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

(Solution 1)
 
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==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) = -1</math> and the estimation becomes <math>x_{1}=-1+{\frac {1}{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)
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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}</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)
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2018|ab=A|num-b=20|num-a=22}}
 
{{AMC12 box|year=2018|ab=A|num-b=20|num-a=22}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 00:46, 15 February 2020

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

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)

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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