Let \(A={(\alpha ,\beta )\in \mathrm{R}\times \mathrm{R}:|\alpha -1|\leq 4\text{ and }|\beta -5|\leq 6}\) and \(B=\left{…

Set A:

\(A=\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}:|\alpha-1| \leq 4\) and \(|\beta-5| \leq 6\}\)

This represents a rectangle with:

\(−3\leq \alpha \leq 5\) (from \(\alpha −1\in [−4,4]\))

\(−1\leq \beta \leq 11\) (from \(\beta −5\in [−6,6]\))

Set B:

\(B=\left\{(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}: 16(\alpha-2)^2+9(\beta-6)^2 \leq 144\right\}\)

This is an ellipse centered at \((2,6)\).

Standard form: \(\frac{(\alpha-2)^2}{9}+\frac{(\beta-6)^2}{16} \leq 1\)

So, it has:

Horizontal semi-axis \(= 3\)

Vertical semi-axis \(= 4\)

Hence, the ellipse spans:

\(\alpha \in [2−3,2+3]=[−1,5]\)

\(\beta \in [6−4,6+4]=[2,10]\)

Now compare sets:

Rectangle A: \(\alpha \in [−3,5],\beta \in [−1,11]\)

Ellipse B: completely lies within bounds of A

So, B is a subset of A, but A is not a subset of B (e.g., point \((−3,−1)\) is in A but not in B)