If f(x)=Sgn x and g(x)=1+x-[x], then for all x, f(g(x)) is equal to :

Options

A. 0
B. 1
C. x
D. f(x)

Answer

To solve the problem, we need to evaluate \( f(g(x)) \) where \( f(x) = \text{sgn}(x) \) and \( g(x) = 1 + x - [x] \).

### Step-by-Step Solution:

1. **Understanding the Functions**:
- The function \( f(x) = \text{sgn}(x) \) is defined as:
\[ \text{sgn}(x) = \begin{cases} 1 & \text{if } x > 0 \\ 0 & \text{if } x = 0 \\ -1 & \text{if } x < 0 \end{cases} \]
- The function \( g(x) = 1 + x - [x] \) can be simplified. Here, \( [x] \) denotes the greatest integer less than or equal to \( x \). The term \( x - [x] \) represents the fractional part of \( x \), which is always between 0 and 1. Therefore, \( g(x) \) can be rewritten as:
\[ g(x) = 1 + (x - [x]) = 1 + \{x\} \]
where \( \{x\} \) is the fractional part of \( x \).

2. **Range of \( g(x) \)**:
- Since \( \{x\} \) is always between 0 and 1, \( g(x) \) will always be:
\[ g(x) = 1 + \{x\} \in (1, 2) \]
- This means that \( g(x) \) is always greater than 1 for any real number \( x \).

3. **Evaluating \( f(g(x)) \)**:
- Now, we need to find \( f(g(x)) \). Since \( g(x) > 1 \) for all \( x \), we can substitute \( g(x) \) into \( f(x) \):
\[ f(g(x)) = f(1 + \{x\}) \]
- Since \( 1 + \{x\} > 0 \), we have:
\[ f(g(x)) = \text{sgn}(g(x)) = 1 \]

4. **Conclusion**:
- Therefore, for all \( x \), \( f(g(x)) = 1 \).

### Final Answer:
\[ f(g(x)) = 1 \quad \text{for all } x \]