Difference between revisions of "2018 AMC 12A Problems/Problem 21"

Latest revision as of 13:07, 11 August 2025

1 Problem
2 Solution 1 (Intermediate Value Theorem, Inequalities, Graphs)
3 Solution 2 (Similar to Solution 1)
4 Solution 3 (Simpler Inequalities)
5 Solution 4 (Similar to Solution 1)
6 Solution 5
7 Solution 6 (Monotonic Function)
8 Solution 7 (Proof by Contradiction)
9 Solution 8 (Calculus)
10 Solution 9 (Calculus)
11 Solution 10 (very fast)
12 Video Solution by Richard Rusczyk
13 Video Solution by grogg007 (5 mins, simple solution)
14 See Also

Problem

Which of the following polynomials has the greatest real root?

$\textbf{(A) } x^{19}+2018x^{11}+1 \qquad \textbf{(B) } x^{17}+2018x^{11}+1 \qquad \textbf{(C) } x^{19}+2018x^{13}+1 \qquad \textbf{(D) } x^{17}+2018x^{13}+1 \qquad \textbf{(E) } 2019x+2018$

Solution 1 (Intermediate Value Theorem, Inequalities, Graphs)

Denote the polynomials in the answer choices by $A(x),B(x),C(x),D(x),$ and $E(x),$ respectively.

Note that $A(x),B(x),C(x),D(x),$ and $E(x)$ are strictly increasing functions with range $(-\infty,\infty).$ So, each polynomial has exactly one real root. The real root of $E(x)$ is $x=-\frac{2018}{2019}\approx-1.000.$ On the other hand, since $A(-1)=B(-1)=C(-1)=D(-1)=-2018$ and $A(0)=B(0)=C(0)=D(0)=1,$ we conclude that the real root for each of $A(x),B(x),C(x),$ and $D(x)$ must satisfy $x\in(-1,0)$ by the Intermediate Value Theorem (IVT).

We analyze the polynomials for $x\in(-1,0):$

We have $\begin{align*} B(x)-A(x)=D(x)-C(x)&=x^{17}-x^{19} \\ &=x^{17}\left(1-x^2\right) \\ &<0. \end{align*}$ As the graph of $y=A(x)$ is always above the graph of $y=B(x)$ in this interval, we deduce that $B(x)$ has a greater real root than $A(x)$ does. By the same reasoning, $D(x)$ has a greater real root than $C(x)$ does.

We have $\begin{align*} B(x)-D(x)&=2018x^{11}-2018x^{13} \\ &=2018x^{11}\left(1-x^2\right) \\ &<0, \end{align*}$ from which $B(x)$ has a greater real root than $D(x)$ does.

Now, we are left with comparing the real roots of $B(x)$ and $E(x).$ Since $B\left(-\frac{1}{\sqrt2}\right)<0<B(0),$ it follows that the real root of $B(x)$ must satisfy $x\in\left(-\frac{1}{\sqrt2},0\right)$ by the IVT. Clearly, $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$ 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 $(-1,0)$ . From there, note that $x^a < x^b$ if $a$ , $b$ are odd positive integers if $a<b$ , so hence it can only either be $\textbf{(B)}$ or $\textbf{(E)}$ (as all of the other polynomials will be larger than the polynomial $\textbf{(B)}$ ). Observe that $\textbf{(E)}$ gives the solution $x=-\frac{2018}{2019}$ . We can approximate the root for $\textbf{(B)}$ by using $x=-\frac 12$ : $\left(-\frac{1}{2}\right)^{17} - \frac{2018}{2048} + 1 \approx 0.$ Therefore, the root for $\textbf{(B)}$ is approximately $-\frac 12$ . The answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$ .

~cpma213

Solution 3 (Simpler Inequalities)

The real roots must lie in the interval $(-1, 0).$ The root of choice E is around $-1$ which is the smallest value on this interval, so we can safely eliminate this.

