If `f(x) = {{:("sin"(pix)/(2)",",x lt 1),([x]",",x ge 1):}`, where [x] denotes the greatest integer function, then

Options

A. f(x) is continuous at x = 1
B. f(x) is discontinuous at x = 1
C. `f(1^(+)) = 0`
D. `f(1^(-)) = -1`

Answer

To determine the continuity of the function \( f(x) \) at \( x = 1 \), we will follow these steps:

### Step 1: Define the function
The function is given as:
\[ f(x) = \begin{cases} \frac{\sin(\pi x)}{2} & \text{if } x < 1 \\ |x| & \text{if } x \geq 1 \end{cases} \]

### Step 2: Find \( f(1) \)
Since \( 1 \geq 1 \), we use the second case of the function:
\[ f(1) = |1| = 1 \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 1
For \( x \) approaching 1 from the right (i.e., \( x \to 1^+ \)), we use the second case of the function:
\[ \lim_{x \to 1^+} f(x) = |1| = 1 \]

### Step 4: Calculate the left-hand limit as \( x \) approaches 1
For \( x \) approaching 1 from the left (i.e., \( x \to 1^- \)), we use the first case of the function:
\[ \lim_{x \to 1^-} f(x) = \frac{\sin(\pi \cdot 1)}{2} = \frac{\sin(\pi)}{2} = \frac{0}{2} = 0 \]

### Step 5: Compare the limits and function value
Now we compare the left-hand limit, right-hand limit, and the function value at \( x = 1 \):
- \( f(1) = 1 \)
- \( \lim_{x \to 1^+} f(x) = 1 \)
- \( \lim_{x \to 1^-} f(x) = 0 \)

### Step 6: Conclusion on continuity
Since the left-hand limit \( (0) \) does not equal the right-hand limit \( (1) \) and does not equal \( f(1) \), we conclude that the function \( f(x) \) is not continuous at \( x = 1 \).

### Final Answer
The function \( f(x) \) is not continuous at \( x = 1 \).

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