NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
116807
For any two real numbers \(\theta\) and \(\phi\), we define \(\theta R \phi\), if and only if \(\sec ^2 \theta-\tan ^2 \phi=1\). The relation \(R\) is
1 reflexive but not transitive
2 symmetric but not reflexive
3 both reflexive and symmetric but not transitive
4 an equivalence relation
Explanation:
D Given, The relation is \(\theta R \phi \Rightarrow \sec ^2 \theta-\tan ^2 \phi=1\) For reflexive:- \(\quad \theta R \theta \Rightarrow \operatorname{Sec}^2 \theta-\tan ^2 \theta=1\) \(1=1\), Which is true \(\therefore \quad\) It is reflexive. For symmetric: \(\theta \mathrm{R} \phi \Rightarrow \sec ^2 \theta-\tan ^2 \phi=1\) \(\left(1+\tan ^2 \theta\right)-\left(\sec ^2 \phi-1\right)=1\) \(1+\tan ^2 \theta-\sec ^2 \phi+1=1\) \(2+\tan ^2 \theta-\sec ^2 \phi=1\) \(\tan ^2 \theta-\sec ^2 \phi=-1\) \(\sec ^2 \phi-\tan ^2 \theta=1\) \(\phi \mathrm{R} \theta\) \(\text { It is symmetric. }\) For transitive:- Let \(\theta R \phi\) and \(\phi R \psi\), then- \(\sec ^2 \theta-\tan ^2 \phi=1\) \(\text { and, } \sec ^2 \phi-\tan ^2 \psi=1\) \(\therefore \theta \mathrm{R} \psi \Rightarrow \sec ^2 \theta-\tan ^2 \psi=1\) \(\Rightarrow \sec ^2 \theta-\tan ^2 \psi+1=1+1\) \(\Rightarrow \sec ^2 \theta-\tan ^2 \psi+\sec ^2 \phi-\tan ^2 \phi=1+1\) \(\Rightarrow \theta R \phi \text { and } \phi \mathrm{R} \psi\) Then, it is transitive. So, it is an equivalence relation.
WB JEE-2014
Sets, Relation and Function
116808
Let the number of elements of the sets \(A\) and \(B\) be \(p\) and \(q\), respectively. Then, the number of relations from the set \(A\) to the set \(B\) is
1 \(2^{\mathrm{p}+\mathrm{q}}\)
2 \(2^{\mathrm{pq}}\)
3 \(\mathrm{p}+\mathrm{q}\)
4 \(\mathrm{pq}\)
Explanation:
B Given, the sets A and B . And, number of elements of the set \(A=p\) number of elements of the set \(B=q\) Then, the cartesian product of \(A\) and \(B\) is - \(\mathrm{A} \times \mathrm{B}=\{(\mathrm{a}, \mathrm{b}):(\mathrm{a} \in \mathrm{A}) \text { and }(\mathrm{b} \in \mathrm{B})\}\) \(\therefore\) Number of elements in \(|\mathrm{A} \times \mathrm{B}|=|\mathrm{A}| \cdot|\mathrm{B}|=\mathrm{pq}\) Then, any relation from \(A\) to \(B\) is a subset of \(A \times B\). So, the number of relations from \(\mathrm{A}\) to \(\mathrm{B}\) is the number of subsets of \(A \times B\) is- \(=2^{\mid \mathrm{A} \times \mathrm{B|}}=2^{\mathrm{pq}}\)
WB JEE-2014
Sets, Relation and Function
116809
A relation \(P\) on the set of real number \(R\) is defined as \(\{x P y: x y>0\}\). Then, which of the following is/are true?
