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

(→‎Solution 4)
(→‎Solution 4)
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Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath>
 
Since we are given <math>f(f(x))=x</math>, we have that <cmath>x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}</cmath>
 
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath>
 
<cmath>cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)</cmath>
−
All the quadratic terms, linear terms, and constant terms must be equal for this to be a true statement so we have that <math>a = -d</math>.
+
All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that <math>a = -d</math>.
  
 
This solution follows in the same manner as the last paragraph of the first solution.
 
This solution follows in the same manner as the last paragraph of the first solution.

Revision as of 17:14, 24 December 2010

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 $\lim_{x\to \infty}f(x) = \frac {a}{c}$. 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 

\[{\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 .\] (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{align*} 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{align*}

Clearly we can discard the positive root, so $e = 58$.

Solution 3

We first note (as before) that the number not in the range of \[f(x) = \frac{ax+b}{cx+ d} = \frac{a}{c} + \frac{b - ad/c}{cx+d}\] is $a/c$, as $\frac{b-ad/c}{cx+d}$ is evidently never 0 (otherwise, $f$ would be a constant function, violating the condition $f(19) \neq f(97)$).

We may represent the real number $x/y$ as $\begin{pmatrix}x \\ y\end{pmatrix}$, with two such column vectors considered equivalent if they are scalar multiples of each other. Similarly, we can represent a function $F(x) = \frac{Ax + B}{Cx + D}$ as a matrix $\begin{pmatrix} A & B\\ C& D \end{pmatrix}$. Function composition and evaluation then become matrix multiplication.

Now in general, \[f^{-1} = \begin{pmatrix} a & b\\ c&d \end{pmatrix}^{-1} = \frac{1}{\det(f)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} .\] In our problem $f^2(x) = x$. It follows that \[\begin{pmatrix} a & b \\ c& d \end{pmatrix} = K \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} ,\] for some nonzero real $K$. Since \[\frac{a}{d} = \frac{b}{-b} = K,\] it follows that $a = -d$. (In fact, this condition condition is equivalent to the condition that $f(f(x)) = x$ for all $x$ in the domain of $f$.)

We next note that the function \[g(x) = x - f(x) = \frac{c x^2 + (d-a) x - b}{cx + d}\] evaluates to 0 when $x$ equals 19 and 97. Therefore \[\frac{c x^2 + (d-a) x - b}{cx+d} = g(x) = \frac{c(x-19)(x-97)}{cx+d}.\] Thus $-19 - 97 = \frac{d-a}{c} = -\frac{2a}{c}$, so $a/c = (19+97)/2 = 58$, our answer. $\blacksquare$

Solution 4

Any number that is not in the domain of the inverse of $f(x)$ cannot be in the range of $f(x)$. Starting with $f(x) = \frac{ax+b}{cx+d}$, we rearrange some things to get $x = \frac{b-f(x)d}{f(x)c-a}$. Clearly, $\frac{a}{c}$ is the number that is outside the range of $f(x)$.


Since we are given $f(f(x))=x$, we have that \[x = \frac{a\frac{ax+b}{cx+d}+b}{c\frac{ax+b}{cx+d}+d} = \frac{a^2x +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{x(bc+a^2)+b(a+d)}{cx(a+d)+(bc+d^2)}\] \[cx^2(a+d)+x(bc+d^2) = x(bc+a^2) + b(a+d)\] All the quadratic terms, linear terms, and constant terms must be equal on both sides for this to be a true statement so we have that $a = -d$.

This solution follows in the same manner as the last paragraph of the first solution.

See also

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
All AIME Problems and Solutions
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