Difference between revisions of "1997 AIME Problems/Problem 12"
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=== Solution 6=== | === Solution 6=== | ||
First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy | First, notice that <math>f(0)=\frac{b}{d}</math>, and <math>f(f(0))=0</math>, so <math>f(\frac{b}{d})=0</math>. Now for <math>f(\frac{b}{d})</math> to be <math>0</math>, <math>a(\frac{b}{d})+b</math> must be <math>0</math>. After some algebra, we find that <math>a=-d</math>. Our function could now be simplified into <math>f(x)=\frac{-dx+b}{cx+d}</math>. Using <math>f(19)=19</math>, we have that <math>b-19d=361c+19d</math>, so <math>b=361c+38d</math>. Using similar process on <math>f(97)=97</math> we have that <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. now our function becomes <math>f(x)=\frac{58cx+b}{cx-58c}</math>. From there, we plug <math>d=-58c</math> back into one of <math>b=361c+38d</math> or <math>b=9409c+194d</math>, and we immediately realize that <math>b</math> must be equal to the product of <math>c</math> and some odd integer, which makes it impossible to achieve a value of <math>58</math> since for <math>f(x)</math> to be 58, <math>58cx+b=58cx-(58^2)c</math> and <math>\frac{b}{c}+58^2=0</math>, which is impossible when <math>\frac{b}{c}</math> is odd. The answer is <math>\boxed{058}</math> - mathleticguyyy | ||
+ | |||
+ | === Solution 7=== | ||
+ | Begin by finding the inverse function of <math>f(X)</math> | ||
== See also == | == See also == |
Revision as of 13:55, 23 March 2019
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
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.
Solution 5
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
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
Solution 7
Begin by finding the inverse function of
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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