Let `f(x)=(1)/(x) and g(x)=(1)/(sqrt(x))`. Then,

Options

A. f(g(x)) and g(f(x)) have different domains
B. f(g(x)) and g(f(x)) have same domain
C. g(f(x)) is a bijective mapping
D. f(g(x)) is neither odd or even.

Answer

To solve the problem, we need to analyze the functions \( f(x) = \frac{1}{x} \) and \( g(x) = \frac{1}{\sqrt{x}} \), and then evaluate the compositions \( f(g(x)) \) and \( g(f(x)) \).

### Step 1: Find \( f(g(x)) \)

1. **Substitute \( g(x) \) into \( f(x) \)**:
\[ f(g(x)) = f\left(\frac{1}{\sqrt{x}}\right) = \frac{1}{\frac{1}{\sqrt{x}}} = \sqrt{x} \]

### Step 2: Find \( g(f(x)) \)

2. **Substitute \( f(x) \) into \( g(x) \)**:
\[ g(f(x)) = g\left(\frac{1}{x}\right) = \frac{1}{\sqrt{\frac{1}{x}}} = \sqrt{x} \]

### Step 3: Determine the domains of \( f(g(x)) \) and \( g(f(x)) \)

3. **Domain of \( f(g(x)) = \sqrt{x} \)**:
- The function \( g(x) = \frac{1}{\sqrt{x}} \) is defined for \( x > 0 \) (since \( \sqrt{x} \) must be real and positive).
- Therefore, \( f(g(x)) \) is defined for \( x > 0 \).

4. **Domain of \( g(f(x)) = \sqrt{x} \)**:
- The function \( f(x) = \frac{1}{x} \) is defined for \( x \neq 0 \).
- Since \( g(f(x)) \) is also \( \sqrt{x} \), it is defined for \( x > 0 \).

### Step 4: Check if \( g(f(x)) \) is a bijective function

5. **Check if \( g(f(x)) = \sqrt{x} \) is one-to-one**:
- Let \( g(f(x_1)) = g(f(x_2)) \).
- This implies \( \sqrt{x_1} = \sqrt{x_2} \), which leads to \( x_1 = x_2 \). Thus, \( g(f(x)) \) is one-to-one.

6. **Check if \( g(f(x)) \) is onto**:
- The range of \( g(f(x)) = \sqrt{x} \) is \( [0, \infty) \).
- The co-domain is also \( [0, \infty) \), so it is onto.

### Step 5: Check if \( f(g(x)) \) is odd or even

7. **Check if \( f(g(x)) = \sqrt{x} \) is odd or even**:
- A function \( h(x) \) is even if \( h(-x) = h(x) \) and odd if \( h(-x) = -h(x) \).
- Since \( \sqrt{-x} \) is not defined for \( x > 0 \), \( f(g(x)) \) is neither odd nor even.

### Conclusion

- The function \( f(g(x)) = \sqrt{x} \) is defined for \( x > 0 \) and is neither odd nor even.
- The function \( g(f(x)) = \sqrt{x} \) is a bijective function.

### Summary of Results

- \( f(g(x)) = \sqrt{x} \) is neither odd nor even.
- \( g(f(x)) = \sqrt{x} \) is a bijective function.