Let x=2 be a root of the equation x^2+p x+q=0 and f(x)= {(1- cos (x^2-4 p x+q^2+8 q+16))/((x-2 p)^4), x ≠ 2 p; 0, x=2 p} Then lim_x → 2 p^+ [f(x)],…

Let $\displaystyle x=2$ be a root of the equation $\displaystyle x^2+p x+q=0$ and $$f(x)=\left\{\begin{array}{cl} \frac{1-\cos \left(x^2-4 p x+q^2+8 q+16\right)}{(x-2 p)^4} & , x \neq 2 p \\ 0, & x=2 p \end{array}\right. $$ Then $\displaystyle \lim _{x \rightarrow 2 p^{+}}[f(x)]$, where $\displaystyle [\cdot]$ denotes greatest integer function, is