JEE Exam > JEE Notes > Mathematics (Maths) Main & Advanced > JEE Main Previous Year Questions (2025): Set, Relations and Functions
JEE Main Previous Year Questions (2025): Set, Relations and Functions
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JEE Main Previous Year Questions
(2025): Set, Relations and Functions
Q1: Let ?? = {?? , ?? , ?? }. The number of relations on ?? , containing ( 1,2 ) and ( 2,3 ), which
are reflexive and transitive but not symmetric, is ____ ?
JEE Main 2025 (Online) 22nd January Evening Shift
Ans: 3
Solution:
Transitivity
( 1,2)? R, ( 2,3)? R ? ( 1,3)? R
For reflexive ( 1,1) , ( 2,2) ( 3,3)? ??
Now ( 2,1) , ( 3,2) , ( 3,1)
( 3,1) cannot be taken
(1) ( 2,1) taken and ( 3,2) not taken
(2) ( 3,2) taken and ( 2,1) not taken
(3) Both not taken
therefore 3 relations are possible.
Q2: Let ?? = {?? ?? , ?? ?? … , ?? ????
} be the set of first ten prime numbers. Let ?? = ?? ? ?? ,
where ?? is the set of all possible products of distinct elements of ?? . Then the number
of all ordered pairs ( ?? , ?? ) , ?? ? ?? , ?? ? ?? , such that ?? divides ?? , is ____ .
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 5120
Solution:
Let
y
x
= ??
y = ?? x
= 10 × (
9
C
0
+
9
C
1
+
9
C
2
+
9
C
3
+ ? +
9
C
9
)
= 10 × ( 2
9
)
10 × 512
5120
Q3: For ?? = ?? , let ?? ?? denote the set of all subsets of {?? , ?? , … , ?? } with no two
consecutive numbers. For example {?? , ?? , ?? } ? ?? ?? , but {?? , ?? , ?? } ? ?? ?? . Then ?? ( ?? ?? ) is
equal to ____
JEE Main 2025 (Online) 7th April Morning Shift
Page 2
JEE Main Previous Year Questions
(2025): Set, Relations and Functions
Q1: Let ?? = {?? , ?? , ?? }. The number of relations on ?? , containing ( 1,2 ) and ( 2,3 ), which
are reflexive and transitive but not symmetric, is ____ ?
JEE Main 2025 (Online) 22nd January Evening Shift
Ans: 3
Solution:
Transitivity
( 1,2)? R, ( 2,3)? R ? ( 1,3)? R
For reflexive ( 1,1) , ( 2,2) ( 3,3)? ??
Now ( 2,1) , ( 3,2) , ( 3,1)
( 3,1) cannot be taken
(1) ( 2,1) taken and ( 3,2) not taken
(2) ( 3,2) taken and ( 2,1) not taken
(3) Both not taken
therefore 3 relations are possible.
Q2: Let ?? = {?? ?? , ?? ?? … , ?? ????
} be the set of first ten prime numbers. Let ?? = ?? ? ?? ,
where ?? is the set of all possible products of distinct elements of ?? . Then the number
of all ordered pairs ( ?? , ?? ) , ?? ? ?? , ?? ? ?? , such that ?? divides ?? , is ____ .
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 5120
Solution:
Let
y
x
= ??
y = ?? x
= 10 × (
9
C
0
+
9
C
1
+
9
C
2
+
9
C
3
+ ? +
9
C
9
)
= 10 × ( 2
9
)
10 × 512
5120
Q3: For ?? = ?? , let ?? ?? denote the set of all subsets of {?? , ?? , … , ?? } with no two
consecutive numbers. For example {?? , ?? , ?? } ? ?? ?? , but {?? , ?? , ?? } ? ?? ?? . Then ?? ( ?? ?? ) is
equal to ____
JEE Main 2025 (Online) 7th April Morning Shift
Ans: 13
Solution:
To find ?? ( ?? 5
) , which is the number of subsets of {1,2,3,4,5} with no consecutive numbers, we
start by enumerating these subsets.
