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

(Reformatted page. About to add a comprehensive solution as Sol 1.)
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<math>\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 </math>
 
<math>\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 </math>
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== Solution 1 ==
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Label the polynomials in the answer choices as <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively.
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Note that <math>A(x),B(x),C(x),D(x),</math> and <math>E(x)</math> are increasing functions with range <math>(-\infty,\infty).</math> So, each polynomial has exactly one real root. Since <math>A(-1)=B(-1)=C(-1)=D(-1)=E(-1)=-1</math> and <math>A(x),B(x),C(x),D(x),E(x)>0</math> for all <math>x\in[0,\infty),</math> we conclude that the real root of each polynomial satisfies <math>x\in(-1,0)</math> by the Intermediate Value Theorem (IVT).
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We analyze the polynomials in the interval <math>x\in(-1,0):</math>
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<ol style="margin-left: 1.5em;">
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  <li>We have
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<cmath>\begin{align*}
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B(x)-A(x)=D(x)-C(x)&=x^{17}-x^{19} \\
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&=x^{17}\left(1-x^2\right) \\
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&<0.
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\end{align*}</cmath>
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As the graph of <math>A(x)</math> is always above the graph of <math>B(x),</math> we deduce that <math>B(x)</math> has a greater real root than <math>A(x)</math> does. By the same reasoning, <math>D(x)</math> has a greater real root than <math>C(x)</math> does.</li><p>
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  <li>We have
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<cmath>\begin{align*}
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B(x)-D(x)&=2018x^{11}-2018x^{13} \\
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&=2018x^{11}\left(1-x^2\right) \\
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&<0,
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\end{align*}</cmath>
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from which <math>B(x)</math> has a greater real root than <math>D(x)</math> does.</li><p>
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</ol>
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Now, we are left with comparing the real roots of <math>B(x)</math> and <math>E(x).</math>
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<b>AFK (away from keyboard). Will finish by today. No edits please.</b>
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 18:17, 19 August 2021

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

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

Note that $A(x),B(x),C(x),D(x),$ and $E(x)$ are increasing functions with range $(-\infty,\infty).$ So, each polynomial has exactly one real root. Since $A(-1)=B(-1)=C(-1)=D(-1)=E(-1)=-1$ and $A(x),B(x),C(x),D(x),E(x)>0$ for all $x\in[0,\infty),$ we conclude that the real root of each polynomial satisfies $x\in(-1,0)$ by the Intermediate Value Theorem (IVT).

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

  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 $A(x)$ is always above the graph of $B(x),$ 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 $E(x).$

AFK (away from keyboard). Will finish by today. No edits please.

Solution 2

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 (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 on 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 4 (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 5

Let the real solution to $B$ be $a.$ It is easy to see that when $a$ is plugged in to $A,$ since $-1 < a < 0,$ $a^{19} < a^{17}$ 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

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

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