Problem:
Find the number of integer values of in the closed interval for which the equation has exactly one real solution.
Solution:
The given equation is equivalent to the equation , with the restrictions and . When , the restrictions require . Then Descartes' Rule of Signs shows that the related polynomial has either or positive roots unless the polynomial has a repeated positive root, as it does exactly when . When , the restrictions require that . On that interval decreases from to while increases from to . This shows that the equation has exactly one real solution for each . Thus the equation has exactly one real solution for each in the set , which has elements.
The problems on this page are the property of the MAA's American Mathematics Competitions