Explanation

(d)
$\because f(x) = x^{3} - 6x^{2} + ax + b$
On differentiating w.r.t. $x$, we get
$f^{'}(x) = 3x^{2} - 12x + a$
By the definition of Rolle’s theorem
$f^{'}(c) = 0\ \ \ \Rightarrow \ \ \ f^{'}\left( 2 + \frac{1}{\sqrt{3}} \right) = 0$
$\Rightarrow \ \ 3\left( 2 + \frac{1}{\sqrt{3}} \right)^{2} = - 12\left( 2 + \frac{1}{\sqrt{3}} \right) + a = 0$
$\Rightarrow \ \ 3\left( 4 + \frac{1}{3} + \frac{4}{\sqrt{3}} \right) - 12\left( 2 + \frac{1}{\sqrt{3}} \right) + a = 0$
$\Rightarrow \ \ \ 12 + 1 + 4\sqrt{3} - 24 - 4\sqrt{3} + a = 0$
$\Rightarrow \ \ \ a = 11$