Explanation

(b)
Here, $\alpha + \beta = - \frac{b}{a},\ \ \alpha\beta = \frac{c}{a}$ ...(i)
The quadratic equation whose roots are $\frac{1 - \alpha}{\alpha}$ and $\frac{1 - \beta}{\beta}$, is
$x^{2} - \left( \frac{1 - \alpha}{\alpha} + \frac{1 - \beta}{\beta} \right)x + \frac{1 - \alpha}{\alpha}.\frac{1 - \beta}{\beta} = 0$
$\Rightarrow x^{2} - \left( \frac{\beta - \alpha\beta + \alpha - \alpha\beta}{\alpha\beta} \right)x + \frac{1 - \beta - \alpha + \alpha\beta}{\alpha\beta} = 0$
$\Rightarrow x^{2} - \left( \frac{- \frac{b}{a},\ - 2.\frac{c}{a}}{\frac{c}{a}} \right)x + \frac{1 - \left( - \frac{b}{a} \right) + \frac{c}{a}}{\frac{c}{a}} = 0$ [from Eq. (i)]
$\Rightarrow x^{2} - \frac{( - b - 2c)x}{c} + \frac{a + b + c}{c} = 0$
$\Rightarrow cx^{2} + (b + 2c)x + (a + b + c) = 0$
On comparing with $px^{2} + qx + r = 0$, we get
$r = q + b + c$