QBManager

Sets, Relation and Function

116815 Let $R$ be the set of real numbers and let $G \subseteq R^{2}$ be a relation defined by $G = {[(a, b), (c, d)] [b -$ $a = d - c]}$ then $G$ is

1 reflexive only

2 symmetric only

3 transitive only

4 an equivalence relation

Explanation:

D Given the relation -
$G = {[(a, b), (c, d)] ∣ b - a = d - c}$
For reflexive -
Let $(x, y) \in R^{2}$
$\Rightarrow y - x = y - x$
$∴ [(x, y), (x, y)] \in G \forall (x, y) \in R^{2}$
$∴ G$ is reflexive
For symmetric -
Let $[(a, b) (c, d)] \in G$
$\Rightarrow b - a = d - c \Rightarrow d - c = b - a$
$\Rightarrow [(c, d), (a, b)] \in G$
$∴ G$ is symmetric -
For transitive -
Let $[(a, b), (c, d)] \in G$
And $[(c, d), (x, y)] \in G$
$\Rightarrow b - a = d - c$
$\Rightarrow d - c = y - x$
(form equation (i))
$\Rightarrow b - a = y - x$
(form equation (ii))
$\Rightarrow [(a, b), (x, y)] \in G$
$∴ G$ is transitive.
Hence, $G$ is an equivalence relation.

J and K CET-2015

Sets, Relation and Function

116823 The relations $R$ defined in the set ${1, 2, 3, 4, 5$ , $6}$ as $R = {(a, b) : b = a + 1}$ is

1 reflexive

2 symmetric

3 transitive

4 None of these

Explanation:

D Consider, $A = {1, 2, 3, 4, 5, 6}$
Given, as $R = {(a, b)} : b = a + 1}$ .
Then, check relation is -
(a) Reflexive relation :-
Then, $_{a} R_{a} \neq_{a} R_{a + 1} \neq_{a} R_{a},_{1} R_{2}, a \in A$
So, it is not reflexive relation.
(b) Symmetric relation
$_{a} R_{b} =_{b} R_{a}$
Then $_{a} R_{a + 1} \neq_{a + 1} R_{a}$ is not defined.
So, it is not a symmetric relation.
(c) Transitive Relation : -
If $a, b, c \in R$
Then, $aRb, bRc \Rightarrow aRc$
It is not transitive because -
$_{5} R_{6},_{6} R_{7} ⧸_{5} R_{7}$
is not defined because 7 A.
So, it is not a transitive relation.
Hence, the relation $R$ is not a reflexive not a symmetric and not a transitive relation.

BCECE-2013

Sets, Relation and Function

116817 Let $A = {1, 3, 4, 6, 9)$ and $B = {2, 4, 5, 8, 10}$ . Let $R$ be a relation defined on $A \times B$ such that $R =$ ${((a_{1}, b_{1}), (a_{2}, b_{2})) : a_{1} \leq b_{2}$ and $b_{1} \leq a_{2}}$ .
Then the number of elements in the set $R$ is

1 26

2 160

3 180

4 52

Explanation:

B Given set,
and
$A = {1, 3, 4, 6, 9)$
$B = {2, 4, 5, 8, 10}$
$R = A \times B \Rightarrow {(a_{1} b_{1}) (a_{2} b_{2}) : a_{1} \leq b_{2} b_{1} \leq a_{2}}$
Let,
{lll}
| $a_{1} = 1$ | then | $b_{2}$ has 5 choices |
|---|---|---|
| $a_{1} = 4$ | then | $b_{2}$ has 4 choices |
| $a_{1} = 6$ | then | $b_{2}$ has 2 choices |
| $a_{1} = 9$ | then | $b_{2}$ has 1 choices
Now,
original image
So, total number of element
$R = 160$

