Explanation

(d)
Since, $\alpha + \beta = - \frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$
Also $\alpha^{2} + \beta^{2} = (\alpha + \beta)^{2} - 2\alpha\beta = \frac{b^{2} - 2ac}{a^{2}}$
Now, $\frac{\alpha}{a\beta + b} + \frac{\beta}{a\alpha + b}$
$= \frac{a\left( \alpha^{2} + \beta^{2} \right) + b(\alpha + \beta)}{\alpha\beta a^{2} + ab(\alpha + \beta) + b^{2}}$
$= \frac{a\left( \frac{b^{2} - 2ac}{a^{2}} \right) + b\left( - \frac{b}{a} \right)}{\left( \frac{c}{a} \right)a^{2} + ab\left( - \frac{b}{a} \right) + b^{2}} = - \frac{2}{a}$