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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| + | === Video Solution by Richard Rusczyk === | ||
| + | |||
| + | https://artofproblemsolving.com/videos/amc/2018amc12a/471 | ||
| + | |||
| + | ~ dolphin7 | ||
==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}} | ||
Revision as of 13:22, 15 May 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 
. ![\[(- \frac {1}{2}) ^{17} - \frac{2018}{2048} + 1 \approx 0\]](http://latex.artofproblemsolving.com/d/d/7/dd7b7d6266b64c094a1a82157b969e4db66b464a.png)
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 ![$[-1,0]$](http://latex.artofproblemsolving.com/3/1/d/31d4af7eb18014c3073c9fa861cdbfc4afb367a0.png)
, 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)
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. 
