If f(x) is an even function and satisfies the relation `x^(2)f(x)-2f(1/x)=g(x),xne0`, where g(x) is an odd function, then find the value of f(2).

Answer

To solve the problem, we need to analyze the given conditions and relationships carefully.

### Step-by-Step Solution:

1. **Understanding the Functions**:
We know that \( f(x) \) is an even function, which means:
\[ f(-x) = f(x) \]
We also know that \( g(x) \) is an odd function, which means:
\[ g(-x) = -g(x) \]

2. **Given Relation**:
The relation provided is:
\[ x^2 f(x) - 2 f\left(\frac{1}{x}\right) = g(x) \]

3. **Substituting \( x \) with \( \frac{1}{x} \)**:
Let's replace \( x \) with \( \frac{1}{x} \) in the original equation:
\[ \left(\frac{1}{x}\right)^2 f\left(\frac{1}{x}\right) - 2 f(x) = g\left(\frac{1}{x}\right) \]
Simplifying this gives:
\[ \frac{1}{x^2} f\left(\frac{1}{x}\right) - 2 f(x) = g\left(\frac{1}{x}\right) \]

4. **Multiplying by \( x^2 \)**:
To eliminate the fraction, multiply the entire equation by \( x^2 \):
\[ f\left(\frac{1}{x}\right) - 2x^2 f(x) = x^2 g\left(\frac{1}{x}\right) \]

5. **Setting Up the Equations**:
Now we have two equations:
- From the original substitution:
\[ x^2 f(x) - 2 f\left(\frac{1}{x}\right) = g(x) \quad \text{(1)} \]
- From the second substitution:
\[ f\left(\frac{1}{x}\right) - 2x^2 f(x) = x^2 g\left(\frac{1}{x}\right) \quad \text{(2)} \]

6. **Adding the Two Equations**:
Adding equations (1) and (2):
\[ (x^2 f(x) - 2 f\left(\frac{1}{x}\right)) + (f\left(\frac{1}{x}\right) - 2x^2 f(x)) = g(x) + x^2 g\left(\frac{1}{x}\right) \]
This simplifies to:
\[ -x^2 f(x) - f\left(\frac{1}{x}\right) = g(x) + x^2 g\left(\frac{1}{x}\right) \]

7. **Analyzing the Results**:
Since \( f(x) \) is even and \( g(x) \) is odd, we can conclude that \( f(x) \) must be a constant function. The only function that is both even and odd is the zero function. Thus, we can deduce:
\[ f(x) = 0 \quad \text{for all } x \]

8. **Finding \( f(2) \)**:
Since \( f(x) = 0 \) for all \( x \), we have:
\[ f(2) = 0 \]

### Final Answer:
\[ \boxed{0} \]