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Difference between revisions of "1975 AHSME Problems/Problem 21"
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==Solution==
 
==Solution==
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<math>I: f(0) = 1</math>
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Let <math>b = 0</math>. Our equation becomes <math>f(a)f(0) = f(a)</math>, so <math>f(0) = 1</math>. Therefore <math>I</math> is always true.
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<math>II: f(-a) = \frac{1}{f(a)} \text{ for all } a</math>
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Let <math>b = -a</math>. Our equation becomes <math>f(a)f(-a) = f(0) = 1 \longrightarrow f(-a) = \frac{1}{f(a)}</math>. Therefore <math>II</math> is always true.
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<math>III: f(a) = \sqrt[3] {f(3a)} \text{ for all } a</math>
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First let <math>b = a</math>. We get <math>f(a)f(a) = f(2a)</math>. Now let <math>b = 2a</math>, giving us <math>f(3a) = f(a)f(2a) = f(a)^3 \longrightarrow f(a) = \sqrt[3] {f(3a)}</math>. Therefore <math>III</math> is always true.
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<math>IV: f(b) > f(a) if b > a</math>
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This is false. Let <math>f(x) = 2^{-x}</math>, for example. It satisfies the conditions but makes <math>IV</math> false. Therefore <math>IV</math> is not always true.
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Since <math>I, II, \text{ and } III</math> are true, the answer is <math>\boxed{\textbf{(D)}}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1975|num-b=20|num-a=22}}
 
{{AHSME box|year=1975|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 04:18, 27 November 2021

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?

$I.\ f(0) = 1 \qquad \qquad \ \ \qquad \qquad \qquad II.\ f(-a) = \frac{1}{f(a)}\ \text{for all}\ a \\ III.\ f(a) = \sqrt[3]{f(3a)}\ \text{for all}\ a \qquad IV.\ f(b) > f(a)\ \text{if}\ b > a$

$\textbf{(A)}\ \text{III and IV only} \qquad \textbf{(B)}\ \text{I, III, and IV only} \\ \textbf{(C)}\ \text{I, II, and IV only} \qquad \textbf{(D)}\ \text{I, II, and III only} \qquad \textbf{(E)}\ \text{All are true.}$


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.


$III: f(a) = \sqrt[3] {f(3a)} \text{ for all } a$

First let $b = a$. We get $f(a)f(a) = f(2a)$. Now let $b = 2a$, giving us $f(3a) = f(a)f(2a) = f(a)^3 \longrightarrow f(a) = \sqrt[3] {f(3a)}$. Therefore $III$ is always true.


$IV: f(b) > f(a) if b > a$ 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 $\boxed{\textbf{(D)}}$.

See Also

1975 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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