Shift-II

Sets, Relation and Function

116818 Let $R$ be the relation in the set ${x : x \in N, x \leq 4}$ given by $R = {(1, 1), (2, 2), (3, 3)}$ then, $R$ is

1 Reflexive and symmetric but not transitive

2 Symmetric and transitive but not reflexive

3 Reflexive and transitive but not symmetric

4 An equivalence relation.

Explanation:

B ${x : x \in N, x \leq 4}$
Let
$A = {1, 2, 3, 4}$
$R = {(1, 1), (2, 2), (3, 3)}$
For symmetry -
$(a, b) \in R \Rightarrow (b, a) \in R$
$(3, 3) \in R$
$(3, 3) \in R$
So, $R$ is symmetry
For transitive -
$(a, b), (b, a) \in R$
Then, $(a, a) \in R$
$(2, 2), (2, 2) \in R$
Also $(2, 2) \in R$
So, $R$ is transitive.
For reflexive -
For all $x \in A$
$(x, x) \in R$
But here
$(4, 4) \in A$
and $(4, 4) \notin R$
So, $R$ is not reflexive.
Hence, this relation is symmetric and transitive but not reflexive.

GUJCET-2021

Sets, Relation and Function

116815 Let $R$ be the set of real numbers and let $G \subseteq R^{2}$ be a relation defined by $G = {[(a, b), (c, d)] [b -$ $a = d - c]}$ then $G$ is

1 reflexive only

2 symmetric only

3 transitive only

4 an equivalence relation

Explanation:

D Given the relation -
$G = {[(a, b), (c, d)] ∣ b - a = d - c}$
For reflexive -
Let $(x, y) \in R^{2}$
$\Rightarrow y - x = y - x$
$∴ [(x, y), (x, y)] \in G \forall (x, y) \in R^{2}$
$∴ G$ is reflexive
For symmetric -
Let $[(a, b) (c, d)] \in G$
$\Rightarrow b - a = d - c \Rightarrow d - c = b - a$
$\Rightarrow [(c, d), (a, b)] \in G$
$∴ G$ is symmetric -
For transitive -
Let $[(a, b), (c, d)] \in G$
And $[(c, d), (x, y)] \in G$
$\Rightarrow b - a = d - c$
$\Rightarrow d - c = y - x$
(form equation (i))
$\Rightarrow b - a = y - x$
(form equation (ii))
$\Rightarrow [(a, b), (x, y)] \in G$
$∴ G$ is transitive.
Hence, $G$ is an equivalence relation.

J and K CET-2015

Sets, Relation and Function

116823 The relations $R$ defined in the set ${1, 2, 3, 4, 5$ , $6}$ as $R = {(a, b) : b = a + 1}$ is

1 reflexive

2 symmetric

3 transitive

4 None of these

Explanation:

D Consider, $A = {1, 2, 3, 4, 5, 6}$
Given, as $R = {(a, b)} : b = a + 1}$ .
Then, check relation is -
(a) Reflexive relation :-
Then, $_{a} R_{a} \neq_{a} R_{a + 1} \neq_{a} R_{a},_{1} R_{2}, a \in A$
So, it is not reflexive relation.
(b) Symmetric relation
$_{a} R_{b} =_{b} R_{a}$
Then $_{a} R_{a + 1} \neq_{a + 1} R_{a}$ is not defined.
So, it is not a symmetric relation.
(c) Transitive Relation : -
If $a, b, c \in R$
Then, $aRb, bRc \Rightarrow aRc$
It is not transitive because -
$_{5} R_{6},_{6} R_{7} ⧸_{5} R_{7}$
is not defined because 7 A.
So, it is not a transitive relation.
Hence, the relation $R$ is not a reflexive not a symmetric and not a transitive relation.

