QBManager

Sets, Relation and Function

116795 Let $R$ be a relation from the set ${1, 2$ , 3.........,60 to itself such that $R = {(a, b) : b =$ pq, where $p, q \geq 3$ are prime numbers $}$ . Then the number of elements in $R$ is :

1 600

2 660

3 540

4 720

Explanation:

B Given set,
$A = {1, 2, 3, 4 \dots 60}$
And, function $R = {(a, b) : b = pq}$
$1 \leq pq \geq 60$
Number of possible values of $a = 60$ for $b = p q$
$\begin{array}{ll} If & p = 3, q = 3, 5, 7, 11, 13, 17, 19 \\ If & p = 5, q = 5, 7, 11 \\ If & p = 7, q = 7 \\ a = 60 b = 11 \\ a . b = 60 \times 11 \end{array}$ So, the number of elements in $R$ is $= 660$ .

Shift-I

Sets, Relation and Function

116796 Let $P (S)$ denote the power set of $S = {1, 2$ , 3....,10 . Define the relation $R_{1}$ and $R_{2}$ on $P (S)$ as $A R_{1} B$ if $(A \cap B^{C}) \cup (B \cap A^{C}) = ϕ$ and $A R_{2} B$ if $A \cup B^{C} = B \cup A^{C}, \forall A, B \in P (S)$ . Then

1 both

R_{1}

and

R_{2}

are not equivalence relations

2 only

R_{2}

is an equivalence relation

3 only

R_{1}

is an equivalence relation

4 both

R_{1}

and

R_{2}

are equivalence relations

Explanation:

D P (S) = power set $S$
$S = {1, 2, 3, \dots 10}$
Given, ${AR}_{1} B \Rightarrow (A \cap B^{c}) \cup (B \cap A^{c}) = ϕ$
$\Rightarrow A \cap B^{c} = ϕ$ and $(B \cap A^{c}) = ϕ$
$\Rightarrow A = B$
$∴ {AR}_{1} B$ is an equivalence relation.
$A R_{2} B \Rightarrow A \cup B = B \cup A^{c}$
$\Rightarrow A B$
$∴ {AR}_{2} B$ is an equivalence relation.
Hence, $R_{1}$ and $R_{2}$ are equivalence relation.

Shift-II

Sets, Relation and Function

116797 Let a relation $R$ in the set $N$ of natural numbers be defined as $(x, y) \Leftrightarrow x^{2} - 4 x y + 3 y^{2} = 0 \forall x, y \in$ N. The relation $R$ is

1 reflexive

2 symmetric

3 transitive

4 an equivalence relation

Explanation:

A We have -
$R = {(x, y) \Leftrightarrow x^{2} - 4 x y + 3 y^{2} = 0 \forall x, y \in N}$
For reflexive -
Let, $x \in N$
$x^{2} - 4 x x$
$(x, x) \in R$
So, $R$ is reflective
For symmetric -
$\begin{array}{ll} Let (x, y) = (3, 1) \Rightarrow (3)^{2} - 4 (3) (1) + 3 (1)^{2} \\ \Rightarrow 9 - 12 + 3 = 0 \\ ∴ & (3, 1) \in R \\ But & (1, 3) \Rightarrow (1)^{2} - 4 (3) (1) + 3 (3)^{2} \\ = 1 - 12 + 27 = 16 \\ (1, 3) \notin R \end{array}$
Hence, $R$ is not symmetric so given relation $R$ is reflexive.

AMU-2009

Sets, Relation and Function

116798 Let $r$ be a relation from $R$ (set of real numbers) to $R$ defined by $r = {(a, b) ∣ a, b \in R$ and $a - b +$ $\sqrt{3}$ is an irrational number $}$ . The relation $r$ is

1 an equivalence relation

2 reflexive only

3 symmetric only

4 transitive only

Explanation:

A Given,
$r = {a, b ∣ a, b \in R}$
And, $r \Rightarrow a - b + \sqrt{3}$ is an irrational number.
For reflexive relation -
Then, $aRa = a - a + \sqrt{3}$
$\Rightarrow aRa = \sqrt{3}$
And, $bRb = b - b + \sqrt{3} \Rightarrow bRb = \sqrt{3}$
Therefore $r$ is reflexive.
For symmetric relation -
Let, $a, b \in R$
$a - b + \sqrt{3} = b - a + \sqrt{3}$ is an irrational number
b, $a \in R$
Therefore $r$ is symmetric.
For transitive relation -
Let $(a, b) \in R$ and $(b, c) \in R$
$a - b + \sqrt{3} = b - c + \sqrt{3}$ is an irrational number
Now, $a - c + 2 \sqrt{3}$ is an also irrational number $∴ (a, c) \in R$
Thus $r$ is transitive relation
Hence, $r$ is an equivalence relation.

