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JEE Mains Previous Year Questions 
(2021-2024): Sets, Relations and Functions 
2024 
Q1: Let ?? ={?? ,?? ,?? ,…,???? }. Let ?? ?? and ?? ?? two relation on ?? such that 
?? ?? ={(?? ,?? ):?? is divisible by ?? } 
?? ?? ={(?? ,?? ):?? is an integral multiple of ?? }. 
Then, number of elements in ?? ?? -?? ?? is equal to    [JEE Main 2024 (Online) 1st February Morning 
Shift] 
Ans:  46 
To determine the number of elements in ?? 1
-?? 2
 let's first articulate the meaning of both relations on 
set ?? ={1,2,3,…,20} : 
?? 1
 includes pairs (?? ,?? ) where ?? is divisible by ?? . This includes pairs like (1,1),(1,2),…,(1,20) for 1 ; 
similar series for 2 up to (2,20) (excluding odd numbers); for 3 up to (3,18); and so on, reflecting the 
divisibility condition. 
?? 1
:
{
 
 
 
 
 
 
 
 
(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),(6,12),(6,18),(7,7),(7,14),
(8,8),(8,16),(9,9),(9,18)(10,10),
(10,20),(11,11),(12,12)…,(20,20)}
 
 
 
 
 
 
 
 
?? (?? 1
)=66
 
?? 2
 consists of pairs (?? ,?? ) where ?? is an integral multiple of ?? . Essentially, this relationship is the reverse 
of ?? 1
. However, for the essence of ?? 1
-?? 2
, the key overlap comes with pairs where ?? =?? , since those 
are the pairs that clearly reflect both conditions identically (i.e., a number is always divisible by itself, 
and it is always a multiple of itself). There are 20 such pairs corresponding to each integer in ?? from 1 to 
20 , resulting in the common elements between ?? 1
 and ?? 2
 (the intersection ?? 1
n?? 2
 ) being 20 pairs. 
R
1
nR
2
={(1,1),(2,2),…(20,20)}
n(R
1
nR
2
)=20
 
The difference ?? 1
-?? 2
 seeks elements present in ?? 1
 but not in ?? 2
. Given that ?? 1
 and ?? 2
 share 20 
elements that are identical, to find ?? 1
-?? 2
 we subtract these 20 common elements from the total in 
?? 1
, resulting in 66-20=46 pairs. 
Page 2


JEE Mains Previous Year Questions 
(2021-2024): Sets, Relations and Functions 
2024 
Q1: Let ?? ={?? ,?? ,?? ,…,???? }. Let ?? ?? and ?? ?? two relation on ?? such that 
?? ?? ={(?? ,?? ):?? is divisible by ?? } 
?? ?? ={(?? ,?? ):?? is an integral multiple of ?? }. 
Then, number of elements in ?? ?? -?? ?? is equal to    [JEE Main 2024 (Online) 1st February Morning 
Shift] 
Ans:  46 
To determine the number of elements in ?? 1
-?? 2
 let's first articulate the meaning of both relations on 
set ?? ={1,2,3,…,20} : 
?? 1
 includes pairs (?? ,?? ) where ?? is divisible by ?? . This includes pairs like (1,1),(1,2),…,(1,20) for 1 ; 
similar series for 2 up to (2,20) (excluding odd numbers); for 3 up to (3,18); and so on, reflecting the 
divisibility condition. 
?? 1
:
{
 
 
 
 
 
 
 
 
(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),(6,12),(6,18),(7,7),(7,14),
(8,8),(8,16),(9,9),(9,18)(10,10),
(10,20),(11,11),(12,12)…,(20,20)}
 
 
 
 
 
 
 
 
?? (?? 1
)=66
 
?? 2
 consists of pairs (?? ,?? ) where ?? is an integral multiple of ?? . Essentially, this relationship is the reverse 
of ?? 1
. However, for the essence of ?? 1
-?? 2
, the key overlap comes with pairs where ?? =?? , since those 
are the pairs that clearly reflect both conditions identically (i.e., a number is always divisible by itself, 
and it is always a multiple of itself). There are 20 such pairs corresponding to each integer in ?? from 1 to 
20 , resulting in the common elements between ?? 1
 and ?? 2
 (the intersection ?? 1
n?? 2
 ) being 20 pairs. 
R
1
nR
2
={(1,1),(2,2),…(20,20)}
n(R
1
nR
2
)=20
 
