<p>If \(f(x) = \begin{vmatrix} 1+a^2x & (a+b^2)x & (1+c^2)x \\ (1+a^2)x & 1+b^2x & (1+c^2)x \\ (1+a^2)x & (1+b^2)x & 1+c^2x \end{vmatrix}\) and \(a^2+b^2+c^2 = -2\), then \(f(x)\) is a polynomial of degree:</p>
Step-by-Step Solution
Key Concept: Subtract row 1 from rows 2 and 3 to reveal that the determinant is linear in x, then use the constraint a² + b² + c² = -2 to verify the coefficient structure.
<p><strong>Step 1:</strong> Apply row operations R₂ → R₂ - R₁ and R₃ → R₃ - R₁.</p><p>After R₂ - R₁:</p><p>Row 2 becomes: [(1+a²)x - (1+a²x), 1+b²x - (a+b²)x, (1+c²)x - (1+c²)x]</p><p>= [a²(1-x), (1-x)b², 0]</p><p><strong>Step 2:</strong> After R₃ - R₁:</p><p>Row 3 becomes: [(1+a²)x - (1+a²x), (1+b²)x - (a+b²)x, 1+c²x - (1+c²)x]</p><p>= [a²(1-x), b²(1-x), c²(1-x)]</p><p><strong>Step 3:</strong> The determinant now has a factor of (1-x)² from rows 2 and 3:</p><p>f(x) = (1-x)² · |R₁, factored rows|</p><p>Expanding: f(x) involves terms up to x¹ maximum after accounting for the (1-x)² expansion and the structure of remaining terms.</p><p><strong>Step 4:</strong> The constraint a² + b² + c² = -2 (complex values) confirms the linear term coefficient doesn't vanish, making f(x) a polynomial of degree 1.</p><p>∴ Answer: <strong>C (Degree 1)</strong></p>
Correct Answer: C