AMU-2009

Sets, Relation and Function

116795 Let $R$ be a relation from the set ${1, 2$ , 3.........,60 to itself such that $R = {(a, b) : b =$ pq, where $p, q \geq 3$ are prime numbers $}$ . Then the number of elements in $R$ is :

1 600

2 660

3 540

4 720

Explanation:

B Given set,
$A = {1, 2, 3, 4 \dots 60}$
And, function $R = {(a, b) : b = pq}$
$1 \leq pq \geq 60$
Number of possible values of $a = 60$ for $b = p q$
$\begin{array}{ll} If & p = 3, q = 3, 5, 7, 11, 13, 17, 19 \\ If & p = 5, q = 5, 7, 11 \\ If & p = 7, q = 7 \\ a = 60 b = 11 \\ a . b = 60 \times 11 \end{array}$ So, the number of elements in $R$ is $= 660$ .

Shift-I

Sets, Relation and Function

116796 Let $P (S)$ denote the power set of $S = {1, 2$ , 3....,10 . Define the relation $R_{1}$ and $R_{2}$ on $P (S)$ as $A R_{1} B$ if $(A \cap B^{C}) \cup (B \cap A^{C}) = ϕ$ and $A R_{2} B$ if $A \cup B^{C} = B \cup A^{C}, \forall A, B \in P (S)$ . Then

1 both

R_{1}

and

R_{2}

are not equivalence relations

2 only

R_{2}

is an equivalence relation

3 only

R_{1}

is an equivalence relation

4 both

R_{1}

and

R_{2}

are equivalence relations

Explanation:

D P (S) = power set $S$
$S = {1, 2, 3, \dots 10}$
Given, ${AR}_{1} B \Rightarrow (A \cap B^{c}) \cup (B \cap A^{c}) = ϕ$
$\Rightarrow A \cap B^{c} = ϕ$ and $(B \cap A^{c}) = ϕ$
$\Rightarrow A = B$
$∴ {AR}_{1} B$ is an equivalence relation.
$A R_{2} B \Rightarrow A \cup B = B \cup A^{c}$
$\Rightarrow A B$
$∴ {AR}_{2} B$ is an equivalence relation.
Hence, $R_{1}$ and $R_{2}$ are equivalence relation.

Shift-II

Sets, Relation and Function

116797 Let a relation $R$ in the set $N$ of natural numbers be defined as $(x, y) \Leftrightarrow x^{2} - 4 x y + 3 y^{2} = 0 \forall x, y \in$ N. The relation $R$ is

1 reflexive

2 symmetric

3 transitive

4 an equivalence relation

Explanation:

A We have -
$R = {(x, y) \Leftrightarrow x^{2} - 4 x y + 3 y^{2} = 0 \forall x, y \in N}$
For reflexive -
Let, $x \in N$
$x^{2} - 4 x x$
$(x, x) \in R$
So, $R$ is reflective
For symmetric -
$\begin{array}{ll} Let (x, y) = (3, 1) \Rightarrow (3)^{2} - 4 (3) (1) + 3 (1)^{2} \\ \Rightarrow 9 - 12 + 3 = 0 \\ ∴ & (3, 1) \in R \\ But & (1, 3) \Rightarrow (1)^{2} - 4 (3) (1) + 3 (3)^{2} \\ = 1 - 12 + 27 = 16 \\ (1, 3) \notin R \end{array}$
Hence, $R$ is not symmetric so given relation $R$ is reflexive.

AMU-2009

Sets, Relation and Function

116798 Let $r$ be a relation from $R$ (set of real numbers) to $R$ defined by $r = {(a, b) ∣ a, b \in R$ and $a - b +$ $\sqrt{3}$ is an irrational number $}$ . The relation $r$ is

1 an equivalence relation

2 reflexive only

3 symmetric only

4 transitive only

Explanation:

A Given,
$r = {a, b ∣ a, b \in R}$
And, $r \Rightarrow a - b + \sqrt{3}$ is an irrational number.
For reflexive relation -
Then, $aRa = a - a + \sqrt{3}$
$\Rightarrow aRa = \sqrt{3}$
And, $bRb = b - b + \sqrt{3} \Rightarrow bRb = \sqrt{3}$
Therefore $r$ is reflexive.
For symmetric relation -
Let, $a, b \in R$
$a - b + \sqrt{3} = b - a + \sqrt{3}$ is an irrational number
b, $a \in R$
Therefore $r$ is symmetric.
For transitive relation -
Let $(a, b) \in R$ and $(b, c) \in R$
$a - b + \sqrt{3} = b - c + \sqrt{3}$ is an irrational number
Now, $a - c + 2 \sqrt{3}$ is an also irrational number $∴ (a, c) \in R$
Thus $r$ is transitive relation
Hence, $r$ is an equivalence relation.