Let's denote the set {1,2,3,4,5} as ?? . The subsets of ?? that meet the criteria are:
The empty set: }
Single-element sets: {1}, {2}, {3}, {4}, {5}
Two-element sets with no consecutive numbers: {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}
Three-element set with no consecutive numbers: {1,3,5}
Counting these subsets, we have:
1 subset with zero elements
5 subsets with one element
6 subsets with two elements
1 subset with three elements
Adding these counts, there are 1 + 5 + 6 + 1 = 13 subsets in total.
Thus, ?? ( ?? 5
)= 13.
Q4: The number of relations on the set ?? = {?? , ?? , ?? }, containing at most 6 elements
including ( ?? , ?? ) , which are reflexive and transitive but not symmetric, is ____.
Ans: 6
Solution:
Since relation needs to be reflexive the ordered pairs ( 1,1) , ( 2,2) , ( 3,3) need to be there and
( 1,2) is also to be included.
Let's call ?? 0
= {( 1,1) , ( 2,2) , ( 3,3) , ( 1,2) } the base relation.
? ?? × ?? contain 3 × 3 = 9 ordered pairs, remaining 5 ordered are
2,1) , ( 1,3) , ( 3,1) , ( 2,3) , ( 3,2)
We have to add at most two ordered pairs to ?? 0
such that resulting relation is reflexive,
transitive but not symmetric.
Following are the only possibilities.
?? = ?? 0
?? {( 1,3) }
OR ?? 0
?? {( 3,2) }
OR ?? 0
?? {( 1,3) , ( 3,1) }
OR ?? 0
?? {( 1,3) , ( 3,2) }
OR ?? 0
?? {( 3,1) , ( 3,2) }
Q5: The number of non-empty equivalence relations on the set {1,2,3} is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
Page 3
JEE Main Previous Year Questions
(2025): Set, Relations and Functions
Q1: Let ?? = {?? , ?? , ?? }. The number of relations on ?? , containing ( 1,2 ) and ( 2,3 ), which
are reflexive and transitive but not symmetric, is ____ ?
JEE Main 2025 (Online) 22nd January Evening Shift
Ans: 3
Solution:
Transitivity
( 1,2)? R, ( 2,3)? R ? ( 1,3)? R
For reflexive ( 1,1) , ( 2,2) ( 3,3)? ??
Now ( 2,1) , ( 3,2) , ( 3,1)
( 3,1) cannot be taken
(1) ( 2,1) taken and ( 3,2) not taken
(2) ( 3,2) taken and ( 2,1) not taken
(3) Both not taken
therefore 3 relations are possible.
Q2: Let ?? = {?? ?? , ?? ?? … , ?? ????
} be the set of first ten prime numbers. Let ?? = ?? ? ?? ,
where ?? is the set of all possible products of distinct elements of ?? . Then the number
of all ordered pairs ( ?? , ?? ) , ?? ? ?? , ?? ? ?? , such that ?? divides ?? , is ____ .
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 5120
Solution:
Let
y
x
= ??
y = ?? x
= 10 × (
9
C
0
+
9
C
1
+
9
C
2
+
9
C
3
+ ? +
9
C
9
)
= 10 × ( 2
9
)
10 × 512
5120
Q3: For ?? = ?? , let ?? ?? denote the set of all subsets of {?? , ?? , … , ?? } with no two
consecutive numbers. For example {?? , ?? , ?? } ? ?? ?? , but {?? , ?? , ?? } ? ?? ?? . Then ?? ( ?? ?? ) is
equal to ____
JEE Main 2025 (Online) 7th April Morning Shift
Ans: 13
Solution:
To find ?? ( ?? 5
) , which is the number of subsets of {1,2,3,4,5} with no consecutive numbers, we
start by enumerating these subsets.
