Difference between revisions of "2018 AMC 12A Problems/Problem 21"
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from which <math>B(x)</math> has a greater real root than <math>D(x)</math> does.</li><p> | from which <math>B(x)</math> has a greater real root than <math>D(x)</math> does.</li><p> | ||
</ol> | </ol> | ||
− | Now, we are left with comparing the real roots of <math>B(x)</math> and <math>E(x).</math> Since <math>B\left(-\frac{1}{\sqrt2}\right)<0<B\left(-\frac{1}{2}\right),</math> it follows that the real root of <math>B(x)</math> must satisfy <math>x\in\left(-\frac{1}{\sqrt2},-\frac{1}{2}\right)</math> by IVT. Clearly, <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math> has the greatest real root. | + | Now, we are left with comparing the real roots of <math>B(x)</math> and <math>E(x).</math> Since <math>B\left(-\frac{1}{\sqrt2}\right)<0<B\left(-\frac{1}{2}\right),</math> it follows that the real root of <math>B(x)</math> must satisfy <math>x\in\left(-\frac{1}{\sqrt2},-\frac{1}{2}\right)</math> by the IVT. Clearly, <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math> has the greatest real root. |
~MRENTHUSIASM | ~MRENTHUSIASM | ||
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~fidgetboss_4000 | ~fidgetboss_4000 | ||
− | ==Solution 4 (Proof by Contradiction)== | + | ==Solution 4== |
+ | |||
+ | Denote the polynomials in the answer choices by <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively. | ||
+ | |||
+ | The real root of <math>E(x)</math> is <math>x=-\frac{2018}{2019}\approx-1.000.</math>, therefore, we can eliminate this polynomial first. Notice that all of the real roots for all answer choices are in <math>(-1,0)</math>. Therefore, | ||
+ | <cmath>x^{11}<0, \quad x^2<1, \quad x^{13}>x^{11}, \quad x^{19}+2018x^{13}+1>x^{19}+2018x^{11}+1, \quad C(x)>A(x)</cmath> | ||
+ | <cmath>\quad x^{17}+2018x^{13}+1>x^{17}+2018x^{11}+1, \quad D(x)>B(x)</cmath> | ||
+ | <cmath>x^{17}<0, \quad x^{19}>x^{17}, \quad x^{19}+2018x^{11}+1>x^{17}+2018x^{11}+1, \quad A(x)>B(x)</cmath> | ||
+ | |||
+ | Now compare the real roots of <math>A(x)</math> and <math>D(x)</math>. | ||
+ | <cmath>A(x) - D(x) = x^{19}+2018x^{11}+1 - x^{17}-2018x^{13}-1 = x^{17}(x^2-1) - 2018x^{11}(x^2-1) = x^{11}(x^2-1)(x^6-2018)</cmath> | ||
+ | <cmath>\because x^{11}<0, \quad 0<x^2<1, \quad 0<x^6<1, \quad \therefore A(x) - D(x)<0, \quad A(x) < D(x)</cmath> | ||
+ | |||
+ | <cmath>C(x)>D(x)>A(x)>B(x)</cmath> | ||
+ | |||
+ | As <math>B(x)</math> has the smallest value for the same <math>x</math>, and <math>A(x),B(x),C(x),D(x),</math> are all monotonically increasing functions, its real root must be the greatest, thus, the answer is <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 5 (Monotonic Function)== | ||
+ | |||
+ | Denote the polynomials in the answer choices by <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively. | ||
+ | |||
+ | Notice that all of the real roots for all answer choices are in <math>(-1,0)</math>. The real root of <math>E(x)</math> is <math>x=-\frac{2018}{2019}\approx-1.000.</math>, therefore, we can eliminate this polynomial first. | ||
+ | |||
+ | First compare the real roots of <math>A(x)</math> and <math>C(x)</math>: | ||
+ | |||
+ | Let the real root of <math>A(x)</math> be <math>a</math>, and the real root of <math>C(x)</math> be <math>b</math>. | ||
+ | <cmath>a^{19}+2018a^{11}+1 = b^{19}+2018b^{13}+1, \quad a^{19}+2018a^{11} = b^{19}+2018b^{13}</cmath> | ||
+ | <cmath>\because a^2 < 1, \quad a^{11} < 0, \quad \therefore a^{13}>a^{11}</cmath> | ||
+ | <cmath>a^{19}+2018a^{11} < a^{19}+2018a^{13}, \quad b^{19}+2018b^{13} < a^{19}+2018a^{13}</cmath> | ||
+ | Let <math>f(x) = x^{19}+2018x^{13}</math>. Notice that <math>f(x)</math> is a monotonically increasing function. As <math>f(b)<f(a)</math>, <math>b<a</math>, <math>A(x)</math> has a bigger real root than <math>C(x)</math>. | ||
+ | |||
+ | Similarly, <math>B(x)</math> has a bigger real root than <math>D(x)</math>. | ||
+ | |||
+ | To determine the polynomial with the greatest real root, now we only need to compare <math>A(x)</math> and <math>B(x)</math>. | ||
+ | |||
+ | Let the real root of <math>A(x)</math> be <math>a</math>, and the real root of <math>B(x)</math> be <math>b</math>. | ||
+ | <cmath>a^{19}+2018a^{11}+1 = b^{17}+2018b^{11}+1, \quad a^{19}+2018a^{11} = b^{17}+2018b^{11}</cmath> | ||
+ | <cmath>\because a^2 < 1, \quad a^{17} < 0, \quad \therefore a^{19}>a^{17}</cmath> | ||
+ | <cmath>a^{17}+2018a^{11} < a^{19}+2018a^{11}, \quad a^{17}+2018a^{11} < b^{17}+2018b^{11}</cmath> | ||
