Difference between revisions of "1997 AIME Problems/Problem 12"

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== Problem ==
 
== Problem ==
The function <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.
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The [[function]] <math>f</math> defined by <math>f(x)= \frac{ax+b}{cx+d}</math>, where <math>a</math>,<math>b</math>,<math>c</math> and <math>d</math> are nonzero real numbers, has the properties <math>f(19)=19</math>, <math>f(97)=97</math> and <math>f(f(x))=x</math> for all values except <math>\frac{-d}{c}</math>. Find the unique number that is not in the range of <math>f</math>.
  
 
== Solution ==
 
== Solution ==
{{solution}}
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First, we note that <math>e = \frac ac</math> is the horizontal [[asymptote]] of the function, and since this is a linear function over a linear function, the unique number not in the range of <math>f</math> will be <math>e</math>. <math>\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}</math>. [[Without loss of generality]], let <math>c=1</math>, so the function becomes <math>\frac{b- \frac{d}{a}}{x+d} + e</math>. 
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(Considering <math>\infty</math> as a limit) By the given, <math>f(f(\infty)) = \infty</math>. <math>\lim_{x \rightarrow \infty} f(x) = e</math>, so <math>f(e) = \infty</math>. <math>f(x) \rightarrow \infty</math> as <math>x</math> reaches the vertical [[asymptote]], which is at <math>-\frac{d}{c} = -d</math>. Hence <math>e = -d</math>. Substituting the givens, we get
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<cmath>\begin{eqnarray*}
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17 &=& \frac{b - \frac da}{17 - e} + e\\
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97 &=& \frac{b - \frac da}{97 - e} + e\\
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b - \frac da &=& (17 - e)^2 = (97 - e)^2\\
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17 - e &=& \pm (97 - e)
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\end{eqnarray*}</cmath>
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Clearly we can discard the positive root, so <math>e = \box{58}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1997|num-b=11|num-a=13}}
 
{{AIME box|year=1997|num-b=11|num-a=13}}
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[[Category:Intermediate Algebra Problems]]

Revision as of 21:30, 23 November 2007

Problem

The function $f$ defined by $f(x)= \frac{ax+b}{cx+d}$, where $a$,$b$,$c$ and $d$ are nonzero real numbers, has the properties $f(19)=19$, $f(97)=97$ and $f(f(x))=x$ for all values except $\frac{-d}{c}$. Find the unique number that is not in the range of $f$.

Solution

First, we note that $e = \frac ac$ is the horizontal asymptote of the function, and since this is a linear function over a linear function, the unique number not in the range of $f$ will be $e$. $\frac{ax+b}{cx+d} = \frac{b-\frac{cd}{a}}{cx+d} + \frac{a}{c}$. Without loss of generality, let $c=1$, so the function becomes $\frac{b- \frac{d}{a}}{x+d} + e$.

(Considering $\infty$ as a limit) By the given, $f(f(\infty)) = \infty$. $\lim_{x \rightarrow \infty} f(x) = e$, so $f(e) = \infty$. $f(x) \rightarrow \infty$ as $x$ reaches the vertical asymptote, which is at $-\frac{d}{c} = -d$. Hence $e = -d$. Substituting the givens, we get

\begin{eqnarray*} 17 &=& \frac{b - \frac da}{17 - e} + e\\ 97 &=& \frac{b - \frac da}{97 - e} + e\\ b - \frac da &=& (17 - e)^2 = (97 - e)^2\\ 17 - e &=& \pm (97 - e) \end{eqnarray*}

Clearly we can discard the positive root, so $e = \box{58}$ (Error compiling LaTeX. ! Missing number, treated as zero.).

See also

1997 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 13
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All AIME Problems and Solutions
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