Let's denote the set {1,2,3,4,5} as ?? . The subsets of ?? that meet the criteria are:
The empty set: }
Single-element sets: {1}, {2}, {3}, {4}, {5}
Two-element sets with no consecutive numbers: {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}
Three-element set with no consecutive numbers: {1,3,5}
Counting these subsets, we have:
1 subset with zero elements
5 subsets with one element
6 subsets with two elements
1 subset with three elements
Adding these counts, there are 1 + 5 + 6 + 1 = 13 subsets in total.
Thus, ?? ( ?? 5
)= 13.
Q4: The number of relations on the set ?? = {?? , ?? , ?? }, containing at most 6 elements
including ( ?? , ?? ) , which are reflexive and transitive but not symmetric, is ____.
Ans: 6
Solution:
Since relation needs to be reflexive the ordered pairs ( 1,1) , ( 2,2) , ( 3,3) need to be there and
( 1,2) is also to be included.
Let's call ?? 0
= {( 1,1) , ( 2,2) , ( 3,3) , ( 1,2) } the base relation.
? ?? × ?? contain 3 × 3 = 9 ordered pairs, remaining 5 ordered are
2,1) , ( 1,3) , ( 3,1) , ( 2,3) , ( 3,2)
We have to add at most two ordered pairs to ?? 0
such that resulting relation is reflexive,
transitive but not symmetric.
Following are the only possibilities.
?? = ?? 0
?? {( 1,3) }
OR ?? 0
?? {( 3,2) }
OR ?? 0
?? {( 1,3) , ( 3,1) }
OR ?? 0
?? {( 1,3) , ( 3,2) }
OR ?? 0
?? {( 3,1) , ( 3,2) }
Q5: The number of non-empty equivalence relations on the set {1,2,3} is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 7
B. 4
C. 5
D. 6
Ans: C
Solution:
An equivalence relation on a finite set is uniquely determined by its partition into equivalence
classes. Hence, counting the number of equivalence relations on a set is equivalent to counting
the number of ways to partition that set.
Step: Counting partitions of {1,2,3}
We want all possible ways to split the set {1,2,3} into nonempty subsets (its "blocks").
3 blocks (each element in its own block)
{{1}, {2}, {3}}.
2 blocks
{{1,2}, {3}}
{{1,3}, {2}}
{{2,3}, {1}}
1 block (all elements together)
{{1,2,3}}.
Counting these, there are a total of ?? distinct partitions, and thus ?? equivalence relations on the
set {1,2,3}.
All equivalence relations are automatically nonempty (they include at least ( 1,1) , ( 2,2) , ( 3,3)
because they are reflexive), so the answer to "the number of nonempty equivalence relations" is
also 5 .
Ans: Option C (5)
Q6: Let ?? = {?? , ?? , ?? , … , ???? } and ?? = {
?? ?? : ?? , ?? ? ?? , ?? < ?? and ?????? ( ?? , ?? )= ?? }. Then
?? ( ?? ) is equal to :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 29
B. 31
C. 37
D. 36
Ans: B
Page 4
JEE Main Previous Year Questions
(2025): Set, Relations and Functions
Q1: Let ?? = {?? , ?? , ?? }. The number of relations on ?? , containing ( 1,2 ) and ( 2,3 ), which
are reflexive and transitive but not symmetric, is ____ ?
JEE Main 2025 (Online) 22nd January Evening Shift
Ans: 3
Solution:
Transitivity
( 1,2)? R, ( 2,3)? R ? ( 1,3)? R
For reflexive ( 1,1) , ( 2,2) ( 3,3)? ??
Now ( 2,1) , ( 3,2) , ( 3,1)
( 3,1) cannot be taken
(1) ( 2,1) taken and ( 3,2) not taken
(2) ( 3,2) taken and ( 2,1) not taken
(3) Both not taken
therefore 3 relations are possible.
Q2: Let ?? = {?? ?? , ?? ?? … , ?? ????
} be the set of first ten prime numbers. Let ?? = ?? ? ?? ,
where ?? is the set of all possible products of distinct elements of ?? . Then the number
of all ordered pairs ( ?? , ?? ) , ?? ? ?? , ?? ? ?? , such that ?? divides ?? , is ____ .
