JEE Main 2023 (Online) 8th April Evening ShiftQuestion 1
If $\alpha>\beta>0$ are the roots of the equation $ax^{2}+bx+1=0$, and $\lim_{x\to\frac{1}{\alpha}}\left(\frac{1-\cos\left(x^{2}+bx+a\right)}{2(1-\alpha x)^{2}}\right)^{\frac{1}{2}}=\frac{1}{k}\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$, then $k$ is equal to
Limits & Continuity
Solution
Roots of $ax^2+bx+1=0$ are $\alpha,\beta$, so $y^2+by+a=0$ has roots $\frac1\alpha,\frac1\beta$, i.e. $x^2+bx+a=(x-\frac1\alpha)(x-\frac1\beta)$. Near $x=\frac1\alpha$: $1-\cos\theta\sim\frac{\theta^2}2$ gives the inside $\sim\frac{(x-\frac1\alpha)^2(x-\frac1\beta)^2}{4\alpha^2(x-\frac1\alpha)^2}\to\frac{(\frac1\alpha-\frac1\beta)^2}{4\alpha^2}$ (using $1-\alpha x=-\alpha(x-\frac1\alpha)$). Square root: $\frac{|1/\alpha-1/\beta|}{2\alpha}=\frac1{2\alpha}\left(\frac1\beta-\frac1\alpha\right)$ since $\alpha>\beta>0$. So $k=2\alpha$. Option (C).