Consider a function defined in `[-2,2]`
`f (x)={{:({x}, -2 le x lt -1),( |sgn x|, -1 le x le 1),( {-x}, 1 lt x le 2):},` where {.} denotes the fractional part function.
The total number of points of discontinuity of `f (x)` for `x in[-2,2]` is:

Options

A. 0
B. 1
C. 2
D. 4

Answer

To determine the total number of points of discontinuity of the function \( f(x) \) defined in the interval \([-2, 2]\), we need to analyze the function piece by piece. The function is defined as follows:

\[ f(x) = \begin{cases} \{x\} & \text{for } -2 \leq x < -1 \\ \text{sgn}(x) & \text{for } -1 \leq x \leq 1 \\ \{-x\} & \text{for } 1 < x \leq 2 \end{cases} \]

Where \(\{x\}\) denotes the fractional part of \(x\) and \(\text{sgn}(x)\) is the signum function.

### Step 1: Analyze the function on the intervals

1. **Interval \([-2, -1)\)**:
- Here, \( f(x) = \{x\} = x + 2 \) since \( x \) is in the range \([-2, -1)\).
- The function is continuous in this interval.

2. **At \( x = -1 \)**:
- Left-hand limit: \( \lim_{x \to -1^-} f(x) = \{-1\} = 0 \)
- Right-hand limit: \( \lim_{x \to -1^+} f(x) = \text{sgn}(-1) = -1 \)
- Since the left-hand limit (0) does not equal the right-hand limit (-1), \( f(x) \) is discontinuous at \( x = -1 \).

3. **Interval \([-1, 1]\)**:
- Here, \( f(x) = \text{sgn}(x) \).
- The function is continuous in this interval except at \( x = 0 \).

4. **At \( x = 0 \)**:
- Left-hand limit: \( \lim_{x \to 0^-} f(x) = \text{sgn}(0) = 0 \)
- Right-hand limit: \( \lim_{x \to 0^+} f(x) = \text{sgn}(0) = 0 \)
- The function is continuous at \( x = 0 \).

5. **At \( x = 1 \)**:
- Left-hand limit: \( \lim_{x \to 1^-} f(x) = \text{sgn}(1) = 1 \)
- Right-hand limit: \( \lim_{x \to 1^+} f(x) = \{-1\} = 0 \)
- Since the left-hand limit (1) does not equal the right-hand limit (0), \( f(x) \) is discontinuous at \( x = 1 \).

6. **Interval \( (1, 2]\)**:
- Here, \( f(x) = \{-x\} = 2 - x \) since \( x \) is in the range \( (1, 2] \).
- The function is continuous in this interval.

### Step 2: Summary of discontinuities

From the analysis, we found discontinuities at:
- \( x = -1 \)
- \( x = 1 \)

### Conclusion

Thus, the total number of points of discontinuity of \( f(x) \) for \( x \in [-2, 2] \) is **2**.