Explanation

(d)
Since, $\alpha$ and $\beta$ are the roots of
$\lambda\ x^{2} + (1 - \lambda)x + 5 = 0$
$\therefore\ \ \alpha + \beta = \frac{\lambda - 1}{\lambda},\ \alpha\beta = \frac{5}{\lambda}$
Since, $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{4}{5}$
$\Rightarrow \ \frac{(\alpha + \beta)^{2} - 2\alpha\beta}{\alpha\beta} = \frac{4}{5}$
$\Rightarrow \ \frac{(\lambda - 1)^{2} - 10\lambda}{5\lambda} = \frac{4}{5}$
$\Rightarrow \ \lambda^{2} - 16\lambda + 1 = 0$
Now, $\lambda_{1} + \lambda_{2} = 16$ and $\lambda_{1}.\lambda_{2} = 1$
$\therefore\frac{\lambda_{1}}{\lambda_{2}^{2}} + \frac{\lambda_{2}}{\lambda_{1}^{2}} = \frac{\lambda_{1}^{3} + \lambda_{2}^{3}}{\left( \lambda_{1}\lambda_{2} \right)^{2}}$
$= \frac{\left( \lambda_{1} + \lambda_{2} \right)^{3} - 3\lambda_{1}\lambda_{2}(\lambda_{1} + \lambda_{2})}{(1)^{2}}$
$= (16)^{3} - 3 \times 1(16)$
$= 4048$