Class JEE Mathematics Sets, Relations, and Functions Q #995
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4 Marks 2025 JEE Main 2025 (Online) 7th April Evening Shift MCQ SINGLE
Let $A = { (\alpha, \beta ) \in R \times R : |\alpha - 1| \leq 4$ and $|\beta - 5| \leq 6 }$

and $B = { (\alpha, \beta ) \in R \times R : 16(\alpha - 2)^{2}+ 9(\beta - 6)^{2} \leq 144 }$.

Then
(A) A $A \subset B$
(B) B $B \subset A$
(C) C neither $A \subset B$ nor $B \subset A$
(D) D $A \cup B = { (x, y) : -4 \leqslant x \leqslant 4, -1 \leqslant y \leqslant 11 }$
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Correct Answer: B
Explanation
$A: |x-1| \leq 4$ and $|y-5| \leq 6$
$\Rightarrow -4 \leq x-1 \leq 4 \Rightarrow -6 \leq y-5 \leq 6$
$\Rightarrow -3 \leq x \leq 5 \Rightarrow -1 \leq y \leq 11$
$B : 16(x-2)^{2} + 9(y-6)^{2} \leq 144$
$B : \frac{(x-2)^{2}}{9} + \frac{(y-6)^{2}}{16} \leq 1$

From Diagram $B \subset A$

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