Difference between revisions of "1997 AIME Problems/Problem 12"
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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{ax^2 +ab+bcx+bd}{acx+bc+cdx+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{ax^2 +ab+bcx+bd}{acx+bc+cdx+d^2} = \frac{ax^2 +bcx+b(a+d)}{cx(a+d)+b(a+d)}</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 for this to be a true statement so we have that <math>a = -d</math>. |
Revision as of 18:10, 24 December 2010
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 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 |