Question

Let $ A = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6\} $ and $ B = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144\}. $ Then:

• A. \( B \subset A \)
• B. \( A \cup B = \{(x, y) : -4 \leq x \leq 4, -1 \leq y \leq 11 \} \)
• C. neither \( A \subset B \) nor \( B \subset A \)
• D. \( A \subset B \)

Hint:

When comparing sets like these, visualize the shapes and check if one is completely contained within the other.

Answer 1

The Correct Option is A

To solve the given problem, let's analyze the sets \( A \) and \( B \) and determine the relationship between them.

1. Start with Set \( A \): \(A = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6\}\).
   • The condition \(|\alpha - 1| \leq 4\) implies that \(-4 \leq \alpha - 1 \leq 4\), or \(-3 \leq \alpha \leq 5\).
   • The condition \(|\beta - 5| \leq 6\) implies that \(-6 \leq \beta - 5 \leq 6\), or \(-1 \leq \beta \leq 11\).
   • Thus, the set \( A \) represents a rectangle with vertices at \((-3, -1)\), \((5, -1)\), \((5, 11)\), and \((-3, 11)\).
2. Next, consider Set \( B \): \(B = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144\}\).
   • This inequality represents an ellipse centered at \((2, 6)\).
   • The equation \(\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \leq 1\) shows the semi-major axis along the \( \beta \)-axis with length \(8\) and the semi-minor axis along the \( \alpha \)-axis with length \(3\).
3. Let's determine the boundaries of each set:
   • The ellipse \( B \) lies within the rectangle \( A \) because:
     • For an ellipse centered at \((2, 6)\) with a semi-horizontal axis of length \(3\), the \( \alpha \)-range is \([-1, 5]\), which is within \([-3, 5]\).
     • For a vertical semi-axis of \(8\), the \( \beta \)-range is \([-2, 14]\), mostly within \([-1, 11]\).
     • Thus, every point of the ellipse \( B \) lies within the rectangle \( A \), confirming \(B \subset A\).

Thus, after analyzing both shapes and their boundaries, it is clear that the correct answer is \( B \subset A \).

Answer 2

We are given two sets \( A \) and \( B \) defined by: \[ A = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6\} \] This defines a rectangular region where \( \alpha \) lies between \( -3 \) and \( 5 \), and \( \beta \) lies between \( -1 \) and \( 11 \). \[ B = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144\} \] This defines an ellipse with center \( (2, 6) \), semi-major axis 4 along the \( \beta \)-axis, and semi-minor axis 3 along the \( \alpha \)-axis. 
We see that the ellipse \( B \) fits entirely within the rectangle \( A \), meaning that \( B \subset A \).