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

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Then, <math>f(x)=f(-x)</math> when <math>\frac{2}{x}-x=\frac{2}{-x}+x</math> or
 
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</math>, so <math>x=\pm{\sqrt2}</math>
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<math>\frac{2}{x}=x</math>, so <math>x=\pm{\sqrt2}</math> are the two real solutions and the answer is \boxed{B}.

Revision as of 00:26, 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

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

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