Let f be a function from a set X to X, such that (f(f(x)) = x, for all `x in X`, then

Options

A. f is one-one
B. f is onto
C. f is many one
D. f is into

Answer

To solve the problem, we need to analyze the function \( f \) given the condition \( f(f(x)) = x \) for all \( x \in X \).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We are given that \( f(f(x)) = x \) for all \( x \in X \). This means that applying the function \( f \) twice returns the original element \( x \).

2. **Proving that \( f \) is One-to-One (Injective)**:
- Assume \( f(a) = f(b) \) for some \( a, b \in X \).
- Applying \( f \) to both sides, we get \( f(f(a)) = f(f(b)) \).
- By the given condition, this simplifies to \( a = b \).
- Since \( a \) and \( b \) were arbitrary, we conclude that \( f \) is injective (one-to-one).

3. **Proving that \( f \) is Onto (Surjective)**:
- To show that \( f \) is onto, we need to demonstrate that for every \( y \in X \), there exists an \( x \in X \) such that \( f(x) = y \).
- Let \( y \) be an arbitrary element in \( X \).
- If we take \( x = f(y) \), then applying \( f \) gives us \( f(x) = f(f(y)) \).
- By the given condition, this simplifies to \( f(f(y)) = y \).
- Thus, for every \( y \in X \), we have found an \( x \) (specifically \( x = f(y) \)) such that \( f(x) = y \).
- Therefore, \( f \) is surjective (onto).

4. **Conclusion**:
Since \( f \) is both injective and surjective, we conclude that \( f \) is a bijection (a one-to-one correspondence).

### Final Answer:
The function \( f \) is both one-to-one (injective) and onto (surjective).

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