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Joint Entrance Examination
Mathematics
Sets, Relations and Functions
Question 39
Question 39 of 161
Easy
MCQ
Let
A
=
1
,
2
,
3
,
4
,
5
,
6
,
7
A={1,2,3,4,5,6,7}
A
=
1
,
2
,
3
,
4
,
5
,
6
,
7
. Then the relation
R
=
(
x
,
y
)
∈
A
×
A
:
x
+
y
=
7
R={(x,y)\in A\times A:x+y=7}
R
=
(
x
,
y
)
∈
A
×
A
:
x
+
y
=
7
is :
A
reflexive but neither symmetric nor transitive
B
transitive but neither symmetric nor reflexive
C
symmetric but neither reflexive nor transitive
D
an equivalence relation
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Official solution
Here,
A
=
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
A=\{1,2,3,4,5,6,7\}
A
=
{
1
,
2
,
3
,
4
,
5
,
6
,
7
}
Since,
x
+
y
=
7
⇒
y
=
7
−
x
x+y=7 \Rightarrow y=7-x
x
+
y
=
7
⇒
y
=
7
−
x
So,
R
=
{
(
1
,
6
)
,
(
2
,
5
)
,
(
3
,
4
)
,
(
4
,
3
)
,
(
5
,
2
)
,
(
6
,
1
)
}
\mathrm{R}=\{(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\}
R
=
{(
1
,
6
)
,
(
2
,
5
)
,
(
3
,
4
)
,
(
4
,
3
)
,
(
5
,
2
)
,
(
6
,
1
)}
∵
(
a
,
b
)
∈
R
⇒
(
b
,
a
)
∈
R
\because(a, b) \in \mathrm{R} \Rightarrow(b, a) \in \mathrm{R}
∵
(
a
,
b
)
∈
R
⇒
(
b
,
a
)
∈
R
∴
R
\therefore \mathrm{R}
∴
R
is symmetric only.
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