Solved: Let \[ A = \{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R} : |\alpha

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

Correct Answer: \(B \subset A\) Explanation: For set \(A\): \[ |x - 1| \leq 4 \;\; \Rightarrow \;\; -3 \leq x \leq 5 \] \[ |y - 5| \leq 6 \;\; \Rightarrow \;\; -1 \leq y \leq 11 \] So \(A\) represents a rectangle with vertices \((-3,-1), (5,-1), (5,11), (-3,11)\). For set \(B\): \[ 16(x - 2)^2 + 9(y - 6)^2 \leq 144 \] \[ \Rightarrow \;\; \frac{(x-2)^2}{9} + \frac{(y-6)^2}{16} \leq 1 \] This is an ellipse centered at \((2,6)\) with semi-axes \(3\) (along \(x\)-axis) and \(4\) (along \(y\)-axis). Clearly, the ellipse \(B\) lies entirely inside the rectangle \(A\). \[ \therefore \quad B \subset A \]

Class: JEE Subject: Math Topic: Relations and Functions Year: 2025

Question:

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

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

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