Problem:
Consider the set of all equations x3+a2x2+a1x+a0=0, where a2,a1,a0 are real constants and ∣ai∣⩽2 for i=0,1,2. Let r be the largest positive real number which satisfies at least one of these equations. Then
Answer Choices:
A. 1⩽r<23
B. 23⩽r<2
C. 2⩽r<25
D. 25⩽r<3
E. 3⩽r<27
Solution:
Let g(x)=x3+a2x2+a1x+a0 be an arbitrary cubic with constants of the specified form. Because x3 dominates the other terms for large enough x, g(x)>0 for all x greater than the largest real root of g. Thus we seek a particular g in which the terms a2x2+a1x+a0 "hold down" g(x) as much as possible, so that the value of the largest real root is as large as possible. This suggests that the answer to the problem is the largest root of f(x) =x3−2x2−2x−2. Call this root r0. To verify this conjecture, note that for x⩾0,−2x2⩽a2x2,−2x⩽a1x, and −2⩽a0.
Summing these inequalities, and adding x3 to both sides, gives f(x)⩽g(x) for all x⩾0. Thus for all x>r0,0<f(x)⩽g(x). That is, no g has a root larger than r0, so r0 is the r of the problem.
A sketch of f shows that it is a typical S-shaped cubic, with largest root a little less than 3 . In fact, f(2)=−6 and f(3)=1. To be absolutely sure the answer is (D), not (C), compute f(25) to see if it is negative. Indeed, f(25)=−831.