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

Difference between revisions of "2018 AMC 12A Problems/Problem 21"
m (added solutions header)
Line 1: Line 1:
−
==Problem==
+
== Problem ==
 
Which of the following polynomials has the greatest real root?
 
Which of the following polynomials has the greatest real root?
 +
 
<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==
+
== Solutions ==
−
 
+
=== 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 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)
 
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)
  
−
==Solution 2 (Calculus version of solution 1)==
+
=== Solution 2 (Calculus version of solution 1) ===
−
 
 
 
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>.
 
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>.
  
−
==Solution 3 (Alternate Calculus Version)==
+
=== Solution 3 (Alternate Calculus Version) ===
 
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)
 
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==
+
=== Solution 4 ===
−
 
 
 
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>
 
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>
 
-fidgetboss_4000
 
-fidgetboss_4000
  
−
=== Video Solution by Richard Rusczyk ===
+
== Video Solution by Richard Rusczyk ==
−
 
 
 
https://artofproblemsolving.com/videos/amc/2018amc12a/471
 
https://artofproblemsolving.com/videos/amc/2018amc12a/471
  

Revision as of 13:20, 19 January 2021

Problem

Which of the following polynomials has the greatest real root?

$\textbf{(A) } x^{19}+2018x^{11}+1 \qquad \textbf{(B) } x^{17}+2018x^{11}+1 \qquad \textbf{(C) } x^{19}+2018x^{13}+1 \qquad \textbf{(D) } x^{17}+2018x^{13}+1 \qquad \textbf{(E) } 2019x+2018$

Solutions

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 B or E(as all of the other polynomials will be larger than the polynomial B). E gives the solution $x=-\frac{2018}{2019}$. We can approximate the root for B by using $x=-\frac 12$. \[(- \frac {1}{2}) ^{17} - \frac{2018}{2048} + 1 \approx 0\] therefore the root for B is approximately $-\frac 12$. The answer is $\fbox{B}$. (cpma213)

Solution 2 (Calculus version of solution 1)

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 $\fbox{B}$.

Solution 3 (Alternate Calculus Version)

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 A to D, $f(-1) = -2018$ and the estimation becomes $x_{1}=-1+{\frac {2018}{f'(-1)}}\,.$ 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 $\fbox{B}$. (Qcumber)

Solution 4

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 $\frac{(-2018)^{17}}{2019^{17}} + 2018\frac{(-2018)^{11}}{2019^{11}} + 1,$ 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 $(\frac{2018}{2019})^{17}$ and $(\frac{2018}{2019}^{11})$ are both much greater than $(\frac{2018}{2019})^{2019} \approx \frac{1}{e},$ 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 $\boxed{B}.$ -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

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=142773"