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To solve this problem, we need to ensure that the function \( f(x) \) is continuous at \( x = 2 \). For a function to be continuous at a point, the following conditions must be met:

1. \( f(2) \) must exist.
2. The left-hand limit as \( x \to 2^- \) must equal the right-hand limit as \( x \to 2^+ \).
3. The left-hand and right-hand limits must equal \( f(2) \).

### The function is defined as:
\[
f(x) =
\begin{cases} 
\frac{1 - 2a}{x}, & \text{jika } x \leq 2 \\
a^2 x^2, & \text{jika } x > 2 
\end{cases}
\]

#### Step 1: Calculate \( f(2) \).
Since \( f(x) = \frac{1 - 2a}{x} \) when \( x \leq 2 \), we can substitute \( x = 2 \) directly into this expression:
\[
f(2) = \frac{1 - 2a}{2}
\]

#### Step 2: Calculate the left-hand limit as \( x \to 2^- \).
The left-hand limit is based on \( \frac{1 - 2a}{x} \), so:
\[
\lim_{x \to 2^-} f(x) = \frac{1 - 2a}{2}
\]

#### Step 3: Calculate the right-hand limit as \( x \to 2^+ \).
For \( x > 2 \), we use \( f(x) = a^2 x^2 \), so the right-hand limit is:
\[
\lim_{x \to 2^+} f(x) = a^2 (2^2) = 4a^2
\]

#### Step 4: Set the left-hand limit equal to the right-hand limit.
To ensure continuity at \( x = 2 \), the left-hand and right-hand limits must be equal:
\[
\frac{1 - 2a}{2} = 4a^2
\]

#### Step 5: Solve the equation.
Now, solve for \( a \) from the equation \( \frac{1 - 2a}{2} = 4a^2 \):
\[
1 - 2a = 8a^2
\]
\[
8a^2 + 2a - 1 = 0
\]
This is a quadratic equation in \( a \). We can solve it using the quadratic formula:
\[
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( b = 2 \), \( a = 8 \), and \( c = -1 \).

Substitute these values into the quadratic formula:
\[
a = \frac{-2 \pm \sqrt{(2)^2 - 4(8)(-1)}}{2(8)}
\]
\[
a = \frac{-2 \pm \sqrt{4 + 32}}{16}
\]
\[
a = \frac{-2 \pm \sqrt{36}}{16}
\]
\[
a = \frac{-2 \pm 6}{16}
\]

Thus, we have two solutions:
\[
a = \frac{-2 + 6}{16} = \frac{4}{16} = \frac{1}{4}
\]
or
\[
a = \frac{-2 - 6}{16} = \frac{-8}{16} = \frac{-1}{2}
\]

### Conclusion:
The values of \( a \) that make \( f(x) \) continuous at \( x = 2 \) are \( a = \frac{1}{4} \) or \( a = \frac{-1}{2} \).

Would you like a more detailed explanation of any of these steps?

### Follow-up Questions:
1. What are the conditions for a function to be continuous at a point?
2. How is the quadratic formula applied to solve quadratic equations?
3. What is the significance of left-hand and right-hand limits in continuity?
4. Can a function be discontinuous even if it is defined at a specific point?
5. How would the solution change if the function were defined differently for \( x > 2 \)?

### Tip:
When solving quadratic equations, always check both roots, as both may be valid depending on the context of the problem.

Tentukan nilai a agar f(x) kontinu di x = 2 dengan f(x) = { (1 - 2a)/x jika x ≤ 2; a^2x^2 jika x > 2 }.

Learn how to determine the value of a for which the function f(x) is continuous at x = 2. This problem covers key concepts in continuity, limits, and solving quadratic equations, with detailed step-by-step solutions. Suitable for students in Grades 11-12.

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