If `f : R -> R` are defined by `f(x) = x - [x]` and `g(x) = [x]` for `x in R`, where [x] is the greatest integer not ex-ceeding x, then for every` x in R, f(g(x))=`

Options

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

Answer

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

- \( f(x) = x - [x] \) (the fractional part of \( x \))
- \( g(x) = [x] \) (the greatest integer not exceeding \( x \))

### Step-by-Step Solution:

1. **Identify the Functions:**
- The function \( g(x) \) gives us the greatest integer less than or equal to \( x \), denoted as \( [x] \).
- The function \( f(x) \) calculates the fractional part of \( x \), which is \( x - [x] \).

2. **Substitute \( g(x) \) into \( f(x) \):**
- We need to find \( f(g(x)) \).
- Since \( g(x) = [x] \), we substitute this into \( f \):
\[ f(g(x)) = f([x]) \]

3. **Evaluate \( f([x]) \):**
- Now, we apply the function \( f \) to \( [x] \):
\[ f([x]) = [x] - [[x]] \]
- Since \( [x] \) is an integer, we have:
\[ f([x]) = [x] - [x] = 0 \]

4. **Conclusion:**
- Therefore, for every \( x \in \mathbb{R} \):
\[ f(g(x)) = 0 \]

### Final Answer:
\[ f(g(x)) = 0 \quad \text{for all } x \in \mathbb{R} \]