Given function is $f\left(x\right)=\frac{{\cos }^{-1}\sqrt{x^2-x+1}}{\sqrt{{\sin }^{-1}\left({\displaystyle \frac{2x-1}{2}}\right)}}$

For, ${\sin }^{-1}x \& {\cos }^{1}x$ to be defined $-1\le x\le 1$ and for $\sqrt{f\left(x\right)}$ to be defined $f\left(x\right)\ge 0$ and the denominator of any expression can never be zero.

Thus, to define the domain of the given function, we have 

$-1\le \sqrt{x^2-x+1}\le 1, x^2-x+1\ge 0$ and ${\sin }^{-1}\left(\frac{2x-1}{2}\right)>0$

Taking the common values and using the range of ${\sin }^{-1}x$ i.e. for positive values $0<{\sin }^{-1}x\le \frac{\mathrm{\pi }}{2},$ we get

$0\le x^2-x+1\le 1$ and $0<{\sin }^{-1}\left(\frac{2x-1}{2}\right)\le \frac{\pi }{2}$

$\Rightarrow x^2-x\le 0$ and $0<\frac{2x-1}{2}\le 1$

$\Rightarrow x\left(x-1\right)\le 0$ and $0<2x-1\le 2$

$\Rightarrow x\in [0, 1]$ and $1<2x\le 3$

$\Rightarrow x\in \left[0, 1\right]$ and $\frac{1}{2}<x\le \frac{3}{2}$

Now, taking intersection, we get $x\in \left(\frac{1}{2}, 1\right]$

Given, domain of the function is $\left(\alpha , \beta \right]=\left(\frac{1}{2}, 1\right]$

$\Rightarrow \alpha =\frac{1}{2}, \beta =1$

$\Rightarrow \alpha +\beta =\frac{3}{2}.$