Correct option is (B) 1 – c
f(x) = \(x^2-p(x+1)-c\)
\(=x^2-px-(p+c)\)
\(\therefore\) \(\alpha+\beta\) \(=\frac{-(-p)}1\) = p
\(\alpha\beta\) \(=\frac{-(p+c)}1\) = -(p+c)
Now, \((\alpha+1)(\beta+1)\) \(=\alpha\beta+\alpha+\beta+1\)
\(=-(p+c)+p+1\)
= 1 - c