<p>Let \[\begin{vmatrix} x & 2 & x \\ x^2 & x & 6 \\ x & x & 6 \end{vmatrix} = Ax^4 + Bx^3 + Cx^2 + Dx + E\]. Then the value of \(5A + 4B + 3C + 2D + E\) is equal to</p>
Step-by-Step Solution
Key Concept: Expand the determinant as a polynomial in x, then use the substitution property: if f(x) = Ax⁴ + Bx³ + Cx² + Dx + E, then 5A + 4B + 3C + 2D + E = f'(1) (derivative evaluated at x=1).
<p><strong>Step 1:</strong> Recognize that the expression 5A + 4B + 3C + 2D + E is the derivative f'(1) where f(x) = Ax⁴ + Bx³ + Cx² + Dx + E.</p><p>This is because: d/dx(Ax⁴ + Bx³ + Cx² + Dx + E) = 4Ax³ + 3Bx² + 2Cx + D, and at x=1 this equals 4A + 3B + 2C + D. But we need 5A + 4B + 3C + 2D + E = f'(1) + E, which means we evaluate d/dx[f(x)] at x=1 plus the constant term.</p><p><strong>Step 2:</strong> More directly: 5A + 4B + 3C + 2D + E = f(1) + f'(1) where f(x) is the determinant polynomial.</p><p>Alternatively, note that if we denote Δ(x) as the determinant, then: 5A + 4B + 3C + 2D + E = Δ'(1) + coefficient structure evaluation.</p><p><strong>Step 3:</strong> Calculate Δ(1): Substitute x=1 into the determinant:</p><p>Δ(1) = |1 2 1| = 1(6-6) - 2(6-6) + 1(x-x) = 0</p><p> |1 1 6|</p><p> |1 1 6|</p><p><strong>Step 4:</strong> The determinant can be computed as: Δ(x) = x⁴ - 2x³ - 5x² + 6x (by expanding along row 1 or using cofactors).</p><p>So A=1, B=-2, C=-5, D=6, E=0</p><p><strong>Step 5:</strong> Calculate: 5(1) + 4(-2) + 3(-5) + 2(6) + 0 = 5 - 8 - 15 + 12 + 0 = <strong>-6</strong></p><p>∴ Answer: <strong>-6</strong></p>
Correct Answer: D