Problem:
Polynomials and each have degree and leading coefficient , and their roots are all elements of . The function has the property that there exist real numbers such that the set of all real numbers with consists of the closed interval together with the open interval . How many functions are possible?
Answer Choices:
A.
B.
C.
D.
E.
Note:
The correct answer is , which is not a provided answer choice. This is because it is possible for multiple pairs to correspond to the exact same function.
Abridged Version
This is not a rigorous proof, but one that would be more feasible on a real life examination.
Let be the roots of and be the roots of . Now, consider the points where changes sign ( need not be defined at those points). As and are polynomials, they're smooth, so they only change sign at - thus we expect the same behavior from .
This means that are amongst our roots. Note and must be roots of , as otherwise wouldn't be defined there. Note and must be roots of , as otherwise, they would be roots of and not , and would be there (and the interval wouldn't be fully open).
Clearly some of the 's and 's must overlap by the . If for some (as , not all can overlap), then wouldn't change sign at , unless some . But we have established and can not be roots of , which leaves us in an impossible situation - we know roots unique to , the last must be shared with , but two must overlap - absurd.
So has unique roots. If doesn't have a double root, it shares a root with . Note that we can just test the values (we know , so we pick the common root, then assign ). It is clear that sharing or works, for solutions.
Otherwise, if does have a double root, we can once again pick in ways. Once done, has options (as or ). However, in either case, must share the double root. To see why, suppose (the other case is symmetric). Then, does not change sign at , so unless does, can not. This means that in this case, which leaves possibility for . Thus, there are a total of solutions here.
However, these solutions only produce distinct functions. This is because the pairs and correspond to the same function . Hence, we have more functions, for a total of .
Formal Solution
First, note that is positive for and . Then for each root at located at a point in switches signs, with multiple roots at a point toggling the sign of once for each root.
Next, notice that is undefined for every zero of , which means and must be roots of .
Furthermore and must be roots of , as the sign of must switch at these locations while keeping defined.
Let be the roots of and be the roots of (not necessarily distinct). We will work with these to get a better understanding of our functions.
First, note that as , and must have at least distinct roots between them. We claim that . Note that as is a rational function, it is continuous on its domain . In particular, we claim that . For the sake of contradiction, assume it is not, and choose arbitrary . As is continuous, we know that
but this implies there exists some such that for any , . However, this would imply that , or that for all . However, this would imply that , absurd as and . Thus, .
By symmetry, as well (the same argument applies, but with the limit from the positive side, noting that ). As the zeroes of are a subset of the zeroes of , we get that and are two distinct roots of .
Similarly, we will prove . For the sake of contradiction, assume that . Thus is defined at , and note (otherwise , a contradiction). This means that , so we can choose arbitrary . Then, we repeat a similar argument: as is continuous on its domain, we have
Thus there exists some (if , we can "shrink" by taking ) such that for any , , or . This would mean is not in , absurd. Thus, , and by symmetry, .
Note that and must share a root. If not, one of the polynomials has a double root. If has a double root, either or . Without loss of generality, assume the first case (the second will follow by symmetry). Then, we claim that indeed - otherwise, if , as is continuous, there exists such that for any , , or . This implies has the same sign for any , but as (as ), we must have with the same nonzero sign, absurd. Thus, , or they share the same root. The same argument applies if , or if has the double root, by symmetry.
If also had unique roots, note that there is exactly way to order the rest of the numbers (we have numbers to assign to ), and we discussed. Finally, this shared root must be odd, as if it were even, our solution set would either be with root or with root (including some endpoints, which won't be specified), but this is impossible as we want only the union of intervals. Thus, if has unique roots, we get solutions.
Otherwise, we have a double root in . Suppose first . Then, by our above paragraph, as well. Thus, we can choose any elements from our set from through , and order them. Then, our roots for would be and . Similarly, if was our double root for , we would have ways to choose the elements and way to make the polynomials. This gives us solutions.
As before, however, these solutions only produce distinct functions. This is because the pairs and actually correspond to the same function . Hence, we have more functions.
This comes out to a total of .
The problems on this page are the property of the MAA's American Mathematics Competitions