If Δ = | x - 2, 2x - 3, 3x - 4; 2x - 3, 3x - 4, 4x - 5; 3x - 5, 5x - 8, 10x - 17 | =
Ax³ + Bx² + Cx + D, then B + C is equal to :

Solution

Maths sir · 5 steps

Do not expand a cubic head-on: subtract neighbouring columns until two zeros appear, then factor.

1. Subtract columns. First apply C₃ → C₃ - C₂, then C₂ → C₂ - C₁. Every entry drops to a simple linear term, and the top two rows of C₂ and C₃ both become x-1.

   Δ = |x-2, x-1, x-1; 2x-3, x-1, x-1; 3x-5, 2x-3, 5x-9| = |x-2, x-1, 0; 2x-3, x-1, 0; 3x-5, 2x-3, 3x-6|

2. Two zeros. The second form comes from C₃ → C₃ - C₂ once more. Expanding along C₃ leaves one term: Δ = (3x-6)[(x-2)(x-1) - (x-1)(2x-3)].
3. Factor. Take out (x-1): the bracket is (x-1)[(x-2)-(2x-3)] = (x-1)(1-x) = -(x-1)². So Δ = -3(x-2)(x-1)².
4. Expand. (x-2)(x²-2x+1) = x³ - 4x² + 5x - 2, so Δ = -3x³ + 12x² - 15x + 6. This gives A=-3, B=12, C=-15, D=6.
5. So. B + C = 12 + (-15) = -3, which is option (b).

The idea to keep. When determinant entries step up by a pattern, subtract adjacent rows or columns first. Making zeros and factoring is far safer than expanding a cubic term by term.

Answer from the NTA final answer key. Solution by PrepWiser.