Difference between revisions of "2018 AMC 12A Problems/Problem 21"
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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) = -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) | ||
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+ | ==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 to <math>B.</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 than the solution to <math>B,</math> our answer is <math>\boxed{B}.</math> | ||
+ | -fidgetboss_4000 | ||
=== Video Solution by Richard Rusczyk === | === Video Solution by Richard Rusczyk === |
Revision as of 21:37, 30 June 2020
Contents
Problem
Which of the following polynomials has the greatest real root?
Solution 1
We can see that our real solution has to lie in the open interval . From there, note that if , are odd positive integers if , 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 . We can approximate the root for B by using . therefore the root for B is approximately . The answer is . (cpma213)
Solution 2 (Calculus version of solution 1)
Note that and . Calculating the definite integral for each function on the interval , we see that 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 .
Solution 3 (Alternate Calculus Version)
Newton's Method is used to approximate the zero of any real valued function given an estimation for the root : After looking at all the options, gives a reasonable estimate. For options A to D, and the estimation becomes 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 . (Qcumber)
Solution 4
Let the real solution to be It is easy to see that when is plugged in to since thus making the real solution to more "negative", or smaller than Similarly we can assert that Now to compare and we can use the same method to what we used before to compare to in which it is easy to see that the smaller exponent "wins". Now, the only thing left is for us to compare and Plugging (or the solution to ) into we obtain which is intuitively close to much smaller than the solution to (For a more rigorous proof, one can note that and are both much greater than by the limit definition of Since is still much smaller than the solution to our answer is -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.