Domain of
`f(x) = sqrt(2x-1) + sqrt(13) cos^(-1) ((2x-1)/2)` is

Options

A. `[1/2, 3/2]`
B. `{1/2}`
C. `[1/2,1]`
D. `[1/2, sqrt(3)/2]`

Answer

To find the domain of the function \( f(x) = \sqrt{2x - 1} + \sqrt{13} \cos^{-1} \left( \frac{2x - 1}{2} \right) \), we need to ensure that both components of the function are defined and return real values.

### Step 1: Determine the condition for the square root
The first part of the function is \( \sqrt{2x - 1} \). For the square root to be defined and return a real value, the expression inside the square root must be non-negative:
\[ 2x - 1 \geq 0 \]
Solving this inequality:
\[ 2x \geq 1 \\ x \geq \frac{1}{2} \]

### Step 2: Determine the condition for the inverse cosine
The second part of the function is \( \cos^{-1} \left( \frac{2x - 1}{2} \right) \). The argument of the inverse cosine function must be in the range \([-1, 1]\):
\[ -1 \leq \frac{2x - 1}{2} \leq 1 \]
We will solve this compound inequality in two parts.

#### Part A: Solve the left side of the inequality
\[ -1 \leq \frac{2x - 1}{2} \]
Multiplying both sides by 2:
\[ -2 \leq 2x - 1 \\ -1 \leq 2x \\ -\frac{1}{2} \leq x \]

#### Part B: Solve the right side of the inequality
\[ \frac{2x - 1}{2} \leq 1 \]
Multiplying both sides by 2:
\[ 2x - 1 \leq 2 \\ 2x \leq 3 \\ x \leq \frac{3}{2} \]

### Step 3: Combine the conditions
Now we have two conditions from the inequalities:
1. \( x \geq \frac{1}{2} \)
2. \( x \leq \frac{3}{2} \)

To find the domain, we take the intersection of these two conditions:
\[ \frac{1}{2} \leq x \leq \frac{3}{2} \]

### Conclusion
Thus, the domain of the function \( f(x) \) is:
\[ \boxed{\left[ \frac{1}{2}, \frac{3}{2} \right]} \]