The range of `f(x)=(2+x-[x])/(1-x+[x])`.where [ ] denotes the greatest integer function is

Options

A. [0,1)
B. `[2,oo)`
C. `[0,1) cup (1,2]`
D. `R^(+)`

Answer

To find the range of the function \( f(x) = \frac{2 + x - [x]}{1 - x + [x]} \), where \([x]\) denotes the greatest integer function, we can follow these steps:

### Step 1: Rewrite the function using the fractional part
We know that \( x - [x] \) is the fractional part of \( x \), denoted as \( \{x\} \). Therefore, we can rewrite the function as:

\[ f(x) = \frac{2 + \{x\}}{1 - x + [x]} \]

### Step 2: Simplify the denominator
The term \( 1 - x + [x] \) can be rewritten as:

\[ 1 - x + [x] = 1 - (x - [x]) = 1 - \{x\} \]

Thus, the function becomes:

\[ f(x) = \frac{2 + \{x\}}{1 - \{x\}} \]

### Step 3: Identify the range of the fractional part
The fractional part \( \{x\} \) lies in the range:

\[ 0 \leq \{x\} < 1 \]

### Step 4: Analyze the function
Now we need to analyze the function \( f(x) = \frac{2 + \{x\}}{1 - \{x\}} \) as \( \{x\} \) varies from 0 to just below 1.

1. When \( \{x\} = 0 \):
\[ f(0) = \frac{2 + 0}{1 - 0} = 2 \]

2. As \( \{x\} \) approaches 1 (but does not equal 1):
\[ f(\{x\} \to 1) = \frac{2 + 1}{1 - 1} \to \infty \]

### Step 5: Determine the range
From the analysis, as \( \{x\} \) moves from 0 to just below 1, \( f(x) \) moves from 2 to \( \infty \). Therefore, the range of \( f(x) \) is:

\[ [2, \infty) \]

### Conclusion
The range of the function \( f(x) = \frac{2 + x - [x]}{1 - x + [x]} \) is \( [2, \infty) \).

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