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Maths Question 1 – JEE-MAIN 2025

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Let A={(α,β)∈R×R:|α−1|≤4 and |β−5|≤6} and B={(α,β)∈R×R:16(α−2)2+9(β−6)2<144}. Then

First, clearly define the regions represented by set A and set B based on their given inequalities.

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Ninja Strategy•Boundary and Range Comparison

Quickly determine the rectangular bounds for A and the extreme bounds for the ellipse B, then compare these intervals to identify the correct subset relationship and union.

Step 1: Analyze Set A✦ Active

Set A is defined by absolute value inequalities:

|α−1|≤4⟹−4≤α−1≤4⟹−3≤α≤5
|β−5|≤6⟹−6≤β−5≤6⟹−1≤β≤11

Thus, A is the rectangular region A=[−3,5]×[−1,11].

💡 Teacher's Secret Hint

Remember that |x−c|≤r defines an interval [c−r,c+r].

Step 2: Analyze Set B○ Expand

Set B is defined by an elliptical inequality:

16(α−2)2+9(β−6)2<144

Divide by 144 to get the standard form of an ellipse:

(α−2)29+(β−6)216<1

This is the interior of an ellipse centered at (2,6) with semi-minor axis a=3 (along α) and semi-major axis b=4 (along β). The α-range for B is (2−3,2+3)=(−1,5). The β-range for B is (6−4,6+4)=(2,10).

💡 Teacher's Secret Hint

The inequality `<` means the interior of the ellipse, excluding the boundary.

Step 3: Compare A and B for Subset Relationships○ Expand

For A⊂B: Consider a point in A, e.g., (−3,5). For this point to be in B, its α-coordinate must satisfy −1<α<5. Since −3≯−1, the point (−3,5) is in A but not in B. Thus, A⊄B.

For B⊂A: Any point (α,β) in B satisfies −1<α<5 and 2<β<10. Comparing these ranges with A's ranges: [−3,5] for α and [−1,11] for β. The interval (−1,5) is contained in [−3,5], and (2,10) is contained in [−1,11]. Since the entire elliptical region B lies within the rectangle defined by these ranges, B⊂A.

💡 Teacher's Secret Hint

Visually sketching the regions can help confirm the subset relationship. The rectangle A is larger and encompasses the ellipse B.

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