If α and β are the roots of the equation x2−a(x+1)−b=0, then (α+1)(β+1)=
b
-b
1-b
b-1
Given equation x2−a(x+1)−b=0
⇒ x2−ax−a−b ⇒ α + β = α, αβ = -(a + b)
Now (α+1)(β+1)=αβ+α+β+1=−(a+b)+a+1=1−b