Mark the correct alternative in the following:

Let be an identify in x, were a, b, c, d, e are independent of x. Then the value of e is


We are given that,


We need to find the value of e.


We know that,


Determinant of 3 × 3 matrix is given as,




So,




⇒ (x2 + 3x)[-2x × 3x – (x – 4)(x + 4)] – (x – 1)[(x + 1) × 3x – (x – 4)(x – 3)] + (x + 3)[(x + 1)(x + 4) – (-2x)(x – 3)] = ax4 + bx3 + cx2 + dx + e


⇒ (x2 + 3x)[-6x – (x2 – 16)] – (x – 1)[3x(x + 1) – (x2 – 3x – 4x + 12)] + (x + 3)[x2 + x + 4x + 4 + 2x(x – 3)] = ax4 + bx3 + cx2 + dx + e


⇒ (x2 + 3x)[-6x – x2 + 16] – (x – 1)[3x2 + 3x – x2 + 7x – 12] + (x + 3)[x2 + 5x + 4 + 2x2 – 6x] = ax4 + bx3 + cx2 + dx + e


⇒ -x4 – 6x3 + 16x2 – 3x3 – 18x2 + 48x – (x – 1)[2x2 + 10x – 12] + (x + 3)[3x2 – x + 4] = ax4 + bx3 + cx2 + dx + e


⇒ -x4 – 9x3 – 2x2 + 48x – (2x3 – 2x2 + 10x2 – 10x – 12x + 12) + 3x3 + 9x2 – x2 – 3x + 4x + 12 = ax4 + bx3 + cx2 + dx + e


⇒ -x4 – 9x3 – 2x2 + 48x – 2x3 + 2x2 – 10x2 + 10x + 12x – 12 + 3x3 + 9x2 – x2 – 3x + 4x + 12 = ax4 + bx3 + cx2 + dx + e


⇒ -x4 – 9x3 – 2x3 + 3x3 – 2x2 + 2x2 + 9x2 – x2 + 48x + 10x + 12x – 3x + 4x – 12 + 12 = ax4 + bx3 + cx2 + dx + e


⇒ -x4 – 8x3 + 8x2 + 23x + 0 = ax4 + bx3 + cx2 + dx + e


Comparing left hand side and right-hand side of the equation, we get


e = 0


Thus, e = 0.

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