AMU-2009

Sets, Relation and Function

116795 Let $R$ be a relation from the set ${1, 2$ , 3.........,60 to itself such that $R = {(a, b) : b =$ pq, where $p, q \geq 3$ are prime numbers $}$ . Then the number of elements in $R$ is :

1 600

2 660

3 540

4 720

Explanation:

B Given set,
$A = {1, 2, 3, 4 \dots 60}$
And, function $R = {(a, b) : b = pq}$
$1 \leq pq \geq 60$
Number of possible values of $a = 60$ for $b = p q$
$\begin{array}{ll} If & p = 3, q = 3, 5, 7, 11, 13, 17, 19 \\ If & p = 5, q = 5, 7, 11 \\ If & p = 7, q = 7 \\ a = 60 b = 11 \\ a . b = 60 \times 11 \end{array}$ So, the number of elements in $R$ is $= 660$ .

Shift-I

Sets, Relation and Function

116796 Let $P (S)$ denote the power set of $S = {1, 2$ , 3....,10 . Define the relation $R_{1}$ and $R_{2}$ on $P (S)$ as $A R_{1} B$ if $(A \cap B^{C}) \cup (B \cap A^{C}) = ϕ$ and $A R_{2} B$ if $A \cup B^{C} = B \cup A^{C}, \forall A, B \in P (S)$ . Then

1 both

R_{1}

and

R_{2}

are not equivalence relations

2 only

R_{2}

is an equivalence relation

3 only

R_{1}

is an equivalence relation

4 both

R_{1}

and

R_{2}

are equivalence relations

Explanation:

D P (S) = power set $S$
$S = {1, 2, 3, \dots 10}$
Given, ${AR}_{1} B \Rightarrow (A \cap B^{c}) \cup (B \cap A^{c}) = ϕ$
$\Rightarrow A \cap B^{c} = ϕ$ and $(B \cap A^{c}) = ϕ$
$\Rightarrow A = B$
$∴ {AR}_{1} B$ is an equivalence relation.
$A R_{2} B \Rightarrow A \cup B = B \cup A^{c}$
$\Rightarrow A B$
$∴ {AR}_{2} B$ is an equivalence relation.
Hence, $R_{1}$ and $R_{2}$ are equivalence relation.

Shift-II

Sets, Relation and Function

116797 Let a relation $R$ in the set $N$ of natural numbers be defined as $(x, y) \Leftrightarrow x^{2} - 4 x y + 3 y^{2} = 0 \forall x, y \in$ N. The relation $R$ is

1 reflexive

2 symmetric

3 transitive

4 an equivalence relation

Explanation:

A We have -
$R = {(x, y) \Leftrightarrow x^{2} - 4 x y + 3 y^{2} = 0 \forall x, y \in N}$
For reflexive -
Let, $x \in N$
$x^{2} - 4 x x$
$(x, x) \in R$
So, $R$ is reflective
For symmetric -
$\begin{array}{ll} Let (x, y) = (3, 1) \Rightarrow (3)^{2} - 4 (3) (1) + 3 (1)^{2} \\ \Rightarrow 9 - 12 + 3 = 0 \\ ∴ & (3, 1) \in R \\ But & (1, 3) \Rightarrow (1)^{2} - 4 (3) (1) + 3 (3)^{2} \\ = 1 - 12 + 27 = 16 \\ (1, 3) \notin R \end{array}$
Hence, $R$ is not symmetric so given relation $R$ is reflexive.

AMU-2009

Sets, Relation and Function

116798 Let $r$ be a relation from $R$ (set of real numbers) to $R$ defined by $r = {(a, b) ∣ a, b \in R$ and $a - b +$ $\sqrt{3}$ is an irrational number $}$ . The relation $r$ is

1 an equivalence relation

2 reflexive only

3 symmetric only

4 transitive only

Explanation:

A Given,
$r = {a, b ∣ a, b \in R}$
And, $r \Rightarrow a - b + \sqrt{3}$ is an irrational number.
For reflexive relation -
Then, $aRa = a - a + \sqrt{3}$
$\Rightarrow aRa = \sqrt{3}$
And, $bRb = b - b + \sqrt{3} \Rightarrow bRb = \sqrt{3}$
Therefore $r$ is reflexive.
For symmetric relation -
Let, $a, b \in R$
$a - b + \sqrt{3} = b - a + \sqrt{3}$ is an irrational number
b, $a \in R$
Therefore $r$ is symmetric.
For transitive relation -
Let $(a, b) \in R$ and $(b, c) \in R$
$a - b + \sqrt{3} = b - c + \sqrt{3}$ is an irrational number
Now, $a - c + 2 \sqrt{3}$ is an also irrational number $∴ (a, c) \in R$
Thus $r$ is transitive relation
Hence, $r$ is an equivalence relation.