1 \(P\) is reflexive and symmetric
2 \(\mathrm{P}\) is symmetric but not reflexive
3 \(P\) is symmetric and transitive
4 \(\mathrm{P}\) is an equivalence relation
Explanation:
A Given, a relation \(\mathrm{P}\) on the set of real number \(\mathrm{R}\) is defined as - \(\{\mathrm{xPy}: \mathrm{xy}>0\}\) For reflexive:- \(\mathrm{x} \cdot \mathrm{x}=\mathrm{x}^2 \geq 0 \forall \mathrm{x} \in \mathrm{R}\) \(\therefore \quad P \text { is reflexive. }\) For symmetric :- Let, \(\quad \mathrm{x}, \mathrm{y} \in \mathrm{R}\) such that \(\mathrm{xy} \geq 0\) \(\Rightarrow \quad \mathrm{yx} \geq 0, \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}\) \(\Rightarrow \quad \mathrm{yPx}\) \(\Rightarrow \quad \mathrm{P}\) is symmetric For transitive:- \((1,0),(0,-2) \in \mathrm{P}\) but, \(\quad(1,-2) \notin \mathrm{P}\) \(\therefore \quad \mathrm{P}\) is not transitive. So, \(\mathrm{P}\) is reflexive and symmetric but not transitive.
WB JEE-2015
Sets, Relation and Function
116810
For any two real numbers \(a\) and \(b\), we define \(a\) \(R b\) if and only if \(\sin ^2 a+\cos ^2 b=1\). The relation \(R\) is
1 reflexive but not symmetric
2 symmetric but not transitive
3 transitive but not reflexive
4 an equivalence relation
Explanation:
D Given, for any two real number a and b. We define \(a R b \Leftrightarrow \sin ^2 a+\cos ^2 b=1\) For reflexive : \(-\mathrm{aRa} \Rightarrow \sin ^2 \mathrm{a}+\cos ^2 \mathrm{a}=1 \forall \mathrm{a} \in \mathrm{R}\). \(\therefore \quad\) It is reflexive relation. For symmetric:- \(\text { aRb } \Rightarrow \sin ^2 a+\cos ^2 b=1\) \(\Rightarrow 1-\cos ^2 a+1-\sin ^2 b=1\) \(\Rightarrow \quad 2-\cos ^2 \mathrm{a}-\sin ^2 \mathrm{~b}=1\) \(\Rightarrow \quad \sin ^2 \mathrm{~b}+\cos ^2 \mathrm{a}=1\) \(\Rightarrow \quad \mathrm{bRa} \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\) \(\therefore \text { It is symmetric relation. }\) \(\text { For transitive:- }\) \(\quad \quad \quad \mathrm{aRb} \text { and } \mathrm{bRc} \Rightarrow \mathrm{aRc}\) \(\Rightarrow \quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{~b}=1 \text { and } \sin ^2 \mathrm{~b}+\cos ^2 \mathrm{c}=1\) \(\therefore \text { Adding these two equation we get- }\) \(\quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{~b}+\sin ^2 \mathrm{~b}+\cos ^2 \mathrm{c}=2\) \(\Rightarrow \quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{c}=1\) \(\Rightarrow \quad \text { aRc }\) \(\therefore \text { It is transitive relation. }\) \(\text { So, } \mathrm{R} \text { is an equivalence relation. }\) \(\therefore\) It is transitive relation. So, \(\mathrm{R}\) is an equivalence relation.
116807
For any two real numbers \(\theta\) and \(\phi\), we define \(\theta R \phi\), if and only if \(\sec ^2 \theta-\tan ^2 \phi=1\). The relation \(R\) is
1 reflexive but not transitive
2 symmetric but not reflexive
3 both reflexive and symmetric but not transitive
4 an equivalence relation
Explanation:
D Given, The relation is \(\theta R \phi \Rightarrow \sec ^2 \theta-\tan ^2 \phi=1\) For reflexive:- \(\quad \theta R \theta \Rightarrow \operatorname{Sec}^2 \theta-\tan ^2 \theta=1\) \(1=1\), Which is true \(\therefore \quad\) It is reflexive. For symmetric: \(\theta \mathrm{R} \phi \Rightarrow \sec ^2 \theta-\tan ^2 \phi=1\) \(\left(1+\tan ^2 \theta\right)-\left(\sec ^2 \phi-1\right)=1\) \(1+\tan ^2 \theta-\sec ^2 \phi+1=1\) \(2+\tan ^2 \theta-\sec ^2 \phi=1\) \(\tan ^2 \theta-\sec ^2 \phi=-1\) \(\sec ^2 \phi-\tan ^2 \theta=1\) \(\phi \mathrm{R} \theta\) \(\text { It is symmetric. }\) For transitive:- Let \(\theta R \phi\) and \(\phi R \psi\), then- \(\sec ^2 \theta-\tan ^2 \phi=1\) \(\text { and, } \sec ^2 \phi-\tan ^2 \psi=1\) \(\therefore \theta \mathrm{R} \psi \Rightarrow \sec ^2 \theta-\tan ^2 \psi=1\) \(\Rightarrow \sec ^2 \theta-\tan ^2 \psi+1=1+1\) \(\Rightarrow \sec ^2 \theta-\tan ^2 \psi+\sec ^2 \phi-\tan ^2 \phi=1+1\) \(\Rightarrow \theta R \phi \text { and } \phi \mathrm{R} \psi\) Then, it is transitive. So, it is an equivalence relation.
