JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Inverse Trigonometric Functions

The inverse cosine needs $0\le \sqrt{x^2-x+1}\le 1$. Since $x^2-x+1={\left(x-\frac{1}{2}\right)}^2+\frac{3}{4}>0$ always, the binding requirement is $x^2-x+1\le 1$, that is $x(x-1)\le 0$, so $0\le x\le 1$. The inverse sine needs $-1\le \frac{2x-1}{2}\le 1$, giving $-\frac{1}{2}\le x\le \frac{3}{2}$. The outer square root sits in the denominator, so its argument must be strictly positive: ${\sin }^{-1}\left(\frac{2x-1}{2}\right)>0$ requires $\frac{2x-1}{2}>0$, that is $x>\frac{1}{2}$. Intersecting the three conditions leaves $\left(\frac{1}{2},1\right]$, so $\alpha =\frac{1}{2}$, $\beta =1$ and $\alpha +\beta =\frac{3}{2}$.