Question

Sol. [Key Concept: The limit is finite so the numerator $\rightarrow 0$ at $x = 2$, forcing $r = 1$ then $p = 3$; the roots of $x^{2} - 3x + q = 0$ lie in $(0,2)$ iff $D \geq 0,\ f(0) > 0,\ f(2) > 0$, bounding $q$.]

$${\lim_{x \rightarrow 2}\mspace{2mu}\frac{tan(x - 2)}{x - 2} \cdot \frac{\left\lbrack {rx}^{2} + (p - 2)x - 2p \right\rbrack}{x - 2} = 5 }{\Rightarrow \lim_{x \rightarrow 2}\mspace{2mu}\frac{{rx}^{2} - 2x + p(x - 2)}{x - 2} = 5 }{\because D^{r} \rightarrow 0\therefore{\text{ }N}^{r} \rightarrow 0 \Rightarrow r = 1 }{\Rightarrow \lim_{x \rightarrow 2}\mspace{2mu}\frac{x(x - 2) + p(x - 2)}{x - 2} = 5 }{\Rightarrow 2 + p = 5 }{\Rightarrow p = 3 }$$Now, quadratic equation is $x^{2} - 3x + q = 0$ its both roots lie in $(0,2)$

$$D \geq 0,\ 0 < \frac{- b}{2a} < 2,\ f(0) > 0,\ f(2) > 0 $$$\Rightarrow 9 - 4q \geq 0$ and $q > 2$

$${\Rightarrow q \in \left( 2,\frac{9}{4} \right\rbrack }{\alpha = 2,\beta = \frac{9}{4} \Rightarrow \left( 2 + \frac{9}{4} \right) = 17 }$$