Find Value of Expression Involving Interval Bounds for Quadratic Roots

To find the required value, we first evaluate the given limit to determine the real constants $r$ and $p$.

The given limit is: $\lim_{x \rightarrow 2} \frac{(\tan(x - 2))(rx^2 + (p - 2)x - 2p)}{(x - 2)^2} = 5$

We can rewrite the limit as: $\lim_{x \rightarrow 2} \left( \frac{\tan(x - 2)}{x - 2} \right) \cdot \left( \frac{rx^2 + (p - 2)x - 2p}{x - 2} \right) = 5$

Since $\lim_{x \rightarrow 2} \frac{\tan(x - 2)}{x - 2} = 1$, the limit simplifies to: $\lim_{x \rightarrow 2} \frac{rx^2 + (p - 2)x - 2p}{x - 2} = 5$

For this limit to exist and be finite, the numerator must be equal to $0$ at $x = 2$: $r(2)^2 + (p - 2)(2) - 2p = 0$ $4r + 2p - 4 - 2p = 0 \implies 4r - 4 = 0 \implies r = 1$

Substituting $r = 1$ back into the numerator gives: $x^2 + (p - 2)x - 2p = (x - 2)(x + p)$

Now, substituting this factorization back into the limit: $\lim_{x \rightarrow 2} \frac{(x - 2)(x + p)}{x - 2} = \lim_{x \rightarrow 2} (x + p) = 2 + p$

Given that the value of the limit is $5$: $2 + p = 5 \implies p = 3$

Thus, we have $r = 1$ and $p = 3$.

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Now, we substitute $r = 1$ and $p = 3$ into the quadratic equation $rx^2 - px + q = 0$: $x^2 - 3x + q = 0$

Let $f(x) = x^2 - 3x + q$. We are given that both roots of $f(x) = 0$ lie in the open interval $(0, 2)$. For both roots of a quadratic function $f(x) = ax^2 + bx + c$ with $a > 0$ to lie in $(0, 2)$, the following conditions must be satisfied:

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

2. Vertex position condition ($0 < -\frac{b}{2a} < 2$): $0 < \frac{3}{2} < 2$ This condition is trivially satisfied.

3. Function values at interval boundaries ($f(0) > 0$ and $f(2) > 0$): $f(0) = q > 0$ $f(2) = 2^2 - 3(2) + q > 0 \implies 4 - 6 + q > 0 \implies q > 2$

Combining all the inequalities for $q$:

• $q \le \frac{9}{4}$
• $q > 0$
• $q > 2$

Taking the intersection of these conditions, we get: $q \in \left( 2, \frac{9}{4} \right]$

Comparing this with the given interval $(\alpha, \beta]$, we find: $\alpha = 2 \quad \text{and} \quad \beta = \frac{9}{4}$

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Finally, we calculate $4(\alpha + \beta)$: $4(\alpha + \beta) = 4 \left( 2 + \frac{9}{4} \right) = 4 \times \frac{17}{4} = 17$

Hence, the correct option is C.