Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Sets and Relations question

2023 · 13 Apr · Shift 2 · Q36
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Sets and Relations
  5. /2023 · 13 Apr · Shift 2 · Q36

Sets and Relations question

2023 · 13 Apr · Shift 2 · Q36

JEE MainMathematicsSets and RelationsNumerical+4 / −1
Let A={−4,−3,−2,0,1,3,4}\mathrm{A}=\{-4,-3,-2,0,1,3,4\}A={−4,−3,−2,0,1,3,4} and R={(a,b)∈A×A:b=∣a∣\mathrm{R}=\left\{(a, b) \in \mathrm{A} \times \mathrm{A}: b=|a|\right.R={(a,b)∈A×A:b=∣a∣ or b2=a+1}\left.b^{2}=a+1\right\}b2=a+1} be a relation on A\mathrm{A}A. Then the minimum number of elements, that must be added to the relation R\mathrm{R}R so that it becomes reflexive and symmetric, is ‾\underline{\hspace{2cm}}​
Numerical answer
View written solutionFree

Correct answer: 7

  1. Given set and relation

We have A={−4,−3,−2,0,1,3,4}A=\{-4,-3,-2,0,1,3,4\}A={−4,−3,−2,0,1,3,4} and R={(a,b)∈A×A:b=∣a∣ or b2=a+1}.R=\{(a,b)\in A\times A: b=|a|\ \text{or}\ b^2=a+1\}.R={(a,b)∈A×A:b=∣a∣ or b2=a+1}.

We must find the minimum number of ordered pairs to add so that RRR becomes reflexive and symmetric.

  1. List all pairs in RRR

We use the two conditions separately.

Condition 1: b=∣a∣b=|a|b=∣a∣

Check each a∈Aa\in Aa∈A:

  • a=−4⇒b=4∈A⇒(−4,4)a=-4 \Rightarrow b=4 \in A \Rightarrow (-4,4)a=−4⇒b=4∈A⇒(−4,4)
  • a=−3⇒b=3∈A⇒(−3,3)a=-3 \Rightarrow b=3 \in A \Rightarrow (-3,3)a=−3⇒b=3∈A⇒(−3,3)
  • a=−2⇒b=2∉Aa=-2 \Rightarrow b=2 \notin Aa=−2⇒b=2∈/A
  • a=0⇒b=0∈A⇒(0,0)a=0 \Rightarrow b=0 \in A \Rightarrow (0,0)a=0⇒b=0∈A⇒(0,0)
  • a=1⇒b=1∈A⇒(1,1)a=1 \Rightarrow b=1 \in A \Rightarrow (1,1)a=1⇒b=1∈A⇒(1,1)
  • a=3⇒b=3∈A⇒(3,3)a=3 \Rightarrow b=3 \in A \Rightarrow (3,3)a=3⇒b=3∈A⇒(3,3)
  • a=4⇒b=4∈A⇒(4,4)a=4 \Rightarrow b=4 \in A \Rightarrow (4,4)a=4⇒b=4∈A⇒(4,4)

So from this condition: (−4,4),(−3,3),(0,0),(1,1),(3,3),(4,4).(-4,4),(-3,3),(0,0),(1,1),(3,3),(4,4).(−4,4),(−3,3),(0,0),(1,1),(3,3),(4,4).

Condition 2: b2=a+1b^2=a+1b2=a+1

This gives a=b2−1.a=b^2-1.a=b2−1. Now test possible b∈Ab\in Ab∈A:

  • b=−4⇒a=16−1=15∉Ab=-4 \Rightarrow a=16-1=15\notin Ab=−4⇒a=16−1=15∈/A
  • b=−3⇒a=9−1=8∉Ab=-3 \Rightarrow a=9-1=8\notin Ab=−3⇒a=9−1=8∈/A
  • b=−2⇒a=4−1=3∈A⇒(3,−2)b=-2 \Rightarrow a=4-1=3\in A \Rightarrow (3,-2)b=−2⇒a=4−1=3∈A⇒(3,−2)
  • b=0⇒a=0−1=−1∉Ab=0 \Rightarrow a=0-1=-1\notin Ab=0⇒a=0−1=−1∈/A
  • b=1⇒a=1−1=0∈A⇒(0,1)b=1 \Rightarrow a=1-1=0\in A \Rightarrow (0,1)b=1⇒a=1−1=0∈A⇒(0,1)
  • b=3⇒a=9−1=8∉Ab=3 \Rightarrow a=9-1=8\notin Ab=3⇒a=9−1=8∈/A
  • b=4⇒a=16−1=15∉Ab=4 \Rightarrow a=16-1=15\notin Ab=4⇒a=16−1=15∈/A

So from this condition: (3,−2),(0,1).(3,-2),(0,1).(3,−2),(0,1).

