Difference between revisions of "2018 AMC 12A Problems/Problem 21"
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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>. | ||
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| + | ==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) | ||
==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 20:28, 22 July 2018
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 so 
, 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 
into B gives a negative, and so 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)
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. 
