
$A,B,C$ and $D$ are four sets such that $A\cap B=C\cap D=\varnothing $. Consider the following:
1. $A\cup C$ and $B\cup D$ are always disjoint
2. $A\cap C$ and $B\cap D$ are always disjoint
Which of the above statement is/are correct.
A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2
Answer
164.1k+ views
Hint: In this question, we are to find the correct statement from the given statements for the given sets $A,B,C$ and $D$. To do this, consider these sets with any of the elements with the given condition, then by evaluating the given statements we get the correct one.
Formula Used:Set: A collection of objects in which it is possible to decide whether a given object belongs to the collection or not is said to be a set. Those objects are nothing but the elements in the set.
Sets are represented in two ways: The roaster method and the Set builder form
Roaster method:
The set of vowels – \[\left\{ a,\text{ }e,\text{ }i,\text{ }o,\text{ }u \right\}\]
The set of natural numbers – \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots. \right\}\]
Set builder form:
The set of vowels – \[\left\{ x:x\text{ }is\text{ }a\text{ }vowel\text{ }in\text{ }English\text{ }alphabet \right\}\]
The set of natural numbers – \[\left\{ x:x\text{ }is\text{ }a\text{ }natural\text{ }number \right\}\]
Some of the important mathematical sets:
$N$ - the set of natural numbers - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
Complete step by step solution:The given sets are $A,B,C$, and $D$.
Consider them as
$\begin{align}
& A=\{a,b\} \\
& B=\{c,d,e\} \\
& C=\{e,f,g\} \\
& D=\{h,i\} \\
\end{align}$
Since we have given $A\cap B=C\cap D=\varnothing $, the above consideration is valid.
Then,
$A\cup C=\{a,b\}\cup \{e,f,g\}=\{a,b,e,f,g\}$
And
$B\cup D=\{c,d,e\}\cup \{h,i\}=\{c,d,e,h,i\}$
So,
$(A\cup C)\cap (B\cup D)=\{a,b,e,f,g\}\cap \{c,d,e,h,i\}=\{e\}$
Therefore, they are not always disjoint. So, statement (1) is incorrect.
Now,
$A\cap C=\{a,b\}\cap \{e,f,g\}=\varnothing $
And
$B\cap D=\{c,d,e\}\cap \{h,i\}=\varnothing $
So, they are always disjoint. Hence, statement (2) is correct.
Option ‘B’ is correct
Note: Here we need to assume the given sets as per the condition $A\cap B=C\cap D=\varnothing $. Then we can able to calculate the given statements.
Formula Used:Set: A collection of objects in which it is possible to decide whether a given object belongs to the collection or not is said to be a set. Those objects are nothing but the elements in the set.
Sets are represented in two ways: The roaster method and the Set builder form
Roaster method:
The set of vowels – \[\left\{ a,\text{ }e,\text{ }i,\text{ }o,\text{ }u \right\}\]
The set of natural numbers – \[\left\{ 1,\text{ }2,\text{ }3,\text{ }\ldots. \right\}\]
Set builder form:
The set of vowels – \[\left\{ x:x\text{ }is\text{ }a\text{ }vowel\text{ }in\text{ }English\text{ }alphabet \right\}\]
The set of natural numbers – \[\left\{ x:x\text{ }is\text{ }a\text{ }natural\text{ }number \right\}\]
Some of the important mathematical sets:
$N$ - the set of natural numbers - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
Complete step by step solution:The given sets are $A,B,C$, and $D$.
Consider them as
$\begin{align}
& A=\{a,b\} \\
& B=\{c,d,e\} \\
& C=\{e,f,g\} \\
& D=\{h,i\} \\
\end{align}$
Since we have given $A\cap B=C\cap D=\varnothing $, the above consideration is valid.
Then,
$A\cup C=\{a,b\}\cup \{e,f,g\}=\{a,b,e,f,g\}$
And
$B\cup D=\{c,d,e\}\cup \{h,i\}=\{c,d,e,h,i\}$
So,
$(A\cup C)\cap (B\cup D)=\{a,b,e,f,g\}\cap \{c,d,e,h,i\}=\{e\}$
Therefore, they are not always disjoint. So, statement (1) is incorrect.
Now,
$A\cap C=\{a,b\}\cap \{e,f,g\}=\varnothing $
And
$B\cap D=\{c,d,e\}\cap \{h,i\}=\varnothing $
So, they are always disjoint. Hence, statement (2) is correct.
Option ‘B’ is correct
Note: Here we need to assume the given sets as per the condition $A\cap B=C\cap D=\varnothing $. Then we can able to calculate the given statements.
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