Let's pick a number between $0$ and $-1,$ say $-1/2.$ We'll plug this in for $x.$ Note that plugging in $-1/2$ here will give us positive values for all polynomials. This means their roots are less than $-1/2$ since the graph increases from $x = -1$ to $0.$ Therefore, the polynomials giving the smallest values when $-1/2$ is plugged in for $x$ have roots closer to $-1/2$ (closer to the positive end of the interval) which means those roots are larger.

Comparing choices A and B:

$(-1/2)^{19} + 2018(-1/2)^{11} + 1 > (-1/2)^{17} + 2018(-1/2)^{11} + 1$

Comparing choices C and D:

$(-1/2)^{19} + 2018(-1/2)^{13} + 1 > (-1/2)^{17} + 2018(-1/2)^{13} + 1$

We're left with choices B and D since they gave the smallest values:

$(-1/2)^{17} + 2018(-1/2)^{13} + 1 > (-1/2)^{17} + 2018(-1/2)^{11} + 1$

Since choice B gave a smaller value than choice D, $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$ has the largest roots.

~grogg007

Desmos (you’ll need to zoom in): https://www.desmos.com/calculator/bgibeqbiku

Solution 4 (Similar to Solution 1)

Let the real solution to $B$ be $a.$ It is easy to see that when $a$ is plugged in to $A,$ since $-1 < a < 0,$ it follows that $a^{19} < a^{17}$ thus making the real solution to $A$ more "negative", or smaller than $B.$ Similarly we can assert that $D > C.$ Now to compare $B$ and $D,$ we can use the same method to what we used before to compare $B$ to $A,$ in which it is easy to see that the smaller exponent $(11)$ "wins". Now, the only thing left is for us to compare $B$ and $E.$ Plugging $\frac{-2018}{2019}$ (or the solution to $E$ ) into $B$ we obtain $\frac{(-2018)^{17}}{2019^{17}} + 2018\cdot\frac{(-2018)^{11}}{2019^{11}} + 1,$ which is intuitively close to $-1 - 2018 + 1 = -2018,$ much smaller than the solution the required $0.$ (For a more rigorous proof, one can note that $\left(\frac{2018}{2019}\right)^{17}$ and $\left(\frac{2018}{2019}\right)^{11}$ are both much greater than $\left(\frac{2018}{2019}\right)^{2019} \approx \frac{1}{e},$ by the limit definition of $e.$ Since $- \frac{1}{e} - 2018 \cdot \frac{1}{e} + 1$ is still much smaller the required $0$ for the solution to $B$ to be a solution, our answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}.$

~fidgetboss_4000

Solution 5

Denote the polynomials in the answer choices by $A(x),B(x),C(x),D(x),$ and $E(x),$ respectively.

The real root of $E(x)$ is $x=-\frac{2018}{2019}\approx-1.000.$ , therefore, we can eliminate this polynomial first. Notice that all of the real roots for all answer choices are in $(-1,0)$ . Therefore, $x^{11}<0, \quad x^2<1, \quad x^{13}>x^{11}, \quad x^{19}+2018x^{13}+1>x^{19}+2018x^{11}+1, \quad C(x)>A(x)$ $\quad x^{17}+2018x^{13}+1>x^{17}+2018x^{11}+1, \quad D(x)>B(x)$ $x^{17}<0, \quad x^{19}>x^{17}, \quad x^{19}+2018x^{11}+1>x^{17}+2018x^{11}+1, \quad A(x)>B(x)$

Now compare the real roots of $A(x)$ and $D(x)$ . $A(x) - D(x) = x^{19}+2018x^{11}+1 - x^{17}-2018x^{13}-1 = x^{17}(x^2-1) - 2018x^{11}(x^2-1) = x^{11}(x^2-1)(x^6-2018)$ $\because x^{11}<0, \quad 0<x^2<1, \quad 0<x^6<1, \quad \therefore A(x) - D(x)<0, \quad A(x) < D(x)$

$C(x)>D(x)>A(x)>B(x)$

As $B(x)$ has the smallest value for the same $x$ , and $A(x),B(x),C(x),D(x),$ are all monotonically increasing functions, its real root must be the greatest, thus, the answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$

~isabelchen

Solution 6 (Monotonic Function)

Denote the polynomials in the answer choices by $A(x),B(x),C(x),D(x),$ and $E(x),$ respectively.

