Difference between revisions of "1981 AHSME Problems/Problem 17"
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==Solution== | ==Solution== | ||
| − | Substitute <math>x</math> with <math>\frac{1}{x}</math> | + | Substitute <math>x</math> with <math>\frac{1}{x}</math>: |
| − | <math>f(\frac{1}{x})+2f(x)=\frac{3}{x}</math> | + | <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 | Adding this to <math>f(x)+2f\left(\dfrac{1}x\right)=3x</math>, we get | ||
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<math>3f(x)+3f\left(\dfrac{1}x\right)=3x+\frac{3}{x}</math>, or | <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> | + | <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 | Subtracting this from <math>f(\frac{1}{x})+2f(x)=\frac{3}{x}</math>, we have | ||
| − | <math>f(x)=\frac{2}{x}-x</math> | + | <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 | Then, <math>f(x)=f(-x)</math> when <math>\frac{2}{x}-x=\frac{2}{-x}+x</math> or | ||
| − | <math>\frac{2}{x}= | + | <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 
is not defined for 
, but, for all non-zero real numbers 
, 
. The equation 
is satisfied by
Solution
Substitute with 
:
.
Adding this to , we get
, or
.
Subtracting this from , we have
.
Then, when 
or
, so 
are the two real solutions and the answer is 
.
