<p>If \(a^2 + b^2 + c^2 = -2\) and \[f(x) = \begin{vmatrix} (1+a^2)x & (1+b^2)x & (1+c^2)x \\ (1+a^2)x & (1+b^2)x & (1+c^2)x \\ (1+a^2)x & (1+b^2)x & (1+c^2)x \end{vmatrix}\] then \(f(x)\) is a polynomial of degree</p>
Step-by-Step Solution
Key Concept: Since all three rows of the determinant are identical, the determinant equals zero for all values of x, making f(x) a polynomial of degree -∞ (or undefined/null polynomial). The constraint a² + b² + c² = -2 is impossible for real numbers, indicating this is a trick question testing whether students recognize identical rows immediately.
<p><strong>Step 1:</strong> Observe the structure of the determinant. All three rows are identical:</p><p>Row 1 = Row 2 = Row 3 = [(1+a²)x, (1+b²)x, (1+c²)x]</p><p><strong>Step 2:</strong> Apply the fundamental property: A determinant with two or more identical rows equals zero.</p><p><strong>Step 3:</strong> Therefore, f(x) = 0 for all values of x.</p><p><strong>Step 4:</strong> The zero polynomial (f(x) = 0) is defined to have degree -∞ or is sometimes stated as having no degree.</p><p><strong>Step 5:</strong> If the question asks for a finite degree, the answer is that f(x) is a constant polynomial (degree 0, since f(x) = 0 is the zero constant).</p><p>∴ Answer: D (The degree is -∞, or f(x) is the zero polynomial with undefined degree, or if forced to choose among positive integers, degree 0 as a degenerate case)</p>
Correct Answer: D