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The problem asks to find the value of \( a \) such that the function \( f(x) \) is continuous at \( x = 2 \).

The piecewise function is given as:

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

### Step-by-step solution:

To ensure that the function is continuous at \( x = 2 \), the following condition must be satisfied:

\[
\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)
\]

#### 1. Left-hand limit \( \lim_{x \to 2^-} f(x) \):

For \( x \leq 2 \), \( f(x) = \frac{1 - 2a}{x} \). Substituting \( x = 2 \):

\[
\lim_{x \to 2^-} f(x) = \frac{1 - 2a}{2}
\]

#### 2. Right-hand limit \( \lim_{x \to 2^+} f(x) \):

For \( x > 2 \), \( f(x) = a^2x^2 \). Substituting \( x = 2 \):

\[
\lim_{x \to 2^+} f(x) = a^2(2^2) = 4a^2
\]

#### 3. Continuity at \( x = 2 \):

For continuity, the left-hand limit must equal the right-hand limit:

\[
\frac{1 - 2a}{2} = 4a^2
\]

#### 4. Solve for \( a \):

Multiply both sides of the equation by 2 to eliminate the denominator on the left side:

\[
1 - 2a = 8a^2
\]

Rearrange the equation:

\[
8a^2 + 2a - 1 = 0
\]

This is a quadratic equation in \( a \). Solve using the quadratic formula:

\[
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Where \( A = 8 \), \( B = 2 \), and \( C = -1 \). Substituting these values:

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

\[
a = \frac{-2 \pm 6}{16}
\]

Thus, the two possible solutions are:

\[
a = \frac{-2 + 6}{16} = \frac{4}{16} = \frac{1}{4}
\]
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 \( \frac{1}{4} \) or \( \frac{-1}{2} \).

---

Would you like further details or explanations?

Here are five related questions you might find useful:

1. What are the conditions for a function to be continuous at a point?
2. How does the quadratic formula help in solving non-linear equations?
3. Can you have a function that is continuous but not differentiable at a point?
4. What are piecewise functions and how are their limits calculated?
5. How do you check for continuity for different types of functions?

**Tip:** When solving for continuity at a point, always check both the left-hand and right-hand limits.

Tentukan nilai a agar f(x) kontiniu di x = 2 dengan f(x) = \frac{1 - 2a}{x} jika x \leq 2 dan f(x) = a^2x^2 jika x > 2.

Find the value of a to ensure the function f(x) is continuous at x = 2. This problem involves calculating limits from both sides of x = 2 and solving a quadratic equation for the parameter a. Suitable for high school or early college students.

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