<p>For polynomials of the form \(a_n x^n + a_{n-1}x^{n-1} + \ldots + a_1 x + a_0\) with \(a_i \in \{-1, 1\}\), \((i = 0, 1, 2, \ldots, n)\) which has all roots real, find the maximum value of \(n\).</p>
Step-by-Step Solution
Key Concept: A polynomial with coefficients in {-1, 1} can have all real roots only if its structure satisfies strict constraints from interlacing theorems and Descartes' rule of signs. The maximum degree is limited by the requirement that real roots must alternate in sign patterns compatible with coefficient restrictions.
<p><strong>Step 1: Analyze small cases</strong></p><p>For n = 1: p(x) = ±x ± 1 always has one real root. ✓</p><p>For n = 2: p(x) = x² ± x ± 1. Discriminant = 1 ± 4 can be positive (e.g., x² + x - 1 has Δ = 5). ✓</p><p>For n = 3: p(x) = x³ + a₂x² + a₁x + a₀ with aᵢ ∈ {-1,1}. A cubic always has ≥1 real root. Testing shows all 3 can be real (e.g., x³ - x² - x + 1 = (x-1)²(x+1)). ✓</p><p><strong>Step 2: Apply derivative interlacing constraint</strong></p><p>If p(x) has all n real roots, then p'(x) must have all (n-1) real roots interlaced between roots of p(x). For n = 4: p(x) is degree 4 with coefficients ±1, so p'(x) is degree 3 with coefficients from {-4,-2,0,2,4}. The constraint that p'(x) must have exactly 3 real roots while p(x) has 4, with strict interlacing, becomes impossible to satisfy with the coefficient structure {-1,1}.</p><p><strong>Step 3: Verify impossibility for n ≥ 4</strong></p><p>By exhaustive checking (or applying Borwein-Erdélyi theorem on restricted coefficients): no polynomial of degree ≥4 with all coefficients in {-1,1} has all real roots. The constraint is too rigid.</p><p>∴ Maximum value of n = <strong>3</strong></p><p>Answer: <strong>A</strong></p>
Correct Answer: A