Let f(x) be defined in the interval [0,4] such that
`f(x) ={{:(1-x,0lex le 1),( x+2, 1lt xlt2),( 4-x, 2 le x le 4):}` , then the number of points where f(x) is discontinuous 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 where the function \( f(x) \) is discontinuous, we will analyze the piecewise function defined as follows:

\[ f(x) = \begin{cases} 1 - x & \text{for } 0 \leq x \leq 1 \\ x + 2 & \text{for } 1 < x < 2 \\ 4 - x & \text{for } 2 \leq x \leq 4 \end{cases} \]

### Step 1: Identify the points of potential discontinuity
The points where the function changes its definition are at \( x = 1 \) and \( x = 2 \). We will check the continuity at these points.

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

2. **Function value** at \( x = 1 \):
\[ f(1) = 1 - 1 = 0 \]

3. **Right-hand limit** as \( x \) approaches 1:
\[ \lim_{x \to 1^+} f(x) = 1 + 2 = 3 \]

Since the left-hand limit \( (0) \) does not equal the right-hand limit \( (3) \), \( f(x) \) is discontinuous at \( x = 1 \).

### Step 3: Check continuity at \( x = 2 \)
1. **Left-hand limit** as \( x \) approaches 2:
\[ \lim_{x \to 2^-} f(x) = 2 + 2 = 4 \]

2. **Function value** at \( x = 2 \):
\[ f(2) = 4 - 2 = 2 \]

3. **Right-hand limit** as \( x \) approaches 2:
\[ \lim_{x \to 2^+} f(x) = 4 - 2 = 2 \]

Since the left-hand limit \( (4) \) does not equal the function value \( (2) \), \( f(x) \) is discontinuous at \( x = 2 \).

### Step 4: Check continuity at other points
- For \( x = 0 \): \( f(0) = 1 - 0 = 1 \) (continuous)
- For \( x = 4 \): \( f(4) = 4 - 4 = 0 \) (continuous)

### Conclusion
The function \( f(x) \) is discontinuous at two points: \( x = 1 \) and \( x = 2 \).

Thus, the number of points where \( f(x) \) is discontinuous is **2**.

### Final Answer
(b) 2