BCECE-2013

Sets, Relation and Function

116817 Let $A = {1, 3, 4, 6, 9)$ and $B = {2, 4, 5, 8, 10}$ . Let $R$ be a relation defined on $A \times B$ such that $R =$ ${((a_{1}, b_{1}), (a_{2}, b_{2})) : a_{1} \leq b_{2}$ and $b_{1} \leq a_{2}}$ .
Then the number of elements in the set $R$ is

1 26

2 160

3 180

4 52

Explanation:

B Given set,
and
$A = {1, 3, 4, 6, 9)$
$B = {2, 4, 5, 8, 10}$
$R = A \times B \Rightarrow {(a_{1} b_{1}) (a_{2} b_{2}) : a_{1} \leq b_{2} b_{1} \leq a_{2}}$
Let,
{lll}
| $a_{1} = 1$ | then | $b_{2}$ has 5 choices |
|---|---|---|
| $a_{1} = 4$ | then | $b_{2}$ has 4 choices |
| $a_{1} = 6$ | then | $b_{2}$ has 2 choices |
| $a_{1} = 9$ | then | $b_{2}$ has 1 choices
Now,
original image
So, total number of element
$R = 160$

Shift-II

Sets, Relation and Function

116818 Let $R$ be the relation in the set ${x : x \in N, x \leq 4}$ given by $R = {(1, 1), (2, 2), (3, 3)}$ then, $R$ is

1 Reflexive and symmetric but not transitive

2 Symmetric and transitive but not reflexive

3 Reflexive and transitive but not symmetric

4 An equivalence relation.

Explanation:

B ${x : x \in N, x \leq 4}$
Let
$A = {1, 2, 3, 4}$
$R = {(1, 1), (2, 2), (3, 3)}$
For symmetry -
$(a, b) \in R \Rightarrow (b, a) \in R$
$(3, 3) \in R$
$(3, 3) \in R$
So, $R$ is symmetry
For transitive -
$(a, b), (b, a) \in R$
Then, $(a, a) \in R$
$(2, 2), (2, 2) \in R$
Also $(2, 2) \in R$
So, $R$ is transitive.
For reflexive -
For all $x \in A$
$(x, x) \in R$
But here
$(4, 4) \in A$
and $(4, 4) \notin R$
So, $R$ is not reflexive.
Hence, this relation is symmetric and transitive but not reflexive.

GUJCET-2021

Sets, Relation and Function

116815 Let $R$ be the set of real numbers and let $G \subseteq R^{2}$ be a relation defined by $G = {[(a, b), (c, d)] [b -$ $a = d - c]}$ then $G$ is

1 reflexive only

2 symmetric only

3 transitive only

4 an equivalence relation

Explanation:

D Given the relation -
$G = {[(a, b), (c, d)] ∣ b - a = d - c}$
For reflexive -
Let $(x, y) \in R^{2}$
$\Rightarrow y - x = y - x$
$∴ [(x, y), (x, y)] \in G \forall (x, y) \in R^{2}$
$∴ G$ is reflexive
For symmetric -
Let $[(a, b) (c, d)] \in G$
$\Rightarrow b - a = d - c \Rightarrow d - c = b - a$
$\Rightarrow [(c, d), (a, b)] \in G$
$∴ G$ is symmetric -
For transitive -
Let $[(a, b), (c, d)] \in G$
And $[(c, d), (x, y)] \in G$
$\Rightarrow b - a = d - c$
$\Rightarrow d - c = y - x$
(form equation (i))
$\Rightarrow b - a = y - x$
(form equation (ii))
$\Rightarrow [(a, b), (x, y)] \in G$
$∴ G$ is transitive.
Hence, $G$ is an equivalence relation.

J and K CET-2015

Sets, Relation and Function

116823 The relations $R$ defined in the set ${1, 2, 3, 4, 5$ , $6}$ as $R = {(a, b) : b = a + 1}$ is

1 reflexive

2 symmetric

3 transitive

4 None of these

Explanation:

D Consider, $A = {1, 2, 3, 4, 5, 6}$
Given, as $R = {(a, b)} : b = a + 1}$ .
Then, check relation is -
(a) Reflexive relation :-
Then, $_{a} R_{a} \neq_{a} R_{a + 1} \neq_{a} R_{a},_{1} R_{2}, a \in A$
So, it is not reflexive relation.
(b) Symmetric relation
$_{a} R_{b} =_{b} R_{a}$
Then $_{a} R_{a + 1} \neq_{a + 1} R_{a}$ is not defined.
So, it is not a symmetric relation.
(c) Transitive Relation : -
If $a, b, c \in R$
Then, $aRb, bRc \Rightarrow aRc$
It is not transitive because -
$_{5} R_{6},_{6} R_{7} ⧸_{5} R_{7}$
is not defined because 7 A.
So, it is not a transitive relation.
Hence, the relation $R$ is not a reflexive not a symmetric and not a transitive relation.

