Question

If $\alpha ,\beta$ 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

• A. $4$
• B. $0$
• C. $1$
• D. $-1$

Answer 1

The correct option is C $1$

$x^2-p(x+1)-c=0\Rightarrow x^2-px-p-c=0$

Sum and product of the roots is,
$\alpha +\beta =p,\alpha \beta =-(p+c)$

$(\alpha +1)(\beta +1)=\alpha \beta +(\alpha +\beta )+1=1-c$

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

Putting the value$1-c=(\alpha +1)(\beta +1)$ , we get
$=\frac{\alpha +1}{\alpha -\beta }+\frac{\beta +1}{\beta -\alpha }=\frac{\alpha +1-\beta -1}{\alpha -\beta }=1$