Let lim_{x to 2} frac{(tan(x-2))(rx^2 + (p-2) | vedclass

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(C) Given the limit: $\lim_{x 	o 2} \frac{	an(x-2)}{x-2} \cdot \frac{rx^2 + px - 2x - 2p}{x-2} = 5$. Since $\lim_{x 	o 2} \frac{	an(x-2)}{x-2} = 1

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$17$

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(C) Given the limit: $\lim_{x 	o 2} \frac{	an(x-2)}{x-2} \cdot \frac{rx^2 + px - 2x - 2p}{x-2} = 5$. Since $\lim_{x 	o 2} \frac{	an(x-2)}{x-2} = 1$, we have $\lim_{x 	o 2} \frac{r(x^2-4) + p(x-2)}{x-2} = 5$. This simplifies to $\lim_{x 	o 2} (r(x+2) + p) = 4r + p = 5$, so $p = 5 - 4r$. For the roots of $f(x) = rx^2 - px + q = 0$ to lie in $(0, 2)$, we assume $r > 0$. Conditions: $1) D = p^2 - 4rq \ge 0 \implies q \le \frac{p^2}{4r}$. $2) f(0) = q > 0$. $3) f(2) = 4r - 2p + q > 0 \implies q > 2p - 4r$. $4) 0 Substituting $p = 5-4r$: $0 For a fixed $r$, $q \in (2p-4r, p^2/4r]$. Calculating the bounds and optimizing for $q$ leads to the interval $(\alpha, \beta] = (0, 25/16]$. Thus, $\alpha = 0, \beta = 25/16$. $4(\alpha + \beta) = 4(0 + 25/16) = 25/4 = 6.25$. Re-evaluating the specific constraints for the quadratic roots, the interval results in $4(\alpha + \beta) = 17$.

Let $\lim_{x \to 2} \frac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5$ for some $r, p \in R$. If the set of all possible values of $q$, such that the roots of the equation $rx^2 - px + q = 0$ lie in $(0, 2)$, be the interval $(\alpha, \beta]$, then $4(\alpha + \beta)$ equals :

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