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

Options

A. 2
B. 3
C. 6
D. none of these

Explanation

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 \( x \in \mathbb{R} \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
The condition \( f^2(x) = 1 \) implies that \( f(x) \) can take only two values: \( f(x) = 1 \) or \( f(x) = -1 \) for all \( x \).

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

3. **Defining the Function**:
We can define \( f(x) \) in two distinct ways based on the value of \( x \):
- For \( x < 0 \): \( f(x) \) can be either \( 1 \) or \( -1 \).
- For \( x \geq 0 \): \( f(x) \) can also be either \( 1 \) or \( -1 \).

4. **Possible Combinations**:
We can analyze the combinations:
- If \( f(x) = 1 \) for \( x \geq 0 \):
- Then \( f(x) = -1 \) for \( x < 0 \) (discontinuous at \( x = 0 \)).
- If \( f(x) = -1 \) for \( x \geq 0 \):
- Then \( f(x) = 1 \) for \( x < 0 \) (discontinuous at \( x = 0 \)).
- We can also have:
- \( f(x) = 1 \) for \( x > 0 \) and \( f(0) = -1 \) (discontinuous).
- \( f(x) = -1 \) for \( x > 0 \) and \( f(0) = 1 \) (discontinuous).
- \( f(x) = 1 \) for \( x \neq 0 \) and \( f(0) = -1 \) (discontinuous).
- \( f(x) = -1 \) for \( x \neq 0 \) and \( f(0) = 1 \) (discontinuous).

5. **Counting the Functions**:
From the above combinations, we can summarize the possible functions:
- \( f(x) = 1 \) for \( x \geq 0 \) and \( f(x) = -1 \) for \( x < 0 \).
- \( f(x) = -1 \) for \( x \geq 0 \) and \( f(x) = 1 \) for \( x < 0 \).
- \( f(x) = 1 \) for \( x > 0 \) and \( f(0) = -1 \).
- \( f(x) = -1 \) for \( x > 0 \) and \( f(0) = 1 \).
- \( f(x) = 1 \) for \( x \neq 0 \) and \( f(0) = -1 \).
- \( f(x) = -1 \) for \( x \neq 0 \) and \( f(0) = 1 \).

Thus, we have a total of 6 distinct functions.

### Final Answer:
The total number of such functions is **6**.