Key Concepts and Formulas

• Absolute Value Inequalities: The inequality $|x - c| < r$ represents all points $x$ within a distance $r$ from $c$, which translates to $c - r < x < c + r$.
• Geometric Interpretation of Inequalities: Understanding how inequalities in two variables ($a$, $b$) define regions in the Cartesian plane (e.g., open squares, closed ellipses).
• Subset Definition: A set $X$ is a subset of set $Y$ ($X \subset Y$) if every element in $X$ is also an element in $Y$.

Step-by-Step Solution

Step 1: Analyze and interpret Set A. Set A is defined by the conditions $|a - 5| < 1$ and $|b - 5| < 1$. The inequality $|a - 5| < 1$ means $5 - 1 < a < 5 + 1$, which simplifies to $4 < a < 6$. The inequality $|b - 5| < 1$ means $5 - 1 < b < 5 + 1$, which simplifies to $4 < b < 6$. Therefore, Set A represents all points $(a, b)$ in the Cartesian plane such that $4 < a < 6$ and $4 < b < 6$. This is an open square with vertices at $(4, 4), (6, 4), (6, 6), (4, 6)$. The center of this square is $(5, 5)$.

Step 2: Analyze and interpret Set B. Set B is defined by the inequality $4(a - 6)^2 + 9(b - 5)^2 \le 36$. To better understand this region, we can divide the inequality by 36: $\frac{4(a - 6)^2}{36} + \frac{9(b - 5)^2}{36} \le \frac{36}{36}$ $\frac{(a - 6)^2}{9} + \frac{(b - 5)^2}{4} \le 1$ This is the standard form of an inequality representing a closed elliptical region. The center of the ellipse is $(h, k) = (6, 5)$. The semi-major axis squared is $a^2 = 9$, so the semi-major axis is $a = 3$ (along the horizontal direction). The semi-minor axis squared is $b^2 = 4$, so the semi-minor axis is $b = 2$ (along the vertical direction). The ellipse extends horizontally from $6 - 3 = 3$ to $6 + 3 = 9$ and vertically from $5 - 2 = 3$ to $5 + 2 = 7$.

Step 3: Compare Set A and Set B to determine subset relationships. We need to check if every point in A is in B, and if every point in B is in A.

• Is A $\subset$ B? This means checking if every point $(a, b)$ satisfying $4 < a < 6$ and $4 < b < 6$ also satisfies $\frac{(a - 6)^2}{9} + \frac{(b - 5)^2}{4} \le 1$. Let's consider a point within the open square A, for example, the center of the square $(5, 5)$. For $(5, 5)$: $4 < 5 < 6$ and $4 < 5 < 6$. So $(5, 5) \in A$. Now check if $(5, 5) \in B$: $\frac{(5 - 6)^2}{9} + \frac{(5 - 5)^2}{4} = \frac{(-1)^2}{9} + \frac{0^2}{4} = \frac{1}{9} + 0 = \frac{1}{9}$ Since $\frac{1}{9} \le 1$, the point $(5, 5)$ is in Set B. This single point does not prove $A \subset B$, but it shows that the center of A is within B.

  Let's consider a point near the boundary of A, for example, $(6, 5)$. While A is an open square and does not include its boundary, we can consider points arbitrarily close to it. Let's test a point that is in A, say $(5.9, 5.9)$. For $(5.9, 5.9)$: $4 < 5.9 < 6$ and $4 < 5.9 < 6$. So $(5.9, 5.9) \in A$. Check if $(5.9, 5.9) \in B$: $\frac{(5.9 - 6)^2}{9} + \frac{(5.9 - 5)^2}{4} = \frac{(-0.1)^2}{9} + \frac{(0.9)^2}{4} = \frac{0.01}{9} + \frac{0.81}{4} \approx 0.0011 + 0.2025 = 0.2036$ Since $0.2036 \le 1$, this point is also in B.

  However, let's consider a point in A that is furthest from the center of B $(6, 5)$. The points in A are in the region $4 < a < 6$ and $4 < b < 6$. Let's test a point like $(4.1, 4.1)$ which is in A. For $(4.1, 4.1)$: $4 < 4.1 < 6$ and $4 < 4.1 < 6$. So $(4.1, 4.1) \in A$. Check if $(4.1, 4.1) \in B$: $\frac{(4.1 - 6)^2}{9} + \frac{(4.1 - 5)^2}{4} = \frac{(-1.9)^2}{9} + \frac{(-0.9)^2}{4} = \frac{3.61}{9} + \frac{0.81}{4} \approx 0.4011 + 0.2025 = 0.6036$ Since $0.6036 \le 1$, this point is also in B.