WB JEE-2014
Sets, Relation and Function
116808
Let the number of elements of the sets \(A\) and \(B\) be \(p\) and \(q\), respectively. Then, the number of relations from the set \(A\) to the set \(B\) is
1 \(2^{\mathrm{p}+\mathrm{q}}\)
2 \(2^{\mathrm{pq}}\)
3 \(\mathrm{p}+\mathrm{q}\)
4 \(\mathrm{pq}\)
Explanation:
B Given, the sets A and B . And, number of elements of the set \(A=p\) number of elements of the set \(B=q\) Then, the cartesian product of \(A\) and \(B\) is - \(\mathrm{A} \times \mathrm{B}=\{(\mathrm{a}, \mathrm{b}):(\mathrm{a} \in \mathrm{A}) \text { and }(\mathrm{b} \in \mathrm{B})\}\) \(\therefore\) Number of elements in \(|\mathrm{A} \times \mathrm{B}|=|\mathrm{A}| \cdot|\mathrm{B}|=\mathrm{pq}\) Then, any relation from \(A\) to \(B\) is a subset of \(A \times B\). So, the number of relations from \(\mathrm{A}\) to \(\mathrm{B}\) is the number of subsets of \(A \times B\) is- \(=2^{\mid \mathrm{A} \times \mathrm{B|}}=2^{\mathrm{pq}}\)
WB JEE-2014
Sets, Relation and Function
116809
A relation \(P\) on the set of real number \(R\) is defined as \(\{x P y: x y>0\}\). Then, which of the following is/are true?
1 \(P\) is reflexive and symmetric
2 \(\mathrm{P}\) is symmetric but not reflexive
3 \(P\) is symmetric and transitive
4 \(\mathrm{P}\) is an equivalence relation
Explanation:
A Given, a relation \(\mathrm{P}\) on the set of real number \(\mathrm{R}\) is defined as - \(\{\mathrm{xPy}: \mathrm{xy}>0\}\) For reflexive:- \(\mathrm{x} \cdot \mathrm{x}=\mathrm{x}^2 \geq 0 \forall \mathrm{x} \in \mathrm{R}\) \(\therefore \quad P \text { is reflexive. }\) For symmetric :- Let, \(\quad \mathrm{x}, \mathrm{y} \in \mathrm{R}\) such that \(\mathrm{xy} \geq 0\) \(\Rightarrow \quad \mathrm{yx} \geq 0, \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}\) \(\Rightarrow \quad \mathrm{yPx}\) \(\Rightarrow \quad \mathrm{P}\) is symmetric For transitive:- \((1,0),(0,-2) \in \mathrm{P}\) but, \(\quad(1,-2) \notin \mathrm{P}\) \(\therefore \quad \mathrm{P}\) is not transitive. So, \(\mathrm{P}\) is reflexive and symmetric but not transitive.
