If ` y= (sqrt(x^2 -4))/(cos^(-1) (2-x))` then the domain of y is

Options

A. `[2,4]`
B. `[1,2]`
C. `[2,3]`
D. `[-2,2]`

Answer

To find the domain of the function \( y = \frac{\sqrt{x^2 - 4}}{\cos^{-1}(2 - x)} \), we need to ensure that both the numerator and the denominator are defined and valid.

### Step 1: Analyze the numerator \( \sqrt{x^2 - 4} \)

For the square root to be defined, the expression inside the square root must be non-negative:

\[ x^2 - 4 \geq 0 \]

This can be rearranged to:

\[ x^2 \geq 4 \]

Taking the square root of both sides gives us:

\[ |x| \geq 2 \]

This means:

\[ x \leq -2 \quad \text{or} \quad x \geq 2 \]

### Step 2: Analyze the denominator \( \cos^{-1}(2 - x) \)

The function \( \cos^{-1}(x) \) is defined for \( x \) in the interval \([-1, 1]\). Therefore, we need:

\[ -1 \leq 2 - x \leq 1 \]

This can be split into two inequalities:

1. \( 2 - x \geq -1 \)
2. \( 2 - x \leq 1 \)

#### Solving the first inequality:

\[ 2 - x \geq -1 \implies x \leq 3 \]

#### Solving the second inequality:

\[ 2 - x \leq 1 \implies x \geq 1 \]

### Step 3: Combine the results

Now we have two conditions:

1. From the numerator: \( x \leq -2 \) or \( x \geq 2 \)
2. From the denominator: \( 1 \leq x \leq 3 \)

Now, we need to find the intersection of these conditions:

- The interval \( x \geq 2 \) intersects with \( 1 \leq x \leq 3 \) to give \( 2 \leq x \leq 3 \).
- The interval \( x \leq -2 \) does not intersect with \( 1 \leq x \leq 3 \).

### Final Domain

Thus, the domain of \( y \) is:

\[ \text{Domain of } y = [2, 3] \]