The difference ?? 1
-?? 2
 seeks elements present in ?? 1
 but not in ?? 2
. Given that ?? 1
 and ?? 2
 share 20 
elements that are identical, to find ?? 1
-?? 2
 we subtract these 20 common elements from the total in 
?? 1
, resulting in 66-20=46 pairs. 
n(R
1
-R
2
)=n(R
1
)-n(R
1
nR
2
)
 =n(R
1
)-20
 =66-20
R
1
-R
2
=46 Pair 
 
Q2: Let ?? ={?? ,?? ,?? ,………,?????? }. Let ?? be a relation on ?? defined by (?? ,?? )??? if and only if ?? ?? =
?? ?? . Let ?? ?? be a symmetric relation on ?? such that ?? ??? ?? and the number of elements in ?? ?? is ?? . 
Then, the minimum value of ?? is    [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: 66 
R={(3,2),(6,4),(9,6),(12,8),………(99,66)}
n(R)=33
 ?66
 
Q3: Let ?? ={?? ,?? ,?? ,?? } and ?? ={(?? ,?? ),(?? ,?? ),(?? ,?? )} be a relation on ?? . Let ?? be the equivalence 
relation on ?? such that ?? ??? and the number of elements in ?? is ?? . Then, the minimum value of ?? is   
[JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 16 
?? ={1,2,3,4} 
?? ={(1,2),(2,3),(1,4)} 
?? is equivalence for ?? <?? and reflexive 
{(1,1),(2,2),(3,3),(4,4)} 
for symmetric 
{(2,1),(4,1),(3,2)} 
for transitive 
{(1,3),(3,1),(4,2),(2,4)} 
Now set 
?? ={(1,1),(2,2),(3,3),(4,4),(1,2),(2,3),(1,4),(4,3),(3,4),(2,1),(4,1),(3,2),(1,3),(3,1),(4,2),(2,4)} 
?? (?? )=16 
Q4: The number of symmetric relations defined on the set {?? ,?? ,?? ,?? } which are not reflexive is     [JEE 
Main 2024 (Online) 30th January Evening Shift] 
Ans: 960 
Total number of relation both symmetric and reflexive =2
?? 2
-?? 2
 
Total number of symmetric relation =2
(
n
2
+n
2
)
 
? Then number of symmetric relation which are not reflexive 
Page 3


JEE Mains Previous Year Questions 
(2021-2024): Sets, Relations and Functions 
2024 
Q1: Let ?? ={?? ,?? ,?? ,…,???? }. Let ?? ?? and ?? ?? two relation on ?? such that 
?? ?? ={(?? ,?? ):?? is divisible by ?? } 
?? ?? ={(?? ,?? ):?? is an integral multiple of ?? }. 
Then, number of elements in ?? ?? -?? ?? is equal to    [JEE Main 2024 (Online) 1st February Morning 
Shift] 
Ans:  46 
To determine the number of elements in ?? 1
-?? 2
 let's first articulate the meaning of both relations on 
set ?? ={1,2,3,…,20} : 
?? 1
 includes pairs (?? ,?? ) where ?? is divisible by ?? . This includes pairs like (1,1),(1,2),…,(1,20) for 1 ; 
similar series for 2 up to (2,20) (excluding odd numbers); for 3 up to (3,18); and so on, reflecting the 
divisibility condition. 
?? 1
:
{
 
 
 
 
 
 
 
 
(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),(6,12),(6,18),(7,7),(7,14),
(8,8),(8,16),(9,9),(9,18)(10,10),
(10,20),(11,11),(12,12)…,(20,20)}
 
 
 
 
 
 
 
 
?? (?? 1
)=66
 
?? 2
 consists of pairs (?? ,?? ) where ?? is an integral multiple of ?? . Essentially, this relationship is the reverse 
of ?? 1
. However, for the essence of ?? 1
-?? 2
, the key overlap comes with pairs where ?? =?? , since those 
are the pairs that clearly reflect both conditions identically (i.e., a number is always divisible by itself, 
and it is always a multiple of itself). There are 20 such pairs corresponding to each integer in ?? from 1 to 
20 , resulting in the common elements between ?? 1
 and ?? 2
 (the intersection ?? 1
n?? 2
 ) being 20 pairs. 
R
1
nR
2
={(1,1),(2,2),…(20,20)}
n(R
1
nR
2
)=20
 
The difference ?? 1
-?? 2
 seeks elements present in ?? 1
 but not in ?? 2
. Given that ?? 1
 and ?? 2
 share 20 
elements that are identical, to find ?? 1
-?? 2
 we subtract these 20 common elements from the total in 
?? 1
, resulting in 66-20=46 pairs. 
n(R
1
-R
2
)=n(R
1
)-n(R
1
nR
2
)
 =n(R
1
)-20
 =66-20
R
1
-R
2
=46 Pair 
 