Notice that all of the real roots for all answer choices are in $(-1,0)$ . The real root of $E(x)$ is $x=-\frac{2018}{2019}\approx-1.000.$ , therefore, we can eliminate this polynomial first.

First compare the real roots of $A(x)$ and $C(x)$ :

Let the real root of $A(x)$ be $a$ , and the real root of $C(x)$ be $b$ . $a^{19}+2018a^{11}+1 = b^{19}+2018b^{13}+1, \quad a^{19}+2018a^{11} = b^{19}+2018b^{13}$ $\because a^2 < 1, \quad a^{11} < 0, \quad \therefore a^{13}>a^{11}$ $a^{19}+2018a^{11} < a^{19}+2018a^{13}, \quad b^{19}+2018b^{13} < a^{19}+2018a^{13}$ Let $f(x) = x^{19}+2018x^{13}$ . Notice that $f(x)$ is a monotonically increasing function. As $f(b)<f(a)$ , $b<a$ , $A(x)$ has a bigger real root than $C(x)$ .

Similarly, $B(x)$ has a bigger real root than $D(x)$ .

To determine the polynomial with the greatest real root, now we only need to compare $A(x)$ and $B(x)$ .

Let the real root of $A(x)$ be $a$ , and the real root of $B(x)$ be $b$ . $a^{19}+2018a^{11}+1 = b^{17}+2018b^{11}+1, \quad a^{19}+2018a^{11} = b^{17}+2018b^{11}$ $\because a^2 < 1, \quad a^{17} < 0, \quad \therefore a^{19}>a^{17}$ $a^{17}+2018a^{11} < a^{19}+2018a^{11}, \quad a^{17}+2018a^{11} < b^{17}+2018b^{11}$ Let $f(x) = x^{17}+2018x^{11}$ . Notice that $f(x)$ is a monotonically increasing function. As $f(a)<f(b)$ , $a<b$ , $B(x)$ has a bigger real root than $A(x)$ .

Thus, the answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$

~isabelchen

Solution 7 (Proof by Contradiction)

Denote the polynomials in the answer choices by $A(x),B(x),C(x),D(x),$ and $E(x),$ respectively.

Notice that all of the real roots for all answer choices are in $(-1,0)$ . The real root of $E(x)$ is $x=-\frac{2018}{2019}\approx-1.000.$ , therefore, we can eliminate this polynomial first.

First compare the real roots of $A(x)$ and $C(x)$ :

Let the real root of $A(x)$ be $a$ , and the real root of $C(x)$ be $b$ . $a^{19}+2018a^{11}+1 = b^{19}+2018b^{13}+1$ $a^{11}(a^8+2018) = b^{11}(b^8+2018b^2)$ $\frac{ a^{11} }{ b^{11} } = \frac{ b^8+2018b^2 }{ a^8+2018 }$ As $b^2<1$ , $b^8+2018b^2<b^8+2018$ , $\frac{ a^{11} }{ b^{11} } = \frac{ b^8+2018b^2 }{ a^8+2018 } < \frac{ b^8+2018 }{ a^8+2018 }$

Assume that $-1<a<b<0$ . Thus, $-1<a^{11}<b^{11}<0, \quad 1<\frac{ a^{11} }{ b^{11} }, \quad 1< \frac{ b^8+2018 }{ a^8+2018 }$

$\because -1<a<b<0, \quad \therefore 1>a^8>b^8>0, \quad a^8 + 1 > b^8+1, \quad \therefore \frac{b^8+2018}{a^8+2018}<1$

$\text{ Contradicts with } 1<\frac{b^8+2018}{a^8+2018}$

$\therefore$ assumption $-1<a<b<0$ is false, $-1<b<a<0$ , $A(x)$ has a bigger real root than $C(x)$ .

Similarly, $B(x)$ has a bigger real root than $D(x)$ .

