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

1981 AHSME Problems/Problem 17

Problem

The function $f$ is not defined for $x=0$, but, for all non-zero real numbers $x$, $f(x)+2f\left(\dfrac{1}x\right)=3x$. The equation $f(x)=f(-x)$ is satisfied by

$\textbf{(A)}\ \text{exactly one real number} \qquad \textbf{(B)}\ \text{exactly two real numbers} \qquad\textbf{(C)}\ \text{no real numbers}\qquad \\ \textbf{(D)}\ \text{infinitely many, but not all, non-zero real numbers} \qquad\textbf{(E)}\ \text{all non-zero real numbers}$

Solution

Substitute $x$ with $\frac{1}{x}$:

$f(\frac{1}{x})+2f(x)=\frac{3}{x}$.

Adding this to $f(x)+2f\left(\dfrac{1}x\right)=3x$, we get

$3f(x)+3f\left(\dfrac{1}x\right)=3x+\frac{3}{x}$, or

$f(x)+f\left(\dfrac{1}x\right)=x+\frac{1}{x}$.

Subtracting this from $f(\frac{1}{x})+2f(x)=\frac{3}{x}$, we have

$f(x)=\frac{2}{x}-x$.

Then, $f(x)=f(-x)$ when $\frac{2}{x}-x=\frac{2}{-x}+x$ or

$\frac{2}{x}-x=0$, so $x=\pm{\sqrt2}$ are the two real solutions and the answer is $\boxed{(B)}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1981_AHSME_Problems/Problem_17&oldid=211007"