JEE Main 2025 Apr 7 Shift 2, Mathematics Q3: Set

JEE Main2025Apr 7, Shift 2Mathematics

Let

$A = {(α, β) \in R \times R : ∣ α - 1∣ \leq 4 and ∣ β - 5∣ \leq 6}$

and

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

Then

A
$B \subset A$
B
$A \cup B = {(x, y) : - 4 \leq x \leq 4, - 1 \leq y \leq 11}$
C
neither $A \subset B$ nor $B \subset A$
D
$A \subset B$

Solution

Set $A$ : $∣ x - 1∣ \leq 4$ gives $- 3 \leq x \leq 5$ , and $∣ y - 5∣ \leq 6$ gives $- 1 \leq y \leq 11$ . So $A$ is the rectangle $[- 3, 5] \times [- 1, 11]$ .

Set $B$ : dividing $16 (x - 2)^{2} + 9 (y - 6)^{2} \leq 144$ by 144 gives $\frac{( x - 2 ) ^{2}}{9} + \frac{( y - 6 ) ^{2}}{16} \leq 1$ , the interior and boundary of an ellipse centred at $(2, 6)$ with semi-axis 3 along $x$ and 4 along $y$ . So $B$ 's $x$ -range is $[- 1, 5]$ and $y$ -range is $[2, 10]$ .

Since $[- 1, 5] \subseteq [- 3, 5]$ and $[2, 10] \subseteq [- 1, 11]$ , every point of the elliptical region $B$ also satisfies the rectangular bounds of $A$ .

$∴ B \subset A$ .

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JEE Main 2025 Apr 7 Shift 2, Mathematics Q3: Set

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