Let A = {(α, β)∈ R x R: |α - 1| ≤ 4 and |β - 5| ≤ 6} and B = {(α, β) ∈ R x R : 16 (α - 2)^2 + 9 (β - 6)^2 ≤ 144} .

Question

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

(1) \(\mathrm{B} \subset \mathrm{A}\)

(2) \(\mathrm{A} \cup \mathrm{B}=\{(\mathrm{x}, \mathrm{y}):-4 \leq \mathrm{x} \leq 4,-1 \leq \mathrm{y} \leq 11\}\)

(3) neither \(\mathrm{A} \subset \mathrm{B}\) nor \(\mathrm{B} \subset \mathrm{A}\)

(4) \(A \subset B\)

Answer 1

Correct option is: (1) \(\mathrm{B} \subset \mathrm{A}\)  

\(A : |x-1| \leq 4\) and \(|y-5| \leq 6\)

\(\Rightarrow-4 \leq \mathrm{x}-1 \leq 4 \Rightarrow-6 \leq \mathrm{y}-5 \leq 6\)

\(\Rightarrow-3 \leq \mathrm{x} \leq 5 \quad \Rightarrow-1 \leq \mathrm{y} \leq 11\)

B : \(16(x-2)^{2}+9(y-6)^{2} \leq 144\)

B : \(\frac{(x-2)^{2}}{9}+\frac{(y-6)^{2}}{16} \leq 1\)

From Diagram \(\mathrm{B} \subset \mathrm{A}\)