Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎See also)
(→‎Problem)
 
(2 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
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>?
+
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.
  
−
<math>\mathrm{(A)}\ -2
+
== Solution (Alcumus)==
−
\qquad\mathrm{(B)}\ -1
+
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>.
−
\qquad\mathrm{(C)}\ 0
+
 
−
\qquad\mathrm{(D)}\ 1
+
Then <math>b = -a^2</math>. Substituting into the equation <math>ab = 1</math>, we get <math>-a^3 = 1</math>. Then <math>a = -1</math>, so <math>b = -1</math>, and<cmath>f(x)=-x-1.</cmath>Likewise<cmath>f^{-1}(x)=-x-1.</cmath>These are inverses to one another since<cmath>f(f^{-1}(x))=-(-x-1)-1=x+1-1=x.</cmath><cmath>f^{-1}(f(x))=-(-x-1)-1=x+1-1=x.</cmath>Therefore <math>a+b=\boxed{-2}</math>.
−
\qquad\mathrm{(E)}\ 2</math>
 
−
== Solution ==
 
−
By the definition of an [[inverse function]], <math>x = f(f^{-1}(x)) = a(bx+a)+b = abx + a^2 + b</math>. By comparing coefficients, we have <math>ab = 1 \Longrightarrow b = \frac 1a</math> and <math>a^2 + b = a^2 + \frac{1}{a} = 0</math>. Simplifying,
 
−
<cmath>a^3 + 1 = 0</cmath>
 
−
and <math>a = b = -1</math>. Thus <math>a+b = -2 \Rightarrow \mathrm{(A)}</math>.
 
  
 
== See also ==
 
== See also ==

Latest revision as of 19:09, 13 October 2024

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 $a+b=\boxed{-2}$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_13&oldid=229761"