Difference between revisions of "2018 AMC 12A Problems/Problem 21"
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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 a, b are odd positive integers so <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 a, b are odd positive integers so <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) | ||
| − | ==Solution 2 (Calculus | + | ==Solution 2 (Calculus version of solution 1)== |
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>. | ||
Revision as of 00:12, 24 February 2018
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 a, b 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 .
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. 