Q2: Let ?? ={?? ,?? ,?? ,………,?????? }. Let ?? be a relation on ?? defined by (?? ,?? )??? if and only if ?? ?? =
?? ?? . Let ?? ?? be a symmetric relation on ?? such that ?? ??? ?? and the number of elements in ?? ?? is ?? . 
Then, the minimum value of ?? is    [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: 66 
R={(3,2),(6,4),(9,6),(12,8),………(99,66)}
n(R)=33
 ?66
 
Q3: Let ?? ={?? ,?? ,?? ,?? } and ?? ={(?? ,?? ),(?? ,?? ),(?? ,?? )} be a relation on ?? . Let ?? be the equivalence 
relation on ?? such that ?? ??? and the number of elements in ?? is ?? . Then, the minimum value of ?? is   
[JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 16 
?? ={1,2,3,4} 
?? ={(1,2),(2,3),(1,4)} 
?? is equivalence for ?? <?? and reflexive 
{(1,1),(2,2),(3,3),(4,4)} 
for symmetric 
{(2,1),(4,1),(3,2)} 
for transitive 
{(1,3),(3,1),(4,2),(2,4)} 
Now set 
?? ={(1,1),(2,2),(3,3),(4,4),(1,2),(2,3),(1,4),(4,3),(3,4),(2,1),(4,1),(3,2),(1,3),(3,1),(4,2),(2,4)} 
?? (?? )=16 
Q4: The number of symmetric relations defined on the set {?? ,?? ,?? ,?? } which are not reflexive is     [JEE 
Main 2024 (Online) 30th January Evening Shift] 
Ans: 960 
Total number of relation both symmetric and reflexive =2
?? 2
-?? 2
 
Total number of symmetric relation =2
(
n
2
+n
2
)
 
? Then number of symmetric relation which are not reflexive 
 ?2
?? (?? +1)
2
-2
?? (?? -1)
2
 ?2
10
-2
6
 ?1024-64
 =960
 
Q5: Let the set ?? ={(?? ,?? )|?? ?? -?? ?? =???????? ,?? ,?? ?N}. Then ?
(?? ,?? )??? ?(?? +?? ) is equal to      [JEE Main 
2024 (Online) 29th January Evening Shift] 
Ans: 46 
?? 2
-2
?? =2023
 ??? =45,?? =1
 ? ?
(?? ,?? )??? ?(?? +?? )=46.
 
Q6: Consider the relations ?? ?? and ?? ?? defined as ?? ?? ?? ?? ??? ?? +?? ?? =?? for all ?? ,?? ??? and 
(?? ,?? )?? ?? (?? ,?? )??? +?? =?? +?? for all (?? ,?? ),(?? ,?? )??? ×?? . Then : 
A.. ?? ?? and ?? ?? both are equivalence relations 
B. Only ?? ?? is an equivalence relation 
C. Only ?? ?? is an equivalence relation 
D. Neither ?? ?? nor ?? ?? is an equivalence relation    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (c) 
To determine if the given relations ?? 1
 and ?? 2
 are equivalence relations, we need to check whether each 
of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and 
transitivity. 
Let's start by analysing ?? 1
 : 
Reflexivity: A relation ?? on a set ?? is reflexive if every element is related to itself, that is, for every ?? ??? , 
the pair (?? ,?? ) is in ?? . In the case of ?? 1
, for an arbitrary ?? ??? , we need to check if ?? ?? 1
?? holds, which 
translates to checking if ?? 2
+?? 2
=1. This would mean 2?? 2
=1 or ?? 2
=
1
2
. Since this is not true for 
every real number ?? ,?? 1
 is not reflexive. 
Symmetry: A relation ?? is symmetric if whenever ?????? , then ?????? . For ?? 1
, if ?? 2
+?? 2
=1, then it is also 
true that ?? 2
+?? 2
=1. Thus, ?? 1
 is symmetric. 
Transitivity: A relation ?? is transitive if whenever ?????? and ?????? , then ?????? . For ?? 1
, suppose ?? ?? 1
?? and 
?? ?? 1
?? , this means ?? 2
+?? 2
=1 and ?? 2
+?? 2
=1. However, if we add these two, we get ?? 2
+2?? 2
+?? 2
=
2. There is no guarantee that ?? 2
+?? 2
=1. Therefore, ?? 1
 is not transitive. 
Conclusion: Relation ?? 1
 is not reflexive or transitive, and hence it is not an equivalence relation. 
Now let's consider ?? 2
 : 
Reflexivity: For any (?? ,?? )??? ×?? , it's clear that ?? +?? =?? +?? , which is just a reiteration of the 
commutative property of addition. Therefore, (?? ,?? )?? 2
(?? ,?? ), and ?? 2
 is reflexive. 
Symmetry: If (?? ,?? )?? 2
(?? ,?? ), meaning ?? +?? =?? +?? , then by reordering the terms we can similarly have 
?? +?? =?? +?? , which means (?? ,?? )?? 2
(?? ,?? ), so ?? 2
 is symmetric. 
Page 4


