JEE Main 2026 — Limits Question with Solution

Given limit is $\underset{x\to 2}{\lim }\frac{\tan (x-2)}{x-2}\cdot \frac{r{x}^2+(p-2)x-2p}{x-2}=5$

Since $\underset{x\to 2}{\lim }\frac{\tan (x-2)}{x-2}=1$, we have:

$\underset{x\to 2}{\lim }\frac{r{x}^2+(p-2)x-2p}{x-2}=5$

For the limit to exist, the numerator must be zero at $x=2$:

$r(2{)}^2+(p-2)(2)-2p=0$

$4r+2p-4-2p=0\Rightarrow 4r=4\Rightarrow r=1$

Substituting $r=1$ into the limit expression:

$\underset{x\to 2}{\lim }\frac{x^2+(p-2)x-2p}{x-2}=5$

$\underset{x\to 2}{\lim }\frac{(x-2)(x+p)}{x-2}=5$

$\underset{x\to 2}{\lim }(x+p)=5\Rightarrow 2+p=5\Rightarrow p=3$

The quadratic equation is $r{x}^2-px+q=0$, which becomes $x^2-3x+q=0$.

For both roots of $f(x)=x^2-3x+q=0$ to lie in the interval $(0,2)$, the following conditions must hold:

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

2) $f(0)>0\Rightarrow q>0$

3) $f(2)>0\Rightarrow 2^2-3(2)+q>0\Rightarrow q-2>0\Rightarrow q>2$

4) The $x$-coordinate of the vertex must lie in $(0,2)\Rightarrow 0<\frac{-(-3)}{2(1)}<2\Rightarrow 0<1.5<2$, which is true.

Taking the intersection of all conditions, we get $q\in (2,\frac{9}{4}]$.

Comparing this with $(\alpha ,\beta ]$, we have $\alpha =2$ and $\beta =\frac{9}{4}$.

Therefore, $4(\alpha +\beta )=4(2+\frac{9}{4})=8+9=17$.

Answer: $17$