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 .
First, notice that , and , so . Now for to be , must be . After some algebra, we find that . Our function could now be simplified into . Using , we have that , so . Using similar process on we have that . Solving for in terms of leads us to . now our function becomes . From there, we plug back into one of or , and we immediately realize that must be equal to the product of and some odd integer, which makes it impossible to achieve a value of since for to be 58, and , which is impossible when is odd. The answer is - 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 , ,
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