JEE Main 2023MathematicsSets and RelationsEasyNumerical

JEE Main 2023Sets and Relations Question with Solution

JEE Main 2023 (08 Apr Shift 1)

Question

Let A=0,3,4,6,7,8,9,10 and R be the relation defined on A such that Rx,yA×A:x-y is odd positive integer or x-y=2. The minimum number of elements that must be added to the relation R, so that it is a symmetric relation, is equal to _________

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Show full solutionCorrect answer: 19
Correct answer
19

Step-by-step explanation

Given,

Set A=10,9,8,7,6,4,3,0

Now relation x-y is odd or x-y=2 can be given by,

R=(10,9),(10,8),(10,7),(10,3),(9,8),(9,7),(9,6),(9,4),(9,0),(8,7),(8,6),(8,3),(7,6),(7,4),(7,0),(6,4),(6,3),(4,3),(3,0)

So, total there are 19 elements and all the elements of R, (a, b) are of type a > b.

Hence, we need to add total of 19 more elements to R to make in symmetric

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About this question

This is a previous-year question from JEE Main 2023, covering the Sets and Relations chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.