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Difference between revisions of "2018 AMC 12A Problems/Problem 21"
(Rearranged the solutions based on similarity. Also, I prioritized the comprehensive Sol 1. Let me know if you disagree.)
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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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== Solution 1 (Intermediate Value Theorem, Inequalities, Graphs) ==
 
Label the polynomials in the answer choices as <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively.
 
Label the polynomials in the answer choices as <math>A(x),B(x),C(x),D(x),</math> and <math>E(x),</math> respectively.
  
−
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).
+
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 must satisfy <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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We analyze the polynomials for <math>x\in(-1,0):</math>
 
<ol style="margin-left: 1.5em;">
 
<ol style="margin-left: 1.5em;">
 
   <li>We have
 
   <li>We have
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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>
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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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Now, we are left with comparing the real roots of <math>B(x)</math> and <math>E(x).</math>  
  
−
<b>AFK (away from keyboard). Will finish by today. No edits please.</b>
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Since <math>B\left(-\frac{1}{\sqrt2}\right)<0<B\left(\frac{1}{2}\right),</math> we conclude 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. On the other hand, the real root of <math>E(x)</math> is <math>x=-\frac{2018}{2019}\approx-1.</math> Clearly, <math>\boxed{\textbf{(B) }  x^{17}+2018x^{11}+1}</math> has the greatest real root.
  
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== Solution 2 ==
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~MRENTHUSIASM
 +
 
 +
== Solution 2 (Similar to Solution 1) ==
 
We can see that our real solution has to lie in the open interval <math>(-1,0)</math>. From there, note that <math>x^a < x^b</math> if <math>a</math>, <math>b</math> are odd positive integers if <math>a<b</math>, so hence it can only either be <math>\textbf{(B)}</math> or <math>\textbf{(E)}</math> (as all of the other polynomials will be larger than the polynomial <math>\textbf{(B)}</math>). Observe that <math>\textbf{(E)}</math> gives the solution <math>x=-\frac{2018}{2019}</math>. We can approximate the root for <math>\textbf{(B)}</math> by using <math>x=-\frac 12</math>: <cmath>\left(-\frac{1}{2}\right)^{17} - \frac{2018}{2048} + 1 \approx 0.</cmath> Therefore, the root for <math>\textbf{(B)}</math> is approximately <math>-\frac 12</math>. The answer is <math>\boxed{\textbf{(B) }  x^{17}+2018x^{11}+1}</math>.
 
We can see that our real solution has to lie in the open interval <math>(-1,0)</math>. From there, note that <math>x^a < x^b</math> if <math>a</math>, <math>b</math> are odd positive integers if <math>a<b</math>, so hence it can only either be <math>\textbf{(B)}</math> or <math>\textbf{(E)}</math> (as all of the other polynomials will be larger than the polynomial <math>\textbf{(B)}</math>). Observe that <math>\textbf{(E)}</math> gives the solution <math>x=-\frac{2018}{2019}</math>. We can approximate the root for <math>\textbf{(B)}</math> by using <math>x=-\frac 12</math>: <cmath>\left(-\frac{1}{2}\right)^{17} - \frac{2018}{2048} + 1 \approx 0.</cmath> Therefore, the root for <math>\textbf{(B)}</math> is approximately <math>-\frac 12</math>. The answer is <math>\boxed{\textbf{(B) }  x^{17}+2018x^{11}+1}</math>.
  
 
~cpma213
 
~cpma213
  
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== Solution 3 (Calculus) ==
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== Solution 3 (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> <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 (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 on 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 on 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 4 (Calculus) ==
+
== Solution 5 (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>.
  
 
~Qcumber
 
~Qcumber
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−
== Solution 5 ==
 
−
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> <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
 
  
 
== Video Solution by Richard Rusczyk ==
 
== Video Solution by Richard Rusczyk ==

Revision as of 23:17, 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 (Intermediate Value Theorem, Inequalities, Graphs)

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 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 $A(x)$ 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 $E(x).$

Since $B\left(-\frac{1}{\sqrt2}\right)<0<B\left(\frac{1}{2}\right),$ we conclude that the real root of $B(x)$ must satisfy $x\in\left(-\frac{1}{\sqrt2},-\frac{1}{2}\right)$ by IVT. On the other hand, the real root of $E(x)$ is $x=-\frac{2018}{2019}\approx-1.$ 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 $-1 < a < 0,$ $a^{19} < a^{17}$ 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 (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 5 (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

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