Given `f (x) =(1)/( sqrt"" (|x| -x))` , `g (x) = (1)/( sqrt""(x-|x|))` then

Options

A. f(x) has some domain and g(x) has no domain
B. f(x) has no domain and g(x) has some domain
C. f (x) and g(x) have the same domain .
D. f (x) and g(x) do not have any domain

Answer

To solve the problem, we need to determine the domains of the functions \( f(x) \) and \( g(x) \).

### Step 1: Determine the Domain of \( f(x) \)

The function is defined as:
\[ f(x) = \frac{1}{\sqrt{|x| - x}} \]
For \( f(x) \) to be defined, the expression inside the square root must be greater than zero:
\[ |x| - x > 0 \]

#### Case 1: \( x \geq 0 \)
If \( x \geq 0 \), then \( |x| = x \). Thus,
\[ |x| - x = x - x = 0 \]
This does not satisfy the inequality \( |x| - x > 0 \).

#### Case 2: \( x < 0 \)
If \( x < 0 \), then \( |x| = -x \). Thus,
\[ |x| - x = -x - x = -2x \]
We need:
\[ -2x > 0 \implies x < 0 \]
This inequality is satisfied for all \( x < 0 \).

### Conclusion for \( f(x) \)
The domain of \( f(x) \) is:
\[ \text{Domain of } f(x) = (-\infty, 0) \]

### Step 2: Determine the Domain of \( g(x) \)

The function is defined as:
\[ g(x) = \frac{1}{\sqrt{x - |x|}} \]
For \( g(x) \) to be defined, the expression inside the square root must be greater than zero:
\[ x - |x| > 0 \]

#### Case 1: \( x \geq 0 \)
If \( x \geq 0 \), then \( |x| = x \). Thus,
\[ x - |x| = x - x = 0 \]
This does not satisfy the inequality \( x - |x| > 0 \).

#### Case 2: \( x < 0 \)
If \( x < 0 \), then \( |x| = -x \). Thus,
\[ x - |x| = x - (-x) = x + x = 2x \]
We need:
\[ 2x > 0 \implies x > 0 \]
This inequality cannot be satisfied since we are considering \( x < 0 \).

### Conclusion for \( g(x) \)
The domain of \( g(x) \) is:
\[ \text{Domain of } g(x) = \emptyset \quad (\text{no real values of } x) \]

### Final Conclusion
- The domain of \( f(x) \) is \( (-\infty, 0) \).
- The domain of \( g(x) \) is empty (no values of \( x \) make \( g(x) \) defined).

### Answer
Thus, the correct statement is that \( f(x) \) has a domain, while \( g(x) \) does not have a domain.

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