$${R_1}$$ is not reflexive, because $$\left( {q,\,q} \right)\left( {r,\,r} \right)\, \notin \,{R_1}$$
$$\therefore \,{R_1}$$ is not equivalence relation
$${R_2}$$ is not reflexive, because $$\left( {p,\,p} \right)\, \notin \,{R_2}$$
$$\therefore \,{R_2}$$ is not equivalence relation
$${R_3}$$ is reflexive, because $$\left( {p,\,p} \right),\,\left( {q,\,q} \right),\,\left( {r,\,r} \right)\, \in \,{R_3}$$
$${R_3}$$ is not symmetric, because $$\left( {p,\,q} \right)\, \in \,{R_3}$$ but $$\left( {q,\,p} \right)\, \notin \,{R_3}.$$
72.
Let $$A$$ and $$B$$ be two sets such that $$A \cup B = A.$$ Then $$A \cap B$$ is equal to :
A
$$\phi $$
B
$$B$$
C
$$A$$
D
none of these
E
Answer :
$$B$$
$$\eqalign{
& A \cup B = A \cr
& \Rightarrow B \subseteq A \cr
& \Rightarrow A \cap B = B \cr} $$
73.
Consider the following relations:
$$R$$ = {($$x, y$$ ) | $$x, y$$ are real numbers and $$x = w y$$ for some rational number $$w$$} ;
$$S =$$ { $$\left( {\frac{m}{n},\frac{p}{q}} \right)$$ | $$m, n, p$$ and $$q$$ are integers such that $$n, q \ne 0$$ and $$qm = pn$$ }. Then
A
Neither $$R$$ nor $$S$$ is an equivalence relation
B
$$S$$ is an equivalence relation but $$R$$ is not an equivalence relation
C
$$R$$ and $$S$$ both are equivalence relations
D
$$R$$ is an equivalence relation but $$S$$ is not an equivalence relation
Answer :
$$S$$ is an equivalence relation but $$R$$ is not an equivalence relation
$$x R y$$ need not implies $$y R x$$
∴ $$R$$ is not symmetric and hence not an equivalence relation.
$$\eqalign{
& S:\frac{m}{n}S\frac{p}{q} \cr
& {\text{Given}}\,{\text{ }}qm = pn \cr
& \Rightarrow \,\,\frac{p}{q} = \frac{m}{n} \cr
& \therefore \,\,\frac{m}{n}S\frac{m}{n}\left( {{\text{reflexive}}} \right)\frac{m}{n}S\frac{p}{q} \cr
& \Rightarrow \,\,\frac{p}{q}S\frac{m}{n}\left( {{\text{symmetric}}} \right)\frac{m}{n}S\frac{p}{q},\frac{p}{q}S\frac{r}{s} \cr
& \Rightarrow \,qm = pn,ps = rq \cr
& \Rightarrow \,\,\frac{p}{q} = \frac{m}{n} = \frac{r}{s} \cr
& \Rightarrow \,\,ms = rn\left( {{\text{transitive}}} \right). \cr} $$
$$S$$ is an equivalence relation.
74.
Let $$A = \left\{ {1,\,2} \right\},\,B = \left\{ {3,\,4} \right\}.$$ Then, number of subsets of $$A \times B$$ is :
A
$$4$$
B
$$8$$
C
$$18$$
D
$$16$$
Answer :
$$16$$
$$\eqalign{
& n\left( A \right) = 2{\text{ and }}n\left( B \right) = 2 \cr
& n\left( {A \times B} \right) = n\left( A \right) \times n\left( B \right) = 2 \times 2 = 4 \cr
& \therefore \,{\text{Number of subsets of }}A \times B = {2^{n\left( {A \times B} \right)}} = {2^4} = 16 \cr} $$
75.
In a survey of 400 students in a school, 100 were listed as taking apple juice, 150 as taking orange juice and 75 were listed as talking both apple as well as orange juice.
Then which of the following is/are true ? I. 150 students are taking at least one juice. II. 225 students were taking neither apple juice nor orange juice.
