<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>The constant term in \(f(x)\) is</p>
Step-by-Step Solution
Key Concept: The constant term in f(x) is found by setting x=0 in the determinant, which gives a circulant matrix whose determinant can be evaluated using the circulant matrix formula or by recognizing its special structure.
<p><strong>Step 1:</strong> Find the constant term by substituting x = 0 into f(x).</p><p>When x = 0: (1+x)^a → 1 and (1+2x)^b → 1</p><p><strong>Step 2:</strong> The determinant becomes:</p><p>f(0) = |1 1 1|</p><p> |1 1 1|</p><p> |1 1 1|</p><p><strong>Step 3:</strong> This is a circulant matrix where all rows are identical. Since all rows are linearly dependent, the determinant is 0.</p><p>Alternatively, notice that the sum of all rows gives [3, 3, 3] = 3×[1, 1, 1], so the rows are dependent.</p><p>∴ <strong>The constant term is 0</strong></p>
Correct Answer: D