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The question asks for which pair of functions \( (f \circ g)(x) = x \), which means the composition of functions \( f \) and \( g \) results in \( x \). 

We need to check each pair by calculating \( (f \circ g)(x) \), which means applying \( g(x) \) first, then applying \( f \) to the result of \( g(x) \).

Let’s evaluate the options one by one:

### Option A:
- \( f(x) = x^2 \) and \( g(x) = \frac{1}{x} \)
- \( (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{x}\right) = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2} \)
- This is not equal to \( x \).

### Option B:
- \( f(x) = \frac{2}{x} \) and \( g(x) = \frac{2}{x} \)
- \( (f \circ g)(x) = f(g(x)) = f\left(\frac{2}{x}\right) = \frac{2}{\frac{2}{x}} = x \)
- This satisfies \( (f \circ g)(x) = x \).

### Option C:
- \( f(x) = \frac{x - 2}{3} \) and \( g(x) = 2 - 3x \)
- \( (f \circ g)(x) = f(g(x)) = f(2 - 3x) = \frac{(2 - 3x) - 2}{3} = \frac{-3x}{3} = -x \)
- This is not equal to \( x \).

### Option D:
- \( f(x) = \frac{1}{2}x - 2 \) and \( g(x) = \frac{1}{2}x + 2 \)
- \( (f \circ g)(x) = f(g(x)) = f\left(\frac{1}{2}x + 2\right) = \frac{1}{2}\left(\frac{1}{2}x + 2\right) - 2 = \frac{1}{4}x + 1 - 2 = \frac{1}{4}x - 1 \)
- This is not equal to \( x \).

### Conclusion:
The correct pair of functions is **Option B**: 
\[ f(x) = \frac{2}{x} \text{ and } g(x) = \frac{2}{x}. \]

Would you like more details on any of these steps or have further questions?

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Here are five related questions for further exploration:
1. How do we define function composition?
2. What is the domain of a composite function?
3. How do we find the inverse of a function?
4. What are some real-life applications of composite functions?
5. How do you verify if two functions are inverses?

**Tip:** When working with composite functions, always apply the inner function first and then the outer function.

For which pair of functions is (f ° g)(x) = x?

Explore how to solve for the pair of functions where (f ° g)(x) = x. This problem covers the core concept of function composition, using algebraic methods to evaluate different options. Suitable for students in Grades 9-11.

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