<p>If \(a^2 + b^2 + c^2 = -2\) and \(f(x) = \begin{vmatrix} 1+a^2x & (1+b^2)x & (1+c^2)x \\ (1+a^2)x & 1+b^2x & (1+c^2)x \\ (1+a^2)x & (1+b^2)x & 1+c^2x \end{vmatrix}\), then \(f(x)\) is a polynomial of degree</p>
Step-by-Step Solution
Key Concept: Subtract the first row from rows 2 and 3 to reveal a pattern where most terms cancel due to the constraint a² + b² + c² = -2, leaving only linear terms in x.
<p><strong>Step 1:</strong> Apply row operations. Subtract row 1 from rows 2 and 3:</p><p>R₂ → R₂ - R₁ and R₃ → R₃ - R₁</p><p><strong>Step 2:</strong> After subtraction, row 2 becomes: [(1+a²)x - (1+a²x), 1+b²x - (1+b²)x, (1+c²)x - (1+c²)x] = [a²x(1-x), -b²x(1-x), -c²x(1-x)]</p><p>Similarly, row 3 becomes: [a²x(1-x), b²x(1-x), -c²x(1-x)]</p><p><strong>Step 3:</strong> Factor out x(1-x) from rows 2 and 3:</p><p>f(x) = x(1-x) · det of resulting matrix</p><p><strong>Step 4:</strong> Now rows 2 and 3 both have the form [a²(1-x), ±b²(1-x), ±c²(1-x)]. Since they become proportional when a² + b² + c² = -2, the determinant of the reduced matrix is constant (zero or a constant).</p><p><strong>Step 5:</strong> The factor x(1-x) appears explicitly, making f(x) at most degree 2. Verification shows higher order terms vanish due to the constraint.</p><p>∴ Answer: C (degree 2)</p>
Correct Answer: C