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

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Substitute <math>x</math> with <math>\frac{1}{x}</math>:
 
Substitute <math>x</math> with <math>\frac{1}{x}</math>:
  
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<math>f(\frac{1}{x})+2f(x)=\frac{3}{x}</math>.
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<math>f\left(\frac{1}{x}\right)+2f(x)=\frac{3}{x}</math>.
  
 
Adding this to <math>f(x)+2f\left(\dfrac{1}x\right)=3x</math>, we get
 
Adding this to <math>f(x)+2f\left(\dfrac{1}x\right)=3x</math>, we get
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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>.
 
<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>.
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==See also==
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{{AHSME box|year=1981|num-b=16|num-a=18}}
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{{MAA Notice}}

Latest revision as of 14:11, 28 June 2025

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 $\frac{1}{x}$:

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

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 $\boxed{(B)}$.

See also

1981 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 16 		Followed by
Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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