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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
  
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<math> \textbf{(A)}\ \text{exactly one real number}  
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<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==

Revision as of 19:30, 17 June 2021

Problem

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

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

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