WB JEE-2015
Sets, Relation and Function
116810
For any two real numbers \(a\) and \(b\), we define \(a\) \(R b\) if and only if \(\sin ^2 a+\cos ^2 b=1\). The relation \(R\) is
1 reflexive but not symmetric
2 symmetric but not transitive
3 transitive but not reflexive
4 an equivalence relation
Explanation:
D Given, for any two real number a and b. We define \(a R b \Leftrightarrow \sin ^2 a+\cos ^2 b=1\) For reflexive : \(-\mathrm{aRa} \Rightarrow \sin ^2 \mathrm{a}+\cos ^2 \mathrm{a}=1 \forall \mathrm{a} \in \mathrm{R}\). \(\therefore \quad\) It is reflexive relation. For symmetric:- \(\text { aRb } \Rightarrow \sin ^2 a+\cos ^2 b=1\) \(\Rightarrow 1-\cos ^2 a+1-\sin ^2 b=1\) \(\Rightarrow \quad 2-\cos ^2 \mathrm{a}-\sin ^2 \mathrm{~b}=1\) \(\Rightarrow \quad \sin ^2 \mathrm{~b}+\cos ^2 \mathrm{a}=1\) \(\Rightarrow \quad \mathrm{bRa} \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\) \(\therefore \text { It is symmetric relation. }\) \(\text { For transitive:- }\) \(\quad \quad \quad \mathrm{aRb} \text { and } \mathrm{bRc} \Rightarrow \mathrm{aRc}\) \(\Rightarrow \quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{~b}=1 \text { and } \sin ^2 \mathrm{~b}+\cos ^2 \mathrm{c}=1\) \(\therefore \text { Adding these two equation we get- }\) \(\quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{~b}+\sin ^2 \mathrm{~b}+\cos ^2 \mathrm{c}=2\) \(\Rightarrow \quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{c}=1\) \(\Rightarrow \quad \text { aRc }\) \(\therefore \text { It is transitive relation. }\) \(\text { So, } \mathrm{R} \text { is an equivalence relation. }\) \(\therefore\) It is transitive relation. So, \(\mathrm{R}\) is an equivalence relation.
116807
For any two real numbers \(\theta\) and \(\phi\), we define \(\theta R \phi\), if and only if \(\sec ^2 \theta-\tan ^2 \phi=1\). The relation \(R\) is
1 reflexive but not transitive
2 symmetric but not reflexive
3 both reflexive and symmetric but not transitive
4 an equivalence relation
Explanation:
D Given, The relation is \(\theta R \phi \Rightarrow \sec ^2 \theta-\tan ^2 \phi=1\) For reflexive:- \(\quad \theta R \theta \Rightarrow \operatorname{Sec}^2 \theta-\tan ^2 \theta=1\) \(1=1\), Which is true \(\therefore \quad\) It is reflexive. For symmetric: \(\theta \mathrm{R} \phi \Rightarrow \sec ^2 \theta-\tan ^2 \phi=1\) \(\left(1+\tan ^2 \theta\right)-\left(\sec ^2 \phi-1\right)=1\) \(1+\tan ^2 \theta-\sec ^2 \phi+1=1\) \(2+\tan ^2 \theta-\sec ^2 \phi=1\) \(\tan ^2 \theta-\sec ^2 \phi=-1\) \(\sec ^2 \phi-\tan ^2 \theta=1\) \(\phi \mathrm{R} \theta\) \(\text { It is symmetric. }\) For transitive:- Let \(\theta R \phi\) and \(\phi R \psi\), then- \(\sec ^2 \theta-\tan ^2 \phi=1\) \(\text { and, } \sec ^2 \phi-\tan ^2 \psi=1\) \(\therefore \theta \mathrm{R} \psi \Rightarrow \sec ^2 \theta-\tan ^2 \psi=1\) \(\Rightarrow \sec ^2 \theta-\tan ^2 \psi+1=1+1\) \(\Rightarrow \sec ^2 \theta-\tan ^2 \psi+\sec ^2 \phi-\tan ^2 \phi=1+1\) \(\Rightarrow \theta R \phi \text { and } \phi \mathrm{R} \psi\) Then, it is transitive. So, it is an equivalence relation.
