Consider the function `y =f(x)` satisfying the condition `f(x+1/x)=x^2+1/(x^2)(x!=0)`. Then the

Options

A. domain of `f(x)` is R
B. domain of `f " is " R-(-2,2)`
C. range of `f(x)" is " [-2,oo)`
D. range of `f(x) " is " [2,oo)`

Explanation

To solve the problem, we need to analyze the function \( f(x) \) given by the condition:

\[ f\left(x + \frac{1}{x}\right) = x^2 + \frac{1}{x^2} \quad (x \neq 0) \]

### Step 1: Rewrite the right-hand side

We start with the right-hand side of the equation:

\[ x^2 + \frac{1}{x^2} \]

We can rewrite this expression by adding and subtracting 2:

\[ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 \]

This allows us to express the function in a more manageable form.

### Step 2: Introduce a new variable

Let:

\[ t = x + \frac{1}{x} \]

Then, we can rewrite the equation as:

\[ f(t) = t^2 - 2 \]

### Step 3: Determine the domain of \( t \)

Next, we need to find the range of \( t = x + \frac{1}{x} \) for \( x \neq 0 \).

- For \( x > 0 \), using the AM-GM inequality, we have:

\[ x + \frac{1}{x} \geq 2 \]

Thus, \( t \geq 2 \).

- For \( x < 0 \), we can let \( x = -u \) where \( u > 0 \):

\[ t = -u - \frac{1}{u} \quad (u > 0) \]

The expression \( -u - \frac{1}{u} \) approaches \( -\infty \) as \( u \) approaches \( 0 \) and has a maximum value of \( -2 \) when \( u = 1 \).

Thus, for \( x < 0 \), \( t \leq -2 \).

Combining both cases, we find that:

\[ t \in (-\infty, -2] \cup [2, \infty) \]

### Step 4: Find the range of \( f(t) \)

Now, we need to find the range of \( f(t) = t^2 - 2 \).

- For \( t \in (-\infty, -2] \):

\[ f(t) = t^2 - 2 \geq (-2)^2 - 2 = 4 - 2 = 2 \]

- For \( t \in [2, \infty) \):

\[ f(t) = t^2 - 2 \geq 2^2 - 2 = 4 - 2 = 2 \]

Thus, in both cases, the minimum value of \( f(t) \) is 2. As \( t \) increases or decreases without bound, \( f(t) \) approaches infinity.

### Conclusion

The range of the function \( f(x) \) is:

\[ [2, \infty) \]