If `f(x)=(x^(2)-[x^(2)])/(1+x^(2)-[x^(2)])` (where `[.]` represents the greatest integer part of x), then the range of `f(x)` is

Options

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

Answer

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

### Step 1: Understand the function
The term \([x^2]\) represents the greatest integer less than or equal to \(x^2\). Thus, we can express \(x^2\) as:
\[ x^2 = n + \{x^2\} \]
where \(n = [x^2]\) is an integer and \(\{x^2\} = x^2 - n\) is the fractional part of \(x^2\).

### Step 2: Rewrite the function
Substituting \(x^2\) into the function, we get:
\[ f(x) = \frac{\{x^2\}}{1 + \{x^2\}} \]
This is because \(x^2 - [x^2] = \{x^2\}\) and \(1 + x^2 - [x^2] = 1 + \{x^2\}\).

### Step 3: Analyze the fractional part
The fractional part \(\{x^2\}\) varies between \(0\) and \(1\) (inclusive of \(0\) and exclusive of \(1\)). Therefore, we can denote:
\[ \{x^2\} \in [0, 1) \]

### Step 4: Determine the range of \(f(x)\)
Now we need to analyze the function:
\[ f(x) = \frac{y}{1 + y} \]
where \(y = \{x^2\}\) and \(y \in [0, 1)\).

To find the range of \(f(x)\), we will evaluate the limits as \(y\) approaches \(0\) and \(1\):
- When \(y = 0\):
\[ f(x) = \frac{0}{1 + 0} = 0 \]
- As \(y\) approaches \(1\):
\[ f(x) = \frac{y}{1 + y} \to \frac{1}{2} \quad \text{(as } y \to 1 \text{, but it never actually reaches 1)} \]

### Step 5: Conclusion on the range
Thus, the range of \(f(x)\) is:
\[ [0, \frac{1}{2}) \]

### Final Answer
The range of \(f(x)\) is \([0, \frac{1}{2})\).