Difference between revisions of "2014 AMC 12A Problems/Problem 18"
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<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\lim_{x\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}&=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^0}}}\\ | \lim_{x\to\infty}\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^{\left(\frac12\right)^x}}}}&=\left(\frac1{16}\right)^{16^{\left(\frac14\right)^{4^0}}}\\ | ||
− | &= | + | &= \left(\frac1{16}\right)^{16^{\left(\frac14\right)}}\\ |
&= \left(\frac1{16}\right)^2\\ | &= \left(\frac1{16}\right)^2\\ | ||
&= \frac{1}{256}. | &= \frac{1}{256}. |
Revision as of 22:54, 7 February 2014
Problem
The domain of the function is an interval of length
, where
and
are relatively prime positive integers. What is
?
Solution
Solution 1
For simplicity, let , and
.
The domain of is
, so
.
Thus,
.
Since
we have
.
Since
, we have
.
Finally, since
,
.
The length of the interval is
and the answer is
.
Solution 2
The domain of is the range of the inverse function
. Now
can be seen to be strictly decreasing, since
is decreasing, so
is decreasing, so
is increasing, so
is increasing, therefore
is decreasing.
Therefore, the range of is an open interval and can be found by taking the limits as
approaches
and
. In fact, the range is
. We find:
Similarly,
Hence the domain of
is
and the answer is
.
See Also
2014 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.