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>. | ||
Revision as of 12:56, 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 
, and![\[f(x)=-x-1.\]](http://latex.artofproblemsolving.com/2/f/3/2f37fa6edb13824dc4671c96b8445db70e5f3eb7.png)
Likewise![\[f^{-1}(x)=-x-1.\]](http://latex.artofproblemsolving.com/1/8/1/1815a3c4b4cb8e4706593728a29e2cac78334fa3.png)
These are inverses to one another since![\[f(f^{-1}(x))=-(-x-1)-1=x+1-1=x.\]](http://latex.artofproblemsolving.com/f/9/a/f9a2ba6c9e59db581ac637bb4fab9f38894cbea0.png)
![\[f^{-1}(f(x))=-(-x-1)-1=x+1-1=x.\]](http://latex.artofproblemsolving.com/b/3/f/b3f20bd3fb5d755851e3caac26d81ad24804ac15.png)
Therefore 
.
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. 
