for all } a$

Let $b = -a$. Our equation becomes $f(a)f(-a) = f(0) = 1 \longrightarrow f(-a) = \frac{1}{f(a)}$. Therefore $II$ is always true. $III: f(a) = \sqrt {f(3a)} \text{ for all } a$

First let $b = a$. We get $f(a)f(a) = f(2a)$. Now let $b = 2a$, giving us $f(3a) = f(a)f(2a) = f(a)^3 \longrightarrow f(a) = \sqrt {f(3a)}$. Therefore $III$ is always true. $IV: f(b) > f(a) \text{ if } b > a$ This is false. Let $f(x) = 2^{-x}$, for example. It satisfies the conditions but makes $IV$ false. Therefore $IV$ is not always true.

Since $I, II, \text{ and } III$ are true, the answer is $\boxed{\textbf{(D)}}$.

- mako17

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