Question

If $A$ and $B$ are two sets then $(A \cup B)' \cup (A' \cap B)$ is equal to
A. $A'$
B. $A$
C. $B'$
D. $B$

Answer 1

The question is related to set theory. Here we have to simplify the given term where it contains the union, intersection and complement. So by using the properties related to these operations and on simplification we can determine the solution for the given question.

Complete step by step answer:
A set is a well defined collection of objects. Georg Cantor, the founder of set theory.
The main operations in the set are
1. $\cup$- Union
2. $\cap$- Intersection
The complement of a set is represented as $A'$and it is defined as if $A$ is a set, then the complement of $A$ is the set of elements not in $A$.

Now consider the given question.
$\Rightarrow (A \cup B)' \cup (A' \cap B)$
As we know that the De-Morgan’s law is given as
$(A \cup B)' = A' \cap B'$------(1)
$(A \cap B)' = A' \cup B'$ ------(2)
From the first De-Morgan’s law the above inequality can be written as
$\Rightarrow (A' \cap B') \cup (A' \cap B)$
As we know, the distributive property is given by $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$.

On applying the distributive law for the above inequality, we have
$\Rightarrow A' \cap (B' \cup B)$
As we know, the union of the complement of a set and the set is the universal set.
$\Rightarrow A' \cap U$
As we know that the intersection of a complement of a set and the universal set is the set only. So, now the above inequality can be written as $A$.

Hence option B is the correct answer.

Note: To simplify the terms in the set theory, one should know the properties in set theory.
-Commutative property - $A \cup B = B \cup A$, $A \cap B = B \cap A$
-Associative property - $A \cup (B \cup C) = (A \cup B) \cup C$, $A \cap (B \cap C) = (A \cap B) \cap C$
-Distributive property - $A \cup (B \cap C) = (A \cup B) \cap (A \cup C)$, $A \cap (B \cup C) = (A \cap B) \cup (A \cap C)$
-De-Morgan’s law - $\left( {A \cup B} \right)' = A' \cap B'$, $\left( {A \cap B} \right)' = A' \cup B'$
-$A \cup \phi = A$, $A \cap \phi = \phi$
-$A \cup A' = U$, $A \cap A' = \phi$