The function `f(x){{:(x-1",",xlt2),(2x-3",",xgt2):}` is continuous function

Options

A. at x=2 only
B. for all real values of x
C. for all real values of x except 2
D. for all integral values of x only

Answer

To determine if the function \( f(x) \) is continuous at \( x = 2 \), we need to check the following conditions:

1. **Function Value at \( x = 2 \)**: We need to find \( f(2) \).
2. **Left-Hand Limit (LHL)**: We need to calculate \( \lim_{x \to 2^-} f(x) \).
3. **Right-Hand Limit (RHL)**: We need to calculate \( \lim_{x \to 2^+} f(x) \).
4. **Check Continuity**: We need to verify if \( f(2) = \text{LHL} = \text{RHL} \).

### Step-by-Step Solution:

**Step 1: Find \( f(2) \)**
Since \( f(x) = 2x - 3 \) for \( x > 2 \), we will use this expression to find \( f(2) \):
\[ f(2) = 2(2) - 3 = 4 - 3 = 1 \]

**Step 2: Calculate the Left-Hand Limit (LHL)**
The left-hand limit as \( x \) approaches 2 from the left is given by the expression \( f(x) = x - 1 \) for \( x < 2 \):
\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (x - 1) = 2 - 1 = 1 \]

**Step 3: Calculate the Right-Hand Limit (RHL)**
The right-hand limit as \( x \) approaches 2 from the right is given by the expression \( f(x) = 2x - 3 \) for \( x > 2 \):
\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (2x - 3) = 2(2) - 3 = 4 - 3 = 1 \]

**Step 4: Check Continuity**
Now we check if the function is continuous at \( x = 2 \):
\[ f(2) = 1, \quad \text{LHL} = 1, \quad \text{RHL} = 1 \]
Since \( f(2) = \text{LHL} = \text{RHL} = 1 \), the function \( f(x) \) is continuous at \( x = 2 \).

### Conclusion:
The function \( f(x) \) is continuous for all real values of \( x \).