1997 AIME Problems/Problem 12
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 .
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
Alternatively, we could have found out that by using the fact that .
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 .
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.
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.
Since is , it must be symmetric across the line . Also, since , it must touch the line at and . a hyperbola that is a scaled and transformed version of . Write as , and z is our desired answer . Take the basic hyperbola, . The distance between points and is , while the distance between and is , so it is scaled by a factor of . Then, we will need to shift it from to , shifting up by , or , so our answer is . Note that shifting the does not require any change from ; it changes the denominator of the part .
Solution 6 (Short)
From , it is obvious that is the value not in the range. First notice that since , which means so . Using , we have that ; on we obtain . Solving for in terms of leads us to , so the answer is .
- solution by mathleticguyyy
Begin by finding the inverse function of , which turns out to be . Since , , so substituting 19 and 97 yields the system, , and after multiplying each equation out and subtracting equation 1 from 2, and after simplifying, you will get . Coincidentally, then , which is familiar because , and since , . Also, , due to . This simplifies to , , , , and substituting and simplifying, you get , then . Looking at one more time, we get , and substituting, we get , and we are done.
Solution 8 (shorter than solution 6)
Because there are no other special numbers other than and , take the average to get . (Note I solved this problem the solution one way but noticed this and this probably generalizes to all questions like these)
|1997 AIME (Problems • Answer Key • Resources)|
|1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15|
|All AIME Problems and Solutions|