JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Limits

$a$ is the positive root of $2x^2+x-2=0$: $a=\frac{-1+\sqrt{17}}{4}$, so $\frac{1}{a}=\frac{1+\sqrt{17}}{4}$.
The quadratic $2x^2-x-2=0$ has roots $\frac{1\pm \sqrt{17}}{4}$ — note $\frac{1}{a}$ is exactly one of these roots. So

As $x\to \frac{1}{a}$, this argument $\to 0$, so $1-\cos (\cdot )\sim \frac{(\cdot {)}^2}{2}$, giving

Also $(1-ax{)}^2=a^2{\left(x-\frac{1}{a}\right)}^2$. The ${\left(x-\frac{1}{a}\right)}^2$ factors cancel in the limit:

$\frac{1}{a}-r=\frac{(1+\sqrt{17})-(1-\sqrt{17})}{4}=\frac{\sqrt{17}}{2}$, so ${\left(\frac{1}{a}-r\right)}^2=\frac{17}{4}$, and $a^2=\frac{18-2\sqrt{17}}{16}=\frac{9-\sqrt{17}}{8}$.

So $\alpha =153,\beta =17$, and $\alpha +\beta =170$.