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

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==Problem==
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== Problem ==
 
Which of the following polynomials has the greatest real root?
 
Which of the following polynomials has the greatest real root?
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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>
  
==Solution 1==
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== Solution 1 (Intermediate Value Theorem, Inequalities, Graphs) ==
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Denote the polynomials in the answer choices by <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 strictly increasing functions with range <math>(-\infty,\infty).</math> So, each polynomial has exactly one real root. The real root of <math>E(x)</math> is <math>x=-\frac{2018}{2019}\approx-1.000.</math> On the other hand, since <math>A(-1)=B(-1)=C(-1)=D(-1)=-2018</math> and <math>A(0)=B(0)=C(0)=D(0)=1,</math> we conclude that the real root for each of <math>A(x),B(x),C(x),</math> and <math>D(x)</math> must satisfy <math>x\in(-1,0)</math> by the Intermediate Value Theorem (IVT).
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We analyze the polynomials for <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>y=A(x)</math> is always above the graph of <math>y=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> Since <math>B\left(-\frac{1}{\sqrt2}\right)<0<B\left(-\frac{1}{2}\right),</math> it follows 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. Clearly, <math>\boxed{\textbf{(B) }  x^{17}+2018x^{11}+1}</math> has the greatest real root.
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~MRENTHUSIASM
  
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 B or E(as all of the other polynomials will be larger than the polynomial B). E gives the solution <math>x=-\frac{2018}{2019}</math>. We can approximate the root for B by using <math>x=-\frac 12</math>. <cmath> (- \frac {1}{2}) ^{17} - \frac{2018}{2048} + 1 \approx 0</cmath> therefore the root for B is approximately <math>-\frac 12</math>. The answer is <math>\fbox{B}</math>. (cpma213)
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== Solution 2 (Similar to Solution 1) ==
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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>.
  
==Solution 2 (Calculus version of solution 1)==
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~cpma213
  
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>\fbox{B}</math>.
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== Solution 3 (Similar to Solution 1) ==
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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>
  
==Solution 3 (Alternate Calculus Version)==
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~fidgetboss_4000
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}</math>: <math>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 A to D, <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 B. Now after comparing B and E through Newton's method, we see that B has the higher root, so the answer is <math>\fbox{B}</math>. (Qcumber)
 
  
==Solution 4==
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== Solution 4 (Calculus) ==
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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>.
  
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\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>(\frac{2018}{2019})^{17}</math> and <math>(\frac{2018}{2019}^{11})</math> are both much greater than  <math>(\frac{2018}{2019})^{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{B}.</math>
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== Solution 5 (Calculus) ==
-fidgetboss_4000
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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>.
  
=== Video Solution by Richard Rusczyk ===
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~Qcumber
  
 +
== Video Solution by Richard Rusczyk ==
 
https://artofproblemsolving.com/videos/amc/2018amc12a/471
 
https://artofproblemsolving.com/videos/amc/2018amc12a/471
  

Revision as of 01:09, 13 September 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)

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 $y=B(x),$ 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 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 (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 (ProblemsAnswer KeyResources)
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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