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 must be < | + | 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 the original equations, 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 |
== See also == | == See also == |
Revision as of 17:36, 29 January 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
[hide]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 the original equations, 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
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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