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

Revision as of 08:54, 18 December 2023

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 $B\left(-\frac{1}{\sqrt2}\right)<0<B\left(-\frac{1}{2}\right),$ it follows that the real root of $B(x)$ must satisfy $x\in\left(-\frac{1}{\sqrt2},-\frac{1}{2}\right)$ by 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 (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 roots of $A(x)$ and $C(x)$ :

Let the root of $A(x)$ be $a$ , and the 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 root than $C(x)$ .

Similarly, $B(x)$ has a bigger 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 root of $A(x)$ be $a$ , and the 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 root than $A(x)$ .

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

~isabelchen

Solution 5 (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 6 (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

@@ Line 44: / Line 44: @@
 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 will be 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>:

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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