If a function satisfies the condition `f(x+1/x)=x^(2)+1/(x^(2)),xne0`, then domain of f(x) is

Options

A. `[-2,2]`
B. `(-oo,-2]cup[2,oo)`
C. `(0,oo)`
D. none of these

Answer

To find the domain of the function \( f(x) \) given the condition \( f\left(x + \frac{1}{x}\right) = x^2 + \frac{1}{x^2} \) for \( x \neq 0 \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Expression**:
We start with the equation:
\[ f\left(x + \frac{1}{x}\right) = x^2 + \frac{1}{x^2} \]

2. **Rewriting the Right Side**:
We can rewrite \( x^2 + \frac{1}{x^2} \) by adding and subtracting 2:
\[ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \]
So, we have:
\[ f\left(x + \frac{1}{x}\right) = \left(x + \frac{1}{x}\right)^2 - 2 \]

3. **Defining a New Variable**:
Let \( y = x + \frac{1}{x} \). Then, we can express \( f(y) \) as:
\[ f(y) = y^2 - 2 \]

4. **Finding the Range of \( y \)**:
We need to determine the possible values of \( y \). For positive \( x \):
- The minimum value of \( y = x + \frac{1}{x} \) occurs at \( x = 1 \), giving \( y = 2 \).
- As \( x \) approaches 0 from the positive side, \( y \) approaches infinity.
Thus, for positive \( x \), \( y \geq 2 \).

For negative \( x \):
- If \( x < 0 \), then \( y = x + \frac{1}{x} \) will yield values less than or equal to -2.
- The minimum occurs at \( x = -1 \), giving \( y = -2 \).

5. **Combining the Results**:
Therefore, the possible values for \( y \) are:
\[ y \leq -2 \quad \text{or} \quad y \geq 2 \]

6. **Determining the Domain of \( f(x) \)**:
Since \( f(y) \) is defined for \( y \) values derived above, the domain of \( f(x) \) is:
\[ (-\infty, -2] \cup [2, \infty) \]

### Final Answer:
The domain of \( f(x) \) is:
\[ \text{Domain of } f(x) = (-\infty, -2] \cup [2, \infty) \]