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

Options

A. f[g(x)] and g[f(x)] have different domain
B. f[g(x)] and g[f(x)] have the same range
C. f[g(x)] is a one-one
D. g[f(x)] is neither odd nor even

Answer

To solve the problem, we need to find \( f(g(x)) \) and \( g(f(x)) \) for the given functions \( f(x) = \frac{1}{x} \) and \( g(x) = \frac{1}{\sqrt{x}} \).

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

1. Start with the function \( g(x) \):
\[ g(x) = \frac{1}{\sqrt{x}} \]
2. Now substitute \( g(x) \) into \( f(x) \):
\[ f(g(x)) = f\left(\frac{1}{\sqrt{x}}\right) = \frac{1}{\frac{1}{\sqrt{x}}} \]
3. Simplify the expression:
\[ f(g(x)) = \sqrt{x} \]

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

1. Start with the function \( f(x) \):
\[ f(x) = \frac{1}{x} \]
2. Now substitute \( f(x) \) into \( g(x) \):
\[ g(f(x)) = g\left(\frac{1}{x}\right) = \frac{1}{\sqrt{\frac{1}{x}}} \]
3. Simplify the expression:
\[ g(f(x)) = \sqrt{x} \]

### Step 3: Compare \( f(g(x)) \) and \( g(f(x)) \)

From the calculations above, we have:
\[ f(g(x)) = \sqrt{x} \quad \text{and} \quad g(f(x)) = \sqrt{x} \]
Thus, both \( f(g(x)) \) and \( g(f(x)) \) are equal.

### Step 4: Check if \( f(g(x)) \) is a one-to-one function

1. A function is one-to-one if \( f(a) = f(b) \) implies \( a = b \).
2. Let \( f(g(x_1)) = f(g(x_2)) \):
\[ \sqrt{x_1} = \sqrt{x_2} \]
3. Squaring both sides:
\[ x_1 = x_2 \]
4. Therefore, \( f(g(x)) \) is a one-to-one function.

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

1. To check if a function is even, we see if \( f(-x) = f(x) \).
2. Substitute \( -x \) into \( f(g(x)) \):
\[ f(g(-x)) = \sqrt{-x} \]
3. Since \( \sqrt{-x} \) is not defined for real numbers, we cannot classify \( f(g(x)) \) as odd or even.

### Conclusion

- \( f(g(x)) = g(f(x)) = \sqrt{x} \)
- Both functions have the same range.
- \( f(g(x)) \) is a one-to-one function.
- The odd/even classification is not applicable.