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 5120
Solution:
Let
y
x
= ??
y = ?? x
= 10 × (
9
C
0
+
9
C
1
+
9
C
2
+
9
C
3
+ ? +
9
C
9
)
= 10 × ( 2
9
)
10 × 512
5120
Q3: For ?? = ?? , let ?? ?? denote the set of all subsets of {?? , ?? , … , ?? } with no two
consecutive numbers. For example {?? , ?? , ?? } ? ?? ?? , but {?? , ?? , ?? } ? ?? ?? . Then ?? ( ?? ?? ) is
equal to ____
JEE Main 2025 (Online) 7th April Morning Shift
Ans: 13
Solution:
To find ?? ( ?? 5
) , which is the number of subsets of {1,2,3,4,5} with no consecutive numbers, we
start by enumerating these subsets.
Let's denote the set {1,2,3,4,5} as ?? . The subsets of ?? that meet the criteria are:
The empty set: }
Single-element sets: {1}, {2}, {3}, {4}, {5}
Two-element sets with no consecutive numbers: {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}
Three-element set with no consecutive numbers: {1,3,5}
Counting these subsets, we have:
1 subset with zero elements
5 subsets with one element
6 subsets with two elements
1 subset with three elements
Adding these counts, there are 1 + 5 + 6 + 1 = 13 subsets in total.
Thus, ?? ( ?? 5
)= 13.
Q4: The number of relations on the set ?? = {?? , ?? , ?? }, containing at most 6 elements
including ( ?? , ?? ) , which are reflexive and transitive but not symmetric, is ____.
Ans: 6
Solution:
Since relation needs to be reflexive the ordered pairs ( 1,1) , ( 2,2) , ( 3,3) need to be there and
( 1,2) is also to be included.
Let's call ?? 0
= {( 1,1) , ( 2,2) , ( 3,3) , ( 1,2) } the base relation.
? ?? × ?? contain 3 × 3 = 9 ordered pairs, remaining 5 ordered are
2,1) , ( 1,3) , ( 3,1) , ( 2,3) , ( 3,2)
We have to add at most two ordered pairs to ?? 0
such that resulting relation is reflexive,
transitive but not symmetric.
Following are the only possibilities.
?? = ?? 0
?? {( 1,3) }
OR ?? 0
?? {( 3,2) }
OR ?? 0
?? {( 1,3) , ( 3,1) }
OR ?? 0
?? {( 1,3) , ( 3,2) }
OR ?? 0
?? {( 3,1) , ( 3,2) }
Q5: The number of non-empty equivalence relations on the set {1,2,3} is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 7
B. 4
C. 5
D. 6
Ans: C
Solution:
An equivalence relation on a finite set is uniquely determined by its partition into equivalence
classes. Hence, counting the number of equivalence relations on a set is equivalent to counting
the number of ways to partition that set.
Step: Counting partitions of {1,2,3}
We want all possible ways to split the set {1,2,3} into nonempty subsets (its "blocks").
3 blocks (each element in its own block)
{{1}, {2}, {3}}.
2 blocks
{{1,2}, {3}}
{{1,3}, {2}}
{{2,3}, {1}}
1 block (all elements together)
{{1,2,3}}.
Counting these, there are a total of ?? distinct partitions, and thus ?? equivalence relations on the
set {1,2,3}.
All equivalence relations are automatically nonempty (they include at least ( 1,1) , ( 2,2) , ( 3,3)
because they are reflexive), so the answer to "the number of nonempty equivalence relations" is
also 5 .
Ans: Option C (5)
Q6: Let ?? = {?? , ?? , ?? , … , ???? } and ?? = {
?? ?? : ?? , ?? ? ?? , ?? < ?? and ?????? ( ?? , ?? )= ?? }. Then
?? ( ?? ) is equal to :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 29
B. 31
C. 37
D. 36
Ans: B
Solution:
To find the number of elements in set ?? , we consider pairs (
?? ?? ) where ?? , ?? ? ?? with ?? < ?? and
gcd( ?? , ?? )= 1.
