The set of values of x for which the function
`f(x)=(1)/(x)+2^(sin^(-1)x)+(1)/(sqrt(x-2))` exists is

Options

A. R
B. R-{0}
C. `phi`
D. none of these

Answer

To determine the set of values of \( x \) for which the function

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

exists, we need to analyze each term in the function separately and find the intersection of their domains.

### Step 1: Analyze the first term \( f_1(x) = \frac{1}{x} \)

The function \( \frac{1}{x} \) is defined for all real numbers except \( x = 0 \). Therefore, the domain for this term is:

\[ x \in \mathbb{R} \setminus \{0\} \]

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

### Step 2: Analyze the second term \( f_2(x) = 2^{\sin^{-1}(x)} \)

The function \( \sin^{-1}(x) \) (the inverse sine function) is defined for \( x \) in the interval \([-1, 1]\). Thus, the domain for this term is:

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

**Hint:** The inverse sine function has a restricted domain.

### Step 3: Analyze the third term \( f_3(x) = \frac{1}{\sqrt{x-2}} \)

The term \( \sqrt{x-2} \) is defined only when \( x - 2 > 0 \) (since we cannot take the square root of a negative number). Therefore, we have:

\[ x > 2 \]

This means the domain for this term is:

\[ x \in (2, \infty) \]

**Hint:** The expression under the square root must be positive.

### Step 4: Find the intersection of the domains

Now we need to find the intersection of the domains obtained from the three terms:

1. From \( f_1(x) \): \( x \in \mathbb{R} \setminus \{0\} \)
2. From \( f_2(x) \): \( -1 \leq x \leq 1 \)
3. From \( f_3(x) \): \( x > 2 \)

The first domain excludes \( 0 \), the second domain is limited to values between \(-1\) and \(1\), and the third domain only includes values greater than \(2\).

Since there is no overlap among these intervals (the second domain is entirely below \(2\) and the third domain is entirely above \(2\)), we conclude that there are no values of \( x \) that satisfy all three conditions simultaneously.

### Conclusion

Thus, the set of values of \( x \) for which the function \( f(x) \) exists is the null set, denoted as:

\[ \text{The set of values of } x \text{ is } \emptyset \]

### Final Answer

The correct option is \( \text{C: null set} \).

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