Let A={(alpha, beta) in R times R :|alpha-1| | vedclass

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(A) For set $A$,we have $|\alpha-1| \leq 4$ and $|\beta-5| \leq 6$.This implies $-4 \leq \alpha-1 \leq 4$,so $-3 \leq \alpha \leq 5$.And $-6 \leq \bet

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$B \subset A$

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(A) For set $A$,we have $|\alpha-1| \leq 4$ and $|\beta-5| \leq 6$.This implies $-4 \leq \alpha-1 \leq 4$,so $-3 \leq \alpha \leq 5$.And $-6 \leq \beta-5 \leq 6$,so $-1 \leq \beta \leq 11$.Thus,$A$ represents a rectangular region defined by $\alpha \in [-3, 5]$ and $\beta \in [-1, 11]$.For set $B$,we have $16(\alpha-2)^2+9(\beta-6)^2 \leq 144$.Dividing by $144$,we get $\frac{(\alpha-2)^2}{9} + \frac{(\beta-6)^2}{16} \leq 1$.This is an ellipse centered at $(2, 6)$ with semi-major axis $b=4$ (along $\beta$) and semi-minor axis $a=3$ (along $\alpha$).The range of $\alpha$ for $B$ is $[2-3, 2+3] = [-1, 5]$,which is contained in $[-3, 5]$.The range of $\beta$ for $B$ is $[6-4, 6+4] = [2, 10]$,which is contained in $[-1, 11]$.Since the entire elliptical region $B$ lies within the rectangular region $A$,we conclude $B \subset A$.

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

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