Let `f(x)=(x-[x])/(1+x-[x]),RtoA` is onto then find set A. (where {.} and [.] represent fractional part and greatest integer part functions respectively) (a) `(0,1/2]` (b) `[0,1/2]` (c) `[0,1/2)` (d) `(0,1/2)`

Options

A. `(0,1/2]`
B. `[0,1/2]`
C. `[0,1/2]`
D. `(0,1/2)`

Explanation

To solve the problem, we need to analyze the function \( f(x) = \frac{x - [x]}{1 + x - [x]} \), where \( [x] \) is the greatest integer function (floor function) and \( x - [x] \) is the fractional part of \( x \).

### Step 1: Understand the Function
The fractional part of \( x \), denoted as \( \{x\} \), is defined as:
\[ \{x\} = x - [x] \]
This means \( \{x\} \) is always in the range \( [0, 1) \).

### Step 2: Rewrite the Function
Substituting the fractional part into the function, we have:
\[ f(x) = \frac{\{x\}}{1 + \{x\}} \]
Since \( \{x\} \) ranges from \( 0 \) to \( 1 \), we can analyze how \( f(x) \) behaves within this range.

### Step 3: Determine the Range of \( f(x) \)
1. When \( \{x\} = 0 \):
\[ f(x) = \frac{0}{1 + 0} = 0 \]
2. When \( \{x\} \) approaches \( 1 \) (but does not reach it):
\[ f(x) = \frac{1 - \epsilon}{1 + (1 - \epsilon)} = \frac{1 - \epsilon}{2 - \epsilon} \]
As \( \epsilon \) approaches \( 0 \), \( f(x) \) approaches \( \frac{1}{2} \).

### Step 4: Analyze the Range
From the above analysis:
- The minimum value of \( f(x) \) is \( 0 \) (when \( \{x\} = 0 \)).
- The maximum value of \( f(x) \) approaches \( \frac{1}{2} \) (as \( \{x\} \) approaches \( 1 \)).

Thus, the range of \( f(x) \) is:
\[ [0, \frac{1}{2}) \]

### Step 5: Determine Set \( A \)
Since the function \( f(x) \) is onto, the range must equal the co-domain \( A \). Therefore, we conclude:
\[ A = [0, \frac{1}{2}) \]

### Step 6: Identify the Correct Option
Among the given options:
(a) \( (0, \frac{1}{2}] \)
(b) \( [0, \frac{1}{2}] \)
(c) \( [0, \frac{1}{2}) \)
(d) \( (0, \frac{1}{2}) \)

The correct option is:
**(c) \( [0, \frac{1}{2}) \)**.