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JEE Main 20241 FebruaryMorning Shift
Exam
JEE Main
Year
2024
Date
1 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Sets and Relations

JEE Main 2024 · 1 February · Morning Shift · Mathematics · Sets and Relations

Let A = \ 1, 2, 3, . ..20\ . Let R _ 1 and R _ 2 two relation on A...

Let A={1,2,3,...20}A = \left\{1, 2, 3, . ..20\right\}. Let R1{R}_{1} and R2{R}_{2} two relation on AA such that

R1=(a,b):b is  divisible  by a{R}_{1} = { \left(a, b\right): b\text{ is }\text{ divisible }\text{ by } a}

R2=(a,b):a is  an  integral  multiple  of b{R}_{2} = { \left(a, b\right): a\text{ is }\text{ an }\text{ integral }\text{ multiple }\text{ of } b}

Then, number of elements in R1−R2{R}_{1} - {R}_{2} is equal to __________.

Correct answer

0

Step-by-step solution

Given: A={1,2,3,...20}A = \left\{1, 2, 3, . ..20\right\}, R1=(a,b):b is  divisible  by a{R}_{1} = { \left(a, b\right): b\text{ is }\text{ divisible }\text{ by } a} and R2=(a,b):a is  an  integral  multiple  of b{R}_{2} = { \left(a, b\right): a\text{ is }\text{ an }\text{ integral }\text{ multiple }\text{ of } b}.

⇒R1={(1,1),(1,2),...,(1,20),(2,2),(2,4),...,(2,20),(3,3),(3,6),...,(3,18),(4,4),(4,8),...,(4,20),(5,5),(5,10),...,(5,20),(6,6),(6,18),...,(6,18),(7,7),(7,14),(8,8),(8,16),(9,9),(9,18),(10,10),(10,20),(11,11),...,(20,20)}⇒ {R}_{1} = \left\{\left(1, 1\right), \left(1, 2\right), . . . , \left(1, 20\right), \left(2, 2\right), \left(2, 4\right), . . . , \left(2, 20\right), \left(3, 3\right), \left(3, 6\right), . . . , \left(3, 18\right), \left(4, 4\right), \left(4, 8\right), . . . , \left(4, 20\right), \left(5, 5\right), \left(5, 10\right), . . . , \left(5, 20\right), \left(6, 6\right), \left(6, 18\right), . . . , \left(6, 18\right), \left(7, 7\right), \left(7, 14\right), \left(8, 8\right), \left(8, 16\right), \left(9, 9\right), \left(9, 18\right), \left(10, 10\right), \left(10, 20\right), \left(11, 11\right), . . . , \left(20, 20\right)\right\}

⇒n(R1)=20+10+6+5+4+3+2+2+2+2+10=66⇒ n\left({R}_{1}\right) = 20 + 10 + 6 + 5 + 4 + 3 + 2 + 2 + 2 + 2 + 10 = 66

⇒R1∩R2={(1,1),(2,2),(3,3),(4,4),...,(20,20)}⇒ {R}_{1} \cap {R}_{2} = \left\{\left(1, 1\right), \left(2, 2\right), \left(3, 3\right), \left(4, 4\right), . . . , \left(20, 20\right)\right\}

⇒n(R1∩R2)=20⇒ n\left({R}_{1} \cap {R}_{2}\right) = 20

⇒n(R1−R2)=n(R1)−n(R1∩R2)⇒ n\left({R}_{1} - {R}_{2}\right) = n\left({R}_{1}\right) - n\left({R}_{1} \cap {R}_{2}\right)

⇒n(R1−R2)=66−20⇒ n\left({R}_{1} - {R}_{2}\right) = 66 - 20

⇒n(R1−R2)=46⇒ n\left({R}_{1} - {R}_{2}\right) = 46

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