Hence, R={(−4,4),(−3,3),(0,0),(1,1),(3,3),(4,4),(3,−2),(0,1)}.R=\{(-4,4),(-3,3),(0,0),(1,1),(3,3),(4,4),(3,-2),(0,1)\}.R={(−4,4),(−3,3),(0,0),(1,1),(3,3),(4,4),(3,−2),(0,1)}.

  1. Make RRR reflexive

A relation on AAA is reflexive if every (x,x)(x,x)(x,x) for x∈Ax\in Ax∈A is present.

Required diagonal pairs are: (−4,−4),(−3,−3),(−2,−2),(0,0),(1,1),(3,3),(4,4).(-4,-4),(-3,-3),(-2,-2),(0,0),(1,1),(3,3),(4,4).(−4,−4),(−3,−3),(−2,−2),(0,0),(1,1),(3,3),(4,4).

Already present: (0,0),(1,1),(3,3),(4,4).(0,0),(1,1),(3,3),(4,4).(0,0),(1,1),(3,3),(4,4).

Missing: (−4,−4),(−3,−3),(−2,−2).(-4,-4),(-3,-3),(-2,-2).(−4,−4),(−3,−3),(−2,−2).

So for reflexivity, we must add 3 pairs.

  1. 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.

Check each non-diagonal pair:

  • (−4,4)∈R(-4,4)\in R(−4,4)∈R requires (4,−4)(4,-4)(4,−4), which is missing.
  • (−3,3)∈R(-3,3)\in R(−3,3)∈R requires (3,−3)(3,-3)(3,−3), which is missing.
  • (3,−2)∈R(3,-2)\in R(3,−2)∈R requires (−2,3)(-2,3)(−2,3), which is missing.
  • (0,1)∈R(0,1)\in R(0,1)∈R requires (1,0)(1,0)(1,0), which is missing.

Diagonal pairs automatically satisfy symmetry.

So for symmetry, we must add 4 pairs: (4,−4),(3,−3),(−2,3),(1,0).(4,-4),(3,-3),(-2,3),(1,0).(4,−4),(3,−3),(−2,3),(1,0).

  1. Check overlap between reflexive and symmetric additions

Reflexive additions needed: (−4,−4),(−3,−3),(−2,−2).(-4,-4),(-3,-3),(-2,-2).(−4,−4),(−3,−3),(−2,−2).

Symmetric additions needed: (4,−4),(3,−3),(−2,3),(1,0).(4,-4),(3,-3),(-2,3),(1,0).(4,−4),(3,−3),(−2,3),(1,0).

There is no overlap.

Therefore total minimum pairs to add: 3+4=7.3+4=7.3+4=7.

  1. Final answer

The minimum number of elements to be added is 7.\boxed{7}.7​.

PreviousNext

More from Sets and Relations

  • The number of elements in the set {n∈N:10≤n≤100 and 3n−3 is a multiple of 7 } is ​.2023 · Numerical
  • Let A={1,2,3,4} and R be a relation on the set A×A defined by R={((a,b),(c,d)):2a+3b=4c+5d}. Then the number of elements in R is ​.2023 · Numerical
  • The relation R={(a,b):gcd(a,b)=1,2aeb,a,b∈Z} is :2023 · MCQ
  • The minimum number of elements that must be added to the relation R = {(a, b), (b, c), (b, d)} on the set {a, b, c, d} so that it is an equivalence relation, is ​.2023 · Numerical
  • Let S = {1, 2, 3, 5, 7, 10, 11}. The number of non-empty subsets of S that have the sum of all elements a multiple of 3, is ​.2023 · Numerical
  • Let R be a relation defined on N as aRb if 2a+3b is a multiple of 5,a,b∈N. Then R is2023 · MCQ
  • The minimum number of elements that must be added to the relation R={(a,b),(b,c)} on the set {a,b,c} so that it becomes symmetric and transitive is :2023 · MCQ
  • Let R be a relation on N×N defined by (a,b) R (c,d) if and only if ad(b−c)=bc(a−d). Then R is2023 · MCQ