Statement 1 : The domain of the function `f(x)=sqrt(x-[x])" is "R^(+)`
Statement 2 : The domain of the function `sqrt(f(x)) is {x : f (x) ge 0}`.

Options

A. Statement-1 is True, Statement-2 is True and Statement-2 is a correct explanation for Statement-1
B. Statement-1 is True, Statement-2 is True and Statement-2 is NOT a correct explanation for Statement-1
C. Statement-1 is True, Statement-2 is False
D. Statement-1 is False, Statement-2 is True

Answer

To solve the problem, we need to analyze both statements regarding the function \( f(x) = \sqrt{x - [x]} \).

### Step 1: Analyze Statement 1
**Statement 1**: The domain of the function \( f(x) = \sqrt{x - [x]} \) is \( \mathbb{R}^{+} \).

1. **Understanding the function**: The expression \( [x] \) denotes the greatest integer less than or equal to \( x \). Therefore, \( x - [x] \) represents the fractional part of \( x \), which is always non-negative and less than 1 for any real number \( x \).

2. **Condition for the square root**: For the square root to be defined, we need \( x - [x] \geq 0 \). This condition is satisfied for all \( x \) because the fractional part is always non-negative.

3. **Finding the domain**: Since \( x - [x] \) can take values from \( 0 \) (when \( x \) is an integer) to just below \( 1 \) (when \( x \) is not an integer), the function is defined for all real numbers \( x \). Thus, the domain of \( f(x) \) is \( \mathbb{R} \) (all real numbers), not just \( \mathbb{R}^{+} \).

### Conclusion for Statement 1
**Statement 1 is false** because the domain is \( \mathbb{R} \), not \( \mathbb{R}^{+} \).

### Step 2: Analyze Statement 2
**Statement 2**: The domain of the function \( \sqrt{f(x)} \) is \( \{ x : f(x) \geq 0 \} \).

1. **Understanding \( f(x) \)**: As established, \( f(x) = \sqrt{x - [x]} \) is non-negative for all \( x \) since the square root function outputs non-negative values.

2. **Condition for the square root**: The statement correctly identifies that the domain of \( \sqrt{f(x)} \) requires \( f(x) \geq 0 \). Since \( f(x) \) is always non-negative, this condition holds for all \( x \).

### Conclusion for Statement 2
**Statement 2 is true** because the domain of \( \sqrt{f(x)} \) is indeed \( \{ x : f(x) \geq 0 \} \), which is all real numbers.

### Final Conclusion
- **Statement 1**: False
- **Statement 2**: True

### Answer
The correct option is that Statement 1 is false and Statement 2 is true.