Difference between revisions of "2020 AMC 12B Problems/Problem 22"
(→Solution1) |
(→Solution2) |
||
Line 31: | Line 31: | ||
Therefore, we can assume that <math>\frac{log_2{x}}{x}</math> equals to <math>\frac{1}{6}</math> when <math>x</math> is somewhere between 1 and 2 (at least), which means that the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> is <math>\textbf{(C)}\ \frac{1}{12}</math>. | Therefore, we can assume that <math>\frac{log_2{x}}{x}</math> equals to <math>\frac{1}{6}</math> when <math>x</math> is somewhere between 1 and 2 (at least), which means that the maximum value of <math>\frac{(2^t-3t)t}{4^t}</math> is <math>\textbf{(C)}\ \frac{1}{12}</math>. | ||
+ | |||
+ | ==Solution 3== | ||
+ | Take the derivative of this function and let the derivative equals to 0, then this gives you <math>2^t=6t</math>. Substitute it into the original function you can get <math>\boxed{C}</math>. |
Revision as of 01:29, 8 February 2020
Contents
[hide]Problem 22
What is the maximum value of for real values of
Solution1
Set . Then the expression in the problem can be written as
It is easy to see that
is attained for some value of
between
and
, thus the maximal value of
is
.
Solution2
First, substitute so that
Notice that
When seen as a function, is a synthesis function that has
as its inner function.
If we substitute , the given function becomes a quadratic function that has a maximum value of
when
.
Now we need to check that can have the value of
in the range of real numbers.
In the range of (positive) real numbers, function is a continuous function whose value gets infinitely smaller as
gets closer to 0 (as
also diverges toward negative infinity in the same condition). When
,
, which is larger than
.
Therefore, we can assume that equals to
when
is somewhere between 1 and 2 (at least), which means that the maximum value of
is
.
Solution 3
Take the derivative of this function and let the derivative equals to 0, then this gives you . Substitute it into the original function you can get
.