Question

If $a{x}^2+bx+c=0$ has roots $\alpha$ and $\beta$ then-$a{\alpha }^2+b\alpha +c=0$ and $a{\beta }^2+b\beta +c=0$
If $\alpha ,\beta$ are the roots of $x^2-p(x+1)-c=0,c\ne 1$, then $\frac{{\alpha }^2+2\alpha +1}{{\alpha }^2+2\alpha +c}+\frac{{\beta }^2+2\beta +1}{\beta 2+2\beta +c}=$

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

Answer 1

The correct option is B $1$

If $\alpha ,\beta$ are roots of $x^2-p(x+1)-c=0$ then $\alpha +\beta =p,\alpha \beta =-p-c----\left(1\right)$

$\frac{{\alpha }^2+2\alpha +1}{{\alpha }^2+2\alpha +c}+\frac{{\beta }^2+2\beta +1}{{\beta }^2+2\beta +c}=\frac{(\alpha +1{)}^2}{(\alpha +1{)}^2+c-1}+\frac{(\beta +1{)}^2}{(\beta +1{)}^2+c-1}$

$=\frac{\alpha (\alpha +1{)}^2+(c-1\left)\right[(\alpha +1{)}^2+(\beta +1{)}^2]}{(\alpha +1{)}^2(\beta +1{)}^2+(c-1)\left(\right(\alpha +1{)}^2+(\beta +1{)}^2)+(c-1{)}^2}-----\left(2\right)$

Now, $(\alpha +1)(\beta +1)=\alpha \beta +(\alpha +\beta )+1=-p-c+p+1=1-c$ [from $1$]

Substituting in $\left(2\right)\frac{{\alpha }^2+2\alpha +1}{{\alpha }^2+2\alpha +c}\frac{{\beta }^2+2\beta +1}{{\beta }^2+2\beta +c}=\frac{2(\alpha +1{)}^2(\beta +1{)}^2+(c-1)\left[\right(\alpha +1{)}^2+(\beta +1{)}^2]}{2(\alpha +1{)}^2(\beta +1{)}^2+(c-1)\left[\right(\alpha +1{)}^2+(\beta +1{)}^2]}$

$=[\because c-1=-(\alpha +1\left)\right(\beta +1\left)\right]$

$\therefore$ Option $B$ is correct.