Explanation

(c)
Given, $f(x) = 2x^{3} - 9ax^{2} + 12a^{2}\ x + 1$ attains maximum and minimum at $p$ and $q$ respectively.
$\therefore\ \ \ \ \ f^{'}(p) = 0,\ \ \ f^{'}(q) = 0$
$f^{''}(p) < 0$ and $f^{''}(q) > 0$
Since, $f^{'}(p) = 0$ and $f^{'}(q) = 0$
$\Rightarrow \ \ \ \ \ \ \ 6p^{2} - 18ap + 12a^{2} = 0$
And $6q^{2} - 18aq + 12a^{2} = 0$
$\Rightarrow \ \ \ \ \ \ p^{2} - 3ap + 2a^{2} = 0$
And $q^{2} - 3aq + 2a^{2} = 0$
$\Rightarrow \ \ \ \ p = a,\ 2a;q = a,\ 2a$ ...(i)
Now, $f^{''}(p) < 0$
$\Rightarrow \ \ \ \ 12p - 18a < 0\ \ \ \ \Rightarrow \ \ \ \ p < \frac{3}{2}a$ ...(ii)
And $f^{''}(q) > 0\ \ \ \ \Rightarrow \ \ \ \ \ 12q - 18a > 0\ \ \ \ \Rightarrow \ \ \ \ q > \frac{3}{2}a$ ...(iii)
From Eqs. (i), (ii), (iii), we get
$p = a,\ \ \ q = 2a$
Now, $p^{2} = q$
$\Rightarrow \ \ \ \ \ a^{2} = 2a\ \ \ \ \Rightarrow \ \ \ \ a = 0,2$
But for $a = 0,\ f(x) = 2x^{3} + 1$ which does not attains a maximum or minimum for any value of $x$.
Hence, $a = 2$