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Question

Let Δ(x)=∣ ∣ ∣(x2)(x1)2x3(x1)x2(x+1)3x(x+1)2(x+2)3∣ ∣ ∣ Then the coefficient of x in Δ(x) is k . Find k.

A
-1
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B
2
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C
-2
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D
1
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Solution

The correct option is C -2
Given Δ(x)=∣ ∣ ∣(x2)(x1)2x3(x1)x2(x+1)3x(x+1)2(x+2)3∣ ∣ ∣
Let Δ(x)=a0+a1x+a2x2+a3x3+.....
∣ ∣ ∣(x2)(x1)2x3(x1)x2(x+1)3x(x+1)2(x+2)3∣ ∣ ∣=a0+a1x+a2x2+a3x3+..... .....(1)
Put x=0 in (1)
∣ ∣210101018∣ ∣=a0
a0=2(1)1(8)=10
So, Δ(x)=∣ ∣ ∣(x2)(x1)2x3(x1)x2(x+1)3x(x+1)2(x+2)3∣ ∣ ∣=10+a1x+a2x2+a3x3+..... ....(2)
Differentiating (2) w.r.t x, we get
∣ ∣ ∣12(x1)3x2(x1)x2(x+1)3x(x+1)2(x+2)3∣ ∣ ∣+∣ ∣ ∣(x2)(x1)2x312x3(x+1)2x(x+1)2(x+2)3∣ ∣ ∣+∣ ∣ ∣(x2)(x1)2x3(x1)x2(x+1)312(x+1)3(x+2)2∣ ∣ ∣=a1+2a2x+3a3x2+.....
Put x=0 in above equation
∣ ∣120101018∣ ∣+∣ ∣210103018∣ ∣+∣ ∣2101011212∣ ∣=a1
a1=172+17=2

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