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1981 AMC 12 Problem 28

Attempt Problem 28 of the 1981 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1981 AMC 12 solutions, or check the answer key.

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28.

Consider the set of all equations x3+a2x2+a1x+a0=0,x^3+a_2x^2+a_1x+a_0=0, where a2,a_2, a1,a_1, a0a_0 are real constants and ∣ai∣≤2|a_i|\le2 for i=0,i=0, 1,1, 2.2. Let rr be the largest positive real number which satisfies at least one of these equations. Then

1≤r<321\le r\lt\frac32

32≤r<2\frac32\le r\lt2

2≤r<522\le r\lt\frac52

52≤r<3\frac52\le r\lt3

3≤r<723\le r\lt\frac72

Answer: D
Concepts:polynomialinequalitybounding to limit cases
Difficulty rating: 2210
Small Hint:

For positive x,x, the coefficients that push a root farthest right take their lower bounds

Big Hint:

Bracket the largest root of x3−2x2−2x−2x^3-2x^2-2x-2 at two consecutive choice endpoints

Solution:

Write an allowed polynomial as g.g. For x≥0,x\ge0, g(x)=x3+a2x2+a1x+a0≥f(x), \begin{aligned} g(x)&=x^3+a_2x^2\\ &\quad+a_1x+a_0\\ &\ge f(x), \end{aligned} where f(x)=x3−2x2−2x−2.f(x)=x^3-2x^2-2x-2. Thus no positive root can exceed the largest root ρ\rho of f.f. Taking a2=a1=a0=−2a_2=a_1=a_0=-2 gives g=f,g=f, so ρ\rho itself is attained. Since f(52)=−318<0,f(3)=1>0, \begin{gathered} f\left(\frac52\right)=-\frac{31}{8}\lt0,\\ f(3)=1\gt0, \end{gathered} we have 52<ρ<3.\frac{5}{2}\lt\rho\lt3.

Therefore, the correct answer is D.

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