`if f(x) ={{:(x+2 ,if xlt0),(-x^(2)-2 ,if 0le xlt1),( x ,if xge1):}`
then the number of points of discontinuity of |f(x)|is (a) 1 (b) 2 (c) 3 (d) none of these

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To determine the number of points of discontinuity of the function \( |f(x)| \), we first need to analyze the function \( f(x) \) given by:

\[ f(x) = \begin{cases} x + 2 & \text{if } x < 0 \\ -x^2 - 2 & \text{if } 0 \leq x < 1 \\ x & \text{if } x \geq 1 \end{cases} \]

### Step 1: Identify the points of interest
The function \( f(x) \) has potential points of discontinuity at the boundaries of the piecewise segments, which are \( x = 0 \) and \( x = 1 \).

### Step 2: Check continuity at \( x = 0 \)
1. **Left-hand limit as \( x \to 0^- \)**:
\[ \lim_{x \to 0^-} f(x) = 0 + 2 = 2 \]
Therefore, \( |f(0^-)| = |2| = 2 \).

2. **Right-hand limit as \( x \to 0^+ \)**:
\[ \lim_{x \to 0^+} f(x) = -0^2 - 2 = -2 \]
Therefore, \( |f(0^+)| = |-2| = 2 \).

3. **Value of \( f(0) \)**:
\[ f(0) = -0^2 - 2 = -2 \quad \Rightarrow \quad |f(0)| = |-2| = 2 \]

Since the left-hand limit, right-hand limit, and the value of the function at \( x = 0 \) are all equal, \( |f(x)| \) is continuous at \( x = 0 \).

### Step 3: Check continuity at \( x = 1 \)
1. **Left-hand limit as \( x \to 1^- \)**:
\[ \lim_{x \to 1^-} f(x) = -1^2 - 2 = -3 \]
Therefore, \( |f(1^-)| = |-3| = 3 \).

2. **Right-hand limit as \( x \to 1^+ \)**:
\[ \lim_{x \to 1^+} f(x) = 1 \]
Therefore, \( |f(1^+)| = |1| = 1 \).

3. **Value of \( f(1) \)**:
\[ f(1) = 1 \quad \Rightarrow \quad |f(1)| = |1| = 1 \]

Since the left-hand limit \( |f(1^-)| = 3 \) is not equal to the right-hand limit \( |f(1^+)| = 1 \), \( |f(x)| \) is discontinuous at \( x = 1 \).

### Conclusion
The only point of discontinuity for \( |f(x)| \) is at \( x = 1 \). Therefore, the number of points of discontinuity of \( |f(x)| \) is:

\[ \text{Number of points of discontinuity} = 1 \]

Thus, the correct answer is (a) 1.