JEE Mains Previous Year Questions 
(2021-2024): Sets, Relations and Functions 
2024 
Q1: Let ?? ={?? ,?? ,?? ,…,???? }. Let ?? ?? and ?? ?? two relation on ?? such that 
?? ?? ={(?? ,?? ):?? is divisible by ?? } 
?? ?? ={(?? ,?? ):?? is an integral multiple of ?? }. 
Then, number of elements in ?? ?? -?? ?? is equal to    [JEE Main 2024 (Online) 1st February Morning 
Shift] 
Ans:  46 
To determine the number of elements in ?? 1
-?? 2
 let's first articulate the meaning of both relations on 
set ?? ={1,2,3,…,20} : 
?? 1
 includes pairs (?? ,?? ) where ?? is divisible by ?? . This includes pairs like (1,1),(1,2),…,(1,20) for 1 ; 
similar series for 2 up to (2,20) (excluding odd numbers); for 3 up to (3,18); and so on, reflecting the 
divisibility condition. 
?? 1
:
{
 
 
 
 
 
 
 
 
(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),(6,12),(6,18),(7,7),(7,14),
(8,8),(8,16),(9,9),(9,18)(10,10),
(10,20),(11,11),(12,12)…,(20,20)}
 
 
 
 
 
 
 
 
?? (?? 1
)=66
 
?? 2
 consists of pairs (?? ,?? ) where ?? is an integral multiple of ?? . Essentially, this relationship is the reverse 
of ?? 1
. However, for the essence of ?? 1
-?? 2
, the key overlap comes with pairs where ?? =?? , since those 
are the pairs that clearly reflect both conditions identically (i.e., a number is always divisible by itself, 
and it is always a multiple of itself). There are 20 such pairs corresponding to each integer in ?? from 1 to 
20 , resulting in the common elements between ?? 1
 and ?? 2
 (the intersection ?? 1
n?? 2
 ) being 20 pairs. 
R
1
nR
2
={(1,1),(2,2),…(20,20)}
n(R
1
nR
2
)=20
 
The difference ?? 1
-?? 2
 seeks elements present in ?? 1
 but not in ?? 2
. Given that ?? 1
 and ?? 2
 share 20 
elements that are identical, to find ?? 1
-?? 2
 we subtract these 20 common elements from the total in 
?? 1
, resulting in 66-20=46 pairs. 
n(R
1
-R
2
)=n(R
1
)-n(R
1
nR
2
)
 =n(R
1
)-20
 =66-20
R
1
-R
2
=46 Pair 
 
Q2: Let ?? ={?? ,?? ,?? ,………,?????? }. Let ?? be a relation on ?? defined by (?? ,?? )??? if and only if ?? ?? =
?? ?? . Let ?? ?? be a symmetric relation on ?? such that ?? ??? ?? and the number of elements in ?? ?? is ?? . 
Then, the minimum value of ?? is    [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: 66 
R={(3,2),(6,4),(9,6),(12,8),………(99,66)}
n(R)=33
 ?66
 
Q3: Let ?? ={?? ,?? ,?? ,?? } and ?? ={(?? ,?? ),(?? ,?? ),(?? ,?? )} be a relation on ?? . Let ?? be the equivalence 
relation on ?? such that ?? ??? and the number of elements in ?? is ?? . Then, the minimum value of ?? is   
[JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 16 
?? ={1,2,3,4} 
?? ={(1,2),(2,3),(1,4)} 
?? is equivalence for ?? <?? and reflexive 
{(1,1),(2,2),(3,3),(4,4)} 
for symmetric 
{(2,1),(4,1),(3,2)} 
for transitive 
{(1,3),(3,1),(4,2),(2,4)} 
Now set 
?? ={(1,1),(2,2),(3,3),(4,4),(1,2),(2,3),(1,4),(4,3),(3,4),(2,1),(4,1),(3,2),(1,3),(3,1),(4,2),(2,4)} 
?? (?? )=16 
Q4: The number of symmetric relations defined on the set {?? ,?? ,?? ,?? } which are not reflexive is     [JEE 
Main 2024 (Online) 30th January Evening Shift] 
Ans: 960 
Total number of relation both symmetric and reflexive =2
?? 2
-?? 2
 
