The domain of the function `psi(x)=1/x+2^(sin^- 1x) +1/(sqrt(x-2))` is

Options

A. [-1,1]
B. R-{0}
C. `[-1,0) cup (0,1]`
D. none of these

Explanation

To find the domain of the function \( \psi(x) = \frac{1}{x} + 2^{\sin^{-1}(x)} + \frac{1}{\sqrt{x-2}} \), we need to analyze each component of the function separately and determine the restrictions on \( x \).

### Step 1: Analyze \( \frac{1}{x} \)
The function \( \frac{1}{x} \) is defined for all real numbers except \( x = 0 \). Thus, the domain for this part is:
\[ \text{Domain of } \frac{1}{x} = \mathbb{R} \setminus \{0\} \]

**Hint:** Remember that division by zero is undefined.

### Step 2: Analyze \( 2^{\sin^{-1}(x)} \)
The function \( \sin^{-1}(x) \) (the inverse sine function) is defined for \( x \) in the interval \([-1, 1]\). Therefore, the domain for this part is:
\[ \text{Domain of } 2^{\sin^{-1}(x)} = [-1, 1] \]

**Hint:** The inverse sine function has a limited range; ensure \( x \) falls within that range.

### Step 3: Analyze \( \frac{1}{\sqrt{x-2}} \)
The expression \( \sqrt{x-2} \) is defined when \( x - 2 > 0 \), which simplifies to \( x > 2 \). Additionally, since it is in the denominator, we also need to ensure that it is not equal to zero. Thus, the domain for this part is:
\[ \text{Domain of } \frac{1}{\sqrt{x-2}} = (2, \infty) \]

**Hint:** The square root function requires the argument to be positive, and we cannot divide by zero.

### Step 4: Find the intersection of the domains
Now, we need to find the intersection of the three domains:
1. \( \mathbb{R} \setminus \{0\} \)
2. \( [-1, 1] \)
3. \( (2, \infty) \)

The intersection of these three sets can be analyzed:
- The first domain excludes \( 0 \) but includes all other real numbers.
- The second domain is limited to values between \(-1\) and \(1\).
- The third domain only includes values greater than \(2\).

Since there are no values that satisfy all three conditions simultaneously, the intersection is empty.

### Conclusion
Thus, the domain of the function \( \psi(x) \) is:
\[ \text{Domain of } \psi(x) = \emptyset \]

**Final Answer:** The domain of the function \( \psi(x) \) is empty, meaning there are no values of \( x \) for which the function is defined.

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