The number of point of discountinuity of the function `f(x)=|x-1|+|x-2|+sinx,x in[0,4]` is

Options

A. 1
B. 2
C. 3
D. 0

Answer

To find the number of points of discontinuity of the function \( f(x) = |x - 1| + |x - 2| + \sin x \) for \( x \in [0, 4] \), we will analyze the function by breaking it down into intervals based on the points where the absolute value expressions change.

### Step 1: Identify critical points
The critical points for the absolute value functions are \( x = 1 \) and \( x = 2 \). We will consider the intervals created by these points:
- Interval 1: \( [0, 1) \)
- Interval 2: \( [1, 2) \)
- Interval 3: \( [2, 4] \)

### Step 2: Analyze each interval

**Interval 1: \( [0, 1) \)**
- In this interval, both \( |x - 1| \) and \( |x - 2| \) can be simplified:
\[ |x - 1| = 1 - x \quad \text{and} \quad |x - 2| = 2 - x \]
- Therefore, the function becomes:
\[ f(x) = (1 - x) + (2 - x) + \sin x = -2x + 3 + \sin x \]

**Interval 2: \( [1, 2) \)**
- In this interval, we have:
\[ |x - 1| = x - 1 \quad \text{and} \quad |x - 2| = 2 - x \]
- Thus, the function simplifies to:
\[ f(x) = (x - 1) + (2 - x) + \sin x = 1 + \sin x \]

**Interval 3: \( [2, 4] \)**
- In this interval, both absolute values are:
\[ |x - 1| = x - 1 \quad \text{and} \quad |x - 2| = x - 2 \]
- The function simplifies to:
\[ f(x) = (x - 1) + (x - 2) + \sin x = 2x - 3 + \sin x \]

### Step 3: Check continuity at critical points
Next, we need to check the continuity of the function at the critical points \( x = 1 \) and \( x = 2 \).

**At \( x = 1 \)**:
- From Interval 1:
\[ f(1) = -2(1) + 3 + \sin(1) = 1 + \sin(1) \]
- From Interval 2:
\[ f(1) = 1 + \sin(1) \]
- Since both values are equal, \( f(x) \) is continuous at \( x = 1 \).

**At \( x = 2 \)**:
- From Interval 2:
\[ f(2) = 1 + \sin(2) \]
- From Interval 3:
\[ f(2) = 2(2) - 3 + \sin(2) = 1 + \sin(2) \]
- Since both values are equal, \( f(x) \) is continuous at \( x = 2 \).

### Conclusion
Since the function \( f(x) \) is continuous at all points in the intervals and at the critical points \( x = 1 \) and \( x = 2 \), there are no points of discontinuity in the interval \( [0, 4] \).

Thus, the number of points of discontinuity of the function is **0**.