Difference between revisions of "1981 AHSME Problems/Problem 17"
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==Problem== | ==Problem== | ||
− | 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)+ | + | 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} | + | <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> |
− | + | ==Solution== | |
− | \ | + | Substitute <math>x</math> with <math>\frac{1}{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 | |
− | == | + | <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 | ||
+ | |||
+ | <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 | |
− | <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>. |
Latest revision as of 01:28, 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
.