If α > β > 0 are the roots of the equation a x ^ 2 + bx + 1 = 0 ,...

Since, $\alpha$ and $\beta$ are the roots of the equation $a{x}^{2} + bx + 1 = 0$, therefore $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$ would be the roots of ${x}^{2} + bx + a = 0$.

Let

$L = \lim_{x \to \dfrac{1}{\alpha}}\left[\dfrac{1 - \cos \left({x}^{2} + bx + a\right)}{2{\left(1 - \alphax\right)}^{2}}\right]\dfrac{1}{2} \to \dfrac{0}{0}$

$⇒ L = \lim_{x \to \dfrac{1}{\alpha}}\left[\dfrac{2\sin^{2}\left(\dfrac{{x}^{2} + bx + a}{2}\right)}{2{\left(1 - \alphax\right)}^{2}}\right]\dfrac{1}{2}$

$⇒ L = \lim_{x \to \dfrac{1}{\alpha}}\left[\dfrac{\sin^{2}\left(\dfrac{1}{2}\left(x - \dfrac{1}{\alpha}\right)\left(x - \dfrac{1}{\beta}\right)\right)}{{\alpha}^{2}{\left(x - \dfrac{1}{\alpha}\right)}^{2}}\right]\dfrac{1}{2}$

$⇒ L = \lim_{x \to \dfrac{1}{\alpha}}\left[\dfrac{{\left(x - \dfrac{1}{\beta}\right)}^{2}}{4{\alpha}^{2}} \times \dfrac{{\left\{\sin \left(\dfrac{1}{2}\left(x - \dfrac{1}{\alpha}\right)\left(x - \dfrac{1}{\beta}\right)\right)\right\}}^{2}}{{\left(\dfrac{1}{2}\right)}^{2}{\left(x - \dfrac{1}{\alpha}\right)}^{2}{\left(x - \dfrac{1}{\beta}\right)}^{2}}\right]\dfrac{1}{2}$

$⇒ L = \dfrac{\left(\dfrac{1}{\alpha} - \dfrac{1}{\beta}\right)}{2\alpha}$

On comparing this value with the value given in the question we get, $k = 2\alpha$.

Hence this is the correct option.