To determine the polynomial with the greatest real root, now we only need to compare $A(x)$ and $B(x)$ .

Let the real root of $A(x)$ be $a$ , and the real root of $B(x)$ be $b$ . $a^{19}+2018a^{11}+1 = b^{17}+2018b^{11}+1$ $a^{19} - b^{17} = 2018( b^{11} - a^{11} )$

Assume that $-1<b<a<0$ . Thus, $-1<b^{11}<a^{11}<0, \quad b^{11}-a^{11}<0, \quad 2018( b^{11} - a^{11} ) < 0, \quad a^{19} - b^{17} < 0$

$-1<b^{17}<a^{17}<0, \quad a^2<1, \quad a^{17}<a^{19}, \quad b^{17}<a^{19}, \quad a^{19} - b^{17} > 0$

$\text{ Contradicts with }a^{19} - b^{17} < 0$

$\therefore$ assumption $-1<b<a<0$ is false, $-1<a<b<0$ , $B(x)$ has a bigger real root than $A(x)$ .

Thus, the answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$

~isabelchen

Solution 8 (Calculus)

Note that $a(-1)=b(-1)=c(-1)=d(-1) < 0$ and $a(0)=b(0)=c(0)=d(0) > 0$ . Calculating the definite integral for each function in the interval $[-1,0]$ , we see that $B(x)\rvert^{0}_{-1}$ 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 $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$ .

Solution 9 (Calculus)

Newton's Method is used to approximate the zero $x_{1}$ of any real valued function given an estimation for the root $x_{0}: \ x_{1}=x_{0}-{\frac {f(x_{0})}{f'(x_{0})}}.$ After looking at all the options, $x_{0}=-1$ gives a reasonable estimate. For options $\textbf{(A)}$ to $\textbf{(D)},$ we have $f(-1) = -2018$ and the estimation becomes $x_{1}=-1+{\frac {2018}{f'(-1)}}.$ Thus we need to minimize the derivative, giving us $\textbf{(B)}$ . Now after comparing $\textbf{(B)}$ and $\textbf{(E)}$ through Newton's method, we see that $\textbf{(B)}$ has the higher root, so the answer is $\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}$ .

~Qcumber

Solution 10 (very fast)

Obviously, eliminate $(E).$ Note that for answer choices $(A)$ and $(B),$ $x \approx \sqrt[11]{\left(-\frac{1}{2018}\right)} > \sqrt[13]{\left(-\frac{1}{2018}\right)},$ with the RHS being approximately the value for $x$ when considering answer choices $(C)$ or $(D),$ so the answer is either $(A)$ or $(B).$ Now, since $x^{19} < x^{17}$ for all negative reals $x$ (we want to choose the option closer to $0$ because then we will have to decrease $x$ less since we estimated), we have $\boxed{(B)}$ as the clear answer.

~ martianrunner

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2018amc12a/471

~ dolphin7

Video Solution by grogg007 (5 mins, simple solution)

