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Sets and Relations question

2025 · 4 Apr · Shift 2 · Q26
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Sets and Relations question

2025 · 4 Apr · Shift 2 · Q26

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let A={−3,−2,−1,0,1,2,3}\mathrm{A}=\{-3,-2,-1,0,1,2,3\}A={−3,−2,−1,0,1,2,3} and R be a relation on A defined by xRyx \mathrm{R} yxRy if and only if 2x−y∈{0,1}2 x-y \in\{0,1\}2x−y∈{0,1}. Let lll be the number of elements in RRR. 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+\mathrm{m}+\mathrm{n}l+m+n is equal to:
  1. A
    17
  2. B
    18
  3. C
    15
  4. D
    16
View written solutionFree

Correct answer: A

  1. Given relation

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

So for each x∈Ax\in Ax∈A, possible yyy satisfy either 2x−y=0  ⟹  y=2x2x-y=0 \implies y=2x2x−y=0⟹y=2x or 2x−y=1  ⟹  y=2x−1.2x-y=1 \implies y=2x-1.2x−y=1⟹y=2x−1.

But only those pairs count for which both x,y∈Ax,y\in Ax,y∈A.

  1. Find all ordered pairs in RRR

We check each x∈Ax\in Ax∈A.

For x=−3x=-3x=−3

  • y=2(−3)=−6∉Ay=2(-3)=-6\notin Ay=2(−3)=−6∈/A
  • y=2(−3)−1=−7∉Ay=2(-3)-1=-7\notin Ay=2(−3)−1=−7∈/A

No pair.

For x=−2x=-2x=−2

  • y=2(−2)=−4∉Ay=2(-2)=-4\notin Ay=2(−2)=−4∈/A
  • y=2(−2)−1=−5∉Ay=2(-2)-1=-5\notin Ay=2(−2)−1=−5∈/A

No pair.

For x=−1x=-1x=−1

  • y=2(−1)=−2∈A⇒(−1,−2)y=2(-1)=-2\in A \Rightarrow (-1,-2)y=2(−1)=−2∈A⇒(−1,−2)
  • y=2(−1)−1=−3∈A⇒(−1,−3)y=2(-1)-1=-3\in A \Rightarrow (-1,-3)y=2(−1)−1=−3∈A⇒(−1,−3)

For x=0x=0x=0

  • y=0∈A⇒(0,0)y=0\in A \Rightarrow (0,0)y=0∈A⇒(0,0)
  • y=−1∈A⇒(0,−1)y=-1\in A \Rightarrow (0,-1)y=−1∈A⇒(0,−1)

For x=1x=1x=1

  • y=2∈A⇒(1,2)y=2\in A \Rightarrow (1,2)y=2∈A⇒(1,2)
  • y=1∈A⇒(1,1)y=1\in A \Rightarrow (1,1)y=1∈A⇒(1,1)

For x=2x=2x=2

  • y=4∉Ay=4\notin Ay=4∈/A
  • y=3∈A⇒(2,3)y=3\in A \Rightarrow (2,3)y=3∈A⇒(2,3)

For x=3x=3x=3

  • y=6∉Ay=6\notin Ay=6∈/A
  • y=5∉Ay=5\notin Ay=5∈/A

No pair.

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

Therefore number of elements in RRR is l=7.l=7.l=7.

  1. Minimum additions to make RRR reflexive

A relation on AAA is reflexive if all (a,a),a∈A(a,a),\quad a\in A(a,a),a∈A are present.

Since A={−3,−2,−1,0,1,2,3},A=\{-3,-2,-1,0,1,2,3\},A={−3,−2,−1,0,1,2,3}, we need these 7 diagonal pairs: (−3,−3),(−2,−2),(−1,−1),(0,0),(1,1),(2,2),(3,3).(-3,-3),(-2,-2),(-1,-1),(0,0),(1,1),(2,2),(3,3).(−3,−3),(−2,−2),(−1,−1),(0,0),(1,1),(2,2),(3,3).

Already present in RRR are: (0,0),(1,1).(0,0),(1,1).(0,0),(1,1).

Missing are: (−3,−3),(−2,−2),(−1,−1),(2,2),(3,3).(-3,-3),(-2,-2),(-1,-1),(2,2),(3,3).(−3,−3),(−2,−2),(−1,−1),(2,2),(3,3).

So minimum number to be added is m=5.m=5.m=5.

  1. Minimum additions 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.

We inspect each pair:

  • (−1,−2)(-1,-2)(−1,−2) requires (−2,−1)(-2,-1)(−2,−1), which is missing.
  • (−1,−3)(-1,-3)(−1,−3) requires (−3,−1)(-3,-1)(−3,−1), which is missing.
  • (0,0)(0,0)(0,0) already has its reverse.
  • (0,−1)(0,-1)(0,−1) requires (−1,0)(-1,0)(−1,0), which is missing.
  • (1,2)(1,2)(1,2) requires (2,1)(2,1)(2,1), which is missing.
  • (1,1)(1,1)(1,1) already has its reverse.
  • (2,3)(2,3)(2,3) requires (3,2)(3,2)(3,2), which is missing.

All these required reverse pairs are distinct, so number of additions needed is n=5.n=5.n=5.

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

l+m+n=7+5+5=17.l+m+n=7+5+5=17.l+m+n=7+5+5=17.

So the correct option is 17.\boxed{17}.17​.

That is Option A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

Hence they agree.

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