Let $A = \{ (\alpha, \beta) \in R \times R : |\alpha - 1| \leq 4 \text{ and } |\beta - 5| \leq 6 \}$ and $B = \{ (\alpha, \beta) \in R \times R : 16(\alpha - 2)^2 + 9(\beta - 6)^2 \leq 144 \}$. Then
(A)
$A \subset B$
(B)
$B \subset A$
(C)
neither $A \subset B$ nor $B \subset A$
(D)
$A \cup B = \{ (x, y) : -4 \leqslant x \leqslant 4, -1 \leqslant y \leqslant 11 \}$
MEDIUM
Correct Answer:B
Explanation
For set A, we have $|x - 1| \leq 4$ and $|y - 5| \leq 6$. This implies $-4 \leq x - 1 \leq 4$ and $-6 \leq y - 5 \leq 6$. Therefore, $-3 \leq x \leq 5$ and $-1 \leq y \leq 11$. For set B, we have $16(x - 2)^2 + 9(y - 6)^2 \leq 144$, which simplifies to $\frac{(x - 2)^2}{9} + \frac{(y - 6)^2}{16} \leq 1$. This represents an ellipse centered at $(2, 6)$ with semi-major axis 4 and semi-minor axis 3. From the diagram, it's clear that $B \subset A$.