SOLUTION —
Here, $\alpha+\beta=-\frac{b}{a}, \alpha \beta=\frac{c}{a}$ and
$\alpha^2+\beta^2=(\alpha+\beta)^2-2 \alpha \beta=\frac{\left(b^2-2 a c\right)}{a^2}$
$\therefore \frac{\alpha}{a \beta+b}+\frac{\beta}{a \alpha+b}=\frac{\alpha(a \alpha+b)+\beta(a \beta+b)}{(a \beta+b)(a \alpha+b)}$
$=\frac{a\left(\alpha^2+\beta^2\right)+b(\alpha+\beta)}{a^2 \alpha \beta+a b(\alpha+\beta)+b^2}$
$=\frac{a \frac{\left(b^2-2 a c\right)}{a^2}+b\left(-\frac{b}{a}\right)}{a^2\left(\frac{c}{a}\right)+a b\left(-\frac{b}{a}\right)+b^2}=-\frac{2}{a}$
So, The correct option will be (D).