Let `f:A to A and g:A to A` be two functions such that fog(x)=gof (x)=x for all `x in A`
Statement-1:`{x in A: f(x)=g(x)}={x in A: f(x)=x}={x in A: g(x)=x}`
Statement-2: `f:A to A` is bijection.

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze the given statements based on the properties of the functions \( f \) and \( g \).

### Step-by-Step Solution:

1. **Understanding the Functions**:
We have two functions \( f: A \to A \) and \( g: A \to A \) such that:
\[ f(g(x)) = g(f(x)) = x \quad \text{for all } x \in A \]
This means that \( f \) and \( g \) are inverses of each other.

2. **Implication of Inverses**:
Since \( f(g(x)) = x \), it implies that \( g \) is the inverse of \( f \). Similarly, \( g(f(x)) = x \) implies that \( f \) is the inverse of \( g \). Therefore, both functions are bijections (one-to-one and onto).

3. **Analyzing Statement 1**:
We need to show that:
\[ \{ x \in A : f(x) = g(x) \} = \{ x \in A : f(x) = x \} = \{ x \in A : g(x) = x \} \]
- Let \( x \) be an element such that \( f(x) = g(x) \). If we denote this common value as \( y \), we have:
\[ f(x) = g(x) = y \]
- Since \( f \) and \( g \) are inverses, we can substitute \( g(x) \) into \( f(g(x)) = x \):
\[ f(y) = x \quad \text{and} \quad g(y) = x \]
- If \( y = f(x) \) and \( y = g(x) \), then \( f(y) = y \) and \( g(y) = y \) must hold true, meaning \( y \) is a fixed point for both functions.

4. **Conclusion for Statement 1**:
Therefore, the set of points where \( f(x) = g(x) \) corresponds to the set of fixed points of both functions, confirming that:
\[ \{ x \in A : f(x) = g(x) \} = \{ x \in A : f(x) = x \} = \{ x \in A : g(x) = x \} \]
Thus, Statement 1 is true.

5. **Analyzing Statement 2**:
Since we established that \( f \) and \( g \) are inverses of each other and both are functions from \( A \) to \( A \), it follows that:
- \( f \) is one-to-one (injective) because if \( f(a) = f(b) \), then applying \( g \) gives \( g(f(a)) = g(f(b)) \) leading to \( a = b \).
- \( f \) is onto (surjective) because for every \( x \in A \), there exists a \( y \in A \) such that \( f(y) = x \) (specifically, \( y = g(x) \)).
Therefore, \( f \) is a bijection.

6. **Conclusion for Statement 2**:
Since \( f \) is both injective and surjective, Statement 2 is also true.

### Final Conclusion:
Both Statement 1 and Statement 2 are true.