+ | Let <math>f(x) = x^{17}+2018x^{11}</math>. Notice that <math>f(x)</math> is a monotonically increasing function. As <math>f(a)<f(b)</math>, <math>a<b</math>, <math>B(x)</math> has a bigger real root than <math>A(x)</math>. | ||
+ | |||
+ | Thus, the answer is <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
+ | |||
+ | ==Solution 6 (Proof by Contradiction)== | ||
Denote the polynomials in the answer choices by <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively. | Denote the polynomials in the answer choices by <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively. | ||
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Notice that all of the real roots for all answer choices are in <math>(-1,0)</math>. The real root of <math>E(x)</math> is <math>x=-\frac{2018}{2019}\approx-1.000.</math>, therefore, we can eliminate this polynomial first. | Notice that all of the real roots for all answer choices are in <math>(-1,0)</math>. The real root of <math>E(x)</math> is <math>x=-\frac{2018}{2019}\approx-1.000.</math>, therefore, we can eliminate this polynomial first. | ||
− | First compare the roots of <math>A(x)</math> and <math>C(x)</math>: | + | First compare the real roots of <math>A(x)</math> and <math>C(x)</math>: |
− | Let the root of <math>A(x)</math> be <math>a</math>, and the root of <math>C(x)</math> be <math>b</math>. | + | Let the real root of <math>A(x)</math> be <math>a</math>, and the real root of <math>C(x)</math> be <math>b</math>. |
<cmath>a^{19}+2018a^{11}+1 = b^{19}+2018b^{13}+1</cmath> | <cmath>a^{19}+2018a^{11}+1 = b^{19}+2018b^{13}+1</cmath> | ||
<cmath>a^{11}(a^8+2018) = b^{11}(b^8+2018b^2)</cmath> | <cmath>a^{11}(a^8+2018) = b^{11}(b^8+2018b^2)</cmath> | ||
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<cmath> \text{ Contradicts with } 1<\frac{b^8+2018}{a^8+2018}</cmath> | <cmath> \text{ Contradicts with } 1<\frac{b^8+2018}{a^8+2018}</cmath> | ||
− | <math>\therefore</math> assumption <math>-1<a<b<0</math> is false, <math>-1<b<a<0</math>, <math>A(x)</math> has a bigger root than <math>C(x)</math>. | + | <math>\therefore</math> assumption <math>-1<a<b<0</math> is false, <math>-1<b<a<0</math>, <math>A(x)</math> has a bigger real root than <math>C(x)</math>. |
− | Similarly, <math>B(x)</math> has a bigger root than <math>D(x)</math>. | + | Similarly, <math>B(x)</math> has a bigger real root than <math>D(x)</math>. |
To determine the polynomial with the greatest real root, now we only need to compare <math>A(x)</math> and <math>B(x)</math>. | To determine the polynomial with the greatest real root, now we only need to compare <math>A(x)</math> and <math>B(x)</math>. | ||
− | Let the root of <math>A(x)</math> be <math>a</math>, and the root of <math>B(x)</math> be <math>b</math>. | + | Let the real root of <math>A(x)</math> be <math>a</math>, and the real root of <math>B(x)</math> be <math>b</math>. |
<cmath>a^{19}+2018a^{11}+1 = b^{17}+2018b^{11}+1</cmath> | <cmath>a^{19}+2018a^{11}+1 = b^{17}+2018b^{11}+1</cmath> | ||
<cmath>a^{19} - b^{17} = 2018( b^{11} - a^{11} )</cmath> | <cmath>a^{19} - b^{17} = 2018( b^{11} - a^{11} )</cmath> | ||
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<cmath> \text{ Contradicts with }a^{19} - b^{17} < 0</cmath> | <cmath> \text{ Contradicts with }a^{19} - b^{17} < 0</cmath> | ||
− | <math>\therefore</math> assumption <math>-1<b<a<0</math> is false, <math>-1<a<b<0</math>, <math>B(x)</math> has a bigger root than <math>A(x)</math>. | + | <math>\therefore</math> assumption <math>-1<b<a<0</math> is false, <math>-1<a<b<0</math>, <math>B(x)</math> has a bigger real root than <math>A(x)</math>. |
Thus, the answer is <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math> | Thus, the answer is <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math> | ||
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~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ||
− | ==Solution | + | == Solution 7 (Calculus) == |
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Note that <math>a(-1)=b(-1)=c(-1)=d(-1) < 0</math> and <math>a(0)=b(0)=c(0)=d(0) > 0</math>. Calculating the definite integral for each function in the interval <math>[-1,0]</math>, we see that <math>B(x)\rvert^{0}_{-1}</math> 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 <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math>. | Note that <math>a(-1)=b(-1)=c(-1)=d(-1) < 0</math> and <math>a(0)=b(0)=c(0)=d(0) > 0</math>. Calculating the definite integral for each function in the interval <math>[-1,0]</math>, we see that <math>B(x)\rvert^{0}_{-1}</math> 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 <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math>. | ||
