Let :f:[-1,3]to R be defined as
`f(x)={{:(|x|+[x]",", -1lexlt1),(x+|x|",", 1lexlt2),(x+[x]",", 2lexle3):}`
Where [t] denotes the greatest integer less than or equal to t. then, f is continuous at :

Options

A. only one point
B. only two point
C. onlythree point
D. four or more points

Answer

To determine the continuity of the function \( f \) defined on the interval \([-1, 3]\), we need to analyze the function piece by piece and check the continuity at the boundaries of the intervals defined in the piecewise function.

### Step 1: Define the function clearly
The function \( f(x) \) is defined as follows:
1. For \( -1 \leq x < 1 \):
\[ f(x) = |x| + [x] \]
2. For \( 1 \leq x < 2 \):
\[ f(x) = x + |x| \]
3. For \( 2 \leq x \leq 3 \):
\[ f(x) = x + [x] \]

### Step 2: Check continuity at the boundaries
We need to check the continuity of \( f \) at the points where the definition of the function changes, which are \( x = 1 \) and \( x = 2 \).

#### Check continuity at \( x = 1 \):
- Calculate \( f(1) \):
\[ f(1) = 1 + |1| = 1 + 1 = 2 \]
- Calculate the left-hand limit as \( x \) approaches 1:
\[ \lim_{x \to 1^-} f(x) = |1| + [1] = 1 + 1 = 2 \]
- Calculate the right-hand limit as \( x \) approaches 1:
\[ \lim_{x \to 1^+} f(x) = 1 + |1| = 1 + 1 = 2 \]
- Since \( f(1) = 2 \), \( \lim_{x \to 1^-} f(x) = 2 \), and \( \lim_{x \to 1^+} f(x) = 2 \), we conclude that \( f \) is continuous at \( x = 1 \).

#### Check continuity at \( x = 2 \):
- Calculate \( f(2) \):
\[ f(2) = 2 + [2] = 2 + 2 = 4 \]
- Calculate the left-hand limit as \( x \) approaches 2:
\[ \lim_{x \to 2^-} f(x) = 2 + |2| = 2 + 2 = 4 \]
- Calculate the right-hand limit as \( x \) approaches 2:
\[ \lim_{x \to 2^+} f(x) = 2 + [2] = 2 + 2 = 4 \]
- Since \( f(2) = 4 \), \( \lim_{x \to 2^-} f(x) = 4 \), and \( \lim_{x \to 2^+} f(x) = 4 \), we conclude that \( f \) is continuous at \( x = 2 \).

### Step 3: Check continuity at the endpoints
- At \( x = -1 \):
\[ f(-1) = |-1| + [-1] = 1 - 1 = 0 \]
The left limit does not exist as it is the left endpoint.

- At \( x = 3 \):
\[ f(3) = 3 + [3] = 3 + 3 = 6 \]
The right limit does not exist as it is the right endpoint.

### Conclusion
The function \( f \) is continuous for all \( x \in [-1, 3] \) except at \( x = 3 \).

### Final Answer
The function \( f \) is continuous at all points in the interval \([-1, 3]\) except at \( x = 3 \).