Question

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

• A. $0$
• B. $2$
• C. $1$
• D. None of these

Answer 1

The correct option is C $1$

$x^2-px-(p+q)=0$
Now, simplifying the above expression we get
$\frac{(\alpha +1{)}^2}{(\alpha +1{)}^2+q-1}+\frac{(\beta +1{)}^2}{(\beta +1{)}^2+q-1}$
$=\frac{\left[\right(\alpha +1\left)\right(\beta +1){]}^2+(q-1\left)\right[(\alpha +1{)}^2+(\beta +1{)}^2]}{(q-1{)}^2+(q-1\left)\right((\alpha +1{)}^2+(\beta +1{)}^2)+[(\alpha +1)(\beta +1){]}^2]}$
$=\frac{[1+(\alpha +\beta {)}^2+\alpha \beta {]}^2+(q-1)[{\alpha }^2+{\beta }^2+2(\alpha +\beta )+2]}{(q-1{)}^2+(q-1\left)\right[{\alpha }^2+{\beta }^2+2(\alpha +\beta )+2+(1+(\alpha \beta )+(\alpha +\beta ){)}^2}$
$=\frac{(1+p^2-p-q{)}^2+(q-1\left)\right(p^2+2(p+q)+2p+2\left)\right]}{(q-1{)}^2+(q-1\left)\right[p^2+2(p+q)+2p+2+(1-(p+q)+p{)}^2]}$
$=\frac{2(q-1{)}^2+(q-1\left)\right[p^2+2(p+q)+2p+2]}{2(q-1{)}^2+(q-1\left)\right[p^2+2(p+q)+2p+2]}$
$=1$