A
Only I is true
B
Only II is true
C
Both I and II are true
D
none of these
Answer :
Only II is true
Let $$U$$ denote the set of surveyed students and $$X$$ denotes the set of students taking apple juice and $$Y$$ denote the set of students taking orange juice. Then, $$n\left( U \right) = 400,\,\,n\left( X \right) = 100,\,\,n\left( Y \right) = 150$$ and $$n\left( {X \cap Y} \right) = 75$$
$$\eqalign{
& n\left( {X \cup Y} \right) = n\left( X \right) + n\left( Y \right) - n\left( {X \cap Y} \right) \cr
& = 100 + 150 - 75 \cr
& = 175 \cr} $$
$$\therefore \,175$$ students were taking at least one juice.
$$\eqalign{
& n\left( {X' \cap Y'} \right) = n\left( {X \cup Y} \right)' \cr
& = n\left( U \right) - n\left( {X \cup Y} \right) \cr
& = 400 - 175 \cr
& = 225 \cr} $$
Hence, 225 students were taking neither apple juice nor orange juice.
76.
Let $$A = Z \cup \left\{ {\sqrt 2 } \right\}.$$ Define a relation $$R$$ in $$A$$ by $$aRb$$ if and only if $$a + b\, \in \,Z.$$ The relation $$R$$ is :
77.
Let $$A$$ and $$B$$ two sets containing 2 elements and 4 elements respectively. The number of subsets of $$A \times B$$ having 3 or more elements is
78.
Let $$R = \left\{ {\left( {1,\,3} \right),\left( {4,\,2} \right),\left( {2,\,4} \right),\left( {2,\,3} \right),\left( {3,\,1} \right)} \right\}$$ be a relation on the set $$A = \left\{ {1,\,2,\,3,\,4} \right\}.$$ The relation $$R$$ is :
A
reflexive
B
transitive
C
not symmetric
D
a function
Answer :
not symmetric
$$\eqalign{
& \left( {2,\,3} \right)\, \in \,R{\text{ but }}\left( {3,\,2} \right)\, \notin \,R \cr
& \therefore \,R{\text{ is not symmentric}}{\text{.}} \cr} $$
79.
If $$\mu $$ is the universal set and $$P$$ is a subset of $$\mu ,$$ then what is $$P \cap \left\{ {\left( {P - \mu } \right) \cup \left( {\mu - P} \right)} \right\}$$ equal to ?
A
$$\phi $$
B
$$P'$$
C
$$m$$
D
$$P$$
Answer :
$$\phi $$
$$\eqalign{
& {\text{Since }}\mu \,{\text{is universal set and }}P \subseteq \mu ,\,P - \mu = \phi \,\,{\text{and }}\mu - P = P' \cr
& {\text{So, }}\left( {P - \mu } \right) \cup \left( {\mu - P} \right) = \phi \cup P' = P' \cr
& {\text{Now, }}P \cap \left\{ {\left( {P - \mu } \right) \cup \left( {\mu - P} \right)} \right\} = P \cap P' = \phi \cr} $$
80.
Which of the following is/are true ? I. If $$A$$ is a subset of the universal set $$U,$$ then its complement $$A'$$ is also a subset of $$U$$. II. If $$U = \left\{ {1,\,2,\,3,.....,\,10} \right\}$$ and $$A = \left\{ {1,\,3,\,5,\,7,\,9} \right\},$$ then $$\left( {A'} \right)' = A.$$
A
Only I is true
B
Only II is true
C
Both I and II are true
D
None of these
Answer :
Both I and II are true
If $$A$$ is a subset of the universal set $$U,$$ then its complement $$A'$$ is also a subset of $$U.$$
We have, $$A' = \left\{ {2,\,4,\,6,\,8,\,10} \right\}$$
Hence, $$\left( {A'} \right)' = \left\{ {x:x\, \in \,U{\text{ and }}x\, \notin \,A'} \right\} = \left\{ {1,\,3,\,5,\,7,\,9} \right\} = A$$
It is clear from the definition of the complement that for any subset of the universal set $$U,$$ we have $$\left( {A'} \right)' = A$$