Question

If $\alpha ,\beta$ be the roots of $x^2-a(x-1)+b=0$, then the value of $\frac{1}{{\alpha }^2-a\alpha }+\frac{1}{{\beta }^2-a\beta }+\frac{2}{a+b}$ is

• A. $\frac{4}{a+b}$
• B. $\frac{1}{a+b}$
• C. $0$
• D. $-1$

Answer 1

Answer: (C)

$0$

Since, $\alpha$ and $\beta$ are the roots of $x^2-ax+a+b=0$, then
$\alpha +\beta =a$ and $\alpha \beta =a+b$
$\Rightarrow {\alpha }^2+\alpha \beta =a\alpha$
$\Rightarrow {\alpha }^2-a\alpha =-(a+b)$
and $\alpha \beta +{\beta }^2=\alpha \beta$
$\Rightarrow {\beta }^2-\alpha \beta =-(a+b)$
$\therefore \frac{1}{{\alpha }^2-a\alpha }+\frac{1}{{\beta }^2-a\beta }+\frac{2}{a+b}$
$=\frac{1}{-(a+b)}+\frac{1}{-(a+b)}+\frac{2}{(a+b)}=0$