Here's the breakdown for each possible ?? :
For ?? = 1 :
Possible values for ?? are 2,3,4,5,6,7,8,9,10.
Total pairs: 9.
For ?? = 2 :
Possible values for ?? are 3,5,7,9 (since these have gcd( 2, ?? )= 1 ).
Total pairs: 4.
For ?? = 3 :
Possible values for ?? are 4,5,7,8,10.
Total pairs: 5.
For ?? = 4 :
Possible values for ?? are 5,7,9.
Total pairs: 3.
For ?? = 5 :
Possible values for ?? are 6,7,8,9.
Total pairs: 4.
For ?? = 6 :
Possible value for ?? is 7 .
Total pairs: 1.
For ?? = 7 :
Possible values for ?? are 8,9,10.
Total pairs: 3.
For ?? = 8 :
Possible value for ?? is 9 .
Total pairs: 1.
For ?? = 9 :
Possible value for ?? is 10 .
Total pairs: 1.
Adding all these up, the total number of elements in set ?? is:
9 + 4 + 5 + 3 + 4 + 1 + 3 + 1 + 1 = 31
Q7: Let ?? = {( ?? , ?? ) , ( ?? , ?? ) , ( ?? , ?? ) } be a relation defined on the set {?? , ?? , ?? , ?? }. Then the
minimum number of elements, needed to be added in ?? so that ?? becomes an
equivalence relation, is:
JEE Main 2025 (Online) 23rd January Morning Shift
Options:
A. 9
B. 8
Page 5
JEE Main Previous Year Questions
(2025): Set, Relations and Functions
Q1: Let ?? = {?? , ?? , ?? }. The number of relations on ?? , containing ( 1,2 ) and ( 2,3 ), which
are reflexive and transitive but not symmetric, is ____ ?
JEE Main 2025 (Online) 22nd January Evening Shift
Ans: 3
Solution:
Transitivity
( 1,2)? R, ( 2,3)? R ? ( 1,3)? R
For reflexive ( 1,1) , ( 2,2) ( 3,3)? ??
Now ( 2,1) , ( 3,2) , ( 3,1)
( 3,1) cannot be taken
(1) ( 2,1) taken and ( 3,2) not taken
(2) ( 3,2) taken and ( 2,1) not taken
(3) Both not taken
therefore 3 relations are possible.
Q2: Let ?? = {?? ?? , ?? ?? … , ?? ????
} be the set of first ten prime numbers. Let ?? = ?? ? ?? ,
where ?? is the set of all possible products of distinct elements of ?? . Then the number
of all ordered pairs ( ?? , ?? ) , ?? ? ?? , ?? ? ?? , such that ?? divides ?? , is ____ .
JEE Main 2025 (Online) 24th January Morning Shift
Ans: 5120
Solution:
Let
y
x
= ??
y = ?? x
= 10 × (
9
C
0
+
9
C
1
+
9
C
2
+
9
C
3
+ ? +
9
C
9
)
= 10 × ( 2
9
)
10 × 512
5120
Q3: For ?? = ?? , let ?? ?? denote the set of all subsets of {?? , ?? , … , ?? } with no two
consecutive numbers. For example {?? , ?? , ?? } ? ?? ?? , but {?? , ?? , ?? } ? ?? ?? . Then ?? ( ?? ?? ) is
equal to ____
JEE Main 2025 (Online) 7th April Morning Shift
Ans: 13
Solution:
To find ?? ( ?? 5
) , which is the number of subsets of {1,2,3,4,5} with no consecutive numbers, we
start by enumerating these subsets.
Let's denote the set {1,2,3,4,5} as ?? . The subsets of ?? that meet the criteria are:
The empty set: }
Single-element sets: {1}, {2}, {3}, {4}, {5}
Two-element sets with no consecutive numbers: {1,3}, {1,4}, {1,5}, {2,4}, {2,5}, {3,5}
Three-element set with no consecutive numbers: {1,3,5}
Counting these subsets, we have:
1 subset with zero elements
5 subsets with one element
6 subsets with two elements
1 subset with three elements
Adding these counts, there are 1 + 5 + 6 + 1 = 13 subsets in total.
