2004 AMC 12B Problems/Problem 13

Problem

If $f(c)=\frac{3}{2c-3}$ , find $\frac{kn^2}{lm}$ when $f^{-1}(c)\times c \times f(c)$ equals the simplified fraction $\frac{kc+l}{mc+n}$ , where $k,l,m,\text{ and }n$ are integers.

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

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2004 AMC 12B Problems/Problem 13

Problem

Solution (Alcumus)

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