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Sets and Relations question

2018 · Shift 0 · Q32
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  5. /2018 · Shift 0 · Q32

Sets and Relations question

2018 · Shift 0 · Q32

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Two sets A and B are as under : A = {(aaa, b) ∈\in∈ R ×\times× R : |aaa - 5| < 1 and |b - 5| < 1}; B = {(aaa, b) ∈\in∈ R ×\times× R : 4(aaa- 6)2 + 9(b - 5)2 ≤\le≤ 36 }; Then
  1. A
    neither A ⊂\subset⊂ B nor B ⊂\subset⊂ A
  2. B
    B ⊂\subset⊂ A
  3. C
    A ⊂\subset⊂ B
  4. D
    A ∩\cap∩ B = ϕ\phiϕ ( an empty set )
View written solutionFree

Correct answer: C

  1. Interpret the sets

    The set A={(a,b)∈R×R:∣a−5∣<1 and ∣b−5∣<1}A = \{(a,b)\in \mathbb{R}\times \mathbb{R} : |a-5|<1 \text{ and } |b-5|<1\}A={(a,b)∈R×R:∣a−5∣<1 and ∣b−5∣<1} means 4<a<6,4<b<6.4<a<6, \qquad 4<b<6.4<a<6,4<b<6. So, AAA is the interior of the square with vertices near (4,4),(6,4),(6,6),(4,6)(4,4),(6,4),(6,6),(4,6)(4,4),(6,4),(6,6),(4,6).

  2. Rewrite set BBB

    Given B={(a,b)∈R×R:4(a−6)2+9(b−5)2≤36}.B=\{(a,b)\in \mathbb{R}\times \mathbb{R}:4(a-6)^2+9(b-5)^2\le 36\}.B={(a,b)∈R×R:4(a−6)2+9(b−5)2≤36}.

    Divide by 363636: (a−6)29+(b−5)24≤1.\frac{(a-6)^2}{9}+\frac{(b-5)^2}{4}\le 1.9(a−6)2​+4(b−5)2​≤1.

    This is an ellipse centered at (6,5)(6,5)(6,5) with semi-major axis 333 in the aaa-direction and semi-minor axis 222 in the bbb-direction.

  3. Check whether A⊂BA\subset BA⊂B

    Take any (a,b)∈A(a,b)\in A(a,b)∈A. Then 4<a<6  ⟹  −2<a−6<0,4<a<6 \implies -2<a-6<0,4<a<6⟹−2<a−6<0, so ∣a−6∣<2  ⟹  (a−6)2<4.|a-6|<2 \implies (a-6)^2<4.∣a−6∣<2⟹(a−6)2<4.

    Also, 4<b<6  ⟹  −1<b−5<1,4<b<6 \implies -1<b-5<1,4<b<6⟹−1<b−5<1, so ∣b−5∣<1  ⟹  (b−5)2<1.|b-5|<1 \implies (b-5)^2<1.∣b−5∣<1⟹(b−5)2<1.

    Therefore, 4(a−6)2+9(b−5)2<4⋅4+9⋅1=16+9=25<36.4(a-6)^2+9(b-5)^2<4\cdot 4+9\cdot 1=16+9=25<36.4(a−6)2+9(b−5)2<4⋅4+9⋅1=16+9=25<36.

    Hence every point of AAA satisfies the defining inequality of BBB. So, A⊂B.A\subset B.A⊂B.

  4. Check whether B⊂AB\subset AB⊂A

    Consider the point (6,7)(6,7)(6,7). Then 4(6−6)2+9(7−5)2=0+9⋅4=36,4(6-6)^2+9(7-5)^2=0+9\cdot 4=36,4(6−6)2+9(7−5)2=0+9⋅4=36, so (6,7)∈B(6,7)\in B(6,7)∈B.

    But for AAA, we need ∣b−5∣<1|b-5|<1∣b−5∣<1. Here, ∣7−5∣=2≮1,|7-5|=2\not<1,∣7−5∣=2<1, so (6,7)∉A(6,7)\notin A(6,7)∈/A.

    Thus, B⊄A.B\not\subset A.B⊂A.

  5. Check other options

    • Option A: false, because A⊂BA\subset BA⊂B is true.
    • Option B: false, since B⊄AB\not\subset AB⊂A.
    • Option C: true, since A⊂BA\subset BA⊂B.
    • Option D: false, because if A⊂BA\subset BA⊂B, then A∩B=A≠∅A\cap B=A\neq \varnothingA∩B=A=∅.

  6. Final answer

    The correct option is

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