Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1975 AHSME Problems/Problem 21"
m (Small fix to IV Latex $IV: f(b) > f(a) if b > a$ to $IV: f(b) > f(a) \text{ if } b > a$)
 
(3 intermediate revisions by 2 users not shown)
Line 1: Line 1:
−
te==Problem==
+
==Problem==
 
Suppose <math>f(x)</math> is defined for all real numbers <math>x; f(x) > 0</math> for all <math>x;</math> and <math>f(a)f(b) = f(a + b)</math> for all <math>a</math> and <math>b</math>. Which of the following statements are true?
 
Suppose <math>f(x)</math> is defined for all real numbers <math>x; f(x) > 0</math> for all <math>x;</math> and <math>f(a)f(b) = f(a + b)</math> for all <math>a</math> and <math>b</math>. Which of the following statements are true?
  
Line 14: Line 14:
  
 
==Solution==
 
==Solution==
−
nothing yet :(
 
  
 +
<math>I: f(0) = 1</math>
 +
 +
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.
 +
 +
 +
<math>II: f(-a) = \frac{1}{f(a)} \text{ for all } a</math>
 +
 +
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.
 +
 +
 +
<math>III: f(a) = \sqrt[3] {f(3a)} \text{ for all } a</math>
 +
 +
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.
 +
 +
 +
<math>IV: f(b) > f(a) \text{ if } b > a</math>
 +
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.
 +
 +
 +
Since <math>I, II, \text{ and } III</math> are true, the answer is <math>\boxed{\textbf{(D)}}</math>.
 +
 +
 +
- mako17
  
 
==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}}

Latest revision as of 20:24, 12 December 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.


$IV: f(b) > f(a) \text{ if } b > a$ 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1975_AHSME_Problems/Problem_21&oldid=167603"