Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2018 AMC 12A Problems/Problem 21

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 $x\in(-1,0):$

  1. 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.

  2. 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 $B\left(-\frac{1}{\sqrt2}\right)<0<B\left(-\frac{1}{2}\right),$ it follows that the real root of $B(x)$ must satisfy $x\in\left(-\frac{1}{\sqrt2},-\frac{1}{2}\right)$ 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 (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 4

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 5 (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 6 (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 7 (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 8 (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

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_21&oldid=223880"