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

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\qquad\mathrm{(E)}\ 2</math>
 
\qquad\mathrm{(E)}\ 2</math>
 
== Solution ==
 
== Solution ==
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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,
+
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>.
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<cmath>a^3 + 1 = 0</cmath>
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and <math>a = b = -1</math>. Thus <math>a+b = -2 \Rightarrow \mathrm{(A)}</math>.
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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>.
  
 
== See also ==
 
== See also ==

Revision as of 22:34, 28 July 2020

Problem

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

$\mathrm{(A)}\ -2 \qquad\mathrm{(B)}\ -1 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 1 \qquad\mathrm{(E)}\ 2$

Solution

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

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