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Sets and Relations question

2025 · 3 Apr · Shift 2 · Q36
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  5. /2025 · 3 Apr · Shift 2 · Q36

Sets and Relations question

2025 · 3 Apr · Shift 2 · Q36

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let A={−2,−1,0,1,2,3}A=\{-2,-1,0,1,2,3\}A={−2,−1,0,1,2,3}. Let R be a relation on AAA defined by xRyx \mathrm{R} yxRy if and only if y=max⁡{x,1}y=\max \{x, 1\}y=max{x,1}. Let lll be the number of elements in R . Let mmm and nnn be the minimum number of elements required to be added in R to make it reflexive and symmetric relations, respectively. Then l+m+nl+m+nl+m+n is equal to
  1. A
    11
  2. B
    12
  3. C
    14
  4. D
    13
View written solutionFree

Correct answer: B

  1. Given set and relation

We have A={−2,−1,0,1,2,3}A=\{-2,-1,0,1,2,3\}A={−2,−1,0,1,2,3} and the relation RRR on AAA is defined by xRy  ⟺  y=max⁡{x,1}.xRy \iff y=\max\{x,1\}.xRy⟺y=max{x,1}.

This means: for each x∈Ax\in Ax∈A, the corresponding yyy is uniquely determined.

  1. Find all ordered pairs in RRR

Compute y=max⁡{x,1}y=\max\{x,1\}y=max{x,1} for each x∈Ax\in Ax∈A:

  • For x=−2x=-2x=−2: y=max⁡{−2,1}=1y=\max\{-2,1\}=1y=max{−2,1}=1
    So (−2,1)∈R(-2,1)\in R(−2,1)∈R
  • For x=−1x=-1x=−1: y=max⁡{−1,1}=1y=\max\{-1,1\}=1y=max{−1,1}=1
    So (−1,1)∈R(-1,1)\in R(−1,1)∈R
  • For x=0x=0x=0: y=max⁡{0,1}=1y=\max\{0,1\}=1y=max{0,1}=1
    So (0,1)∈R(0,1)\in R(0,1)∈R
  • For x=1x=1x=1: y=max⁡{1,1}=1y=\max\{1,1\}=1y=max{1,1}=1
    So (1,1)∈R(1,1)\in R(1,1)∈R
  • For x=2x=2x=2: y=max⁡{2,1}=2y=\max\{2,1\}=2y=max{2,1}=2
    So (2,2)∈R(2,2)\in R(2,2)∈R
  • For x=3x=3x=3: y=max⁡{3,1}=3y=\max\{3,1\}=3y=max{3,1}=3
    So (3,3)∈R(3,3)\in R(3,3)∈R

Hence, R={(−2,1),(−1,1),(0,1),(1,1),(2,2),(3,3)}.R=\{(-2,1),(-1,1),(0,1),(1,1),(2,2),(3,3)\}.R={(−2,1),(−1,1),(0,1),(1,1),(2,2),(3,3)}.

Therefore, the number of elements in RRR is l=6.l=6.l=6.

  1. Minimum number of pairs to add to make RRR reflexive

A relation on AAA is reflexive if (a,a)∈Rfor every a∈A.(a,a)\in R \quad \text{for every } a\in A.(a,a)∈Rfor every a∈A.

Since A={−2,−1,0,1,2,3},A=\{-2,-1,0,1,2,3\},A={−2,−1,0,1,2,3}, the required diagonal pairs are (−2,−2),(−1,−1),(0,0),(1,1),(2,2),(3,3).(-2,-2),(-1,-1),(0,0),(1,1),(2,2),(3,3).(−2,−2),(−1,−1),(0,0),(1,1),(2,2),(3,3).

Already present in RRR are: (1,1),(2,2),(3,3).(1,1),(2,2),(3,3).(1,1),(2,2),(3,3).

Missing are: (−2,−2),(−1,−1),(0,0).(-2,-2),(-1,-1),(0,0).(−2,−2),(−1,−1),(0,0).

So the minimum number to be added is m=3.m=3.m=3.

  1. Minimum number of pairs to add to make RRR symmetric

A relation is symmetric if whenever (a,b)∈R(a,b)\in R(a,b)∈R, then (b,a)∈R(b,a)\in R(b,a)∈R also.

Check each pair in RRR:

  • (−2,1)∈R(-2,1)\in R(−2,1)∈R requires (1,−2)(1,-2)(1,−2) to be in RRR — missing
  • (−1,1)∈R(-1,1)\in R(−1,1)∈R requires (1,−1)(1,-1)(1,−1) to be in RRR — missing
  • (0,1)∈R(0,1)\in R(0,1)∈R requires (1,0)(1,0)(1,0) to be in RRR — missing
  • (1,1)(1,1)(1,1) is already symmetric with itself
  • (2,2)(2,2)(2,2) is already symmetric with itself
  • (3,3)(3,3)(3,3) is already symmetric with itself

Thus we need to add exactly (1,−2),(1,−1),(1,0).(1,-2),(1,-1),(1,0).(1,−2),(1,−1),(1,0).

So, n=3.n=3.n=3.

  1. Compute l+m+nl+m+nl+m+n

l+m+n=6+3+3=12.l+m+n=6+3+3=12.l+m+n=6+3+3=12.

  1. Compare with given options

The value is 12,12,12, which corresponds to Option B.

  1. Comparison with stored correct answer

Stored correct answer: B
Derived answer: B

So the derived answer agrees with the stored correct answer.

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