Total number of symmetric relation =2
(
n
2
+n
2
)
 
? Then number of symmetric relation which are not reflexive 
 ?2
?? (?? +1)
2
-2
?? (?? -1)
2
 ?2
10
-2
6
 ?1024-64
 =960
 
Q5: Let the set ?? ={(?? ,?? )|?? ?? -?? ?? =???????? ,?? ,?? ?N}. Then ?
(?? ,?? )??? ?(?? +?? ) is equal to      [JEE Main 
2024 (Online) 29th January Evening Shift] 
Ans: 46 
?? 2
-2
?? =2023
 ??? =45,?? =1
 ? ?
(?? ,?? )??? ?(?? +?? )=46.
 
Q6: Consider the relations ?? ?? and ?? ?? defined as ?? ?? ?? ?? ??? ?? +?? ?? =?? for all ?? ,?? ??? and 
(?? ,?? )?? ?? (?? ,?? )??? +?? =?? +?? for all (?? ,?? ),(?? ,?? )??? ×?? . Then : 
A.. ?? ?? and ?? ?? both are equivalence relations 
B. Only ?? ?? is an equivalence relation 
C. Only ?? ?? is an equivalence relation 
D. Neither ?? ?? nor ?? ?? is an equivalence relation    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (c) 
To determine if the given relations ?? 1
 and ?? 2
 are equivalence relations, we need to check whether each 
of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and 
transitivity. 
Let's start by analysing ?? 1
 : 
Reflexivity: A relation ?? on a set ?? is reflexive if every element is related to itself, that is, for every ?? ??? , 
the pair (?? ,?? ) is in ?? . In the case of ?? 1
, for an arbitrary ?? ??? , we need to check if ?? ?? 1
?? holds, which 
translates to checking if ?? 2
+?? 2
=1. This would mean 2?? 2
=1 or ?? 2
=
1
2
. Since this is not true for 
every real number ?? ,?? 1
 is not reflexive. 
Symmetry: A relation ?? is symmetric if whenever ?????? , then ?????? . For ?? 1
, if ?? 2
+?? 2
=1, then it is also 
true that ?? 2
+?? 2
=1. Thus, ?? 1
 is symmetric. 
Transitivity: A relation ?? is transitive if whenever ?????? and ?????? , then ?????? . For ?? 1
, suppose ?? ?? 1
?? and 
?? ?? 1
?? , this means ?? 2
+?? 2
=1 and ?? 2
+?? 2
=1. However, if we add these two, we get ?? 2
+2?? 2
+?? 2
=
2. There is no guarantee that ?? 2
+?? 2
=1. Therefore, ?? 1
 is not transitive. 
Conclusion: Relation ?? 1
 is not reflexive or transitive, and hence it is not an equivalence relation. 
Now let's consider ?? 2
 : 
Reflexivity: For any (?? ,?? )??? ×?? , it's clear that ?? +?? =?? +?? , which is just a reiteration of the 
commutative property of addition. Therefore, (?? ,?? )?? 2
(?? ,?? ), and ?? 2
 is reflexive. 
Symmetry: If (?? ,?? )?? 2
(?? ,?? ), meaning ?? +?? =?? +?? , then by reordering the terms we can similarly have 
?? +?? =?? +?? , which means (?? ,?? )?? 2
(?? ,?? ), so ?? 2
 is symmetric. 
Transitivity: Suppose (?? ,?? )?? 2
(?? ,?? ) and (?? ,?? )?? 2
(?? ,?? ), i.e., ?? +?? =?? +?? and ?? +?? =?? +?? , we want 
to show that (?? ,?? )?? 2
(?? ,?? ). Adding the two equations, we get ?? +?? +?? +?? =?? +?? +?? +?? . By 
rearranging and simplifying, we get ?? +?? =?? +?? , thus, (?? ,?? )?? 2
(?? ,?? ). 
So ?? 2
 is reflexive, symmetric, and transitive, and therefore it is an equivalence relation. 
Conclusion: According to the above examination, only ?? 2
 is an equivalence relation. Thus, the correct 
answer is: 
Option C: Only ?? 2
 is an equivalence relation. 
Q7: If ?? is the smallest equivalence relation on the set {?? ,?? ,?? ,?? } such that {(?? ,?? ),(?? ,?? )}??? , then 
the number of elements in ?? is 
A. 15 
B. 10 
C. 12 
D. 8    [JEE Main 2024 (Online) 29th January Evening Shift] 
Ans: (b) 
Given set {1,2,3,4} 
Minimum order pairs are 
(1,1),(2,2),(3,3),(4,4),(3,1),(2,1),(2,3),(3,2),(1,3),(1,2) 
Thus no. of elements =10 
Q8: Let ?? be a relation on ?? ×?? defined by (?? ,?? )?? (?? ,?? ) if and only if ???? -???? is divisible by 5 . Then 
?? is 
A. Reflexive and transitive but not symmetric 
B. Reflexive and symmetric but not transitive 
C. Reflexive but neither symmetric nor transitive 
D. Reflexive, symmetric and transitive    [JEE Main 2024 (Online) 29th January Morning Shift] 
Ans: (b) 
Explanation: Currently no explanation available 
Q9: Let ?? ={?? ,?? ,?? ,…,???? }. Suppose ?? is the set of all the subsets of ?? , then the relation ?? =
{(?? ,?? ):?? n?? ??? ;?? ,?? ??? } is : 
A symmetric only 
B reflexive only 
C symmetric and reflexive only 
D symmetric and transitive only 
Ans: (a) 
Explanation: Currently no explanation available 
Page 5


