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

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 $\boxed{(B)}$ .

Revision as of 00:27, 17 January 2024 (view source) Alexanderruan (talk \| contribs) (→‎Solution) ← Older edit		Revision as of 00:28, 17 January 2024 (view source) Alexanderruan (talk \| contribs) (→‎Solution) Newer edit →
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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

−	<math>\frac{2}{x}=x</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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Difference between revisions of "1981 AHSME Problems/Problem 17"

Revision as of 00:28, 17 January 2024

Problem

Solution