The domain of the function
`f(x)=sin^(-1)""(1)/abs(x^(2)-1)+1/sqrt(sin^(2)x+sinx+1)` is

Options

A. a) `(-infty, infty)`
B. b) `(-infty,-sqrt(2)] cup [sqrt(2), infty)`
C. c) `(-infty,-sqrt(2)] cup [sqrt(2),infty) cup {0}`
D. d) None of the above

Answer

To find the domain of the function

\[ f(x) = \sin^{-1}\left(\frac{1}{|x^2 - 1|}\right) + \frac{1}{\sqrt{\sin^2 x + \sin x + 1}} \]

we need to ensure that both components of the function are defined.

### Step 1: Analyze the first term \( \sin^{-1}\left(\frac{1}{|x^2 - 1|}\right) \)

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

\[ -\frac{1}{|x^2 - 1|} \leq 1 \quad \text{and} \quad \frac{1}{|x^2 - 1|} \geq 0 \]

Since \(\frac{1}{|x^2 - 1|} \geq 0\) is always true when \( |x^2 - 1| > 0\), we only need to consider:

\[ |x^2 - 1| \geq 1 \]

### Step 2: Solve the inequality \( |x^2 - 1| \geq 1 \)

This leads to two cases:

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

#### Case 1: \( x^2 - 1 \geq 1 \)

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

#### Case 2: \( x^2 - 1 \leq -1 \)

\[ x^2 \leq 0 \implies x = 0 \]

### Step 3: Combine the results from both cases

From Case 1, we have:

\[ x \in (-\infty, -\sqrt{2}] \cup [\sqrt{2}, \infty) \]

From Case 2, we have:

\[ x = 0 \]

### Step 4: Check the second term \( \frac{1}{\sqrt{\sin^2 x + \sin x + 1}} \)

The expression \( \sin^2 x + \sin x + 1 \) is always positive because:

- The minimum value of \( \sin^2 x \) is \( 0 \) (when \( \sin x = 0 \)).
- The minimum value of \( \sin x \) is \( -1 \) (when \( \sin x = -1 \)).

Thus, the minimum value of \( \sin^2 x + \sin x + 1 \) occurs when \( \sin x = -1 \):

\[ \sin^2 x + \sin x + 1 = 0 + (-1) + 1 = 1 > 0 \]

This means \( \frac{1}{\sqrt{\sin^2 x + \sin x + 1}} \) is always defined.

### Final Domain

Combining both parts, the domain of \( f(x) \) is:

\[ (-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty) \cup \{0\} \]

### Conclusion

Thus, the domain of the function \( f(x) \) is:

\[ \boxed{(-\infty, -\sqrt{2}) \cup (\sqrt{2}, \infty) \cup \{0\}} \]