JEE Mains Previous Year Questions 
(2021-2024): Sets, Relations and Functions 
2024 
Q1: Let ?? ={?? ,?? ,?? ,…,???? }. Let ?? ?? and ?? ?? two relation on ?? such that 
?? ?? ={(?? ,?? ):?? is divisible by ?? } 
?? ?? ={(?? ,?? ):?? is an integral multiple of ?? }. 
Then, number of elements in ?? ?? -?? ?? is equal to    [JEE Main 2024 (Online) 1st February Morning 
Shift] 
Ans:  46 
To determine the number of elements in ?? 1
-?? 2
 let's first articulate the meaning of both relations on 
set ?? ={1,2,3,…,20} : 
?? 1
 includes pairs (?? ,?? ) where ?? is divisible by ?? . This includes pairs like (1,1),(1,2),…,(1,20) for 1 ; 
similar series for 2 up to (2,20) (excluding odd numbers); for 3 up to (3,18); and so on, reflecting the 
divisibility condition. 
?? 1
:
{
 
 
 
 
 
 
 
 
(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),(6,12),(6,18),(7,7),(7,14),
(8,8),(8,16),(9,9),(9,18)(10,10),
(10,20),(11,11),(12,12)…,(20,20)}
 
 
 
 
 
 
 
 
?? (?? 1
)=66
 
?? 2
 consists of pairs (?? ,?? ) where ?? is an integral multiple of ?? . Essentially, this relationship is the reverse 
of ?? 1
. However, for the essence of ?? 1
-?? 2
, the key overlap comes with pairs where ?? =?? , since those 
are the pairs that clearly reflect both conditions identically (i.e., a number is always divisible by itself, 
and it is always a multiple of itself). There are 20 such pairs corresponding to each integer in ?? from 1 to 
20 , resulting in the common elements between ?? 1
 and ?? 2
 (the intersection ?? 1
n?? 2
 ) being 20 pairs. 
R
1
nR
2
={(1,1),(2,2),…(20,20)}
n(R
1
nR
2
)=20
 
The difference ?? 1
-?? 2
 seeks elements present in ?? 1
 but not in ?? 2
. Given that ?? 1
 and ?? 2
 share 20 
elements that are identical, to find ?? 1
-?? 2
 we subtract these 20 common elements from the total in 
?? 1
, resulting in 66-20=46 pairs. 
n(R
1
-R
2
)=n(R
1
)-n(R
1
nR
2
)
 =n(R
1
)-20
 =66-20
R
1
-R
2
=46 Pair 
 
Q2: Let ?? ={?? ,?? ,?? ,………,?????? }. Let ?? be a relation on ?? defined by (?? ,?? )??? if and only if ?? ?? =
?? ?? . Let ?? ?? be a symmetric relation on ?? such that ?? ??? ?? and the number of elements in ?? ?? is ?? . 
Then, the minimum value of ?? is    [JEE Main 2024 (Online) 31st January Evening Shift] 
Ans: 66 
R={(3,2),(6,4),(9,6),(12,8),………(99,66)}
n(R)=33
 ?66
 
