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Step-by-step Solution
Step 1: Analyze the expression inside $ \cos^{-1} $
We have $ \cos^{-1}\bigl(\sqrt{x^2 - x + 1}\bigr) $. For $ \cos^{-1}(z) $ to be defined, we need
$ -1 \le z \le 1 $. Here,
$ z = \sqrt{x^2 - x + 1} \ge 0 $.
Hence the relevant condition is:
$ 0 \le x^2 - x + 1 \le 1 $.
Solving the inequality $ x^2 - x + 1 \le 1 $:
$ x^2 - x \le 0 $
$ x(x - 1) \le 0 $
This implies
$ x \in [0, 1] $.
Step 2: Analyze the expression inside $ \sqrt{\sin^{-1}\bigl(\frac{2x-1}{2}\bigr)} $
We need $ \sin^{-1}\bigl(\tfrac{2x-1}{2}\bigr) $ to be in the domain of $ \sin^{-1} $ and also to remain positive for the square-root to be real and non-zero. So:
1) $ \frac{2x - 1}{2} $ must lie within $ [-1, 1] $ for $ \sin^{-1}\bigl(\tfrac{2x - 1}{2}\bigr) $ to be defined.
2) We also need $ \sin^{-1}\bigl(\tfrac{2x - 1}{2}\bigr) > 0 $ for the square-root to be real and non-zero.
From $ \sin^{-1}\bigl(\tfrac{2x - 1}{2}\bigr) > 0 $, we get:
$ \frac{2x - 1}{2} > 0 \quad \Rightarrow \quad 2x - 1 > 0 \quad \Rightarrow \quad 2x > 1 \quad \Rightarrow \quad x > \frac{1}{2} $.
Additionally, $ \frac{2x - 1}{2} \le 1 \quad \Rightarrow \quad 2x - 1 \le 2 \quad \Rightarrow \quad 2x \le 3 \quad \Rightarrow \quad x \le \frac{3}{2} $.
Combining these, from the positivity condition we get:
$ x \in \left(\frac{1}{2}, \frac{3}{2}\right] $.
Step 3: Find the common interval for $ x $
We intersect the two intervals:
1) $ x \in [0, 1] $
2) $ x \in \bigl(\tfrac{1}{2}, \tfrac{3}{2}\bigr] $
The common portion is:
$ x \in \Bigl(\tfrac{1}{2}, 1\Bigr] $.
Therefore, the domain of the function is $ \bigl(\alpha, \beta \bigr] $ where $ \alpha = \tfrac{1}{2} $ and $ \beta = 1 $.
Step 4: Sum of the bounds of the domain
We have
$ \alpha + \beta = \tfrac{1}{2} + 1 = \tfrac{3}{2} $.
Hence, the required value of $ \alpha + \beta $ is
$ \tfrac{3}{2} $.
