Let $$\displaystyle\lim_{x \to 2} \dfrac{(\tan(x - 2))(rx^2 + (p - 2)x - 2p)}{(x - 2)^2} = 5$$ for some $$r, p \in \mathbb{R}$$. 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 :

$$\lim_{x\to2} \frac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5$$

$$\lim_{x\to2} \left[ \frac{\tan(x-2)}{x-2} \right] \cdot \lim_{x\to2} \left[ \frac{rx^2 + (p-2)x - 2p}{x-2} \right] = 5$$

$$\lim_{\theta\to0} \frac{\tan\theta}{\theta} = 1$$

$$\lim_{x\to2} \frac{rx^2 + (p-2)x - 2p}{x-2} = 5$$

For this limit to exist and be finite, the numerator must evaluate to $$0$$ when $$x = 2$$:

$$r(2)^2 + (p-2)(2) - 2p = 0 \implies 4r + 2p - 4 - 2p = 0 \implies 4r = 4 \implies r = 1$$

$$\lim_{x\to2} \frac{x^2 + (p-2)x - 2p}{x-2} = 5$$

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

$$\lim_{x\to2} \frac{(x-2)(x+p)}{x-2} = 5 \implies \lim_{x\to2} (x+p) = 5 \implies 2 + p = 5 \implies p = 3$$

$$x^2 - 3x + q = 0$$

For both roots of a quadratic equation to lie within an open interval $$(d, e)$$:

Discriminant condition ($$D \ge 0$$): $$(-3)^2 - 4(1)(q) \ge 0 \implies 9 - 4q \ge 0 \implies q \le \frac{9}{4}$$

Vertex condition ($$d < -\frac{b}{2a} < e$$): $$0 < -\frac{-3}{2(1)} < 2 \implies 0 < 1.5 < 2 \quad (\text{Always true})$$

Boundary conditions ($$f(d) > 0$$ and $$f(e) > 0$$):

At $$x = 0$$: $$f(0) > 0 \implies 0^2 - 3(0) + q > 0 \implies q > 0$$

$$q \in \left( 2, \frac{9}{4} \right]$$

$$4(\alpha + \beta) = 4\left( 2 + \frac{9}{4} \right) = 8 + 9 = 17$$