JEE Main · 2026 · 06 Apr 2026 · Shift 2 · Limits

For the limit to be finite, $\frac{\tan (x-2)}{x-2}\to 1$ forces $\frac{r{x}^2+(p-2)x-2p}{x-2}$ to have a finite limit, so the numerator must vanish at $x=2$: $4r+2(p-2)-2p=4r-4=0$, giving $r=1$. Then $x^2+(p-2)x-2p=(x-2)(x+p)$, and ${\lim }_{x\to 2}(x+p)=2+p=5$, so $p=3$. The equation $x^2-3x+q=0$ has both roots in $(0,2)$ when the discriminant is non-negative, $9-4q\ge 0\Rightarrow q\le \frac{9}{4}$; when $f(0)=q>0$; and when $f(2)=q-2>0\Rightarrow q>2$ (the vertex $x=\frac{3}{2}$ already lies in the interval). Hence $q\in (2,\frac{9}{4}]$, so $\alpha =2,\beta =\frac{9}{4}$ and $4(\alpha +\beta )=4\cdot \frac{17}{4}=17$.