Relations Test 21

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Relations Test 21

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Weekly Quiz Competition

Question 1
1 / -0

If $$R$$ is a relation from a set $$A$$ to the set $$B$$ and $$S$$ is a relation from $$B$$ to $$C,$$ then the relation $$SoR$$

A
is from $$C$$ to $$A$$

B
does not exist

C
is from $$A$$ to $$C$$

D
None of these

Solution

Since $$R\subseteq A\times B$$ and $$S\subseteq B\times C$$, we have
So $$R\subseteq A\times C$$.
$$\therefore$$ So R is a relation from $$A$$ to $$C.$$
Question 2
1 / -0

What is the Cartesian product of $$A = \left \{1, 2\right \}$$ and $$B = \left \{a, b\right \}$$?

A
$$\left \{(1, a), (1, b), (2, a), (b, b)\right \}$$

B
$$\left \{(1, 1), (2, 2), (a, a), (b, b)\right \}$$

C
$$\left \{(1, a), (2, a), (1, b), (2, b)\right \}$$

D
$$\left \{(1, 1), (a, a), (2, a), (1, b)\right \}$$

Solution

If $$A $$ and $$B$$ are two non empty sets, then the Cartesian product $$A \times B$$ is set of all ordered pairs $$(a,b)$$ such that $$a\in A$$ and $$b\in B$$.

Given $$A =\{1,2\}$$ and $$B = \{a,b\}$$

Hence $$A\times B = \{(1,a),(1,b),(2,a),(2,b)\}$$
Question 3
1 / -0

$$A = \left \{1, 2, 3, 4\right \}$$ and $$B = \left \{a, b, c\right \}$$. The relations from $$A$$ to $$B$$ is

A
$$\left \{(1, 2), (1, 3), (2, 3), (2, 4), (3, 4), (3, 1)\right \}$$

B
$$\left \{(a, b), (a, c), (b, a), (b, c), (c, a)\right \}$$

C
$$\{(1,a),(1,b)(1,c),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c),(4,a),(4,b),(4,c)\}$$

D
$$\left \{(a, 1), (1, 3), (b, 2), (c, 3), (b, 3), (b, 4)\right \}$$

Solution

$$A=\{1,2,3,4\}$$
$$B=\{a,b,c\}$$

Possible Relation $$R:A\to B$$
$$=\{(1,a),(1,b)(1,c),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c),(4,a),(4,b),(4,c)\}$$

Option C contains Relation $$R:A\to B$$
Question 4
1 / -0

What is the first component of an ordered pair $$(1, -1)$$?

A
$$1$$

B
$$-1$$

C
$$2$$

D
$$0$$

Solution

In an ordered pair $$(x,y)$$, the first component is $$x$$ and the second component is $$y$$.
Therefore, in an ordered pair $$(1,-1)$$, the first component is $$1$$.
Question 5
1 / -0

$$R$$ is a relation defined in $$R\times T$$ by $$(a,b) R (c,d)$$ iff $$a-c$$ is an integer and $$b=d$$. The relation $$R$$ is

A
an identity relation

B
an universal relation

C
an equivalence relation

D
None of these

Solution

We have $$R=\left \{((a,b), (c,d)):a-c\in Z\ and \ b=d; a,b,c,d\in R\right \}$$.
Let $$(a,b)\in R\times R$$
$$\therefore (a,b)R(a,b)$$, because $$a-a=0\in Z$$ and $$b=b$$
$$\therefore$$ R is reflexive.
Let $$(a,b) R (c,d).\Rightarrow a-c\in Z$$ and $$b=d$$
$$\Rightarrow c-a\in Z$$ and $$d=b$$
$$\Rightarrow (c,d) R (a,b)\Rightarrow R$$ is symmetric.
Let $$(a,b)R(c,d)$$ and $$(c,d)R(e,f)$$.
$$\Rightarrow a-c\in Z, b=d, c-e\in Z, d=f$$
$$\Rightarrow (a-c)+(c-e)\in z, b=f$$
$$\Rightarrow a-e\in z, b=f\Rightarrow (a,b)R(e,f)$$
$$\Rightarrow R$$ is transitive.
$$\therefore R$$ is an equivalence relation.
$$\therefore$$ the correct answer is (c).
Question 6
1 / -0