BCECE-2013

Sets, Relation and Function

116817 Let $A = {1, 3, 4, 6, 9)$ and $B = {2, 4, 5, 8, 10}$ . Let $R$ be a relation defined on $A \times B$ such that $R =$ ${((a_{1}, b_{1}), (a_{2}, b_{2})) : a_{1} \leq b_{2}$ and $b_{1} \leq a_{2}}$ .
Then the number of elements in the set $R$ is

1 26

2 160

3 180

4 52

Explanation:

B Given set,
and
$A = {1, 3, 4, 6, 9)$
$B = {2, 4, 5, 8, 10}$
$R = A \times B \Rightarrow {(a_{1} b_{1}) (a_{2} b_{2}) : a_{1} \leq b_{2} b_{1} \leq a_{2}}$
Let,
{lll}
| $a_{1} = 1$ | then | $b_{2}$ has 5 choices |
|---|---|---|
| $a_{1} = 4$ | then | $b_{2}$ has 4 choices |
| $a_{1} = 6$ | then | $b_{2}$ has 2 choices |
| $a_{1} = 9$ | then | $b_{2}$ has 1 choices
Now,
original image
So, total number of element
$R = 160$

Shift-II

Sets, Relation and Function

116818 Let $R$ be the relation in the set ${x : x \in N, x \leq 4}$ given by $R = {(1, 1), (2, 2), (3, 3)}$ then, $R$ is

1 Reflexive and symmetric but not transitive

2 Symmetric and transitive but not reflexive

3 Reflexive and transitive but not symmetric

4 An equivalence relation.

Explanation:

B ${x : x \in N, x \leq 4}$
Let
$A = {1, 2, 3, 4}$
$R = {(1, 1), (2, 2), (3, 3)}$
For symmetry -
$(a, b) \in R \Rightarrow (b, a) \in R$
$(3, 3) \in R$
$(3, 3) \in R$
So, $R$ is symmetry
For transitive -
$(a, b), (b, a) \in R$
Then, $(a, a) \in R$
$(2, 2), (2, 2) \in R$
Also $(2, 2) \in R$
So, $R$ is transitive.
For reflexive -
For all $x \in A$
$(x, x) \in R$
But here
$(4, 4) \in A$
and $(4, 4) \notin R$
So, $R$ is not reflexive.
Hence, this relation is symmetric and transitive but not reflexive.

GUJCET-2021

Sets, Relation and Function

116815 Let $R$ be the set of real numbers and let $G \subseteq R^{2}$ be a relation defined by $G = {[(a, b), (c, d)] [b -$ $a = d - c]}$ then $G$ is

1 reflexive only

2 symmetric only

3 transitive only

4 an equivalence relation

Explanation:

D Given the relation -
$G = {[(a, b), (c, d)] ∣ b - a = d - c}$
For reflexive -
Let $(x, y) \in R^{2}$
$\Rightarrow y - x = y - x$
$∴ [(x, y), (x, y)] \in G \forall (x, y) \in R^{2}$
$∴ G$ is reflexive
For symmetric -
Let $[(a, b) (c, d)] \in G$
$\Rightarrow b - a = d - c \Rightarrow d - c = b - a$
$\Rightarrow [(c, d), (a, b)] \in G$
$∴ G$ is symmetric -
For transitive -
Let $[(a, b), (c, d)] \in G$
And $[(c, d), (x, y)] \in G$
$\Rightarrow b - a = d - c$
$\Rightarrow d - c = y - x$
(form equation (i))
$\Rightarrow b - a = y - x$
(form equation (ii))
$\Rightarrow [(a, b), (x, y)] \in G$
$∴ G$ is transitive.
Hence, $G$ is an equivalence relation.

