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

Revision as of 20:00, 11 February 2009

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

Solution 1

First, we use the fact that $f(f(x)) = x$ for all $x$ in the domain. Substituting the function definition, we have ${\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x}$ , which reduces to ${\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} = \frac {ex + f}{gx + h} = x}$ . In order for this fraction to reduce to $x$ , we must have $f = g = 0$ and $e = h\not = 0$ . From $c(a + d) = b(a + d) = 0$ , we get $a = - d$ or $b = c = 0$ . The second cannot be true, since we are given that $a,b,c,d$ are nonzero. This means $a = - d$ , so $f(x) = \frac {ax + b}{cx - a}$ .

The only value that is not in the range of this function is $\lim_{x\to - d/c}f(x) = \frac {a}{c}$ . To find $\frac {a}{c}$ , we use the two values of the function given to us. We get $2(97)a + b = 97^2c$ and $2(19)a + b = 19^2c$ . Subtracting the second equation from the first will eliminate $b$ , and this results in $2(97 - 19)a = (97^2 - 19^2)c$ , so ${\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = \boxed{058}$ (Error compiling LaTeX. Unknown error_msg).

Alternatively, we could have found out that $a = -d$ by using the fact that $f(f(-b/a))=-b/a$ .

Solution 2

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 $e = \boxed{58}$ .

@@ Line 12: / Line 12: @@
 === Solution 2 ===
-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>.
+First, we note that <math>e = \frac ac</math> is the horizontal [[Asymptote (Geometry)|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>.
 (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

1997 AIME (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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