If `a^(3)-3a^(2)+5a-17=0` and `b^(3)-3b^(2)+5b+11=0` are such that `a+b` is a real number, then the value of `a+b` is

Options

A. `-1`
B. `1`
C. `2`
D. `-2`

Answer

C

Explanation

`(c )` Let `a+b=lambdaimpliesb=lambda-a`
So `(lambda-a)^(3)-3(lambda-a)^(2)+5(lambda-a)+11=0`
`lambda^(3)-a^(3)-3lambda^(2)a+3lambdaa^(2)-3lambda^(2)+6lambda a-3a^(2)+5lambda-5a+11=0` ,brgt `implies a^(3)+(3-3lambda)a^(2)+(3lambda^(2)-6lambda+5)a-(lambda^(3)-3lambda^(2)+5lambda+11)=0`
Comparing it with the equation `a^(3)-3a^(2)+5a-17=0`, we get `lambda=2`