Let $$x = 2$$ be a root of the equation $$x^2 + px + q = 0$$ and $$f(x) = \begin{cases} \frac{1-\cos(x^2-4px+q^2+8q+16)}{(x-2p)^4}, & x \neq 2p \\ 0, & x = 2p \end{cases}$$. Then $$\lim_{x \to 2p^+} [f(x)]$$
where $$[.]$$ denotes greatest integer function, is

To find the value of the limit, we first simplify the expression inside the cosine function using the given root condition.

Since $$x = 2$$ is a root of the quadratic equation $$x^2 + px + q = 0$$, substituting $$x = 2$$ gives:

$$2^2 + p(2) + q = 0 \implies 4 + 2p + q = 0 \implies q + 4 = -2p$$

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Step 1: Simplify the argument of the cosine function

Let the algebraic expression inside the cosine be $$g(x)$$:

$$g(x) = x^2 - 4px + q^2 + 8q + 16$$

Recognizing the perfect square formula for the last three terms:

$$q^2 + 8q + 16 = (q + 4)^2$$

Substituting $$(q + 4) = -2p$$ into this perfect square:

$$(q + 4)^2 = (-2p)^2 = 4p^2$$

Now substitute this back into the full expression for $$g(x)$$:

$$g(x) = x^2 - 4px + 4p^2 = (x - 2p)^2$$

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Step 2: Rewrite the function $$f(x)$$ for $$x \neq 2p$$

Substituting the simplified argument back into the definition of $$f(x)$$:

$$f(x) = \frac{1 - \cos((x - 2p)^2)}{(x - 2p)^4}$$

Let $$u = (x - 2p)^2$$. As $$x \to 2p^+$$, we have $$u \to 0^+$$. The function becomes:

$$f(x) = \frac{1 - \cos u}{u^2}$$

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Step 3: Analyze the behavior of $$f(x)$$ near $$u = 0$$

Using the Taylor series expansion for $$\cos u$$ around 0:

$$\cos u = 1 - \frac{u^2}{2!} + \frac{u^4}{4!} - \dots = 1 - \frac{u^2}{2} + \frac{u^4}{24} - \dots$$

Substituting this series into the function expression:

$$f(x) = \frac{1 - \left(1 - \frac{u^2}{2} + \frac{u^4}{24} - \dots\right)}{u^2}$$

$$f(x) = \frac{\frac{u^2}{2} - \frac{u^4}{24} + \dots}{u^2} = \frac{1}{2} - \frac{u^2}{24} + \dots$$

For small positive values of $$u$$ (since $$u = (x-2p)^2 > 0$$ as $$x \to 2p^+$$), the term $$-\frac{u^2}{24}$$ is strictly negative. This means that as $$x$$ approaches $$2p^+$$ from the right, the value of $$f(x)$$ approaches $$\frac{1}{2}$$ from below:

$$0 < f(x) < \frac{1}{2}$$

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Step 4: Evaluate the limit with the greatest integer function

Since $$f(x)$$ lies strictly between 0 and $$\frac{1}{2}$$ in the immediate right-neighborhood of $$2p$$, applying the greatest integer function $$[\cdot]$$ yields:

$$[f(x)] = 0$$

Taking the limit of this constant value:

$$\lim_{x \to 2p^+} [f(x)] = 0$$

Therefore, the value of the limit is equal to 0.