Thus, ?? ( ?? 5
)= 13.
Q4: The number of relations on the set ?? = {?? , ?? , ?? }, containing at most 6 elements
including ( ?? , ?? ) , which are reflexive and transitive but not symmetric, is ____.
Ans: 6
Solution:
Since relation needs to be reflexive the ordered pairs ( 1,1) , ( 2,2) , ( 3,3) need to be there and
( 1,2) is also to be included.
Let's call ?? 0
= {( 1,1) , ( 2,2) , ( 3,3) , ( 1,2) } the base relation.
? ?? × ?? contain 3 × 3 = 9 ordered pairs, remaining 5 ordered are
2,1) , ( 1,3) , ( 3,1) , ( 2,3) , ( 3,2)
We have to add at most two ordered pairs to ?? 0
such that resulting relation is reflexive,
transitive but not symmetric.
Following are the only possibilities.
?? = ?? 0
?? {( 1,3) }
OR ?? 0
?? {( 3,2) }
OR ?? 0
?? {( 1,3) , ( 3,1) }
OR ?? 0
?? {( 1,3) , ( 3,2) }
OR ?? 0
?? {( 3,1) , ( 3,2) }
Q5: The number of non-empty equivalence relations on the set {1,2,3} is :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 7
B. 4
C. 5
D. 6
Ans: C
Solution:
An equivalence relation on a finite set is uniquely determined by its partition into equivalence
classes. Hence, counting the number of equivalence relations on a set is equivalent to counting
the number of ways to partition that set.
Step: Counting partitions of {1,2,3}
We want all possible ways to split the set {1,2,3} into nonempty subsets (its "blocks").
3 blocks (each element in its own block)
{{1}, {2}, {3}}.
2 blocks
{{1,2}, {3}}
{{1,3}, {2}}
{{2,3}, {1}}
1 block (all elements together)
{{1,2,3}}.
Counting these, there are a total of ?? distinct partitions, and thus ?? equivalence relations on the
set {1,2,3}.
All equivalence relations are automatically nonempty (they include at least ( 1,1) , ( 2,2) , ( 3,3)
because they are reflexive), so the answer to "the number of nonempty equivalence relations" is
also 5 .
Ans: Option C (5)
Q6: Let ?? = {?? , ?? , ?? , … , ???? } and ?? = {
?? ?? : ?? , ?? ? ?? , ?? < ?? and ?????? ( ?? , ?? )= ?? }. Then
?? ( ?? ) is equal to :
JEE Main 2025 (Online) 22nd January Morning Shift
Options:
A. 29
B. 31
C. 37
D. 36
Ans: B
Solution:
To find the number of elements in set ?? , we consider pairs (
?? ?? ) where ?? , ?? ? ?? with ?? < ?? and
gcd( ?? , ?? )= 1.
Here's the breakdown for each possible ?? :
For ?? = 1 :
Possible values for ?? are 2,3,4,5,6,7,8,9,10.
Total pairs: 9.
For ?? = 2 :
Possible values for ?? are 3,5,7,9 (since these have gcd( 2, ?? )= 1 ).
Total pairs: 4.
For ?? = 3 :
Possible values for ?? are 4,5,7,8,10.
Total pairs: 5.
For ?? = 4 :
Possible values for ?? are 5,7,9.
Total pairs: 3.
For ?? = 5 :
Possible values for ?? are 6,7,8,9.
Total pairs: 4.
For ?? = 6 :
Possible value for ?? is 7 .
Total pairs: 1.
For ?? = 7 :
Possible values for ?? are 8,9,10.
Total pairs: 3.
For ?? = 8 :
Possible value for ?? is 9 .
Total pairs: 1.
For ?? = 9 :
Possible value for ?? is 10 .
Total pairs: 1.
