If the domain of the function f (x) = ^ - 1 x ^ 2 - x + 1 ^ - 1...

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

For, $\sin^{- 1}x \text{ and } \cos^{1}x$ to be defined $- 1 \le x \le 1 and for$\sqrt{f\left(x\right)}$to be defined$f\left(x\right) \ge 0$and the denominator of any expression can never be zero.</p><p>Thus, to define the domain of the given function, we have </p><p>$- 1 \le \sqrt{{x}^{2} - x + 1} \le 1, {x}^{2} - x + 1 \ge 0$and$\sin^{- 1}\left(\dfrac{2x - 1}{2}\right) > 0$</p><p>Taking the common values and using the range of$\sin^{- 1}x$i.e. for positive values$0 < \sin^{- 1}x \le \dfrac{\pi}{2},$we get</p><p>$0 \le {x}^{2} - x + 1 \le 1$and$0 < \sin^{- 1}\left(\dfrac{2x - 1}{2}\right) \le \dfrac{\pi}{2}$</p><p>$⇒ {x}^{2} - x \le 0$and$0 < \dfrac{2x - 1}{2} \le 1$</p><p>$⇒ x\left(x - 1\right) \le 0$and$0 < 2x - 1 \le 2$</p><p>$⇒ x \in [ 0, 1 ]$and$1 < 2x \le 3$</p><p>$⇒ x \in \left[0, 1\right]$and$\dfrac{1}{2} < x \le \dfrac{3}{2}$</p><p>Now, taking intersection, we get$x \in \left(\dfrac{1}{2}, 1\right]$</p><p>Given, domain of the function is \left(\alpha, \beta\right] = \left(\dfrac{1}{2}, 1\right]$

$⇒ \alpha = \dfrac{1}{2}, \beta = 1$

$⇒ \alpha + \beta = \dfrac{3}{2} .$