Relations Test 32

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Relations Test 32

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Question 1
1 / -0

Let $$R$$ be a relation on $$N$$ defined by $$x+2y=8$$. The domain of $$R$$ is

A
$$\left\{ 2,\ 4,\ 8 \right\} $$

B
$$\left\{ 2,\ 4,\ 6,\ 8 \right\} $$

C
$$\left\{ 2,\ 4,\ 6 \right\} $$

D
$$\left\{ 1,\ 2,\ 3,\ 4 \right\} $$

Solution

R is Relation
$$ R = (x,y) $$
$$ x+2y = 8 $$
$$ R = \left \{ (6,1);(4,2);(2,3) \right \} $$
$$ \therefore $$ Domain of set
$$ = \left \{ 6,4,2 \right \} $$
Question 2
1 / -0

Let $$A\equiv \left\{1,2,3,4\right\},\ B\equiv \left\{a,b,c\right\}$$, then number of function from $$A\rightarrow B$$, which are not onto is

A
$$9$$

B
$$24$$

C
$$45$$

D
$$6$$

Solution

$$A \equiv\{1,2,3,4\} \\$$
$$B \equiv\{a, b, c\}$$
Total function $$(A \rightarrow B)$$

$$\text { Total onto function }(A \rightarrow B)$$

$$=4 c_{3} \times 3 ! \times 3$$
$$=\frac{4 !}{3 ! 1 !} \times 3 ! \times 3$$
$$=41 \times 3=72$$
Hence, no. of $$f^{n}$$ which are not onto
$$=81-72 \\$$
$$=9$$
Question 3
1 / -0

If domain of $$y = f\left( x \right)$$ is $$ \left[ { - 3,\,\,2} \right]$$ then domain of $$y = f\left( {\left| {\left[ x \right]} \right|} \right)$$ is

A
$$[-3, 2]$$

B
$$[-2, 3)$$

C
$$[-3, 3]$$

D
$$[-2, 3]$$

Solution
Question 4
1 / -0

Let $$R = \{ ( 3,3 ) , ( 6,6 ) , ( 9,9 ) , ( 6,12 ) , ( 3,9 ) , ( 3,12 ) , ( 3,6 ) \}$$ be a relation on the set $$A=\{ 3,6,9,12\} .$$ Then the relation $$R ^ { - 1 }$$ is

A
reflexive and transitive.

B
not symmetric

C
transitive

D
all the above

Solution

$$\begin{array}{l} We\, have \\ R=\left\{ { \left( { 3,3 } \right) ,\left( { 6,6 } \right) \left( { 9,9 } \right) \left( { 12,12 } \right) ,\, \left( { 6,12 } \right) ,\left( { 3,12 } \right) ,\, \left( { 3,6 } \right) } \right\} \\ and, \\ A=\left\{ { 3,6,9,12 } \right\} \\ Then \\ The\, relation\, is\, reflexive\, and\, transitive\, only.\, because \\ \left\{ { \left( { 3,3 } \right) ,\left( { 6,6 } \right) \left( { 9,9 } \right) \left( { 12,12 } \right) } \right\} \, \, it\, \, is\, { { Re } }flexive \\ and,\, \left( { 6,12 } \right) ,\left( { 3,12 } \right) ,\, \left( { 3,6 } \right) \left[ { it\, is\, transitive. } \right] \\ \end{array}$$
Hence, option $$A$$ is the correct answer.
Question 5
1 / -0

Consider the following relations :-
$$R=\{ (x,y):x,y$$ are real numbers and $$x =w y$$ for some rational number $$w \}$$ :
$$S = \{ \left( \dfrac { m } { n } , \dfrac { p } { q } \right) : m , n , p$$ and $$q$$ are integers such that $${ n },{ q } \neq 0$$ and $${ qm }={ pn }\}.$$ Then :

A
$$R$$ is an equivalence relation but $$S$$ is not an equivalence relation

B
Neither $$R$$ nor $$S$$ is an equivalence relation

C
$$S$$ is an equivalence relation but $$R$$ is not an equivalence relation

D
$$R$$ and $$S$$ both are equivalence relations

Solution

R: $$x=w y$$ for a function to be a reflexive function
$$ (a, a) \in R \quad \Rightarrow \quad a=\omega a $$
$$w=1$$ (only) $$\therefore R$$ is not a reflexive function.
$$ \begin{aligned} S:\left\{\left(\frac{m}{n}, \frac{p}{q}\right)\right.& m q=n p \\ \Rightarrow \frac{q}{n} &=\frac{p}{m} \Rightarrow \frac{m}{n}=\frac{p}{q} \end{aligned} $$

If $$\left(\frac{a}{b}, \frac{a}{b}\right) \in S$$, then $$S$$ will be reflexive function If $$S \in\left(\frac{a}{b}, \frac{c}{d}\right)$$ and $$\in\left(\frac{c}{d}, \frac{a}{b}\right) \Rightarrow S$$ is symmetric for a function to be transitive,

$$ \begin{aligned} S \in\left(\frac{1}{2}, \frac{2}{4}\right), & S \in\left(\frac{2}{4}, \frac{4}{8}\right) \\ s \in(a, b), S \in(b, c) & \Rightarrow S \in(a, c) \\ \therefore \frac{1}{2}=\frac{4}{8}=\frac{1}{2} \quad & \Rightarrow S \in\left(\frac{1}{2}, \frac{4}{8}\right) \\ & \therefore \text { S is transitive } \end{aligned} $$

$$ \begin{array}{l} \text { Hence, } S \text { is an equivalence relation. } \\ \Rightarrow \text { (c) is the correct option } \end{array} $$
Question 6
1 / -0

If $$A=\left\{1,2,3\right\}$$, the number of symmetric relation in $$A$$ is

A
$$64$$

B
$$8$$

C
$$324$$

D
$$328$$

Solution
Question 7
1 / -0

If $$A=\left\{ 2,3,5 \right\} ,B=\left\{ 4,6,8 \right\} $$, then the relation from $$A$$ into $$B$$ is

A
$$\left\{(2,4),(3,5),(5,6)\right\}$$

B
$$\left\{(2,4),(5,8),(6,5)\right\}$$

C
$$\left\{(2,4),(3,6),(5,6)\right\}$$

D
$$\left\{(2,5),(3,6)\right\}$$
Question 8
1 / -0

If A= {a, b} then possible number of relation on the set A.

A
$$2$$

B
$$4$$

C
$$16$$

D
none of these

Solution
Question 9
1 / -0

If n(A)=4, n(B)=3, $$n(A\times B\times C)=24,then\quad n(C)$$ is equal to

A
288

B
1

C
2

D
12

Solution

$$n\left(A\right)=4\,\,n\left(B\right)=3\,\,n\left(A\times B\times C\right)=24$$
We have $$n\left(A\times B\times C\right)=n\left(A\right)\times n\left(B\right)\times n\left(C\right)$$
$$\Rightarrow 24=4\times 3\times n\left(C\right)$$
$$\therefore n\left(C\right)=\dfrac{24}{12}=2$$
Question 10
1 / -0

If a set $$A$$ has $$n$$ elements then the number of relations defined on $$A$$ is

A
$$2^{ \left( { n }^{ 2 } \right) }$$

B
$$2^{ \left( { n }^{ 2 } \right) }-1$$

C
$$2^{ n }$$

D
None of these.

Solution