Relations Test 19

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Relations Test 19

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Question 1
1 / -0

If $$R$$ is an equivalence relation in a set $$A$$, then $$R^{-1}$$ is

A
Reflexive but not symmetric

B
Symmetric but not transitive

C
An equivalence relation

D
None of these

Solution

Let R be a relation on A i.e. $$R\subseteq A\times A$$
$$R=\{(a,b)|a,b \in A \}$$
Also, given $$R$$ is equivalence relation,

Now, let $$R^{-1}=\{(b,a) | (a,b) \in R\}$$
We will check whether $$R^{-1}$$ is reflexive, symmetric, transitive or an equivalence relation.

Reflexive:
Since, $$R$$ is reflexive
$$\Rightarrow (a,a)\in R$$
$$\Rightarrow (a,a) \in R^{-1}$$ (by def of $$R^{-1}$$)
Hence, $$R^{-1}$$ is reflexive.

Symmetric: Let $$(b,a) \in R^{-1}$$
$$\Rightarrow (a,b) \in R$$ (by def of $$R^{-1}$$)
$$\Rightarrow (b,a) \in R$$ (Since, R is symmetric)
$$\Rightarrow (a,b) \in R^{-1}$$ (by def of $$R^{-1}$$)
Hence, $$R^{-1}$$ is symmetric.

Transitive : Let $$(b,a), (a,c) \in R^{-1}$$
$$\Rightarrow (a,b), (c,a) \in R$$ (by def of $$R^{-1}$$)
or $$(c,a)(a,b) \in R$$
$$\Rightarrow (c,b) \in R$$ (since, R is transitive.)
$$\Rightarrow (b,c) \in R^{-1}$$
Hence, $$R^{-1}$$ is transitive.
Hence, $$R^{-1}$$ is an equivalence relation.
Question 2
1 / -0

If $$\displaystyle A=\left\{ 2,4,5 \right\} ,B=\left\{ 7,8,9 \right\} $$ then $$\displaystyle n\left( A \times B \right) $$ is equal to

A
$$6$$

B
$$9$$

C
$$3$$

D
$$0$$

Solution

$$\displaystyle A=\left\{ 2,4,5 \right\} ,B=\left\{ 7,8,9 \right\} $$

$$\Rightarrow n(A)=3$$ and $$n(B)=3$$

$$\therefore n(A\times B)=n(A)n(B)=9$$

Hence, option B.
Question 3
1 / -0

If $$(3p+q,p-q)=(p-q,3p+q)$$, then:

A
$$p=q=0$$

B
$$p=q$$

C
$$p=2q$$

D
$$p+q=0$$

Solution

As the ordered pairs are equal,
$$ 3p + q = p - q $$
$$ => 2p = -2q $$
$$ => 2p + 2q = 0 $$
$$ => p + q = 0 $$
Question 4
1 / -0

Let $$A = \left\{p,q,r\right\}$$. Which of the following is an equivalence relation in $$A$$?

A
$$R_1 = \left\{(p,q), (q,r), (p,r), (p,p)\right\}$$

B
$$R_2 = \left\{(r,q), (r,p), (r,r), (q,q)\right\}$$

C
$$R_3 = \left\{(p,p), (q,q), (r,r), (p,q)\right\}$$

D
None of these

Solution

Given $$A=\{p,q,r\}$$
$$R_1 = \left\{(p,q), (q,r), (p,r), (p,p)\right\}$$
$$R_1$$ is not reflexive as it does not have $$(q,q),(r,r)$$ . So, $$R_1$$ is not an equivalence relation on A.

$$R_2 = \left\{(r,q), (r,p), (r,r), (q,q)\right\}$$
$$R_2$$ is not reflexive as it does not have $$(p,p)$$ . So, $$R_2$$ is not an equivalence relation on A.

$$R_3 = \left\{(p,p), (q,q), (r,r), (p,q)\right\}$$
$$R_3$$ is reflexive but not symmetric as $$(p,q)$$ is there ,but not $$(q,p)$$
Question 5
1 / -0

