The function f(x) is discontinuous only at x = 0 such that `f^(2)(x)=1 AA x in R`. The total number of such functions is

Answer

To solve the problem, we need to find the total number of functions \( f(x) \) that are discontinuous only at \( x = 0 \) and satisfy the condition \( f^2(x) = 1 \) for all \( x \in \mathbb{R} \). This means that for every \( x \), \( f(x) \) can only take the values \( 1 \) or \( -1 \).

### Step-by-Step Solution:

1. **Identify Possible Values of \( f(x) \)**:
Since \( f^2(x) = 1 \), we can conclude that:
\[ f(x) = 1 \quad \text{or} \quad f(x) = -1 \quad \text{for all } x \in \mathbb{R}. \]

2. **Discontinuity at \( x = 0 \)**:
The function \( f(x) \) must be discontinuous only at \( x = 0 \). This means that the limits from the left and right at \( x = 0 \) must not be equal.

3. **Define Cases for \( f(x) \)**:
We can define \( f(x) \) in the following ways based on the intervals:
- For \( x < 0 \): \( f(x) \) can be either \( 1 \) or \( -1 \).
- For \( x > 0 \): \( f(x) \) can also be either \( 1 \) or \( -1 \).
- At \( x = 0 \): \( f(0) \) can be either \( 1 \) or \( -1 \).

4. **Construct Possible Functions**:
We can create functions based on the combinations of values for \( f(x) \) in the intervals:
- If \( f(x) = 1 \) for \( x < 0 \) and \( f(x) = -1 \) for \( x > 0 \), then \( f(0) \) can be either \( 1 \) or \( -1 \).
- If \( f(x) = -1 \) for \( x < 0 \) and \( f(x) = 1 \) for \( x > 0 \), then again \( f(0) \) can be either \( 1 \) or \( -1 \).

5. **Count the Combinations**:
- For the first case (where \( f(x) = 1 \) for \( x < 0 \) and \( f(x) = -1 \) for \( x > 0 \)), we have:
- \( f(0) = 1 \) → Function 1
- \( f(0) = -1 \) → Function 2
- For the second case (where \( f(x) = -1 \) for \( x < 0 \) and \( f(x) = 1 \) for \( x > 0 \)), we have:
- \( f(0) = 1 \) → Function 3
- \( f(0) = -1 \) → Function 4

6. **Total Functions**:
Thus, we have a total of 4 functions from the above combinations. However, we also need to consider the cases where \( f(x) \) can be constant on either side of \( 0 \):
- \( f(x) = 1 \) for all \( x < 0 \) and \( f(x) = 1 \) for all \( x > 0 \) (discontinuous at \( 0 \)).
- \( f(x) = -1 \) for all \( x < 0 \) and \( f(x) = -1 \) for all \( x > 0 \) (discontinuous at \( 0 \)).

This gives us 2 additional functions.

7. **Final Count**:
Adding these up, we have:
\[ 4 \text{ (from combinations)} + 2 \text{ (constant functions)} = 6. \]

### Conclusion:
The total number of such functions is \( 6 \).