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

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==Problem==
 
==Problem==
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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
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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)+2f\left(\dfrac{1}x\right)=3x</math>. The equation <math>f(x)=f(-x)</math> is satisfied by
  
 
<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>
 
<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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==Solution==
 
==Solution==
  
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Substitute <math>x</math> with <math>\frac{1}{x}</math>.
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Substitute <math>x</math> with <math>\frac{1}{x}</math>:
  
−
<math>f(\frac{1}{x})+2f(x)=\frac{3}{x}</math>
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<math>f(\frac{1}{x})+2f(x)=\frac{3}{x}</math>.
 +
 
 +
Adding this to <math>f(x)+2f\left(\dfrac{1}x\right)=3x</math>, we get
 +
 
 +
<math>3f(x)+3f\left(\dfrac{1}x\right)=3x+\frac{3}{x}</math>, or
 +
 
 +
<math>f(x)+f\left(\dfrac{1}x\right)=x+\frac{1}{x}</math>.
 +
 
 +
Subtracting this from <math>f(\frac{1}{x})+2f(x)=\frac{3}{x}</math>, we have
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 +
<math>f(x)=\frac{2}{x}-x</math>.
 +
 
 +
Then, <math>f(x)=f(-x)</math> when <math>\frac{2}{x}-x=\frac{2}{-x}+x</math> or
 +
 
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<math>\frac{2}{x}-x=0</math>, so <math>x=\pm{\sqrt2}</math> are the two real solutions and the answer is <math>\boxed{(B)}</math>.

Latest revision as of 00:28, 17 January 2024

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 $\boxed{(B)}$.

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