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If =1 and =1+i 2, where i= -1 are two roots of the equation x3+ax2+bx+c=0, a, b, c R, then -11 (x3+ax2+bx+c )dx is equal to :
(3) -8
a, b, c , the third root is =1-i 2.
(x- )(x- )=x2-2x+(1+2)=x2-2x+3
x3+ax2+bx+c=(x-1) (x2-2x+3 )=x3-3x2+5x-3
-11 (x3-3x2+5x-3 )dx
= -11 (-3x2-3 )dx ( x3 and 5x are odd functions)
=2 01 (-3x2-3 )dx=2(-1-3)=-8
Answer: (3)
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