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Difference between revisions of "1983 AHSME Problems/Problem 8"

(Created page with "== Problem 8 == Let <math>f(x) = \frac{x+1}{x-1}</math>. Then for <math>x^2 \neq 1, f(-x)</math> is <math>\textbf{(A)}\ \frac{1}{f(x)}\qquad \textbf{(B)}\ -f(x)\qquad \textb...")
(No difference)

Revision as of 13:02, 7 June 2016

Problem 8

Let $f(x) = \frac{x+1}{x-1}$. Then for $x^2 \neq 1, f(-x)$ is

$\textbf{(A)}\ \frac{1}{f(x)}\qquad \textbf{(B)}\ -f(x)\qquad \textbf{(C)}\ \frac{1}{f(-x)}\qquad \textbf{(D)}\ -f(-x)\qquad \textbf{(E)}\ f(x)$

Solution

$\frac{-x+1}{-x-1}\implies\frac{x-1}{x+1}=\frac{1}{f(x)}$, $\fbox{A}$

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