Let a relation $$R$$ be defined by $$R=\left \{(4,5), (1,4), (4,6), (7,6), (3,7)\right \}$$. The relation $$R^{-1}\circ R$$ is given by

A
$$\left \{(1,1), (4,4), (7,4), (4,7), (7,7)\right \}$$

B
$$\left \{(1,1), (4,4), (4,7), (7,4), (7,7),(3,3)\right \}$$

C
$$\left \{(1,5), (1,6), (3,6)\right \}$$

D
None of these

Solution

We have $$R=\left \{(4,5), (1,4), (4,6), (7,6), (3,7)\right \}$$.
$$\therefore R^{-1}=\left \{(5,4), (4,1), (6,4), (6,7), (7,3)\right \}$$
$$(4,4)\in R^{-1}\circ R$$ because $$(4,5)\in R$$ and $$(5,4)\in R^{-1}$$
$$(1,1)\in R^{-1}\circ R$$ because $$(1,4)\in R$$ and $$(4,1)\in R^{-1}$$
$$(4,4)\in R^{-1}\circ R$$ because $$(4,6)\in R$$ and $$(6,4)\in R^{-1}$$
$$(4,7)\in R^{-1}\circ R$$ because $$(4,6)\in R$$ and $$(6,7)\in R^{-1}$$
$$(7,4)\in R^{-1}\circ R$$ because $$(7,6)\in R$$ and $$(6,4)\in R^{-1}$$
$$(7,7)\in R^{-1}\circ R$$ because $$(7,6)\in R$$ and $$(6,7)\in R^{-1}$$
$$(3,3)\in R^{-1}\circ R$$ because $$(3,7)\in R$$ and $$(7,3)\in R^{-1}$$
$$\therefore R^{-1}\circ R=\left \{(4,4), (1,1), (4,7), (7,4), (7,7), (3,3)\right \}$$.
$$\therefore$$ The correct answer is $$B$$.
Question 7
1 / -0

Given $$(a - 2, b + 3) = (6, 8)$$, are equal ordered pair. Find the value of $$a$$ and $$b$$.

A
$$a = 8$$ and $$b = 5$$

B
$$a = 8$$ and $$b = 3$$

C
$$a = 5$$ and $$b = 5$$

D
$$a = 8$$ and $$b = 6$$

Solution

By equality of ordered pairs, we have
$$(a - 2, b + 3) = (6, 8)$$
On equating we get
$$a - 2 = 6$$
$$a = 8$$
$$b + 3 = 8$$
$$b = 5$$
So, the value of$$ a = 8 $$ and $$ b = 5.$$
Question 8
1 / -0

What is the second component of an ordered pair $$(3, -0.2)$$?

A
$$3$$

B
$$0.2$$

C
$$1$$

D
$$-0.2$$

Solution

In an ordered pair $$(x,y)$$, the first component is $$x$$ and the second component is $$y$$.
Therefore, in an ordered pair $$(3,-0.2)$$, the second component is $$-0.2$$.
Question 9
1 / -0

The relation $$R$$ defined on the set $$A=\left \{1,2,3,4,5\right \}$$ by $$R=\left \{(x,y):|x^2-y^2| < 16\right \}$$ is given by

A
$$\left \{(1,1), (2,1), (3,1), (4,1), (2,3)\right \}$$

B
$$\left \{(2,2), (3,2), (4,2), (2,4)\right \}$$

C
$$\left \{(3,3), (4,3), (5,4), (3,4)\right \}$$

D
None of these

Solution

We have $$R=\left \{(x,y):|x^2-y^2| < 16\right \}$$.

