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

Latest revision as of 23:45, 19 February 2019

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

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

We find $f(-x) = \frac{-x+1}{-x-1} = \frac{x-1}{x+1} = \frac{1}{f(x)}$ , so the answer is $\boxed{\textbf{(A)}}$ .

1983 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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

@@ Line 10: / Line 10: @@
 == Solution ==
-We find <math>f(-x) = \frac{-x+1}{-x-1} = \frac{x-1}{x+1} = \frac{1}{f(x)}</math>, so the answer is <math>\fbox{A}</math>.
+We find <math>f(-x) = \frac{-x+1}{-x-1} = \frac{x-1}{x+1} = \frac{1}{f(x)}</math>, so the answer is <math>\boxed{\textbf{(A)}}</math>.
+==See Also==
+{{AHSME box|year=1983|num-b=7|num-a=9}}
+{{MAA Notice}}