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

Revision as of 16:02, 25 January 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, $f(-1) = -1$ and the estimation becomes $x_{1}=-1+{\frac {1}{f'(-1)}}\,.$ 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)

@@ Line 5: / Line 5: @@
 ==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). Finally, we can see that plugging in the root of <math>2019x+2018</math> into B gives a negative, and so 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)==

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 20	Followed by Problem 22
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