2018 AMC 12A Problems/Problem 21
Contents
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 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 (Calculus)
Note that and . Calculating the definite integral for each function on 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 5 (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 |
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