<p><strong>For Problems 16–18</strong><br>Consider the polynomial function<br>\[f(x) = \begin{vmatrix} (1+x)^a & (1+2x)^b & 1 \\ 1 & (1+x)^a & (1+2x)^b \\ (1+2x)^b & 1 & (1+x)^a \end{vmatrix}\]<br>\(a, b\) being positive integers.<br>Which of the following is true?</p>
<p>All the roots of the equation \(f(x) = 0\) are positive.</p>
<p>All the roots of the equation \(f(x) = 0\) are negative.</p>
<p>At least one of the equation \(f(x) = 0\) is repeating one.</p>
<p>None of these.</p>
Step-by-Step Solution
Key Concept: Recognize that this determinant has a circulant matrix structure where each row is a cyclic shift of the previous row. For circulant matrices, the determinant factors as a product involving roots of unity, making direct expansion unnecessary.
<p><strong>Step 1: Identify the circulant structure.</strong> The matrix has the form where column vectors [c₁, c₂, c₃] → [c₃, c₁, c₂] → [c₂, c₃, c₁], characteristic of a circulant matrix with entries (1+x)^a, (1+2x)^b, and 1.</p><p><strong>Step 2: Apply circulant matrix determinant formula.</strong> For a circulant matrix with first row [p, q, r], the determinant equals (p+q+r)(p+ωq+ω²r)(p+ω²q+ωr), where ω = e^(2πi/3) is a primitive cube root of unity.</p><p><strong>Step 3: Substitute values p=(1+x)^a, q=(1+2x)^b, r=1.</strong> One factor is (1+x)^a + (1+2x)^b + 1. The other factors involve cube roots of unity applied to the polynomial expressions.</p><p><strong>Step 4: Key observation.</strong> Since the determinant must be a polynomial (real coefficients in x), complex conjugate factors combine. The determinant simplifies to: f(x) = [(1+x)^a + (1+2x)^b + 1][(1+x)^(2a) + (1+2x)^(2b) + 1 - (1+x)^a(1+2x)^b - (1+2x)^b - (1+x)^a]</p><p><strong>Step 5: Alternative direct form.</strong> f(x) = (1+x)^(3a) + (1+2x)^(3b) + 1 - 3(1+x)^a(1+2x)^b, which is a key factorization revealing the polynomial's degree as max(3a, 3b).</p><p>∴ Answer: C</p>
Correct Answer: C