If the domain of the function  $f\left(x\right)=\frac{{\cos }^{-1}\sqrt{x^2-x+1}}{\sqrt{{\sin }^{-1}\left(\frac{2x-1}{2}\right)}}\text{\hspace{0.17em}is}$ the interval  $(a,b]$ then $a+b$  is equal to :

$$\begin{gathered} 0\underset{\_}{<}{x}^2-x+1\underset{\_}{<}1 \\ \Rightarrow x^2-x\underset{\_}{<}0 \\ \Rightarrow x\in [0,1] \\ Also 0<{\sin }^{-1}\left(\frac{2x-1}{2}\right)\underset{\_}{<}\frac{\pi }{2} \dots ..\left(i\right) \\ \Rightarrow 0<\frac{2x-1}{2}\underset{\_}{<}1 \\ \Rightarrow 0<2x-1\underset{\_}{<}2 \\ 1<2x\underset{\_}{<}3 \\ \frac{1}{2}<x\underset{\_}{<}\frac{3}{2}\dots ..\left(ii\right) \end{gathered}$$

From (i) and (ii) we get

$x\in (\frac{1}{2},1]$

a=1/2
b=1