Difference between revisions of "2018 AMC 12A Problems/Problem 21"
MRENTHUSIASM (talk | contribs) (Rearranged the solutions based on similarity. Also, I prioritized the comprehensive Sol 1. Let me know if you disagree.) |
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== Solution 3 (Similar to Solution 1) == | == Solution 3 (Similar to Solution 1) == | ||
− | Let the real solution to <math>B</math> be <math>a.</math> It is easy to see that when <math>a</math> is plugged in to <math>A,</math> since <math>-1 < a < 0,</math> <math>a^{19} < a^{17}</math> thus making the real solution to <math>A</math> more "negative", or smaller than <math>B.</math> Similarly we can assert that <math>D > C.</math> Now to compare <math>B</math> and <math>D,</math> we can use the same method to what we used before to compare <math>B</math> to <math>A,</math> in which it is easy to see that the smaller exponent <math>(11)</math> "wins". Now, the only thing left is for us to compare <math>B</math> and <math>E.</math> Plugging <math>\frac{-2018}{2019}</math> (or the solution to <math>E</math>) into <math>B</math> we obtain <math>\frac{(-2018)^{17}}{2019^{17}} + 2018\cdot\frac{(-2018)^{11}}{2019^{11}} + 1,</math> which is intuitively close to <math>-1 - 2018 + 1 = -2018, </math> much smaller than the solution the required <math>0.</math> (For a more rigorous proof, one can note that <math>\left(\frac{2018}{2019}\right)^{17}</math> and <math>\left(\frac{2018}{2019}\right)^{11}</math> are both much greater than <math>\left(\frac{2018}{2019}\right)^{2019} \approx \frac{1}{e},</math> by the limit definition of <math>e.</math> Since <math>- \frac{1}{e} - 2018 \cdot \frac{1}{e} + 1</math> is still much smaller the required <math>0</math> for the solution to <math>B</math> to be a solution, our answer is <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}.</math> | + | Let the real solution to <math>B</math> be <math>a.</math> It is easy to see that when <math>a</math> is plugged in to <math>A,</math> since <math>-1 < a < 0,</math> it follows that <math>a^{19} < a^{17}</math> thus making the real solution to <math>A</math> more "negative", or smaller than <math>B.</math> Similarly we can assert that <math>D > C.</math> Now to compare <math>B</math> and <math>D,</math> we can use the same method to what we used before to compare <math>B</math> to <math>A,</math> in which it is easy to see that the smaller exponent <math>(11)</math> "wins". Now, the only thing left is for us to compare <math>B</math> and <math>E.</math> Plugging <math>\frac{-2018}{2019}</math> (or the solution to <math>E</math>) into <math>B</math> we obtain <math>\frac{(-2018)^{17}}{2019^{17}} + 2018\cdot\frac{(-2018)^{11}}{2019^{11}} + 1,</math> which is intuitively close to <math>-1 - 2018 + 1 = -2018, </math> much smaller than the solution the required <math>0.</math> (For a more rigorous proof, one can note that <math>\left(\frac{2018}{2019}\right)^{17}</math> and <math>\left(\frac{2018}{2019}\right)^{11}</math> are both much greater than <math>\left(\frac{2018}{2019}\right)^{2019} \approx \frac{1}{e},</math> by the limit definition of <math>e.</math> Since <math>- \frac{1}{e} - 2018 \cdot \frac{1}{e} + 1</math> is still much smaller the required <math>0</math> for the solution to <math>B</math> to be a solution, our answer is <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}.</math> |
~fidgetboss_4000 | ~fidgetboss_4000 |
Revision as of 00:46, 20 August 2021
Contents
Problem
Which of the following polynomials has the greatest real root?
Solution 1 (Intermediate Value Theorem, Inequalities, Graphs)
Label the polynomials in the answer choices as and
respectively.
Note that and
are increasing functions with range
So, each polynomial has exactly one real root. Since
and
for all
we conclude that the real root of each polynomial must satisfy
by the Intermediate Value Theorem (IVT).
We analyze the polynomials for
- We have
As the graph of
is always above the graph of
we deduce that
has a greater real root than
does. By the same reasoning,
has a greater real root than
does.
- We have
from which
has a greater real root than
does.
Now, we are left with comparing the real roots of and
Since we conclude that the real root of
must satisfy
by IVT. On the other hand, the real root of
is
Clearly,
has the greatest real root.
~MRENTHUSIASM
Solution 2 (Similar to Solution 1)
We can see that our real solution has to lie in the open interval . From there, note that
if
,
are odd positive integers if
, so hence it can only either be
or
(as all of the other polynomials will be larger than the polynomial
). Observe that
gives the solution
. We can approximate the root for
by using
:
Therefore, the root for
is approximately
. The answer is
.
~cpma213
Solution 3 (Similar to Solution 1)
Let the real solution to be
It is easy to see that when
is plugged in to
since
it follows that
thus making the real solution to
more "negative", or smaller than
Similarly we can assert that
Now to compare
and
we can use the same method to what we used before to compare
to
in which it is easy to see that the smaller exponent
"wins". Now, the only thing left is for us to compare
and
Plugging
(or the solution to
) into
we obtain
which is intuitively close to
much smaller than the solution the required
(For a more rigorous proof, one can note that
and
are both much greater than
by the limit definition of
Since
is still much smaller the required
for the solution to
to be a solution, our answer is
~fidgetboss_4000
Solution 4 (Calculus)
Note that and
. Calculating the definite integral for each function on the interval
, we see that
gives the most negative value. To maximize our real root, we want to maximize the area of the curve under the x-axis, which means we want our integral to be as negative as possible and thus the answer is
.
Solution 5 (Calculus)
Newton's Method is used to approximate the zero of any real valued function given an estimation for the root
After looking at all the options,
gives a reasonable estimate. For options
to
we have
and the estimation becomes
Thus we need to minimize the derivative, giving us
. Now after comparing
and
through Newton's method, we see that
has the higher root, so the answer is
.
~Qcumber
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2018amc12a/471
~ dolphin7
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
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.