Two sets A and B are defined as follows:A = { | vedclass

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(A) The set $A$ is defined by $|a - 5| < 1$ and $|b - 5| < 1$. Let $x = a - 5$ and $y = b - 5$. Then $A$ represents the interior of a square centered

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$A \subset B$

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(A) The set $A$ is defined by $|a - 5| The set $B$ is defined by $4(a - 6)^2 + 9(b - 5)^2 \le 36$. Substituting $x = a - 5$ and $y = b - 5$,we get $4(x - 1)^2 + 9y^2 \le 36$,which simplifies to $\frac{(x - 1)^2}{9} + \frac{y^2}{4} \le 1$. This represents the interior and boundary of an ellipse centered at $(1, 0)$ in the $(x, y)$ plane.The vertices of the square $A$ in the $(x, y)$ plane are $(1, 1), (-1, 1), (-1, -1), (1, -1)$.Checking if these points satisfy the ellipse inequality $\frac{(x - 1)^2}{9} + \frac{y^2}{4} \le 1$:For $(1, 1): \frac{0}{9} + \frac{1}{4} = 0.25 \le 1$ (True)For $(-1, 1): \frac{4}{9} + \frac{1}{4} = \frac{16+9}{36} = \frac{25}{36} \le 1$ (True)For $(-1, -1): \frac{4}{9} + \frac{1}{4} = \frac{25}{36} \le 1$ (True)For $(1, -1): \frac{0}{9} + \frac{1}{4} = 0.25 \le 1$ (True)Since all vertices of the square lie within the ellipse,$A \subset B$.

Two sets $A$ and $B$ are defined as follows:
$A = \{ (a,b) \in R \times R : |a - 5| < 1 \text{ and } |b - 5| < 1 \}$
$B = \{ (a,b) \in R \times R : 4(a - 6)^2 + 9(b - 5)^2 \le 36 \}$
Then:

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Two sets $A$ and $B$ are defined as follows:$A = \{ (a,b) \in R \times R : |a - 5| < 1 \text{ and } |b - 5| < 1 \}$$B = \{ (a,b) \in R \times R : 4(a - 6)^2 + 9(b - 5)^2 \le 36 \}$Then: