1997 AIME Problems/Problem 12

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. Unknown error_msg).

1997 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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1997 AIME Problems/Problem 12

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