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

We need to find the domain of $$f(x) = \frac{\cos^{-1}\sqrt{x^2 - x + 1}}{\sqrt{\sin^{-1}\left(\frac{2x-1}{2}\right)}}$$.

Condition 1 (square root inside cos inverse): We need $$x^2 - x + 1 \geq 0$$. The discriminant is $$1 - 4 = -3 < 0$$ with positive leading coefficient, so this is always positive. Always satisfied.

Condition 2 (argument of cos inverse in $$[-1,1]$$): Since $$\sqrt{x^2 - x + 1} \geq 0$$, we need $$\sqrt{x^2 - x + 1} \leq 1$$, i.e., $$x^2 - x + 1 \leq 1$$, so $$x^2 - x \leq 0$$, giving $$x(x-1) \leq 0$$, i.e., $$0 \leq x \leq 1$$.

Condition 3 (argument of sin inverse in $$[-1,1]$$): $$-1 \leq \frac{2x-1}{2} \leq 1$$ gives $$-\frac{1}{2} \leq x \leq \frac{3}{2}$$.

Condition 4 (denominator positive): $$\sin^{-1}\left(\frac{2x-1}{2}\right) > 0$$. Since $$\sin^{-1}(t) > 0$$ iff $$t > 0$$, we need $$\frac{2x-1}{2} > 0$$, i.e., $$x > \frac{1}{2}$$.

Intersecting all conditions: $$0 \leq x \leq 1$$ and $$x > \frac{1}{2}$$ gives $$\frac{1}{2} < x \leq 1$$, i.e., the interval $$\left(\frac{1}{2}, 1\right]$$.

So $$\alpha = \frac{1}{2}$$ and $$\beta = 1$$, giving $$\alpha + \beta = \frac{3}{2}$$.

The answer is $$\frac{3}{2}$$, which is Option A.