AMU-2009

Sets, Relation and Function

116795 Let $R$ be a relation from the set ${1, 2$ , 3.........,60 to itself such that $R = {(a, b) : b =$ pq, where $p, q \geq 3$ are prime numbers $}$ . Then the number of elements in $R$ is :

1 600

2 660

3 540

4 720

Explanation:

B Given set,
$A = {1, 2, 3, 4 \dots 60}$
And, function $R = {(a, b) : b = pq}$
$1 \leq pq \geq 60$
Number of possible values of $a = 60$ for $b = p q$
$\begin{array}{ll} If & p = 3, q = 3, 5, 7, 11, 13, 17, 19 \\ If & p = 5, q = 5, 7, 11 \\ If & p = 7, q = 7 \\ a = 60 b = 11 \\ a . b = 60 \times 11 \end{array}$ So, the number of elements in $R$ is $= 660$ .

Shift-I

Sets, Relation and Function

116796 Let $P (S)$ denote the power set of $S = {1, 2$ , 3....,10 . Define the relation $R_{1}$ and $R_{2}$ on $P (S)$ as $A R_{1} B$ if $(A \cap B^{C}) \cup (B \cap A^{C}) = ϕ$ and $A R_{2} B$ if $A \cup B^{C} = B \cup A^{C}, \forall A, B \in P (S)$ . Then

1 both

R_{1}

and

R_{2}

are not equivalence relations

2 only

R_{2}

is an equivalence relation

3 only

R_{1}

is an equivalence relation

4 both

R_{1}

and

R_{2}

are equivalence relations

Explanation:

D P (S) = power set $S$
$S = {1, 2, 3, \dots 10}$
Given, ${AR}_{1} B \Rightarrow (A \cap B^{c}) \cup (B \cap A^{c}) = ϕ$
$\Rightarrow A \cap B^{c} = ϕ$ and $(B \cap A^{c}) = ϕ$
$\Rightarrow A = B$
$∴ {AR}_{1} B$ is an equivalence relation.
$A R_{2} B \Rightarrow A \cup B = B \cup A^{c}$
$\Rightarrow A B$
$∴ {AR}_{2} B$ is an equivalence relation.
Hence, $R_{1}$ and $R_{2}$ are equivalence relation.

Shift-II

Sets, Relation and Function

116797 Let a relation $R$ in the set $N$ of natural numbers be defined as $(x, y) \Leftrightarrow x^{2} - 4 x y + 3 y^{2} = 0 \forall x, y \in$ N. The relation $R$ is

1 reflexive

2 symmetric

3 transitive

4 an equivalence relation

Explanation:

A We have -
$R = {(x, y) \Leftrightarrow x^{2} - 4 x y + 3 y^{2} = 0 \forall x, y \in N}$
For reflexive -
Let, $x \in N$
$x^{2} - 4 x x$
$(x, x) \in R$
So, $R$ is reflective
For symmetric -
$\begin{array}{ll} Let (x, y) = (3, 1) \Rightarrow (3)^{2} - 4 (3) (1) + 3 (1)^{2} \\ \Rightarrow 9 - 12 + 3 = 0 \\ ∴ & (3, 1) \in R \\ But & (1, 3) \Rightarrow (1)^{2} - 4 (3) (1) + 3 (3)^{2} \\ = 1 - 12 + 27 = 16 \\ (1, 3) \notin R \end{array}$
Hence, $R$ is not symmetric so given relation $R$ is reflexive.

AMU-2009

Sets, Relation and Function

116798 Let $r$ be a relation from $R$ (set of real numbers) to $R$ defined by $r = {(a, b) ∣ a, b \in R$ and $a - b +$ $\sqrt{3}$ is an irrational number $}$ . The relation $r$ is

1 an equivalence relation

2 reflexive only

3 symmetric only

4 transitive only

Explanation:

A Given,
$r = {a, b ∣ a, b \in R}$
And, $r \Rightarrow a - b + \sqrt{3}$ is an irrational number.
For reflexive relation -
Then, $aRa = a - a + \sqrt{3}$
$\Rightarrow aRa = \sqrt{3}$
And, $bRb = b - b + \sqrt{3} \Rightarrow bRb = \sqrt{3}$
Therefore $r$ is reflexive.
For symmetric relation -
Let, $a, b \in R$
$a - b + \sqrt{3} = b - a + \sqrt{3}$ is an irrational number
b, $a \in R$
Therefore $r$ is symmetric.
For transitive relation -
Let $(a, b) \in R$ and $(b, c) \in R$
$a - b + \sqrt{3} = b - c + \sqrt{3}$ is an irrational number
Now, $a - c + 2 \sqrt{3}$ is an also irrational number $∴ (a, c) \in R$
Thus $r$ is transitive relation
Hence, $r$ is an equivalence relation.

AMU-2009

Question Bank Series

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