Difference between revisions of "2004 AMC 12B Problems/Problem 13"
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− | == Solution == | + | == Solution (Alcumus)== |
Since <math>f(f^{-1}(x))=x</math>, it follows that <math>a(bx+a)+b=x</math>, which implies <math>abx + a^2 +b = x</math>. This equation holds for all values of <math>x</math> only if <math>ab=1</math> and <math>a^2+b=0</math>. | Since <math>f(f^{-1}(x))=x</math>, it follows that <math>a(bx+a)+b=x</math>, which implies <math>abx + a^2 +b = x</math>. This equation holds for all values of <math>x</math> only if <math>ab=1</math> and <math>a^2+b=0</math>. | ||
Latest revision as of 13:56, 2 March 2021
Problem
If and with and real, what is the value of ?
Solution (Alcumus)
Since , it follows that , which implies . This equation holds for all values of only if and .
Then . Substituting into the equation , we get . Then , so , andLikewiseThese are inverses to one another sinceTherefore .
See also
2004 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.