JEE Main 2026 (Online) 6th April Evening ShiftQuestion 1
Let $\lim_{x\to2}\frac{(\tan(x-2))\left(rx^{2}+(p-2)x-2p\right)}{(x-2)^{2}}=5$ for some $r,p\in\mathbf{R}$. If the set of all possible values of $q$, such that the roots of the equation $rx^{2}-px+q=0$ lie in $(0,2)$, be the interval $(\alpha,\beta]$, then $4(\alpha+\beta)$ equals :
Limits & Continuity
Solution
$\tan(x-2)\sim(x-2)$, so $\frac{rx^2+(p-2)x-2p}{x-2}\to5$ needs numerator zero at $x=2$: $4r+2p-4-2p=0\Rightarrow r=1$. Then $x^2+(p-2)x-2p=(x-2)(x+p)$, limit $=2+p=5\Rightarrow p=3$. Roots of $x^2-3x+q=0$ in $(0,2)$: $D\ge0\Rightarrow q\le\frac94$; $f(0)>0\Rightarrow q>0$; $f(2)>0\Rightarrow q>2$; vertex $\frac32\in(0,2)$. So $q\in(2,\frac94]$, $4(\alpha+\beta)=4\cdot\frac{17}{4}=17$: option C.