  Let's consider a point in A that is close to the corner $(6, 6)$. For example, let's consider a point $(5.9, 5.9)$. We already checked this and it's in B. Consider the point $(5.9, 5.9)$. $a = 5.9$, $b = 5.9$. $|a - 5| = |5.9 - 5| = 0.9 < 1$. $|b - 5| = |5.9 - 5| = 0.9 < 1$. So $(5.9, 5.9) \in A$. Now check for B: $4(5.9 - 6)^2 + 9(5.9 - 5)^2 = 4(-0.1)^2 + 9(0.9)^2 = 4(0.01) + 9(0.81) = 0.04 + 7.29 = 7.33$. Since $7.33 \not\le 36$, the point $(5.9, 5.9)$ is NOT in B. Therefore, A is NOT a subset of B.

• Is B $\subset$ A? This means checking if every point $(a, b)$ satisfying $\frac{(a - 6)^2}{9} + \frac{(b - 5)^2}{4} \le 1$ also satisfies $4 < a < 6$ and $4 < b < 6$. Consider the center of the ellipse, which is $(6, 5)$. For $(6, 5)$: $4(6 - 6)^2 + 9(5 - 5)^2 = 4(0)^2 + 9(0)^2 = 0 \le 36$. So $(6, 5) \in B$. Now check if $(6, 5) \in A$: $|a - 5| = |6 - 5| = |1| = 1$. The condition for A is $|a - 5| < 1$, so $1 < 1$ is false. $|b - 5| = |5 - 5| = |0| = 0$. The condition for A is $|b - 5| < 1$, so $0 < 1$ is true. Since $|a - 5| < 1$ is not satisfied, the point $(6, 5)$ is NOT in A. Since we found a point in B that is not in A, B is NOT a subset of A.

Step 4: Evaluate the options based on the subset relationships.

• Option (A) neither A $\subset$ B nor B $\subset$ A. We found this to be true.
• Option (B) B $\subset$ A. We found this to be false.
• Option (C) A $\subset$ B. We found this to be false.
• Option (D) A $\cap$ B = $\phi$. We found points like $(5, 5)$ which are in both A and B, so the intersection is not empty. For example, $(5,5)$ is in A since $|5-5|<1$ and $|5-5|<1$. And $(5,5)$ is in B since $4(5-6)^2 + 9(5-5)^2 = 4(-1)^2 + 9(0)^2 = 4 \le 36$. Thus $A \cap B \ne \phi$.

Our analysis shows that neither set is a subset of the other, and their intersection is not empty.

Common Mistakes & Tips

• Confusing open and closed regions: Pay close attention to strict inequalities ($<, >$) versus non-strict inequalities ($\le, \ge$). Set A is an open square (boundaries not included), while Set B is a closed ellipse (boundary is included).
• Testing only the center: While testing the center of a region is a good starting point, it's crucial to test points near the boundaries or corners to definitively prove or disprove subset relationships.
• Algebraic vs. Geometric Interpretation: Visualizing the shapes (square and ellipse) in the Cartesian plane can greatly aid in understanding their relationship. The center of the square is $(5, 5)$ and the center of the ellipse is $(6, 5)$. The square is defined by $4 < a < 6$ and $4 < b < 6$. The ellipse is centered at $(6, 5)$ and extends from $a=3$ to $a=9$ and $b=3$ to $b=7$.

Summary

We analyzed Set A as an open square region defined by $4 < a < 6$ and $4 < b < 6$, and Set B as a closed elliptical region defined by $\frac{(a - 6)^2}{9} + \frac{(b - 5)^2}{4} \le 1$. By testing specific points, we determined that the point $(5.9, 5.9)$, which is in Set A, is not in Set B, thus disproving $A \subset B$. We also found that the point $(6, 5)$, which is in Set B, is not in Set A, thus disproving $B \subset A$. Since neither set is a subset of the other, and their intersection is not empty, option (A) is the correct conclusion.

The final answer is \boxed{A}.