Explanation

(d)
$\therefore f(x) = - x^{3} + 4ax^{2} + 2x - 5$
$\therefore f^{'}(x) = - 3x^{2} + 8ax + 2$
Since, $f(x)$ is decreasing. $\forall x,$ therefore $f^{'}(x) < 0$
$\Rightarrow - 3^{2} + 8ax + 2 < 0$
$\Rightarrow 3x^{2} - 8ax - 2 > 0$
Now, $D = ( - 8a)^{2} - 4(3)( - 2) < 0$
$\Rightarrow 64a^{2} + 24 < 0$
$\Rightarrow a^{2} < - \frac{3}{8}$
Hence, no value of a exist.