Let g(x) = 1 + x – [x] and `f(x)={:{(-1,if,xlt0),(0,if, x=0),(1,if,x gt0):}` then `Aax, fog(x)` equals
(where [ * ] represents greatest integer function).

Options

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

Answer

To solve the problem, we need to find \( f(g(x)) \) where:

1. \( g(x) = 1 + x - [x] \) (where \([x]\) is the greatest integer function)
2. \( f(x) = \begin{cases} -1 & \text{if } x < 0 \\ 0 & \text{if } x = 0 \\ 1 & \text{if } x > 0 \end{cases} \)

### Step 1: Understand the function \( g(x) \)

The function \( g(x) \) can be rewritten as:
\[ g(x) = 1 + x - [x] \]
The term \( x - [x] \) represents the fractional part of \( x \), denoted as \( \{x\} \). Therefore, we can express \( g(x) \) as:
\[ g(x) = 1 + \{x\} \]
where \( 0 \leq \{x\} < 1 \).

### Step 2: Analyze the range of \( g(x) \)

Since \( \{x\} \) is always between 0 and 1, we can determine the range of \( g(x) \):
\[ g(x) = 1 + \{x\} \implies 1 \leq g(x) < 2 \]
This means \( g(x) \) is always greater than 1.

### Step 3: Substitute \( g(x) \) into \( f(x) \)

Now we need to find \( f(g(x)) \). Since we have established that \( g(x) > 1 \) for all \( x \), we can use the definition of \( f(x) \):
- Since \( g(x) > 1 \), it falls under the case where \( x > 0 \) in the function \( f(x) \).

Thus, we have:
\[ f(g(x)) = 1 \quad \text{(since \( g(x) > 0 \))} \]

### Conclusion

The value of \( f(g(x)) \) is:
\[ \boxed{1} \]