Let x=2 be a root of the equation x^2+px+q=0 | vedclass

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(C) Given $x=2$ is a root of $x^2+px+q=0$,we have $4+2p+q=0$,so $q = -2p-4$. Substituting $q$ into the expression inside the cosine: $x^2-4px+q^2+8q+1

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$0$

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(C) Given $x=2$ is a root of $x^2+px+q=0$,we have $4+2p+q=0$,so $q = -2p-4$. Substituting $q$ into the expression inside the cosine: $x^2-4px+q^2+8q+16 = x^2-4px+(-2p-4)^2+8(-2p-4)+16 = x^2-4px+4p^2+16p+16-16p-32+16 = x^2-4px+4p^2 = (x-2p)^2$. Thus,$f(x) = \frac{1-\cos((x-2p)^2)}{(x-2p)^4}$ for $x 
eq 2p$. Using the limit $\lim_{	heta 	o 0} \frac{1-\cos 	heta}{	heta^2} = \frac{1}{2}$,we have $\lim_{x 	o 2p} f(x) = \lim_{x 	o 2p} \frac{1-\cos((x-2p)^2)}{((x-2p)^2)^2} = \frac{1}{2}$. Since $\lim_{x 	o 2p^+} f(x) = \frac{1}{2}$,the greatest integer function $[f(x)]$ for $x$ near $2p$ (where $0 < f(x) < 1$) is $[f(x)] = 0$.

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 $\lim _{x \rightarrow 2p^{+}}[f(x)]$,where $[.]$ denotes the greatest integer function,is $........$

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