Difference between revisions of "1983 AHSME Problems/Problem 8"
Katzrockso (talk | contribs) (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...") |
Sevenoptimus (talk | contribs) (Fixed formatting and added box at the bottom) |
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== Solution == | == Solution == | ||
| − | <math>\frac{-x+1}{-x-1} | + | 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}} | ||
Latest revision as of 23:45, 19 February 2019
Problem 8
Let 
. Then for 
is
Solution
We find 
, so the answer is 
.
See Also
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
