Difference between revisions of "2018 AMC 12A Problems/Problem 21"
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| − | + | ==Problem== | |
Which of the following polynomials has the greatest real root? | Which of the following polynomials has the greatest real root? | ||
<math>\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 </math> | <math>\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 </math> | ||
| − | + | ==Solution== | |
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) | ||
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| + | ==See Also== | ||
| + | {{AMC12 box|year=2018|ab=A|num-b=20|num-a=22}} | ||
| + | {{MAA Notice}} | ||
Revision as of 14:50, 8 February 2018
Problem
Which of the following polynomials has the greatest real root?
Solution
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)
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. 