− | == Solution | + | == Solution 8 (Calculus) == |
Newton's Method is used to approximate the zero <math>x_{1}</math> of any real valued function given an estimation for the root <math>x_{0}: \ x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}.</math> After looking at all the options, <math>x_{0}=-1</math> gives a reasonable estimate. For options <math>\textbf{(A)}</math> to <math>\textbf{(D)},</math> we have <math>f(-1) = -2018</math> and the estimation becomes <math>x_{1}=-1+{\frac {2018}{f'(-1)}}.</math> Thus we need to minimize the derivative, giving us <math>\textbf{(B)}</math>. Now after comparing <math>\textbf{(B)}</math> and <math>\textbf{(E)}</math> through Newton's method, we see that <math>\textbf{(B)}</math> has the higher root, so the answer is <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math>. | Newton's Method is used to approximate the zero <math>x_{1}</math> of any real valued function given an estimation for the root <math>x_{0}: \ x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}.</math> After looking at all the options, <math>x_{0}=-1</math> gives a reasonable estimate. For options <math>\textbf{(A)}</math> to <math>\textbf{(D)},</math> we have <math>f(-1) = -2018</math> and the estimation becomes <math>x_{1}=-1+{\frac {2018}{f'(-1)}}.</math> Thus we need to minimize the derivative, giving us <math>\textbf{(B)}</math>. Now after comparing <math>\textbf{(B)}</math> and <math>\textbf{(E)}</math> through Newton's method, we see that <math>\textbf{(B)}</math> has the higher root, so the answer is <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math>. | ||
Latest revision as of 06:58, 29 July 2024
Contents
- 1 Problem
- 2 Solution 1 (Intermediate Value Theorem, Inequalities, Graphs)
- 3 Solution 2 (Similar to Solution 1)
- 4 Solution 3 (Similar to Solution 1)
- 5 Solution 4
- 6 Solution 5 (Monotonic Function)
- 7 Solution 6 (Proof by Contradiction)
- 8 Solution 7 (Calculus)
- 9 Solution 8 (Calculus)
- 10 Video Solution by Richard Rusczyk
- 11 See Also
Problem
Which of the following polynomials has the greatest real root?
Solution 1 (Intermediate Value Theorem, Inequalities, Graphs)
Denote the polynomials in the answer choices by and
respectively.
Note that and
are strictly increasing functions with range
So, each polynomial has exactly one real root. The real root of
is
On the other hand, since
and
we conclude that the real root for each of
and
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
in this interval, 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
it follows that the real root of
must satisfy
by the IVT. 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
Denote the polynomials in the answer choices by and
respectively.
The real root of is
, therefore, we can eliminate this polynomial first. Notice that all of the real roots for all answer choices are in
. Therefore,
Now compare the real roots of and
.
As has the smallest value for the same
, and
are all monotonically increasing functions, its real root must be the greatest, thus, the answer is
Solution 5 (Monotonic Function)
Denote the polynomials in the answer choices by and
respectively.
Notice that all of the real roots for all answer choices are in . The real root of
is
, therefore, we can eliminate this polynomial first.
First compare the real roots of and
:
Let the real root of be
, and the real root of
be
.
Let
. Notice that
is a monotonically increasing function. As
,
,
has a bigger real root than
.
Similarly, has a bigger real root than
.
To determine the polynomial with the greatest real root, now we only need to compare and
.
Let the real root of be
, and the real root of
be
.
Let
. Notice that
is a monotonically increasing function. As
,
,
has a bigger real root than
.
Thus, the answer is
Solution 6 (Proof by Contradiction)
Denote the polynomials in the answer choices by and
respectively.
Notice that all of the real roots for all answer choices are in . The real root of
is
, therefore, we can eliminate this polynomial first.
First compare the real roots of and
:
Let the real root of be
, and the real root of
be
.
As
,
,
Assume that . Thus,
assumption
is false,
,
has a bigger real root than
.
Similarly, has a bigger real root than
.
To determine the polynomial with the greatest real root, now we only need to compare and
.
Let the real root of be
, and the real root of
be
.
Assume that . Thus,
assumption
is false,
,
has a bigger real root than
.
Thus, the answer is
Solution 7 (Calculus)
Note that and
. Calculating the definite integral for each function in 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 8 (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.