JEE Main 2023 — Limits Question with Solution

Since, $\alpha$ and $\beta$ are the roots of the equation $a{x}^2+bx+1=0$, therefore $\frac{1}{\alpha }$ and $\frac{1}{\beta }$ would be the roots of $x^2+bx+a=0$.

Let

$L=\underset{x\to \frac{1}{\alpha }}{\lim }{\left[\frac{1-\cos \left(x^2+bx+a\right)}{2{\left(1-\alpha x\right)}^2}\right]}^{\frac{1}{2}}\to \frac{0}{0}$

$\Rightarrow L=\underset{x\to \frac{1}{\alpha }}{\lim }{\left[\frac{2{\sin }^2\left({\displaystyle \frac{x^2+bx+a}{2}}\right)}{2{\left(1-\alpha x\right)}^2}\right]}^{\frac{1}{2}}$

$\Rightarrow L=\underset{x\to \frac{1}{\alpha }}{\lim }{\left[\frac{{\sin }^2\left({\displaystyle \frac{1}{2}\left(x-\frac{1}{\alpha }\right)\left(x-\frac{1}{\beta }\right)}\right)}{{\alpha }^2{\left(x-{\displaystyle \frac{1}{\alpha }}\right)}^2}\right]}^{\frac{1}{2}}$

$\Rightarrow L=\underset{x\to \frac{1}{\alpha }}{\lim }{\left[\frac{{\left(x-\frac{1}{\beta }\right)}^2}{4{\alpha }^2}\times \frac{{\left\{\sin \left({\displaystyle \frac{1}{2}\left(x-\frac{1}{\alpha }\right)\left(x-\frac{1}{\beta }\right)}\right)\right\}}^2}{{\left({\displaystyle \frac{1}{2}}\right)}^2{\left(x-{\displaystyle \frac{1}{\alpha }}\right)}^2{\left(x-\frac{1}{\beta }\right)}^2}\right]}^{\frac{1}{2}}$

$\Rightarrow L=\frac{\left({\displaystyle \frac{1}{\alpha }}-{\displaystyle \frac{1}{\beta }}\right)}{2\alpha }$

On comparing this value with the value given in the question we get, $k=2\alpha$.

Hence this is the correct option.