https://www.youtube.com/watch?v=kjjCdK8lPDY

@@ Line 35: / Line 35: @@
 ~cpma213
-==Solution 2.5 (inequalities)==
+==Solution 3 (Simpler Inequalities)==
-Our real roots must be in the interval <math>(-1, 0).</math> Immediately, we see the root of choice E is around <math>-1</math> which is the smallest possible value on this interval, so we can safely eliminate this.
+The real roots must lie in the interval <math>(-1, 0).</math> The root of choice E is around <math>-1</math> which is the smallest value on this interval, so we can safely eliminate this.
+Let's pick a number between <math>0</math> and <math>-1,</math> say <math>-1/2.</math> We'll plug this in for <math>x.</math> Note that plugging in <math>-1/2</math> here will give us positive values for all polynomials. This means their roots are less than <math>-1/2</math> since the graph increases from <math>x = -1</math> to <math>0.</math> Therefore, the polynomials giving the smallest values when <math>-1/2</math> is plugged in for <math>x</math> have roots closer to <math>-1/2</math> (closer to the positive end of the interval) which means those roots are larger.
-Let's pick a number between <math>0</math> and <math>-1,</math> say <math>-1/2.</math> We'll plug this in for <math>x.</math> Note that plugging in <math>-1/2</math> here will give us positive values for all polynomials. This means their roots are less than <math>-1/2.</math> Therefore, the polynomials giving the smallest values when <math>-1/2</math> is plugged in for <math>x</math> have roots closer to <math>-1/2,</math> which makes those roots larger.
 Comparing choices A and B:
@@ Line 44: / Line 45: @@
 <cmath>(-1/2)^{19} + 2018(-1/2)^{11} + 1 > (-1/2)^{17} + 2018(-1/2)^{11} + 1</cmath>
 Comparing choices C and D:
 <cmath>(-1/2)^{19} + 2018(-1/2)^{13} + 1 > (-1/2)^{17} + 2018(-1/2)^{13} + 1</cmath>
@@ Line 52: / Line 53: @@
 <cmath>(-1/2)^{17} + 2018(-1/2)^{13} + 1 > (-1/2)^{17} + 2018(-1/2)^{11} + 1</cmath>
 Since choice B gave a smaller value than choice D, <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math> has the largest roots.
@@ Line 57: / Line 59: @@
 ~[[User:grogg007|grogg007]]
-Desmos: https://www.desmos.com/calculator/bgibeqbiku
+Desmos (you’ll need to zoom in): https://www.desmos.com/calculator/bgibeqbiku
-== Solution 3 (Similar to Solution 1) ==
+== Solution 4 (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> 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
-==Solution 4==
+==Solution 5==
 Denote the polynomials in the answer choices by <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively.
@@ Line 83: / Line 85: @@
 ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
-==Solution 5 (Monotonic Function)==
+==Solution 6 (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.
@@ Line 111: / Line 113: @@
 ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
-==Solution 6 (Proof by Contradiction)==
+==Solution 7 (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.
@@ Line 156: / Line 158: @@
 ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
-== Solution 7 (Calculus) ==
+== Solution 8 (Calculus) ==
 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 8 (Calculus) ==
+== Solution 9 (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>.
 ~Qcumber
+== Solution 10 (very fast) ==
+Obviously, eliminate <math>(E).</math> Note that for answer choices <math>(A)</math> and <math>(B),</math> <math>x \approx \sqrt[11]{\left(-\frac{1}{2018}\right)} > \sqrt[13]{\left(-\frac{1}{2018}\right)},</math> with the RHS being approximately the value for <math>x</math> when considering answer choices <math>(C)</math> or <math>(D),</math> so the answer is either <math>(A)</math> or <math>(B).</math> Now, since <math>x^{19} < x^{17}</math> for all negative reals <math>x</math> (we want to choose the option closer to <math>0</math> because then we will have to decrease <math>x</math> less since we estimated), we have <math>\boxed{(B)}</math> as the clear answer.
+~ martianrunner
 == Video Solution by Richard Rusczyk ==
@@ Line 168: / Line 175: @@
 ~ dolphin7
+==Video Solution by [[User:grogg007|grogg007]] (5 mins, simple solution)==
+https://www.youtube.com/watch?v=kjjCdK8lPDY
 ==See Also==
 {{AMC12 box|year=2018|ab=A|num-b=20|num-a=22}}
 {{MAA Notice}}

2018 AMC 12A (Problems • Answer Key • Resources)
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Difference between revisions of "2018 AMC 12A Problems/Problem 21"

Latest revision as of 13:07, 11 August 2025

Contents

Problem

Solution 1 (Intermediate Value Theorem, Inequalities, Graphs)

Solution 2 (Similar to Solution 1)

Solution 3 (Simpler Inequalities)

Solution 4 (Similar to Solution 1)

Solution 5

Solution 6 (Monotonic Function)

Solution 7 (Proof by Contradiction)

Solution 8 (Calculus)

Solution 9 (Calculus)

Solution 10 (very fast)

Video Solution by Richard Rusczyk

Video Solution by grogg007 (5 mins, simple solution)

See Also