Difference between revisions of "2004 AMC 12B Problems/Problem 13"

Revision as of 11:11, 26 April 2025

Problem

If $f(x) = ax+b$ and $f^{-1}(x) = bx+a$ with $a$ and $b$ real, what is the value of $a+b$ ?

Solution (Alcumus)

Since $f(f^{-1}(x))=x$ , it follows that $a(bx+a)+b=x$ , which implies $abx + a^2 +b = x$ . This equation holds for all values of $x$ only if $ab=1$ and $a^2+b=0$ .

Then $b = -a^2$ . Substituting into the equation $ab = 1$ , we get $-a^3 = 1$ . Then $a = -1$ , so $b = -1$ , and $f(x)=-x-1.$ Likewise $f^{-1}(x)=-x-1.$ These are inverses to one another since $f(f^{-1}(x))=-(-x-1)-1=x+1-1=x.$ $f^{-1}(f(x))=-(-x-1)-1=x+1-1=x.$ Therefore $a+b=\boxed{-2}$ .

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

@@ Line 1: / Line 1: @@
 == Problem ==
-If <math>f(c)=\frac{3}{2c-3}</math>, find <math>\frac{kn^2}{lm}</math> when <math>f^{-1}(c)\times c \times f(c)</math> equals the simplified fraction<math>\frac{kc+l}{mc+n}</math>, where <math>k,l,m,\text{ and }n</math> are integers.
+If <math>f(x) = ax+b</math> and <math>f^{-1}(x) = bx+a</math> with <math>a</math> and <math>b</math> real, what is the value of <math>a+b</math>?
 == Solution (Alcumus)==

Page

Toolbox

Search

Difference between revisions of "2004 AMC 12B Problems/Problem 13"

Revision as of 11:11, 26 April 2025

Problem

Solution (Alcumus)

See also