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To solve the problem and determine the value of \( a \) such that the function \( f(x) \) is continuous at \( x = 2 \), we need to ensure that the left-hand limit and right-hand limit of \( f(x) \) at \( x = 2 \) are equal, and that they also equal \( f(2) \).

The function \( f(x) \) is given as:

\[
f(x) =
\begin{cases} 
\frac{1 - 2a}{x}, & \text{if } x \leq 2 \\
a^2x^2, & \text{if } x > 2 
\end{cases}
\]

### Continuity condition:
To be continuous at \( x = 2 \), we require:
\[
\lim_{x \to 2^-} f(x) = f(2) = \lim_{x \to 2^+} f(x)
\]

1. **Left-hand limit (as \( x \to 2^- \)):**
   \[
   \lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} \frac{1 - 2a}{x} = \frac{1 - 2a}{2}
   \]

2. **Right-hand limit (as \( x \to 2^+ \)):**
   \[
   \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} a^2x^2 = a^2(2)^2 = 4a^2
   \]

3. **Function value at \( x = 2 \):**
   From the first piece of the function for \( x \leq 2 \), we have:
   \[
   f(2) = \frac{1 - 2a}{2}
   \]

### Setting the limits equal:
For continuity at \( x = 2 \), we equate the left-hand and right-hand limits:
\[
\frac{1 - 2a}{2} = 4a^2
\]

### Solving for \( a \):

1. Multiply both sides by 2 to eliminate the denominator:
   \[
   1 - 2a = 8a^2
   \]

2. Rearrange the equation:
   \[
   8a^2 + 2a - 1 = 0
   \]

3. Solve the quadratic equation using the quadratic formula:
   \[
   a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   Here, \( a = 8 \), \( b = 2 \), and \( c = -1 \).

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

4. This gives two solutions:
   \[
   a_1 = \frac{-2 + 6}{16} = \frac{4}{16} = \frac{1}{4}
   \]
   \[
   a_2 = \frac{-2 - 6}{16} = \frac{-8}{16} = -\frac{1}{2}
   \]

Thus, the values of \( a \) that make the function continuous at \( x = 2 \) are \( a = \frac{1}{4} \) and \( a = -\frac{1}{2} \).

Would you like to go deeper into any part of this explanation or have any questions? 

Here are 5 follow-up questions for you:

1. Why is the condition for continuity important in functions like this?
2. How does the quadratic formula help in solving equations with higher degrees?
3. What is the significance of having two possible values for \( a \)?
4. Can a function be continuous but not differentiable at a point? Why or why not?
5. How would you graph the function \( f(x) \) for different values of \( a \)?

**Tip:** When solving quadratic equations, always simplify the equation first, and double-check your calculations when using the quadratic formula.

Tentukan nilai a agar f(x) kontinu di x = 2 dengan f(x) = \begin{cases} \frac{1 - 2a}{x}, & jika \ x \leq 2 \ \ a^2x^2, & jika \ x > 2 \end{cases}

Learn how to find the value of a such that the function f(x) is continuous at x = 2. This problem involves key concepts like continuity, limits, and quadratic equations, suitable for students in Grades 10-12.

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