JEE Main 2026 · 6 Apr · Shift 2, Mathematics question 16

The limit. $\dfrac{\tan(x-2)}{x-2}\to1$, so the quadratic must vanish at 2: $4r-4=0$ gives $r=1$. Then $x^2+(p-2)x-2p=(x-2)(x+p)$, and the limit is $2+p=5$, so $p=3$.

Roots of $x^2-3x+q=0$ in (0, 2).

• Real roots: $q\le\tfrac94$.
• $f(0)=q>0$ and $f(2)=q-2>0$.
• Vertex $\tfrac32\in(0,2)$.

So $q\in\left(2,\tfrac94\right]$ and $4(\alpha+\beta)=4\cdot\tfrac{17}4=17$.

(C)