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Sets and Relations question

2025 · 7 Apr · Shift 2 · Q40
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  5. /2025 · 7 Apr · Shift 2 · Q40

Sets and Relations question

2025 · 7 Apr · Shift 2 · Q40

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let A = { (α,β\alpha, \betaα,β) ∈R×R\in \mathbb{R} \times \mathbb{R}∈R×R : |α\alphaα- 1|≤4\leq 4≤4 and |β\betaβ- 5|≤6\leq 6≤6 } and B = { (α,β\alpha, \betaα,β) ∈R×R\in \mathbb{R} \times \mathbb{R}∈R×R : 16(α\alphaα-2)22)^22)2+ 9(β\betaβ-6)2≤1446)^2\leq 1446)2≤144 }. Then
  1. A
    A ⊂\subset⊂ B
  2. B
    B ⊂\subset⊂ A
  3. C
    neither A ⊂\subset⊂ B nor B ⊂\subset⊂ A
  4. D
    A∪B={(x,y):−4⩽x⩽4,−1⩽y⩽11}A \cup B=\{(x, y):-4 \leqslant x \leqslant 4,-1 \leqslant y \leqslant 11\}A∪B={(x,y):−4⩽x⩽4,−1⩽y⩽11}
View written solutionFree

Correct answer: B

  1. Interpret set AAA

Given A={(α,β)∈R2:∣α−1∣≤4, ∣β−5∣≤6}.A=\{(\alpha,\beta)\in \mathbb R^2: |\alpha-1|\le 4,\ |\beta-5|\le 6\}.A={(α,β)∈R2:∣α−1∣≤4, ∣β−5∣≤6}.

These inequalities give −4≤α−1≤4  ⟹  −3≤α≤5,-4\le \alpha-1\le 4 \implies -3\le \alpha\le 5,−4≤α−1≤4⟹−3≤α≤5, −6≤β−5≤6  ⟹  −1≤β≤11.-6\le \beta-5\le 6 \implies -1\le \beta\le 11.−6≤β−5≤6⟹−1≤β≤11.

So AAA is the rectangle A=[−3,5]×[−1,11].A=[-3,5]\times[-1,11].A=[−3,5]×[−1,11].

  1. Interpret set BBB

Given B={(α,β)∈R2:16(α−2)2+9(β−6)2≤144}.B=\{(\alpha,\beta)\in \mathbb R^2: 16(\alpha-2)^2+9(\beta-6)^2\le 144\}.B={(α,β)∈R2:16(α−2)2+9(β−6)2≤144}.

Divide by 144144144: (α−2)29+(β−6)216≤1.\frac{(\alpha-2)^2}{9}+\frac{(\beta-6)^2}{16}\le 1.9(α−2)2​+16(β−6)2​≤1.

So BBB is an ellipse centered at (2,6)(2,6)(2,6) with semi-axes:

  • along xxx-direction: 333
  • along yyy-direction: 444

Hence the ellipse lies entirely inside the rectangle 2−3≤α≤2+3  ⟹  −1≤α≤5,2-3\le \alpha\le 2+3 \implies -1\le \alpha\le 5,2−3≤α≤2+3⟹−1≤α≤5, 6−4≤β≤6+4  ⟹  2≤β≤10.6-4\le \beta\le 6+4 \implies 2\le \beta\le 10.6−4≤β≤6+4⟹2≤β≤10.

Thus every point of BBB satisfies −1≤α≤5,2≤β≤10,-1\le \alpha\le 5,\qquad 2\le \beta\le 10,−1≤α≤5,2≤β≤10, which certainly implies −3≤α≤5,−1≤β≤11.-3\le \alpha\le 5,\qquad -1\le \beta\le 11.−3≤α≤5,−1≤β≤11. Therefore, B⊂A.B\subset A.B⊂A.

  1. Check whether A⊂BA\subset BA⊂B

Take the point (−3,−1)∈A(-3,-1)\in A(−3,−1)∈A since it satisfies ∣−3−1∣=4≤4,∣−1−5∣=6≤6.|{-3}-1|=4\le 4,\qquad |{-1}-5|=6\le 6.∣−3−1∣=4≤4,∣−1−5∣=6≤6.

Now test in BBB: 16(−3−2)2+9(−1−6)2=16(25)+9(49)=400+441=841>144.16(-3-2)^2+9(-1-6)^2=16(25)+9(49)=400+441=841>144.16(−3−2)2+9(−1−6)2=16(25)+9(49)=400+441=841>144. So (−3,−1)∉B(-3,-1)\notin B(−3,−1)∈/B.

Hence, A⊄B.A\not\subset B.A⊂B.

  1. Check option DDD

Option DDD says A∪B={(x,y):−4≤x≤4,−1≤y≤11}.A\cup B=\{(x,y):-4\le x\le 4,-1\le y\le 11\}.A∪B={(x,y):−4≤x≤4,−1≤y≤11}.

But since B⊂AB\subset AB⊂A, A∪B=A=[−3,5]×[−1,11],A\cup B=A=[-3,5]\times[-1,11],A∪B=A=[−3,5]×[−1,11], which is not equal to [−4,4]×[−1,11].[-4,4]\times[-1,11].[−4,4]×[−1,11]. So option DDD is false.

  1. Evaluate all options
  • A: A⊂BA\subset BA⊂B — False
  • B: B⊂AB\subset AB⊂A — True
  • C: neither A⊂BA\subset BA⊂B nor B⊂AB\subset AB⊂A — False
  • D: False

Therefore, the correct option is B.\boxed{B}.B​.

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