Difference between revisions of "1997 AIME Problems/Problem 12"
m (→Solution 6) |
(→Solution 6) |
||
Line 74: | Line 74: | ||
=== 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> | + | 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>. 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 obtain <math>b=9409c+194d</math>. Solving for <math>d</math> in terms of <math>c</math> leads us to <math>d=-58c</math>. Rewrite the function as <math>f(x)=\frac{58cx+b}{cx-58c}</math>; from there, we see that either 58 is the only unattainable or 58 is the only attainable value. The second case clearly cannot happen, which leads us to the answer being <math>\boxed{58}</math>. - mathleticguyyy |
=== Solution 7=== | === Solution 7=== |
Revision as of 10:38, 18 October 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
. Using
, we have that
, so
. Using similar process on
we obtain
. Solving for
in terms of
leads us to
. Rewrite the function as
; from there, we see that either 58 is the only unattainable or 58 is the only attainable value. The second case clearly cannot happen, which leads us to the answer being
. - mathleticguyyy
Solution 7
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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.