WB JEE-2014
Sets, Relation and Function
116808
Let the number of elements of the sets \(A\) and \(B\) be \(p\) and \(q\), respectively. Then, the number of relations from the set \(A\) to the set \(B\) is
1 \(2^{\mathrm{p}+\mathrm{q}}\)
2 \(2^{\mathrm{pq}}\)
3 \(\mathrm{p}+\mathrm{q}\)
4 \(\mathrm{pq}\)
Explanation:
B Given, the sets A and B . And, number of elements of the set \(A=p\) number of elements of the set \(B=q\) Then, the cartesian product of \(A\) and \(B\) is - \(\mathrm{A} \times \mathrm{B}=\{(\mathrm{a}, \mathrm{b}):(\mathrm{a} \in \mathrm{A}) \text { and }(\mathrm{b} \in \mathrm{B})\}\) \(\therefore\) Number of elements in \(|\mathrm{A} \times \mathrm{B}|=|\mathrm{A}| \cdot|\mathrm{B}|=\mathrm{pq}\) Then, any relation from \(A\) to \(B\) is a subset of \(A \times B\). So, the number of relations from \(\mathrm{A}\) to \(\mathrm{B}\) is the number of subsets of \(A \times B\) is- \(=2^{\mid \mathrm{A} \times \mathrm{B|}}=2^{\mathrm{pq}}\)
WB JEE-2014
Sets, Relation and Function
116809
A relation \(P\) on the set of real number \(R\) is defined as \(\{x P y: x y>0\}\). Then, which of the following is/are true?
1 \(P\) is reflexive and symmetric
2 \(\mathrm{P}\) is symmetric but not reflexive
3 \(P\) is symmetric and transitive
4 \(\mathrm{P}\) is an equivalence relation
Explanation:
A Given, a relation \(\mathrm{P}\) on the set of real number \(\mathrm{R}\) is defined as - \(\{\mathrm{xPy}: \mathrm{xy}>0\}\) For reflexive:- \(\mathrm{x} \cdot \mathrm{x}=\mathrm{x}^2 \geq 0 \forall \mathrm{x} \in \mathrm{R}\) \(\therefore \quad P \text { is reflexive. }\) For symmetric :- Let, \(\quad \mathrm{x}, \mathrm{y} \in \mathrm{R}\) such that \(\mathrm{xy} \geq 0\) \(\Rightarrow \quad \mathrm{yx} \geq 0, \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}\) \(\Rightarrow \quad \mathrm{yPx}\) \(\Rightarrow \quad \mathrm{P}\) is symmetric For transitive:- \((1,0),(0,-2) \in \mathrm{P}\) but, \(\quad(1,-2) \notin \mathrm{P}\) \(\therefore \quad \mathrm{P}\) is not transitive. So, \(\mathrm{P}\) is reflexive and symmetric but not transitive.
WB JEE-2015
Sets, Relation and Function
116810
For any two real numbers \(a\) and \(b\), we define \(a\) \(R b\) if and only if \(\sin ^2 a+\cos ^2 b=1\). The relation \(R\) is
1 reflexive but not symmetric
2 symmetric but not transitive
3 transitive but not reflexive
4 an equivalence relation
Explanation:
D Given, for any two real number a and b. We define \(a R b \Leftrightarrow \sin ^2 a+\cos ^2 b=1\) For reflexive : \(-\mathrm{aRa} \Rightarrow \sin ^2 \mathrm{a}+\cos ^2 \mathrm{a}=1 \forall \mathrm{a} \in \mathrm{R}\). \(\therefore \quad\) It is reflexive relation. For symmetric:- \(\text { aRb } \Rightarrow \sin ^2 a+\cos ^2 b=1\) \(\Rightarrow 1-\cos ^2 a+1-\sin ^2 b=1\) \(\Rightarrow \quad 2-\cos ^2 \mathrm{a}-\sin ^2 \mathrm{~b}=1\) \(\Rightarrow \quad \sin ^2 \mathrm{~b}+\cos ^2 \mathrm{a}=1\) \(\Rightarrow \quad \mathrm{bRa} \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\) \(\therefore \text { It is symmetric relation. }\) \(\text { For transitive:- }\) \(\quad \quad \quad \mathrm{aRb} \text { and } \mathrm{bRc} \Rightarrow \mathrm{aRc}\) \(\Rightarrow \quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{~b}=1 \text { and } \sin ^2 \mathrm{~b}+\cos ^2 \mathrm{c}=1\) \(\therefore \text { Adding these two equation we get- }\) \(\quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{~b}+\sin ^2 \mathrm{~b}+\cos ^2 \mathrm{c}=2\) \(\Rightarrow \quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{c}=1\) \(\Rightarrow \quad \text { aRc }\) \(\therefore \text { It is transitive relation. }\) \(\text { So, } \mathrm{R} \text { is an equivalence relation. }\) \(\therefore\) It is transitive relation. So, \(\mathrm{R}\) is an equivalence relation.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Sets, Relation and Function
116807
For any two real numbers \(\theta\) and \(\phi\), we define \(\theta R \phi\), if and only if \(\sec ^2 \theta-\tan ^2 \phi=1\). The relation \(R\) is
1 reflexive but not transitive
2 symmetric but not reflexive
3 both reflexive and symmetric but not transitive
4 an equivalence relation
Explanation:
D Given, The relation is \(\theta R \phi \Rightarrow \sec ^2 \theta-\tan ^2 \phi=1\) For reflexive:- \(\quad \theta R \theta \Rightarrow \operatorname{Sec}^2 \theta-\tan ^2 \theta=1\) \(1=1\), Which is true \(\therefore \quad\) It is reflexive. For symmetric: \(\theta \mathrm{R} \phi \Rightarrow \sec ^2 \theta-\tan ^2 \phi=1\) \(\left(1+\tan ^2 \theta\right)-\left(\sec ^2 \phi-1\right)=1\) \(1+\tan ^2 \theta-\sec ^2 \phi+1=1\) \(2+\tan ^2 \theta-\sec ^2 \phi=1\) \(\tan ^2 \theta-\sec ^2 \phi=-1\) \(\sec ^2 \phi-\tan ^2 \theta=1\) \(\phi \mathrm{R} \theta\) \(\text { It is symmetric. }\) For transitive:- Let \(\theta R \phi\) and \(\phi R \psi\), then- \(\sec ^2 \theta-\tan ^2 \phi=1\) \(\text { and, } \sec ^2 \phi-\tan ^2 \psi=1\) \(\therefore \theta \mathrm{R} \psi \Rightarrow \sec ^2 \theta-\tan ^2 \psi=1\) \(\Rightarrow \sec ^2 \theta-\tan ^2 \psi+1=1+1\) \(\Rightarrow \sec ^2 \theta-\tan ^2 \psi+\sec ^2 \phi-\tan ^2 \phi=1+1\) \(\Rightarrow \theta R \phi \text { and } \phi \mathrm{R} \psi\) Then, it is transitive. So, it is an equivalence relation.
WB JEE-2014
Sets, Relation and Function
116808
Let the number of elements of the sets \(A\) and \(B\) be \(p\) and \(q\), respectively. Then, the number of relations from the set \(A\) to the set \(B\) is
1 \(2^{\mathrm{p}+\mathrm{q}}\)
2 \(2^{\mathrm{pq}}\)
3 \(\mathrm{p}+\mathrm{q}\)
4 \(\mathrm{pq}\)
Explanation:
B Given, the sets A and B . And, number of elements of the set \(A=p\) number of elements of the set \(B=q\) Then, the cartesian product of \(A\) and \(B\) is - \(\mathrm{A} \times \mathrm{B}=\{(\mathrm{a}, \mathrm{b}):(\mathrm{a} \in \mathrm{A}) \text { and }(\mathrm{b} \in \mathrm{B})\}\) \(\therefore\) Number of elements in \(|\mathrm{A} \times \mathrm{B}|=|\mathrm{A}| \cdot|\mathrm{B}|=\mathrm{pq}\) Then, any relation from \(A\) to \(B\) is a subset of \(A \times B\). So, the number of relations from \(\mathrm{A}\) to \(\mathrm{B}\) is the number of subsets of \(A \times B\) is- \(=2^{\mid \mathrm{A} \times \mathrm{B|}}=2^{\mathrm{pq}}\)
WB JEE-2014
Sets, Relation and Function
116809
A relation \(P\) on the set of real number \(R\) is defined as \(\{x P y: x y>0\}\). Then, which of the following is/are true?