Q3: Let ?? ={?? ,?? ,?? ,?? } and ?? ={(?? ,?? ),(?? ,?? ),(?? ,?? )} be a relation on ?? . Let ?? be the equivalence 
relation on ?? such that ?? ??? and the number of elements in ?? is ?? . Then, the minimum value of ?? is   
[JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: 16 
?? ={1,2,3,4} 
?? ={(1,2),(2,3),(1,4)} 
?? is equivalence for ?? <?? and reflexive 
{(1,1),(2,2),(3,3),(4,4)} 
for symmetric 
{(2,1),(4,1),(3,2)} 
for transitive 
{(1,3),(3,1),(4,2),(2,4)} 
Now set 
?? ={(1,1),(2,2),(3,3),(4,4),(1,2),(2,3),(1,4),(4,3),(3,4),(2,1),(4,1),(3,2),(1,3),(3,1),(4,2),(2,4)} 
?? (?? )=16 
Q4: The number of symmetric relations defined on the set {?? ,?? ,?? ,?? } which are not reflexive is     [JEE 
Main 2024 (Online) 30th January Evening Shift] 
Ans: 960 
Total number of relation both symmetric and reflexive =2
?? 2
-?? 2
 
Total number of symmetric relation =2
(
n
2
+n
2
)
 
? Then number of symmetric relation which are not reflexive 
 ?2
?? (?? +1)
2
-2
?? (?? -1)
2
 ?2
10
-2
6
 ?1024-64
 =960
 
Q5: Let the set ?? ={(?? ,?? )|?? ?? -?? ?? =???????? ,?? ,?? ?N}. Then ?
(?? ,?? )??? ?(?? +?? ) is equal to      [JEE Main 
2024 (Online) 29th January Evening Shift] 
Ans: 46 
?? 2
-2
?? =2023
 ??? =45,?? =1
 ? ?
(?? ,?? )??? ?(?? +?? )=46.
 
