Let `f : I - {-1,0,1} to [-pi, pi]` be defined as `f(x) = 2 tan^(-1) x - tan^(-1)((2x)/(1 -x^(2)))`, then which of the following statements (s) is (are) correct ?

Options

A. f(x) is bijective
B. f(x) is injective but not surjective
C. f(x) is neither injective nor surjective
D. f(x) is an odd function

Answer

To solve the problem, we need to analyze the function \( f(x) = 2 \tan^{-1}(x) - \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \) and determine the properties of this function based on the given statements.

### Step 1: Understanding the Function
The function is defined as:
\[ f(x) = 2 \tan^{-1}(x) - \tan^{-1}\left(\frac{2x}{1 - x^2}\right) \]
We know from trigonometric identities that:
\[ \tan^{-1}\left(\frac{2x}{1 - x^2}\right) = 2 \tan^{-1}(x) \quad \text{for } x \in (-1, 1) \]
This implies that within the interval \( (-1, 1) \):
\[ f(x) = 2 \tan^{-1}(x) - 2 \tan^{-1}(x) = 0 \]

### Step 2: Evaluating the Function Outside the Interval
For \( x < -1 \) and \( x > 1 \), we need to analyze the behavior of \( f(x) \):
- For \( x < -1 \):
\[ f(x) = 2 \tan^{-1}(x) - \left(\pi + 2 \tan^{-1}(x)\right) = -\pi \]
- For \( x > 1 \):
\[ f(x) = 2 \tan^{-1}(x) - \left(-\pi + 2 \tan^{-1}(x)\right) = \pi \]

### Step 3: Summary of Function Values
- For \( x \in (-1, 1) \), \( f(x) = 0 \)
- For \( x < -1 \), \( f(x) = -\pi \)
- For \( x > 1 \), \( f(x) = \pi \)

### Step 4: Analyzing Injectivity and Surjectivity
1. **Injectivity**: A function is injective if it gives distinct outputs for distinct inputs. Since \( f(x) = 0 \) for all \( x \in (-1, 1) \), it is not injective.
2. **Surjectivity**: A function is surjective if its range covers the entire codomain. The range of \( f(x) \) is \( \{-\pi, 0, \pi\} \), which does not cover the entire interval \( [-\pi, \pi] \). Hence, it is not surjective.

### Step 5: Checking if \( f(x) \) is an Odd Function
To check if \( f(x) \) is odd, we need to verify if:
\[ f(-x) = -f(x) \]
Calculating \( f(-x) \):
- For \( x \in (-1, 1) \), \( f(-x) = 0 = -0 \)
- For \( x < -1 \), \( f(-x) = \pi \) and \( -f(x) = -(-\pi) = \pi \)
- For \( x > 1 \), \( f(-x) = -\pi \) and \( -f(x) = -\pi \)

Thus, \( f(x) \) is an odd function.

### Conclusion
Based on the analysis:
- **Option A**: Incorrect (not bijective)
- **Option B**: Incorrect (not injective)
- **Option C**: Correct (neither injective nor surjective)
- **Option D**: Correct (is an odd function)

### Final Answer
The correct statements are:
- Option C: \( f(x) \) is neither injective nor surjective.
- Option D: \( f(x) \) is an odd function.