Mathematics Question from JEE Main 2023 · 29 January, Shift 1

Let $x = 2$ be a root of the equation $x^2 + px + q = 0$ and

$$f(x) = \begin{cases} \dfrac{1 - \cos\left(x^2 - 4px + q^2 + 8q + 16\right)}{(x - 2p)^4}, & x \ne 2p, \\ 0, & x = 2p. \end{cases}$$

Then $\lim_{x \to 2p^{+}} [f(x)]$, where $[\cdot]$ denotes the greatest integer function, is: