
If μ is the universal set and P is a subset of μ, then what is \[P \cap \{ \left( {P - \mu } \right) \cup \left( {\mu - P} \right)\} \;\] equal to?
A) \[\phi \]
B) \[P'\]
C) m
D) P
Answer
163.5k+ views
Hint: In this question, we have to find the value of given equation of set. In order to find this apply algebra of sets. After that, apply the concept that intersection of any subset with its complement is null sets.
Formula used: Let X is a universal set and A is a subset of X then
\[A - X\]Gives null set
Complete step by step solution: Given: \[P \cap \{ \left( {P - \mu } \right) \cup \left( {\mu - P} \right)\} \;\]
We know that if X is a universal set and A is a subset of X then
\[A - X\] Gives null set
\[P \cap \{ \left( {P - \mu } \right) \cup \left( {\mu - P} \right)\} \; = P \cap \{ \phi \cup P'\} \;\]
\[P \cap \{ \phi \cup P'\} \; = P \cap P'\]
Now apply the concept that intersection of any subset with its complement is null sets.
\[P \cap P' = \phi \]
Thus, Option (A) is correct.
Note: Here we must remember the algebra used in Venn diagram.
Some important properties of Sets are given below:
A. Idempotent Law is given as
(i) Union of two same sets \[A{\rm{ }} \cup {\rm{ }}A{\rm{ }} = {\rm{ }}A\]
(ii) Intersection of two same sets \[A{\rm{ }} \cap {\rm{ }}A{\rm{ }} = {\rm{ }}A\]
B. Associative Law is given as
(i) \[\left( {A{\rm{ }} \cup {\rm{ }}B} \right){\rm{ }} \cup {\rm{ }}C{\rm{ }} = {\rm{ }}A{\rm{ }} \cup {\rm{ }}\left( {B{\rm{ }} \cup {\rm{ }}C} \right)\]
(ii) \[\left( {A{\rm{ }} \cap {\rm{ }}B} \right){\rm{ }} \cap {\rm{ }}C{\rm{ }} = {\rm{ }}A{\rm{ }} \cap {\rm{ }}\left( {B{\rm{ }} \cap {\rm{ }}C} \right)\]
C. Commutative Law is given as
(i) \[A{\rm{ }} \cup {\rm{ }}B{\rm{ }} = {\rm{ }}B{\rm{ }} \cup {\rm{ }}A\]
(ii) \[A{\rm{ }} \cap {\rm{ }}B{\rm{ }} = {\rm{ }}B{\rm{ }} \cap {\rm{ }}A\]
D. Distributive law is given as
(i) \[A{\rm{ }} \cup {\rm{ }}\left( {B{\rm{ }} \cap {\rm{ }}C} \right){\rm{ }} = {\rm{ }}\left( {A{\rm{ }} \cup {\rm{ }}B} \right){\rm{ }} \cap {\rm{ }}\left( {A{\rm{ }} \cup {\rm{ }}C} \right)\]
(ii) \[A{\rm{ }} \cap {\rm{ }}\left( {B{\rm{ }} \cup {\rm{ }}C} \right){\rm{ }} = \left( {A{\rm{ }} \cap {\rm{ }}B} \right){\rm{ }} \cup {\rm{ }}\left( {A{\rm{ }} \cap {\rm{ }}C} \right)\]
Where A, B, C are set or subset of any universal set
E. De Morgan’s law is given as
(i) \[{\left( {A{\rm{ }} \cup B} \right)^c} = {A^c} \cap {\rm{ }}{B^c}\]
(ii) \[{\left( {A{\rm{ }} \cap B} \right)^c} = {A^c} \cup {\rm{ }}{B^c}\]
Where, \[{A^c},{B^c}\] is complement of set A and B respectively
Formula used: Let X is a universal set and A is a subset of X then
\[A - X\]Gives null set
Complete step by step solution: Given: \[P \cap \{ \left( {P - \mu } \right) \cup \left( {\mu - P} \right)\} \;\]
We know that if X is a universal set and A is a subset of X then
\[A - X\] Gives null set
\[P \cap \{ \left( {P - \mu } \right) \cup \left( {\mu - P} \right)\} \; = P \cap \{ \phi \cup P'\} \;\]
\[P \cap \{ \phi \cup P'\} \; = P \cap P'\]
Now apply the concept that intersection of any subset with its complement is null sets.
\[P \cap P' = \phi \]
Thus, Option (A) is correct.
Note: Here we must remember the algebra used in Venn diagram.
Some important properties of Sets are given below:
A. Idempotent Law is given as
(i) Union of two same sets \[A{\rm{ }} \cup {\rm{ }}A{\rm{ }} = {\rm{ }}A\]
(ii) Intersection of two same sets \[A{\rm{ }} \cap {\rm{ }}A{\rm{ }} = {\rm{ }}A\]
B. Associative Law is given as
(i) \[\left( {A{\rm{ }} \cup {\rm{ }}B} \right){\rm{ }} \cup {\rm{ }}C{\rm{ }} = {\rm{ }}A{\rm{ }} \cup {\rm{ }}\left( {B{\rm{ }} \cup {\rm{ }}C} \right)\]
(ii) \[\left( {A{\rm{ }} \cap {\rm{ }}B} \right){\rm{ }} \cap {\rm{ }}C{\rm{ }} = {\rm{ }}A{\rm{ }} \cap {\rm{ }}\left( {B{\rm{ }} \cap {\rm{ }}C} \right)\]
C. Commutative Law is given as
(i) \[A{\rm{ }} \cup {\rm{ }}B{\rm{ }} = {\rm{ }}B{\rm{ }} \cup {\rm{ }}A\]
(ii) \[A{\rm{ }} \cap {\rm{ }}B{\rm{ }} = {\rm{ }}B{\rm{ }} \cap {\rm{ }}A\]
D. Distributive law is given as
(i) \[A{\rm{ }} \cup {\rm{ }}\left( {B{\rm{ }} \cap {\rm{ }}C} \right){\rm{ }} = {\rm{ }}\left( {A{\rm{ }} \cup {\rm{ }}B} \right){\rm{ }} \cap {\rm{ }}\left( {A{\rm{ }} \cup {\rm{ }}C} \right)\]
(ii) \[A{\rm{ }} \cap {\rm{ }}\left( {B{\rm{ }} \cup {\rm{ }}C} \right){\rm{ }} = \left( {A{\rm{ }} \cap {\rm{ }}B} \right){\rm{ }} \cup {\rm{ }}\left( {A{\rm{ }} \cap {\rm{ }}C} \right)\]
Where A, B, C are set or subset of any universal set
E. De Morgan’s law is given as
(i) \[{\left( {A{\rm{ }} \cup B} \right)^c} = {A^c} \cap {\rm{ }}{B^c}\]
(ii) \[{\left( {A{\rm{ }} \cap B} \right)^c} = {A^c} \cup {\rm{ }}{B^c}\]
Where, \[{A^c},{B^c}\] is complement of set A and B respectively
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