JEE MAIN - Mathematics (2025 - 7th April Evening Shift - No. 15)
Let A = { ($\alpha, \beta$) $\in \mathbb{R} \times \mathbb{R}$ : |$\alpha$ - 1| $\leq 4$ and |$\beta$ - 5| $\leq 6$ }
and B = { ($\alpha, \beta$) $\in \mathbb{R} \times \mathbb{R}$ : 16($\alpha$ - $2)^2 $+ 9($\beta$ - $6)^2$ $\leq 144$ }.
Then
and B = { ($\alpha, \beta$) $\in \mathbb{R} \times \mathbb{R}$ : 16($\alpha$ - $2)^2 $+ 9($\beta$ - $6)^2$ $\leq 144$ }.
Then
A $\subset$ B
B $\subset$ A
neither A $\subset$ B nor B $\subset$ A
$A \cup B=\{(x, y):-4 \leqslant x \leqslant 4,-1 \leqslant y \leqslant 11\}$
Explanation
$$\begin{aligned} & \text { A: }|x-1| \leq 4 \text { and }|y-5| \leq 6 \\ & \Rightarrow-4 \leq x-1 \leq 4 \Rightarrow-6 \leq y-5 \leq 6 \\ & \Rightarrow-3 \leq x \leq 5 \quad \Rightarrow-1 \leq y \leq 11 \\ & \text { B : } 16(x-2)^2+9(y-6)^2 \leq 144 \\ & \text { B : } \frac{(x-2)^2}{9}+\frac{(y-6)^2}{16} \leq 1 \end{aligned}$$
_7th_April_Evening_Shift_en_15_1.png)
From Diagram $\mathrm{B} \subset \mathrm{A}$
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