Adding all these up, the total number of elements in set ?? is:
9 + 4 + 5 + 3 + 4 + 1 + 3 + 1 + 1 = 31
Q7: Let ?? = {( ?? , ?? ) , ( ?? , ?? ) , ( ?? , ?? ) } be a relation defined on the set {?? , ?? , ?? , ?? }. Then the
minimum number of elements, needed to be added in ?? so that ?? becomes an
equivalence relation, is:
JEE Main 2025 (Online) 23rd January Morning Shift
Options:
A. 9
B. 8
C. 7
D. 10
Ans: C
Solution:
A = {1,2,3,4}
For relation to be reflexive
R = {( 1,2) , ( 2,3) , ( 3,3) }
Minimum elements added will be
( 1,1) , ( 2,2) , ( 4,4) ( 2,1) ( 3,2) ( 3,2) ( 3,1) ( 1,3)
? Minimum number of elements = 7
Q8: Let ?? = {( ?? , ?? )? ?? × ?? : |?? + ?? | ? ?? } and ?? = {( ?? , ?? )? ?? × ?? : |?? | + |?? | = ?? }.If
?? = {( ?? , ?? )? ?? n ?? : ?? = ?? or ?? = ?? }, then ?
( ?? ,?? ) ??? ?|?? + ?? | is :
JEE Main 2025 (Online) 23rd January Evening Shift
Options:
A. 18
B. 24
C. 15
D. 12
Ans: D
Solution:
C = {( 3,0) , ( -3,0) , ( 0,3) , ( 0, -3) }
S|x + y| = 12
Read MoreFAQs on JEE Main Previous Year Questions (2025): Set, Relations and Functions
| 1. What are sets and how are they defined in mathematics? |  |
Ans. A set is a well-defined collection of distinct objects, which can be anything from numbers to letters or even other sets. Sets are typically denoted by curly braces, for example, {1, 2, 3} is a set containing the numbers 1, 2, and 3. The elements of a set are called its members or elements, and a set can be finite or infinite. For example, the set of natural numbers is infinite.
| 2. What is the difference between a relation and a function in mathematics? | |
Ans. A relation is a set of ordered pairs, typically defined between two sets. For example, if A = {1, 2} and B = {3, 4}, a relation R from A to B could be R = {(1, 3), (2, 4)}. A function, on the other hand, is a special type of relation where each element in the domain (first set) is associated with exactly one element in the codomain (second set). This means that no two ordered pairs in a function can have the same first element.
| 3. How can you determine whether a relation is a function? | |
Ans. To determine if a relation is a function, you can use the vertical line test. If any vertical line intersects the graph of the relation at more than one point, then the relation is not a function. In terms of ordered pairs, for a relation to be a function, each input must map to one and only one output. For example, if you have the pairs (1, 2), (1, 3), and (2, 4), the relation is not a function because the input 1 corresponds to two different outputs.
| 4. What are the different types of relations in mathematics? | |
Ans. There are several types of relations in mathematics, including: 1. <b>Reflexive Relation</b>: A relation R on a set A is reflexive if every element is related to itself. For example, for any a in A, (a, a) is in R. 2. <b>Symmetric Relation</b>: A relation R is symmetric if, whenever (a, b) is in R, then (b, a) is also in R. 3. <b>Transitive Relation</b>: A relation R is transitive if, whenever (a, b) and (b, c) are in R, then (a, c) must also be in R. 4. <b>Anti-symmetric Relation</b>: A relation R is anti-symmetric if, whenever (a, b) and (b, a) are both in R, then a must equal b.
| 5. What is the significance of the Cartesian product in set theory? | |
Ans. The Cartesian product is a fundamental operation in set theory that combines two sets to form a new set of ordered pairs. If A and B are two sets, their Cartesian product, denoted as A × B, consists of all possible ordered pairs (a, b) where a is an element of A and b is an element of B. This operation is essential in defining relations and functions, as it establishes the framework for how elements from one set can be paired with elements from another set. For example, if A = {1, 2} and B = {x, y}, then A × B = {(1, x), (1, y), (2, x), (2, y)}.
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