JEE Main 2026 — 6 April, Evening Shift — Question 35

$\lim _{\mathrm{x} \rightarrow 2} \frac{\tan (\mathrm{x}-2)}{\mathrm{x}-2} \cdot \frac{\left[\mathrm{rx}^{2}+(\mathrm{p}-2) \mathrm{x}-2 \mathrm{p}\right]}{\mathrm{x}-2}=5$ $\Rightarrow \lim _{\mathrm{x} \rightarrow 2} \frac{\mathrm{rx}^{2}-2 \mathrm{x}+\mathrm{p}(\mathrm{x}-2)}{\mathrm{x}-2}=5$ $\because \mathrm{D}^{\mathrm{r}} \rightarrow 0 \therefore \mathrm{~N}^{\mathrm{r}} \rightarrow 0 \Rightarrow \mathrm{r}=1$ $\Rightarrow \lim _{\mathrm{x} \rightarrow 2} \frac{\mathrm{x}(\mathrm{x}-2)+\mathrm{p}(\mathrm{x}-2)}{\mathrm{x}-2}=5$ $\Rightarrow 2+\mathrm{p}=5$ $\Rightarrow \mathrm{p}=3$ Now, quadratic equation is $\mathrm{x}^{2}-3 \mathrm{x}+\mathrm{q}=0$ its both roots lie in $(0,2)$ $\mathrm{D} \geq 0, 0<\frac{-\mathrm{b}}{2 \mathrm{a}}<2, \mathrm{f}(0)>0, \mathrm{f}(2)>0$ $\Rightarrow 9-4 \mathrm{q} \geq 0$ and $\mathrm{q}>2$ $\Rightarrow \mathrm{q} \in\left(2, \frac{9}{4}\right]$ $\alpha=2, \beta=\frac{9}{4} \Rightarrow\left(2+\frac{9}{4}\right)=17$