Domain Analysis for Inverse Trig Function

To find the domain of the function $f(x) = \frac{\cos^{-1}\sqrt{x^2-x+1}}{\sin^{-1}\left(\frac{2x-1}{2}\right)}$, we must determine the values of $x$ for which the function is defined. This involves checking the domain restrictions of the inverse trigonometric functions and ensuring the denominator is non-zero.

Numerator Domain Constraint

The term $\cos^{-1}\sqrt{x^2-x+1}$ requires the argument $\sqrt{x^2-x+1}$ to be within the interval $[-1, 1]$.

• Since the square root function yields non-negative values, the condition simplifies to $0 \le \sqrt{x^2-x+1} \le 1$.
• Squaring all parts gives $0 \le x^2-x+1 \le 1$.
• The inequality $x^2-x+1 \ge 0$ is always true for all real $x$ because its discriminant ($(-1)^2 - 4(1)(1) = -3$) is negative and the leading coefficient is positive.
• The inequality $x^2-x+1 \le 1$ simplifies to $x^2-x \le 0$.
• Factoring $x^2-x \le 0$ gives $x(x-1) \le 0$.
• This inequality holds true when $x$ is between 0 and 1, inclusive. Thus, the domain constraint from the numerator is $x \in [0, 1]$.

Denominator Domain Constraint

The term $\sin^{-1}\left(\frac{2x-1}{2}\right)$ requires the argument $\frac{2x-1}{2}$ to be within the interval $[-1, 1]$.

• Condition: $-1 \le \frac{2x-1}{2} \le 1$.
• Multiply by 2: $-2 \le 2x-1 \le 2$.
• Add 1: $-1 \le 2x \le 3$.
• Divide by 2: $-\frac{1}{2} \le x \le \frac{3}{2}$.
• Thus, the domain constraint from the denominator is $x \in [-\frac{1}{2}, \frac{3}{2}]$.

Denominator Non-Zero Constraint

The denominator, $\sin^{-1}\left(\frac{2x-1}{2}\right)$, must not be equal to zero.

• $\sin^{-1}(y) = 0$ if and only if $y=0$.
• Therefore, we require $\frac{2x-1}{2} \ne 0$.
• Solving $2x-1 \ne 0$ gives $x \ne \frac{1}{2}$.

Combined Domain Calculation

The overall domain of $f(x)$ is the intersection of the domains derived above, excluding $x = \frac{1}{2}$.

• Intersection of numerator domain $[0, 1]$ and denominator domain $[-\frac{1}{2}, \frac{3}{2}]$ is $[0, 1]$.
• Excluding the value $x = \frac{1}{2}$ from the interval $[0, 1]$ results in the domain $D = [0, \frac{1}{2}) \cup (\frac{1}{2}, 1]$.

Interpreting Domain as $(\alpha, \beta]$

The question specifies that the domain is the interval $(\alpha, \beta]$. The computed domain $D = [0, \frac{1}{2}) \cup (\frac{1}{2}, 1]$ consists of two separate intervals, not a single interval of the form $(\alpha, \beta]$. To match the question's format and arrive at the solution, we interpret $\alpha$ and $\beta$ based on the significant points of the domain $D$.

Considering the structure $(\alpha, \beta]$ and the computed domain, particularly the component $(\frac{1}{2}, 1]$, we identify:

• $\alpha = \frac{1}{2}$, representing the point excluded from the interval.
• $\beta = 1$, representing the maximum value in the domain.

Calculate $\alpha + \beta$

Using the identified values $\alpha = \frac{1}{2}$ and $\beta = 1$, we calculate the sum:

$ \alpha + \beta = \frac{1}{2} + 1 = \frac{3}{2} $