Key Concepts and Formulas

• Absolute Value Inequality: For a real number $x$ and a non-negative real number $r$, the inequality $|x - c| \leq r$ is equivalent to $c - r \leq x \leq c + r$.
• Standard Equation of an Ellipse: The inequality $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} \leq 1$ represents the region inside and on the boundary of an ellipse centered at $(h, k)$ with semi-major/minor axes of lengths $a$ and $b$ parallel to the x and y axes, respectively.
• Set Inclusion: For two sets $A$ and $B$, $A \subset B$ means that every element of $A$ is also an element of $B$. This implies that the geometric region representing $A$ must be entirely contained within the geometric region representing $B$.

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Step-by-Step Solution

Step 1: Analyze Set A and determine its geometric representation.

Set A is defined as $A = \{ (\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6 \}$. We will analyze the inequalities for $\alpha$ and $\beta$ separately.

• Inequality for $\alpha$: $|\alpha - 1| \leq 4$.

  • Reasoning: This inequality describes all real numbers $\alpha$ whose distance from $1$ is less than or equal to $4$.
  • Calculation: Using the property of absolute value inequalities, we get: $-4 \leq \alpha - 1 \leq 4$ Adding $1$ to all parts of the inequality to isolate $\alpha$: $-4 + 1 \leq \alpha - 1 + 1 \leq 4 + 1$ $-3 \leq \alpha \leq 5$
  • Interpretation: This means $\alpha$ lies in the closed interval $[-3, 5]$.

• Inequality for $\beta$: $|\beta - 5| \leq 6$.

  • Reasoning: This inequality describes all real numbers $\beta$ whose distance from $5$ is less than or equal to $6$.
  • Calculation: Using the property of absolute value inequalities, we get: $-6 \leq \beta - 5 \leq 6$ Adding $5$ to all parts of the inequality to isolate $\beta$: $-6 + 5 \leq \beta - 5 + 5 \leq 6 + 5$ $-1 \leq \beta \leq 11$
  • Interpretation: This means $\beta$ lies in the closed interval $[-1, 11]$.

• Geometric Representation of Set A:

  • Reasoning: Since the conditions on $\alpha$ and $\beta$ are independent, Set A represents a rectangular region in the $\alpha\beta$-plane.
  • Description: $A = \{ (\alpha, \beta) : -3 \leq \alpha \leq 5 \text{ and } -1 \leq \beta \leq 11 \}$. This is a rectangle with vertices at $(-3, -1), (5, -1), (-3, 11), (5, 11)$. The center of this rectangle is at $(1, 5)$.

Step 2: Analyze Set B and determine its geometric representation.

Set B is defined as $B = \{ (\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144 \}$. We need to convert this inequality into the standard form of an ellipse.

• Converting to standard form:

  • Reasoning: Dividing by a positive constant preserves the inequality and helps us identify the standard parameters of an ellipse.
  • Calculation: Divide the entire inequality by $144$: $\frac{16(\alpha - 2)^2}{144} + \frac{9(\beta - 6)^2}{144} \leq \frac{144}{144}$ Simplifying the fractions: $\frac{(\alpha - 2)^2}{9} + \frac{(\beta - 6)^2}{16} \leq 1$
  • Interpretation: This inequality represents the region inside and on the boundary of an ellipse.

• Identifying Ellipse Parameters:

  • Reasoning: Comparing the standard form $\frac{(\alpha-h)^2}{a^2} + \frac{(\beta-k)^2}{b^2} \leq 1$ with our derived inequality.
  • Center $(h, k)$: $(2, 6)$.
  • Semi-axis along $\alpha$ ($a$): $a^2 = 9 \implies a = 3$. The $\alpha$-range of the ellipse is $[h-a, h+a]$.
  • Semi-axis along $\beta$ ($b$): $b^2 = 16 \implies b = 4$. The $\beta$-range of the ellipse is $[k-b, k+b]$.

• Determining the Extents of Set B:

  • Reasoning: The extreme values of $\alpha$ and $\beta$ for points within the ellipse define its bounding box.
  • $\alpha$-range: $h - a \leq \alpha \leq h + a \implies 2 - 3 \leq \alpha \leq 2 + 3 \implies -1 \leq \alpha \leq 5$.
  • $\beta$-range: $k - b \leq \beta \leq k + b \implies 6 - 4 \leq \beta \leq 6 + 4 \implies 2 \leq \beta \leq 10$.
  • Description: Set B is an elliptical region centered at $(2, 6)$, extending from $\alpha = -1$ to $\alpha = 5$ horizontally, and from $\beta = 2$ to $\beta = 10$ vertically.

