Difference between revisions of "1997 AIME Problems/Problem 12"
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First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>{\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x}</math>, which reduces to | First, we use the fact that <math>f(f(x)) = x</math> for all <math>x</math> in the domain. Substituting the function definition, we have <math>{\frac {a\frac {ax + b}{cx + d} + b}{c\frac {ax + b}{cx + d} + d} = x}</math>, which reduces to | ||
<cmath>{\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} = | <cmath>{\frac {(a^2 + bc)x + b(a + d)}{c(a + d)x + (bc + d^2)} = | ||
− | \frac { | + | \frac {px + q}{rx + s} = x}. </cmath> |
− | In order for this fraction to reduce to <math>x</math>, we must have <math> | + | In order for this fraction to reduce to <math>x</math>, we must have <math>q = r = 0</math> and <math>p = s\not = 0</math>. From <math>c(a + d) = b(a + d) = 0</math>, we get <math>a = - d</math> or <math>b = c = 0</math>. The second cannot be true, since we are given that <math>a,b,c,d</math> are nonzero. This means <math>a = - d</math>, so <math>f(x) = \frac {ax + b}{cx - a}</math>. |
The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so | The only value that is not in the range of this function is <math>\frac {a}{c}</math>. To find <math>\frac {a}{c}</math>, we use the two values of the function given to us. We get <math>2(97)a + b = 97^2c</math> and <math>2(19)a + b = 19^2c</math>. Subtracting the second equation from the first will eliminate <math>b</math>, and this results in <math>2(97 - 19)a = (97^2 - 19^2)c</math>, so |
Revision as of 05:44, 29 June 2011
Problem
The function defined by , where ,, and are nonzero real numbers, has the properties , and for all values except . Find the unique number that is not in the range of .
Contents
Solution
Solution 1
First, we use the fact that for all in the domain. Substituting the function definition, we have , which reduces to In order for this fraction to reduce to , we must have and . From , we get or . The second cannot be true, since we are given that are nonzero. This means , so .
The only value that is not in the range of this function is . To find , we use the two values of the function given to us. We get and . Subtracting the second equation from the first will eliminate , and this results in , so
\[{\frac {a}{c} = \frac {(97 - 19)(97 + 19)}{2(97 - 19)} = 58 .\] (Error compiling LaTeX. ! Missing } inserted.)
Alternatively, we could have found out that by using the fact that .
Solution 2
First, we note that 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 will be . . Without loss of generality, let , so the function becomes .
(Considering as a limit) By the given, . , so . as reaches the vertical asymptote, which is at . Hence . Substituting the givens, we get
Clearly we can discard the positive root, so .
Solution 3
We first note (as before) that the number not in the range of is , as is evidently never 0 (otherwise, would be a constant function, violating the condition ).
We may represent the real number as , with two such column vectors considered equivalent if they are scalar multiples of each other. Similarly, we can represent a function as a matrix . Function composition and evaluation then become matrix multiplication.
Now in general, In our problem . It follows that for some nonzero real . Since it follows that . (In fact, this condition condition is equivalent to the condition that for all in the domain of .)
We next note that the function evaluates to 0 when equals 19 and 97. Therefore Thus , so , our answer.
Solution 4
Any number that is not in the domain of the inverse of cannot be in the range of . Starting with , we rearrange some things to get . Clearly, is the number that is outside the range of .
Since we are given , we have that
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 .
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 |