Take $$x=1,$$ we have
$$ |x^2-y^2| < 16\Rightarrow |1-y^2| < 16$$
$$\Rightarrow |y^2-1| < 16\Rightarrow y=1,2,3,4$$.

Take $$x=2,$$ we have
$$ |x^2-y^2| < 16 \Rightarrow |4-y^2| < 16$$
$$\Rightarrow |y^2-4| < 16\Rightarrow y=1,2,3,4$$.

Take $$x=3,$$ we have
$$ |x^2-y^2| < 16\Rightarrow |9-y^2| < 16$$
$$\Rightarrow |y^2-9| < 16\Rightarrow y=1,2,3,4$$.

Take $$x=4,$$ we have
$$ |x^2-y^2| < 16\Rightarrow |16-y^2| < 16$$
$$\Rightarrow |y^2-16| < 16\Rightarrow y=1,2,3,4,5$$.

Take $$x=5,$$
$$|x^2-y^2| < 16 \Rightarrow |25-y^2| < 16$$
$$\Rightarrow |y^2-25|<16 \Rightarrow y = 4,5$$

$$\therefore R=\left \{(1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), \\(3,3), (3,4), (4,1), (4,2), (4,3), (4,4), (4,5), (5,4), (5,5)\right \}$$
$$\therefore$$ The correct answer is $$D$$.
Question 10
1 / -0

Ordered pairs $$(x, y)$$ and $$(3, 6)$$ are equal if $$x = 3$$ and $$y = ?$$

A
$$3$$

B
$$6$$

C
$$-6$$

D
$$-3$$

Solution

Given
$$(x,y)= (3,6)$$
$$x=3$$
$$y=6$$
The value of $$x=3$$ and $$y=6$$.

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Relations Test 21

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Relations Test 21

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Question 1

is from $$C$$ to $$A$$

does not exist

is from $$A$$ to $$C$$

None of these

Solution

Question 2

$$\left \{(1, a), (1, b), (2, a), (b, b)\right \}$$

$$\left \{(1, 1), (2, 2), (a, a), (b, b)\right \}$$

$$\left \{(1, a), (2, a), (1, b), (2, b)\right \}$$

$$\left \{(1, 1), (a, a), (2, a), (1, b)\right \}$$

Solution

Question 3

$$\left \{(1, 2), (1, 3), (2, 3), (2, 4), (3, 4), (3, 1)\right \}$$

$$\left \{(a, b), (a, c), (b, a), (b, c), (c, a)\right \}$$

$$\{(1,a),(1,b)(1,c),(2,a),(2,b),(2,c),(3,a),(3,b),(3,c),(4,a),(4,b),(4,c)\}$$

$$\left \{(a, 1), (1, 3), (b, 2), (c, 3), (b, 3), (b, 4)\right \}$$

Solution

Question 4

$$1$$

$$-1$$

$$2$$

$$0$$

Solution

Question 5

an identity relation

an universal relation

an equivalence relation

None of these

Solution

Question 6

$$\left \{(1,1), (4,4), (7,4), (4,7), (7,7)\right \}$$

$$\left \{(1,1), (4,4), (4,7), (7,4), (7,7),(3,3)\right \}$$

$$\left \{(1,5), (1,6), (3,6)\right \}$$

None of these

Solution

Question 7

$$a = 8$$ and $$b = 5$$

$$a = 8$$ and $$b = 3$$

$$a = 5$$ and $$b = 5$$

$$a = 8$$ and $$b = 6$$

Solution

Question 8

$$3$$

$$0.2$$

$$1$$

$$-0.2$$

Solution

Question 9

$$\left \{(1,1), (2,1), (3,1), (4,1), (2,3)\right \}$$

$$\left \{(2,2), (3,2), (4,2), (2,4)\right \}$$

$$\left \{(3,3), (4,3), (5,4), (3,4)\right \}$$

None of these

Solution

Question 10

$$3$$

$$6$$

$$-6$$

$$-3$$

Solution

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