If $$A = \{1, 2 \}$$ and $$B = \{3, 4\}$$ then find $$A \times B$$

A
$$A \times B = \{(1,3),(1,2),(2,3),(2,4)\}$$

B
$$A \times B = \{(1,3),(1,4),(2,3),(2,4)\}$$

C
$$A \times B = \{(1,3),(1,4),(2,1),(2,4)\}$$

D
$$A \times B = \{(1,3),(1,4),(2,3),(2,1)\}$$

Solution
Question 6
1 / -0

If $$\displaystyle R_{n}=\left \{ x:\frac{-1}{n}< x< \frac{1}{n} \right \}$$ then $$\displaystyle R_{5}\cup R_{15}=$$ ________

A
$$\displaystyle R_{5}$$

B
$$\displaystyle R_{15}$$

C
$$\displaystyle R_{3}$$

D
$$\displaystyle R_{20}$$

Solution

$$ R_5 = $$ { $$ x:-\frac {1}{5} < x < \frac {1}{5} $$ }
And
$$ R_{15} = $$ {$$ x:-\frac {1}{15} < x < \frac {1}{15} $$ }

But, $$ \frac {1}{15} < x < \frac {1}{15} $$ is within the limits of $$ \frac {1}{5} < x < \frac {1}{5} $$

Hence, $$ R_5 \cup R_{15} = \frac {1}{5} < x < \frac {1}{5} = R_5 $$
Question 7
1 / -0

a R b if "a and b are animals in different zoological parks" then R is

A
only reflexive

B
only symmetric

C
only transitive

D
equivalence

Solution

A relation is said to be symmetric if an element a is related to b then b is also related to a.

Since the relation defines about the animals in different parks, it will be symmetric as the animals are same.
Question 8
1 / -0

R is a relation on set A then $$\displaystyle \left [ \left ( R^{-1} \right )^{-1} \right ]^{-1}$$ is_______

A
R

B
$$\displaystyle R^{-1}$$

C
$$\displaystyle A\times A$$

D
None of these

Solution

We know that Inverse of inverse of a Relation is the original Relation itself.

So, $$ \left[ \left( { R }^{ -1 } \right) ^{ -1 } \right] ^{ -1 } = \left[ R \right] ^{ -1 } $$
Question 9
1 / -0

A relation which satisfies reflexive symmetric and transitive is ________ relation

A
an identity

B
a constant

C
an equivalence

D
None of these

Solution

A relation which satisfies reflexive symmetric and transitive is _called an equivalence relation
Question 10
1 / -0

The domain of the function f(x) = $$\displaystyle \frac{\left | x \right |-2}{\left | x \right |-3}$$ is ________

A
R

B
R - {2, 3}

C
R - {2, -2}

D
R - {-3, 3}

Solution

The function has a real value only when denominator of the fraction is not equal to zero.

So, $$\left| x \right| - 3 \neq = 0 $$
$$ => \left| x \right| \neq = 3 $$
$$ => x \neq = 3 ; -3 $$

So, the domain of the function is $$ R - $$ { $$ -3,3 $$ }

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Relations Test 19

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Relations Test 19

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SHARING IS CARING

Question 1

Reflexive but not symmetric

Symmetric but not transitive

An equivalence relation

None of these

Solution

Question 2

$$6$$

$$9$$

$$3$$

$$0$$

Solution

Question 3

$$p=q=0$$

$$p=q$$

$$p=2q$$

$$p+q=0$$

Solution

Question 4

$$R_1 = \left\{(p,q), (q,r), (p,r), (p,p)\right\}$$

$$R_2 = \left\{(r,q), (r,p), (r,r), (q,q)\right\}$$

$$R_3 = \left\{(p,p), (q,q), (r,r), (p,q)\right\}$$

None of these

Solution

Question 5

$$A \times B = \{(1,3),(1,2),(2,3),(2,4)\}$$

$$A \times B = \{(1,3),(1,4),(2,3),(2,4)\}$$

$$A \times B = \{(1,3),(1,4),(2,1),(2,4)\}$$

$$A \times B = \{(1,3),(1,4),(2,3),(2,1)\}$$

Solution

Question 6

$$\displaystyle R_{5}$$

$$\displaystyle R_{15}$$

$$\displaystyle R_{3}$$

$$\displaystyle R_{20}$$

Solution

Question 7

only reflexive

only symmetric

only transitive

equivalence

Solution

Question 8

R

$$\displaystyle R^{-1}$$

$$\displaystyle A\times A$$

None of these

Solution

Question 9

an identity

a constant

an equivalence

None of these

Solution

Question 10

R

R - {2, 3}

R - {2, -2}

R - {-3, 3}

Solution

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