If `f (tan x) = cos 2x , x != (2n + 1) pi/2, n in I` then incorrect statement is

Options

A. f(x) is an even function
B. f(x) is an odd function
C. Range of `f(x) "is" [-1,1]`
D. Domain of `f(x) "is" x in R`

Answer

To solve the problem, we need to analyze the function given by \( f(\tan x) = \cos(2x) \) and derive the correct statements about this function.

### Step-by-Step Solution:

1. **Rewrite the Function**:
We start with the equation:
\[ f(\tan x) = \cos(2x) \]
We can express \(\cos(2x)\) in terms of \(\tan x\):
\[ \cos(2x) = \frac{1 - \tan^2 x}{1 + \tan^2 x} \]
This is derived from the double angle formula for cosine.

2. **Substituting \(t = \tan x\)**:
Let \(t = \tan x\). Then, we can rewrite the equation as:
\[ f(t) = \frac{1 - t^2}{1 + t^2} \]

3. **Finding \(f(x)\)**:
Since we want \(f(x)\) instead of \(f(t)\), we replace \(t\) with \(x\):
\[ f(x) = \frac{1 - x^2}{1 + x^2} \]

4. **Check if the Function is Even or Odd**:
To determine if \(f(x)\) is even or odd, we need to check:
- For even: \(f(-x) = f(x)\)
- For odd: \(f(-x) = -f(x)\)

Calculate \(f(-x)\):
\[ f(-x) = \frac{1 - (-x)^2}{1 + (-x)^2} = \frac{1 - x^2}{1 + x^2} = f(x) \]
Since \(f(-x) = f(x)\), the function is **even**.

5. **Determine the Domain**:
The function \(f(x) = \frac{1 - x^2}{1 + x^2}\) is defined for all real numbers \(x\) since the denominator \(1 + x^2\) is always positive. Thus, the domain is:
\[ \text{Domain} = \mathbb{R} \]

6. **Determine the Range**:
To find the range of \(f(x)\), we analyze the expression:
\[ f(x) = \frac{1 - x^2}{1 + x^2} \]
As \(x\) approaches \(\infty\) or \(-\infty\), \(f(x)\) approaches \(-1\). When \(x = 0\), \(f(0) = 1\). Therefore, the function takes values from \(-1\) to \(1\):
\[ \text{Range} = [-1, 1] \]

7. **Identifying Incorrect Statement**:
Now, we analyze the statements provided in the question:
- A: The function is even (True)
- B: The function is odd (False)
- C: The range of the function is \([-1, 1]\) (True)
- D: The domain of the function is \(\mathbb{R}\) (True)

The incorrect statement is:
\[ \text{B: The function is odd} \]

### Final Answer:
The incorrect statement is **B**.