J and K CET-2015

Sets, Relation and Function

116823 The relations $R$ defined in the set ${1, 2, 3, 4, 5$ , $6}$ as $R = {(a, b) : b = a + 1}$ is

1 reflexive

2 symmetric

3 transitive

4 None of these

Explanation:

D Consider, $A = {1, 2, 3, 4, 5, 6}$
Given, as $R = {(a, b)} : b = a + 1}$ .
Then, check relation is -
(a) Reflexive relation :-
Then, $_{a} R_{a} \neq_{a} R_{a + 1} \neq_{a} R_{a},_{1} R_{2}, a \in A$
So, it is not reflexive relation.
(b) Symmetric relation
$_{a} R_{b} =_{b} R_{a}$
Then $_{a} R_{a + 1} \neq_{a + 1} R_{a}$ is not defined.
So, it is not a symmetric relation.
(c) Transitive Relation : -
If $a, b, c \in R$
Then, $aRb, bRc \Rightarrow aRc$
It is not transitive because -
$_{5} R_{6},_{6} R_{7} ⧸_{5} R_{7}$
is not defined because 7 A.
So, it is not a transitive relation.
Hence, the relation $R$ is not a reflexive not a symmetric and not a transitive relation.

BCECE-2013

Sets, Relation and Function

116817 Let $A = {1, 3, 4, 6, 9)$ and $B = {2, 4, 5, 8, 10}$ . Let $R$ be a relation defined on $A \times B$ such that $R =$ ${((a_{1}, b_{1}), (a_{2}, b_{2})) : a_{1} \leq b_{2}$ and $b_{1} \leq a_{2}}$ .
Then the number of elements in the set $R$ is

1 26

2 160

3 180

4 52

Explanation:

B Given set,
and
$A = {1, 3, 4, 6, 9)$
$B = {2, 4, 5, 8, 10}$
$R = A \times B \Rightarrow {(a_{1} b_{1}) (a_{2} b_{2}) : a_{1} \leq b_{2} b_{1} \leq a_{2}}$
Let,
{lll}
| $a_{1} = 1$ | then | $b_{2}$ has 5 choices |
|---|---|---|
| $a_{1} = 4$ | then | $b_{2}$ has 4 choices |
| $a_{1} = 6$ | then | $b_{2}$ has 2 choices |
| $a_{1} = 9$ | then | $b_{2}$ has 1 choices
Now,
original image
So, total number of element
$R = 160$

Shift-II

Sets, Relation and Function

116818 Let $R$ be the relation in the set ${x : x \in N, x \leq 4}$ given by $R = {(1, 1), (2, 2), (3, 3)}$ then, $R$ is

1 Reflexive and symmetric but not transitive

2 Symmetric and transitive but not reflexive

3 Reflexive and transitive but not symmetric

4 An equivalence relation.

Explanation:

B ${x : x \in N, x \leq 4}$
Let
$A = {1, 2, 3, 4}$
$R = {(1, 1), (2, 2), (3, 3)}$
For symmetry -
$(a, b) \in R \Rightarrow (b, a) \in R$
$(3, 3) \in R$
$(3, 3) \in R$
So, $R$ is symmetry
For transitive -
$(a, b), (b, a) \in R$
Then, $(a, a) \in R$
$(2, 2), (2, 2) \in R$
Also $(2, 2) \in R$
So, $R$ is transitive.
For reflexive -
For all $x \in A$
$(x, x) \in R$
But here
$(4, 4) \in A$
and $(4, 4) \notin R$
So, $R$ is not reflexive.
Hence, this relation is symmetric and transitive but not reflexive.

GUJCET-2021