Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Mathematics
On set A = 1, 2, 3, relations R and S are given by R = (1, 1), (2, 2), (3, 3), (1, 2), (2, 1) S = (1, 1), (2, 2), (3, 3), (1, 3), (3, 1) Then
Q. On set
A
=
1
,
2
,
3
, relations
R
and
S
are given by
R
=
{
(
1
,
1
)
,
(
2
,
2
)
,
(
3
,
3
)
,
(
1
,
2
)
,
(
2
,
1
)
}
S
=
{
(
1
,
1
)
,
(
2
,
2
)
,
(
3
,
3
)
,
(
1
,
3
)
,
(
3
,
1
)
}
Then
2092
213
WBJEE
WBJEE 2017
Relations and Functions - Part 2
Report Error
A
R
∪
S
is an equivalence relation
39%
B
R
∪
S
is reflexive and transitive but not symmetric
14%
C
R
∪
S
is reflexive and symmetric but not transitive
43%
D
R
∪
S
is symmetric and transitive but not reflexive
4%
Solution:
We have,
R
=
{(
1
,
1
)
,
(
2
,
2
)
,
(
3
,
3
)
,
(
1
,
2
)
,
(
2
,
1
)}
,
S
=
{(
1
,
1
)
,
(
2
,
2
)
,
(
3
,
3
)
,
(
1
,
3
)
,
(
3
,
1
)}
.
∴
R
∪
S
=
{(
1
,
1
)
,
(
2
,
2
)
,
(
3
,
3
)
,
(
1
,
2
)
,
(
2
,
1
)
,
(
1
,
3
)
,
(
3
,
1
)}
.
Since,
(
2
,
1
)
∈
R
∪
S
,
(
1
,
3
)
∈
R
∪
S
but
(
2
,
3
)
∈
/
R
∪
S
∴
R
∪
S
is reflexive and symmetric but not transitive.