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The question asks which of the given piecewise functions shows continuity ("Fungsi manakah yang menunjukkan kekontinuan?"). To solve this, we need to check if the function is continuous at the boundary points where the function definition changes. This involves ensuring that the limits from both sides of the boundary are equal and that the function value at the boundary is defined and matches these limits.

Let's analyze the four options for continuity:

1. **First function**:
   \[
   f(x) =
   \begin{cases}
   \frac{x^2 - 1}{x - 1}, & x < 1 \\
   x^2 + 1, & x \geq 1
   \end{cases}
   \]
   The boundary is at \( x = 1 \). We check the limit from the left (\( x \to 1^- \)) and the limit from the right (\( x \to 1^+ \)).

   For \( x < 1 \):
   \[
   f(x) = \frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1} = x + 1 \quad (\text{for } x \neq 1)
   \]
   As \( x \to 1^- \), \( f(x) \to 2 \).

   For \( x \geq 1 \), \( f(x) = x^2 + 1 \), so as \( x \to 1^+ \), \( f(x) = 1^2 + 1 = 2 \).

   The limit from both sides is equal to 2, and \( f(1) = 2 \). Therefore, the function is continuous at \( x = 1 \).

2. **Second function**:
   \[
   f(x) =
   \begin{cases}
   x^2 + 1, & x < 2 \\
   2x - 1, & x \geq 2
   \end{cases}
   \]
   We check at \( x = 2 \).

   For \( x < 2 \), \( f(x) = x^2 + 1 \). As \( x \to 2^- \), \( f(x) = 2^2 + 1 = 5 \).

   For \( x \geq 2 \), \( f(x) = 2x - 1 \). As \( x \to 2^+ \), \( f(x) = 2(2) - 1 = 3 \).

   The limits from both sides are not equal, so the function is **not continuous** at \( x = 2 \).

3. **Third function**:
   \[
   f(x) =
   \begin{cases}
   \frac{x^2 - 1}{x}, & x < 1 \\
   \frac{x^2 - 4}{x - 2}, & x \geq 1
   \end{cases}
   \]
   The boundary is at \( x = 1 \). Let's check the limits.

   For \( x < 1 \), \( f(x) = \frac{x^2 - 1}{x} = x - 1 \). As \( x \to 1^- \), \( f(x) = 1 - 1 = 0 \).

   For \( x \geq 1 \), \( f(x) = \frac{x^2 - 4}{x - 2} = x + 2 \). As \( x \to 1^+ \), \( f(x) = 1 + 2 = 3 \).

   The limits from both sides are not equal, so the function is **not continuous** at \( x = 1 \).

4. **Fourth function**:
   \[
   f(x) =
   \begin{cases}
   x^2 + 3, & x < 2 \\
   \frac{x^2 - 4}{x - 2}, & x \geq 2
   \end{cases}
   \]
   We check at \( x = 2 \).

   For \( x < 2 \), \( f(x) = x^2 + 3 \). As \( x \to 2^- \), \( f(x) = 2^2 + 3 = 7 \).

   For \( x \geq 2 \), \( f(x) = \frac{x^2 - 4}{x - 2} = x + 2 \) (for \( x \neq 2 \)). As \( x \to 2^+ \), \( f(x) = 4 \).

   The limits from both sides are not equal, so the function is **not continuous** at \( x = 2 \).

### Conclusion:
The first function is the only one that is continuous at the boundary point. Therefore, the correct answer is the first option.

Would you like a deeper explanation of any part? Here are some follow-up questions you could explore:

1. What are the conditions for a function to be continuous at a point?
2. How does factoring help simplify expressions to check continuity?
3. What is the significance of limits in determining continuity?
4. How can discontinuities in piecewise functions appear?
5. Can you describe different types of discontinuities?

**Tip**: Always check the limit from both sides and compare with the function value at the boundary to ensure continuity!

Fungsi manakah yang menunjukkan kekontinuan?

This problem focuses on analyzing which piecewise function is continuous by checking the limits and values at the boundary points. It involves understanding the concept of continuity and the limit definition. Suitable for students in Grades 11-12.

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