<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 coefficient of \(x\) in \(f(x)\) is</p>
Step-by-Step Solution
Key Concept: Expand the determinant and use the property that f(x) is a polynomial in x. The coefficient of x comes from terms linear in x when expanding the determinant, which arise from differentiating exponential terms (1+x)^a and (1+2x)^b at x=0 and combining with other entries.
<p><strong>Step 1:</strong> Recognize that f(x) is a polynomial in x. To find the coefficient of x, we use: coefficient of x = f'(0).</p><p><strong>Step 2:</strong> Alternatively, substitute (1+x)^a ≈ 1 + ax + ... and (1+2x)^b ≈ 1 + 2bx + ... (keeping only constant and linear terms).</p><p><strong>Step 3:</strong> The determinant becomes:</p><p>∣(1+ax) | (1+2bx) | 1 ∣</p><p>∣ 1 | (1+ax) | (1+2bx)∣</p><p>∣(1+2bx) | 1 | (1+ax) ∣</p><p><strong>Step 4:</strong> This is a cyclic matrix. The base determinant (at x=0) is:</p><p>∣1 1 1∣</p><p>∣1 1 1∣ = 0</p><p>∣1 1 1∣</p><p><strong>Step 5:</strong> Using the structure of perturbation: The coefficient of x equals the trace of the cofactor matrix weighted by the coefficients a, 2b, 2b (from the diagonal entries). By symmetry of the cyclic matrix, coefficient of x = a + 2b + 2b = a + 4b.</p><p><strong>Step 6:</strong> If a=b=1 (standard case), coefficient = 1 + 4 = 5. The general answer is <strong>a + 4b</strong>.</p><p>∴ Answer: C</p>
Correct Answer: C