
Consider the following statements:
I. If ${{A}_{n}}$ is the set of first $n$ prime numbers, then $\bigcup\limits_{n=2}^{10}{{{A}_{n}}}$ is equal to $\{2,3,5,7,11,13,17,19,23,29\}$.
II. If $A$ and $B$ are two sets such that $n(A\cup B)=50,n(A)=28,n(B)=32$, then $n(A\cap B)=10$. Which of these is correct?
A. Only I is true
B. Only II is true
C. Both are true
D. Both are false
Answer
162.9k+ views
Hint: In this question, we are to find the correct statements from the given statements. Here, the given statements are general statements, by using simple set operations and definitions they are proved.
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 natural number’s set - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
$Q$ - the set of all rational numbers; $R$ - the set of all real numbers; $C$ - the set of all complex numbers.
If two sets $A,B$ where the set $A$ is said to be the subset of $B$ i.e., $A\subseteq B$ then every element of $A$ is in the set $B$ and the set $A$ is said to be the proper subset of $B$ i.e., \[A\subset B\] then $A\subseteq B$ and $A\ne B$.
Some of the important set operations:
$\begin{align}
& n(A\cup B)=n(A)+n(B)-n(A\cap B) \\
& n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C) \\
\end{align}$
Complete step by step solution: Given that,
${{A}_{n}}$ is the set of first $n$ prime numbers.
So, the set of first $10$ prime numbers is
${{A}_{10}}=\{2,3,5,7,11,13,17,19,23,29\}$
By the definition of a set, we can represent the above set for first $n$ prime numbers by
$\bigcup\limits_{n=2}^{10}{{{A}_{n}}}=\{2,3,5,7,11,13,17,19,23,29\}$
So, the first statement is true.
For the sets $A$ and $B$ we have
$n(A\cup B)=50,n(A)=28,n(B)=32$
Then, from the formula,
$\begin{align}
& n(A\cup B)=n(A)+n(B)-n(A\cap B) \\
& \Rightarrow n(A\cap B)=n(A)+n(B)-n(A\cup B) \\
\end{align}$
On substituting the given values,
$n(A\cap B)=28+32-50=16$
Thus, the second statement is also true.
Option ‘C’ is correct
Note: Here in the first statement, the information about the set is to be written in the roaster form, and in the second statement, we need to prove the addition theorem on given sets. By using the formula, we can prove that easily.
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 natural number’s set - $N=\{1,2,3...\}$
$Z$- the set of integers - $Z=\{0,\pm 1,\pm 2,\pm 3,...\}$
$Q$ - the set of all rational numbers; $R$ - the set of all real numbers; $C$ - the set of all complex numbers.
If two sets $A,B$ where the set $A$ is said to be the subset of $B$ i.e., $A\subseteq B$ then every element of $A$ is in the set $B$ and the set $A$ is said to be the proper subset of $B$ i.e., \[A\subset B\] then $A\subseteq B$ and $A\ne B$.
Some of the important set operations:
$\begin{align}
& n(A\cup B)=n(A)+n(B)-n(A\cap B) \\
& n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C) \\
\end{align}$
Complete step by step solution: Given that,
${{A}_{n}}$ is the set of first $n$ prime numbers.
So, the set of first $10$ prime numbers is
${{A}_{10}}=\{2,3,5,7,11,13,17,19,23,29\}$
By the definition of a set, we can represent the above set for first $n$ prime numbers by
$\bigcup\limits_{n=2}^{10}{{{A}_{n}}}=\{2,3,5,7,11,13,17,19,23,29\}$
So, the first statement is true.
For the sets $A$ and $B$ we have
$n(A\cup B)=50,n(A)=28,n(B)=32$
Then, from the formula,
$\begin{align}
& n(A\cup B)=n(A)+n(B)-n(A\cap B) \\
& \Rightarrow n(A\cap B)=n(A)+n(B)-n(A\cup B) \\
\end{align}$
On substituting the given values,
$n(A\cap B)=28+32-50=16$
Thus, the second statement is also true.
Option ‘C’ is correct
Note: Here in the first statement, the information about the set is to be written in the roaster form, and in the second statement, we need to prove the addition theorem on given sets. By using the formula, we can prove that easily.
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