# Difference between revisions of "1981 AHSME Problems/Problem 17"

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

− | <math> \textbf{(A)}\ \text{exactly one real number} | + | <math>\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}</math> |

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− | \qquad \textbf{(B)}\ \text{exactly two real numbers} | ||

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− | \qquad\textbf{(C)}\ \text{no real numbers}\qquad \\ | ||

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− | \textbf{(D)}\ \text{infinitely many, but not all, non-zero real numbers} | ||

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− | \qquad\textbf{(E)}\ \text{all non-zero real numbers} </math> | ||

==Solution== | ==Solution== |

## Latest revision as of 20:30, 17 June 2021

## Problem

The function is not defined for , but, for all non-zero real numbers , . The equation is satisfied by

## Solution

Substitute with .