JEE MAIN - Mathematics Hindi (2025 - 7th April Evening Shift - No. 15)
Let $\mathrm{A}=\{(\alpha, \beta) \in \mathbf{R} \times \mathbf{R}:|\alpha-1| \leqslant 4$ and $|\beta-5| \leqslant 6\}$ and $B=\left\{(\alpha, \beta) \in \mathbf{R} \times \mathbf{R}: 16(\alpha-2)^2+9(\beta-6)^2 \leqslant 144\right\}$. Then
$\mathrm{B} \subset \mathrm{A}$
$A \subset B$
$\mathrm{A} \cup \mathrm{B}=\{(x, y):-4 \leqslant x \leqslant 4,-1 \leqslant y \leqslant 11\}$
neither $\mathrm{A} \subset \mathrm{B}$ nor $\mathrm{B} \subset \mathrm{A}$
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