Question

If $A$ and $B$ are two disjoint sets and $N$ is universal set, then ${A^\circ } \cup \left[ {(A \cup B) \cap {B^\circ }} \right]$ is
A.$\phi $
B.A
C.B
D.N

Answer 1

Hint: Disjoint sets are a pair of sets that have no common element amongst them. That is, the intersection of disjoint sets gives null or empty sets. They are two separate sets who have completely different elements and no elements that they share. This can be illustrated in the following manner for ease of mathematical understanding;
$A \cap B = \phi $
Also the complement of a set $A$ is ${A^\circ }$ which represents the set of all elements in the universal set that are not in the set $P$ .
Identities/Formulae used:
The distributive property of sets gives distribution of union of sets with intersection with another set, that is
$(A \cup B) \cap C = (A \cap C) \cup (B \cap C)$
$(A \cap B) \cup C = (A \cup C) \cap (B \cup C)$
The complement law is applicable with union and intersection of a set with its complement. It can be mathematically represented as
$A \cup {A^\circ } = U$
$A \cap {A^\circ } = \phi $
Hence using these properties of sets, we can now expand it on the basis of our requirements.

Complete step by step solution:Due to the presence of square parenthesis, we will first expand within the square brackets and then solve outside it.
That is,
Separating using distributive property,
${A^\circ } \cup \left[ {(A \cup B) \cap {B^\circ }} \right] = {A^\circ } \cup \left[ {(A \cap {B^\circ }) \cup (B \cap {B^\circ })} \right]$
$ \Rightarrow {A^\circ } \cup \left[ {(A \cap {B^\circ }) \cup (B \cap {B^\circ })} \right]$
$ \Rightarrow {A^\circ } \cup \left[ {(A \cap {B^\circ }) \cup ({\phi ^{}})} \right]$
Since the union with null set gives the initial set itself therefore,
$ \Rightarrow {A^\circ } \cup (A \cap {B^\circ })$
Separating using distributive property,
$ \Rightarrow ({A^\circ } \cup A) \cap ({A^\circ } \cup {B^\circ })$
$
   \Rightarrow N \cup ({A^\circ } \cup {B^\circ }) \\
   \Rightarrow N \\
$

Note:
To solve such questions, it is also possible to draw the Venn diagram for convenience of picturing the question better. You can use different colours and patterns for different areas suggested in the expression until you get the final area, which will also turn out to be the entire universal set, as it is now. The mathematical approach is more viable on paper and hence the one chosen here.