Step 3: Compare Set A and Set B to determine set inclusion.

We have the ranges for Set A: $-3 \leq \alpha \leq 5$ and $-1 \leq \beta \leq 11$. We have the ranges for Set B: $-1 \leq \alpha \leq 5$ and $2 \leq \beta \leq 10$.

• Check for $A \subset B$:

  • Reasoning: For $A \subset B$, every point in the rectangle A must also be in the ellipse B. We can test a point in A that is on the boundary of A but potentially outside the bounding box of B.
  • Test Point: Consider the point $(\alpha, \beta) = (-3, 5)$. This point is within A since $-3 \leq -3 \leq 5$ and $-1 \leq 5 \leq 11$.
  • Check if in B: Substitute $(-3, 5)$ into the inequality for B: $16(-3 - 2)^2 + 9(5 - 6)^2 \leq 144$ $16(-5)^2 + 9(-1)^2 \leq 144$ $16(25) + 9(1) \leq 144$ $400 + 9 \leq 144$ $409 \leq 144$
  • Conclusion: This is false. Therefore, the point $(-3, 5)$ is in A but not in B. So, $A \not\subset B$. Option (A) is incorrect.

• Check for $B \subset A$:

  • Reasoning: For $B \subset A$, every point in the ellipse B must also be in the rectangle A. This means the $\alpha$-range of B must be within the $\alpha$-range of A, and the $\beta$-range of B must be within the $\beta$-range of A.
  • Comparing $\alpha$-ranges: Range for A: $[-3, 5]$. Range for B: $[-1, 5]$. Since $-3 \leq -1$ and $5 \leq 5$, the interval $[-1, 5]$ is contained within $[-3, 5]$. Thus, the $\alpha$-condition for A is satisfied for all points in B.
  • Comparing $\beta$-ranges: Range for A: $[-1, 11]$. Range for B: $[2, 10]$. Since $-1 \leq 2$ and $10 \leq 11$, the interval $[2, 10]$ is contained within $[-1, 11]$. Thus, the $\beta$-condition for A is satisfied for all points in B.
  • Conclusion: Since both the $\alpha$ and $\beta$ ranges of the ellipse B are contained within the respective ranges of the rectangle A, every point in B is also in A. Therefore, $B \subset A$. This corresponds to option (B).

• Check Option (D): $A \cup B=\{(x, y):-4 \leqslant x \leqslant 4,-1 \leqslant y \leqslant 11\}$

  • Reasoning: Since we found $B \subset A$, the union $A \cup B$ is simply equal to $A$.
  • Comparison: Set A is defined by $-3 \leq \alpha \leq 5$ and $-1 \leq \beta \leq 11$. Option (D) proposes a union with $\alpha$ range $[-4, 4]$ and $\beta$ range $[-1, 11]$. The $\alpha$ range does not match. Therefore, option (D) is incorrect.

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Common Mistakes & Tips

• Misinterpreting Absolute Value: Ensure that $|x-c| \leq r$ is correctly translated to $c-r \leq x \leq c+r$.
• Algebraic Errors: Be meticulous with algebraic manipulations, especially when simplifying fractions or solving inequalities.
• Visualizing the Regions: Sketching the rectangle and the ellipse can provide a strong visual confirmation of the set inclusion relationship. The center and extents of both shapes are crucial for this visualization.
• Exhaustive Check for Inclusion: When checking if $A \subset B$, if you find even one point in A that is not in B, you can conclude $A \not\subset B$. Conversely, to prove $B \subset A$, you must show that all points in B satisfy the conditions for A.

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Summary

Set A defines a rectangular region in the $\alpha\beta$-plane with $\alpha \in [-3, 5]$ and $\beta \in [-1, 11]$. Set B defines an elliptical region centered at $(2, 6)$ with semi-axes $a=3$ and $b=4$, meaning its $\alpha$ values range from $-1$ to $5$ and its $\beta$ values range from $2$ to $10$. By comparing these ranges, we see that the $\alpha$-range of B is a subset of A's $\alpha$-range, and the $\beta$-range of B is a subset of A's $\beta$-range. This confirms that the entire elliptical region B is contained within the rectangular region A, leading to the conclusion that $B \subset A$.

The final answer is $\boxed{B}$.