If `f(x)=|[x-2, (x-1)^2, x^3] , [(x-1), x^2, (x+1)^3] , [x,(x+1)^2, (x+2)^3]|` then coefficient of `x` in `f(x)` is

Options

A. `-4`
B. `-2`
C. `-6`
D. `0`

Answer

B

Explanation

`(b)` Note that `Delta(x)` is a polynomial of degree at most `6` in `x`. If `Delta(x)=a_(0)+a_(1)x+a_(2)x^(2)+…+a_(6)x^(6)`, then `Delta'(x)=a_(1)+2a_(2)x+……+6a_(6)x^(5)`
`impliesa_(1)=Delta'(0)`,
Now,
`Delta'(x)=|{:(1,(x-1)^(2),x^(3)),(1,x^(2),(x+1)^(3)),(1,(x+1)^(2),(x+2)^(3)):}|+|{:(x-2,2(x-1),x^(3)),(x-1,2x,(x+1)^(3)),(x,2(x+1),(x+2)^(3)):}|+|{:(x-2,(x-1)^(2),3x^(2)),(x-1,x^(2),3(x+1)^(2)),(x,(x+1)^(2),3(x+2)^(2)):}|`
`impliesDelta'(0)=|{:(1,1,0),(1,0,1),(1,1,8):}|+|{:(-2,-2,0),(-1,0,1),(0,2,8):}|+|{:(-2,1,0),(-1,0,3),(0,1,12):}|`
`=-8-12+18=-2`