Q6: Consider the relations ?? ?? and ?? ?? defined as ?? ?? ?? ?? ??? ?? +?? ?? =?? for all ?? ,?? ??? and 
(?? ,?? )?? ?? (?? ,?? )??? +?? =?? +?? for all (?? ,?? ),(?? ,?? )??? ×?? . Then : 
A.. ?? ?? and ?? ?? both are equivalence relations 
B. Only ?? ?? is an equivalence relation 
C. Only ?? ?? is an equivalence relation 
D. Neither ?? ?? nor ?? ?? is an equivalence relation    [JEE Main 2024 (Online) 1st February Evening Shift] 
Ans: (c) 
To determine if the given relations ?? 1
 and ?? 2
 are equivalence relations, we need to check whether each 
of them satisfies the three defining properties of an equivalence relation: reflexivity, symmetry, and 
transitivity. 
Let's start by analysing ?? 1
 : 
Reflexivity: A relation ?? on a set ?? is reflexive if every element is related to itself, that is, for every ?? ??? , 
the pair (?? ,?? ) is in ?? . In the case of ?? 1
, for an arbitrary ?? ??? , we need to check if ?? ?? 1
?? holds, which 
translates to checking if ?? 2
+?? 2
=1. This would mean 2?? 2
=1 or ?? 2
=
1
2
. Since this is not true for 
every real number ?? ,?? 1
 is not reflexive. 
Symmetry: A relation ?? is symmetric if whenever ?????? , then ?????? . For ?? 1
, if ?? 2
+?? 2
=1, then it is also 
true that ?? 2
+?? 2
=1. Thus, ?? 1
 is symmetric. 
Transitivity: A relation ?? is transitive if whenever ?????? and ?????? , then ?????? . For ?? 1
, suppose ?? ?? 1
?? and 
?? ?? 1
?? , this means ?? 2
+?? 2
=1 and ?? 2
+?? 2
=1. However, if we add these two, we get ?? 2
+2?? 2
+?? 2
=
2. There is no guarantee that ?? 2
+?? 2
=1. Therefore, ?? 1
 is not transitive. 
Conclusion: Relation ?? 1
 is not reflexive or transitive, and hence it is not an equivalence relation. 
Now let's consider ?? 2
 : 
Reflexivity: For any (?? ,?? )??? ×?? , it's clear that ?? +?? =?? +?? , which is just a reiteration of the 
commutative property of addition. Therefore, (?? ,?? )?? 2
(?? ,?? ), and ?? 2
 is reflexive. 
Symmetry: If (?? ,?? )?? 2
(?? ,?? ), meaning ?? +?? =?? +?? , then by reordering the terms we can similarly have 
?? +?? =?? +?? , which means (?? ,?? )?? 2
(?? ,?? ), so ?? 2
 is symmetric. 
Transitivity: Suppose (?? ,?? )?? 2
(?? ,?? ) and (?? ,?? )?? 2
(?? ,?? ), i.e., ?? +?? =?? +?? and ?? +?? =?? +?? , we want 
to show that (?? ,?? )?? 2
(?? ,?? ). Adding the two equations, we get ?? +?? +?? +?? =?? +?? +?? +?? . By 
rearranging and simplifying, we get ?? +?? =?? +?? , thus, (?? ,?? )?? 2
(?? ,?? ). 
So ?? 2
 is reflexive, symmetric, and transitive, and therefore it is an equivalence relation. 
Conclusion: According to the above examination, only ?? 2
 is an equivalence relation. Thus, the correct 
answer is: 
Option C: Only ?? 2
 is an equivalence relation. 
Q7: If ?? is the smallest equivalence relation on the set {?? ,?? ,?? ,?? } such that {(?? ,?? ),(?? ,?? )}??? , then 
the number of elements in ?? is 
A. 15 
B. 10 
C. 12 
D. 8    [JEE Main 2024 (Online) 29th January Evening Shift] 
Ans: (b) 
Given set {1,2,3,4} 
Minimum order pairs are 
(1,1),(2,2),(3,3),(4,4),(3,1),(2,1),(2,3),(3,2),(1,3),(1,2) 
Thus no. of elements =10 
Q8: Let ?? be a relation on ?? ×?? defined by (?? ,?? )?? (?? ,?? ) if and only if ???? -???? is divisible by 5 . Then 
?? is 
A. Reflexive and transitive but not symmetric 
B. Reflexive and symmetric but not transitive 
C. Reflexive but neither symmetric nor transitive 
D. Reflexive, symmetric and transitive    [JEE Main 2024 (Online) 29th January Morning Shift] 
Ans: (b) 
Explanation: Currently no explanation available 
Q9: Let ?? ={?? ,?? ,?? ,…,???? }. Suppose ?? is the set of all the subsets of ?? , then the relation ?? =
{(?? ,?? ):?? n?? ??? ;?? ,?? ??? } is : 
A symmetric only 
B reflexive only 
C symmetric and reflexive only 
D symmetric and transitive only 
Ans: (a) 
Explanation: Currently no explanation available 
2023 
Q1: Let A = {1 , 3 , 4 , 6 , 9} and B = {2 , 4 , 5 , 8 , 10}. Let R be a relation defined on A × B such that R = 
{((a
1
, b
1
) , (a
2 
, b
2
)) : a
1
 = b
2
 and b
1
 = a
2
}. Then the number of elements in the set R is: 
(a) 180 
(b) 26 
(c) 52 
(d) 160 
Ans: (d) 
Given that the sets are A = {1, 3, 4, 6, 9} and B = {2, 4, 5, 8, 10}, for the relation R on the set A × B, we 
need to find the combinations of pairs that satisfy the conditions a
1
 =  b
2
 and b
1
 =  a
2
. 
We find the number of combinations by considering the possible values for b
2
 for each a
1
 and the 
possible values for a
2
 for each b
1
: 
For each a
1
 in A = {1, 3, 4, 6, 9} , the number of valid b
2
 values in B = {2, 4, 5, 8, 10} are: 
- For a
1
 = 1 , there are 5 choices for b
2
. 
- For a
1
 = 3 , there are 4 choices for b
2
. 
- For a
1
 = 4 , there are 4 choices for b
2
. 
- For a
1
 = 6 , there are 2 choices for b
2
. 
-For a
1
 = 9 , there is 1 choice for b
2
. 
This results in a total of 5 + 4 + 4 + 2 + 1 = 16 possible pairs (a
1
 , b
2
).   
Similarly, for each b
1
 in B, the number of valid a
2
 values in A are: 
- For b
1
 = 2 , there are 4 choices for a
2
. 
- For b
1
 = 4 , there are 3 choices for a
2
. 
- For b
1
 = 5 , there are 2 choices for a
2
. 
- For b
1
 = 8 , there is 1 choice for a
2
. 
- For b
1
 = 10 , there are no choices for a
2
. 
This results in a total of 4 + 3 + 2 + 1 + 0 = 10 possible pairs (b
1
 , a
2
). 
Therefore, the total number of elements in the relation R , which satisfies the given conditions, is 16 × 
10 = 160. 
So, the correct answer is 160. 
 
Q2: An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals 
went to total 60 men and only five men got medals in all the three events, then, how many received 
medals in exactly two of three events? 
(a) 10 
(b) 15 
(c) 21 
(d) 9 
Ans: (c) 
We are given the number of medals for events A, B, and C which are 48, 25, and 18 respectively. We are 
also given that the total number of unique medal recipients across all events is 60 and that 5 people 
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