If $\alpha$ and $\beta$ are the roots of $x^2-p(x+1)-c=0$, then the value of $\frac{{\alpha }^2+2\alpha +1}{{\alpha }^2+2\alpha +c}+\frac{{\beta }^2+2\beta +1}{{\beta }^2+2\beta +c}$ is

Solution

1. The equation is $x^2-px-p-c=0$, so $x^2=px+p+c$.
2. Substitute $x^2$ into the denominator: $$\begin{array}{rl}{\alpha }^2+2\alpha +c& =p\alpha +p+c+2\alpha +c\\ & =(p+2)\alpha +p+c\end{array}$$.
3. Alternatively, note that $$\begin{array}{rl}{\alpha }^2+2\alpha +c& =p\alpha +p+2\alpha \\ & =(p+2)(\alpha +1)\end{array}$$.
4. The expression becomes $\frac{(\alpha +1{)}^2}{(p+2)(\alpha +1)}+\frac{(\beta +1{)}^2}{(p+2)(\beta +1)}=\frac{\alpha +1+\beta +1}{p+2}$.
5. From the equation, $\alpha +\beta =p$, so the numerator is $p+2$.
6. Hence the answer is (B).