1 \(P\) is reflexive and symmetric
2 \(\mathrm{P}\) is symmetric but not reflexive
3 \(P\) is symmetric and transitive
4 \(\mathrm{P}\) is an equivalence relation
Explanation:
A Given, a relation \(\mathrm{P}\) on the set of real number \(\mathrm{R}\) is defined as - \(\{\mathrm{xPy}: \mathrm{xy}>0\}\) For reflexive:- \(\mathrm{x} \cdot \mathrm{x}=\mathrm{x}^2 \geq 0 \forall \mathrm{x} \in \mathrm{R}\) \(\therefore \quad P \text { is reflexive. }\) For symmetric :- Let, \(\quad \mathrm{x}, \mathrm{y} \in \mathrm{R}\) such that \(\mathrm{xy} \geq 0\) \(\Rightarrow \quad \mathrm{yx} \geq 0, \forall \mathrm{x}, \mathrm{y} \in \mathrm{R}\) \(\Rightarrow \quad \mathrm{yPx}\) \(\Rightarrow \quad \mathrm{P}\) is symmetric For transitive:- \((1,0),(0,-2) \in \mathrm{P}\) but, \(\quad(1,-2) \notin \mathrm{P}\) \(\therefore \quad \mathrm{P}\) is not transitive. So, \(\mathrm{P}\) is reflexive and symmetric but not transitive.
WB JEE-2015
Sets, Relation and Function
116810
For any two real numbers \(a\) and \(b\), we define \(a\) \(R b\) if and only if \(\sin ^2 a+\cos ^2 b=1\). The relation \(R\) is
1 reflexive but not symmetric
2 symmetric but not transitive
3 transitive but not reflexive
4 an equivalence relation
Explanation:
D Given, for any two real number a and b. We define \(a R b \Leftrightarrow \sin ^2 a+\cos ^2 b=1\) For reflexive : \(-\mathrm{aRa} \Rightarrow \sin ^2 \mathrm{a}+\cos ^2 \mathrm{a}=1 \forall \mathrm{a} \in \mathrm{R}\). \(\therefore \quad\) It is reflexive relation. For symmetric:- \(\text { aRb } \Rightarrow \sin ^2 a+\cos ^2 b=1\) \(\Rightarrow 1-\cos ^2 a+1-\sin ^2 b=1\) \(\Rightarrow \quad 2-\cos ^2 \mathrm{a}-\sin ^2 \mathrm{~b}=1\) \(\Rightarrow \quad \sin ^2 \mathrm{~b}+\cos ^2 \mathrm{a}=1\) \(\Rightarrow \quad \mathrm{bRa} \forall \mathrm{a}, \mathrm{b} \in \mathrm{R}\) \(\therefore \text { It is symmetric relation. }\) \(\text { For transitive:- }\) \(\quad \quad \quad \mathrm{aRb} \text { and } \mathrm{bRc} \Rightarrow \mathrm{aRc}\) \(\Rightarrow \quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{~b}=1 \text { and } \sin ^2 \mathrm{~b}+\cos ^2 \mathrm{c}=1\) \(\therefore \text { Adding these two equation we get- }\) \(\quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{~b}+\sin ^2 \mathrm{~b}+\cos ^2 \mathrm{c}=2\) \(\Rightarrow \quad \sin ^2 \mathrm{a}+\cos ^2 \mathrm{c}=1\) \(\Rightarrow \quad \text { aRc }\) \(\therefore \text { It is transitive relation. }\) \(\text { So, } \mathrm{R} \text { is an equivalence relation. }\) \(\therefore\) It is transitive relation. So, \(\mathrm{R}\) is an equivalence relation.