Difference between revisions of "2018 AMC 12A Problems/Problem 21"
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<cmath>C(x)>D(x)>A(x)>B(x)</cmath> | <cmath>C(x)>D(x)>A(x)>B(x)</cmath> | ||
− | As <math>B(x)</math> has the smallest value for the same <math>x</math>, its real root must be the greatest, thus, the answer is <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math> | + | As <math>B(x)</math> has the smallest value for the same <math>x</math>, and <math>A(x),B(x),C(x),D(x),</math> are all monotonically increasing functions, its real root must be the greatest, thus, the answer is <math>\boxed{\textbf{(B) } x^{17}+2018x^{11}+1}</math> |
~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] | ~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen] |
Revision as of 09:50, 19 December 2023
Contents
[hide]- 1 Problem
- 2 Solution 1 (Intermediate Value Theorem, Inequalities, Graphs)
- 3 Solution 2 (Similar to Solution 1)
- 4 Solution 3 (Similar to Solution 1)
- 5 Solution 4
- 6 Solution 5 (Monotonic Function)
- 7 Solution 6 (Proof by Contradiction)
- 8 Solution 7 (Calculus)
- 9 Solution 8 (Calculus)
- 10 Video Solution by Richard Rusczyk
- 11 See Also
Problem
Which of the following polynomials has the greatest real root?
Solution 1 (Intermediate Value Theorem, Inequalities, Graphs)
Denote the polynomials in the answer choices by and respectively.
Note that and are strictly increasing functions with range So, each polynomial has exactly one real root. The real root of is On the other hand, since and we conclude that the real root for each of and must satisfy by the Intermediate Value Theorem (IVT).
We analyze the polynomials for
- We have As the graph of is always above the graph of in this interval, we deduce that has a greater real root than does. By the same reasoning, has a greater real root than does.
- We have from which has a greater real root than does.
Now, we are left with comparing the real roots of and Since it follows that the real root of must satisfy by IVT. Clearly, has the greatest real root.
~MRENTHUSIASM
Solution 2 (Similar to 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 or (as all of the other polynomials will be larger than the polynomial ). Observe that gives the solution . We can approximate the root for by using : Therefore, the root for is approximately . The answer is .
~cpma213
Solution 3 (Similar to Solution 1)
Let the real solution to be It is easy to see that when is plugged in to since it follows that 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 the required (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 the required for the solution to to be a solution, our answer is
~fidgetboss_4000
Solution 4
Denote the polynomials in the answer choices by and respectively.
The real root of is , therefore, we can eliminate this polynomial first. Notice that all of the real roots for all answer choices are in . Therefore,
Now compare the real roots of and .
As has the smallest value for the same , and are all monotonically increasing functions, its real root must be the greatest, thus, the answer is
Solution 5 (Monotonic Function)
Denote the polynomials in the answer choices by and respectively.
Notice that all of the real roots for all answer choices are in . The real root of is , therefore, we can eliminate this polynomial first.
First compare the real roots of and :
Let the real root of be , and the real root of be . Let . Notice that is a monotonically increasing function. As , , has a bigger real root than .
Similarly, has a bigger real root than .
To determine the polynomial with the greatest real root, now we only need to compare and .
Let the real root of be , and the real root of be . Let . Notice that is a monotonically increasing function. As , , has a bigger real root than .
Thus, the answer is
Solution 6 (Proof by Contradiction)
Denote the polynomials in the answer choices by and respectively.
Notice that all of the real roots for all answer choices are in . The real root of is , therefore, we can eliminate this polynomial first.
First compare the real roots of and :
Let the real root of be , and the real root of be . As , ,
Assume that . Thus,
assumption is false, , has a bigger real root than .
Similarly, has a bigger real root than .
To determine the polynomial with the greatest real root, now we only need to compare and .
Let the real root of be , and the real root of be .
Assume that . Thus,
assumption is false, , has a bigger real root than .
Thus, the answer is
Solution 7 (Calculus)
Note that and . Calculating the definite integral for each function in 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 8 (Calculus)
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 to we have and the estimation becomes Thus we need to minimize the derivative, giving us . Now after comparing and through Newton's method, we see that 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.