Answer
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Hint: First, before proceeding for this, we will be calculating terms for X and Y for the values of natural number 1, 2, 3 and so on to comment on the given question answer. Then, calculate all the values for set X and Y to get a relation between them. Then, we can clearly observe from the two sets values that X is a subset of Y where subset means that X has elements which are present in Y compulsorily, which gives the final result.
Complete step-by-step answer:
In this question, we are supposed to find the value of $ X\cup Y $ when we are given with two sets defined as $ X=\left\{ {{4}^{n}}-3n-1:n\in N \right\} $ and $ Y=\left\{ 9\left( n-1 \right):n\in N \right\} $ where N is the set of natural numbers.
So, before proceeding for this, we will be calculating terms for X and Y for the values of natural numbers 1, 2, 3 and so on to comment on the given question answer.
So, by substituting the value of n as 1 in X as:
$ \begin{align}
& {{4}^{1}}-3\left( 1 \right)-1=4-3-1 \\
& \Rightarrow 0 \\
\end{align} $
So, we get the first value of the set X as 0.
Similarly, by substituting the value of n as 2 in X as:
$ \begin{align}
& {{4}^{2}}-3\left( 2 \right)-1=16-6-1 \\
& \Rightarrow 9 \\
\end{align} $
So, we get the second value of the set X as 9.
Similarly, by substituting the value of n as 3 in X as:
$ \begin{align}
& {{4}^{3}}-3\left( 3 \right)-1=64-9-1 \\
& \Rightarrow 54 \\
\end{align} $
So, we get the third value of the set X as 54.
Then, we get the set X as {0, 9, 54, ....}.
Now, by substituting the value of n as 1 in Y as:
$ \begin{align}
& 9\left( 1-1 \right)=9\left( 0 \right) \\
& \Rightarrow 0 \\
\end{align} $
So, we get the first value of the set Y as 0.
Similarly, by substituting the value of n as 2 in Y as:
$ \begin{align}
& 9\left( 2-1 \right)=9\left( 1 \right) \\
& \Rightarrow 9 \\
\end{align} $
So, we get the second value of the set Y as 9.
Similarly, by substituting the value of n as 3 in Y as:
$ \begin{align}
& 9\left( 3-1 \right)=9\left( 2 \right) \\
& \Rightarrow 18 \\
\end{align} $
So, we get the third value of the set Y as 18.
Similarly, by substituting the value of n as 4 in Y as:
$ \begin{align}
& 9\left( 4-1 \right)=9\left( 3 \right) \\
& \Rightarrow 27 \\
\end{align} $
So, we get the fourth value of the set Y as 27.
Similarly, by substituting the value of n as 5 in Y as:
$ \begin{align}
& 9\left( 5-1 \right)=9\left( 4 \right) \\
& \Rightarrow 36 \\
\end{align} $
So, we get the fifth value of the set Y as 36.
Similarly, by substituting the value of n as 6 in Y as:
$ \begin{align}
& 9\left( 6-1 \right)=9\left( 5 \right) \\
& \Rightarrow 45 \\
\end{align} $
So, we get the sixth value of the set Y as 45.
Similarly, by substituting the value of n as 7 in Y as:
$ \begin{align}
& 9\left( 7-1 \right)=9\left( 6 \right) \\
& \Rightarrow 54 \\
\end{align} $
So, we get the seventh value of the set Y as 54.
Then, we get the set Y as {0, 9, 18, 27, 36, 45, 54, ....}.
So, we can clearly observe from the two sets values that X is a subset of Y where the subset means that X has elements which are present in Y compulsorily.
Now, we need to find the value of union of X and Y which means all the values which are commonly taken as once and remaining values are also considered.
Then, we get $ X\subset Y $ because of it, we get:
$ X\cup Y=Y $
So, the correct answer is “Option (d)”.
Note: Now, to solve these types of the questions we need to know some of the basic rules of the sets which is a series of values defined by some relation. So, the basic rule is that if X is the subset of Y( $ X\subset Y $ ), then $ X\cup Y=Y $ is the result.
Complete step-by-step answer:
In this question, we are supposed to find the value of $ X\cup Y $ when we are given with two sets defined as $ X=\left\{ {{4}^{n}}-3n-1:n\in N \right\} $ and $ Y=\left\{ 9\left( n-1 \right):n\in N \right\} $ where N is the set of natural numbers.
So, before proceeding for this, we will be calculating terms for X and Y for the values of natural numbers 1, 2, 3 and so on to comment on the given question answer.
So, by substituting the value of n as 1 in X as:
$ \begin{align}
& {{4}^{1}}-3\left( 1 \right)-1=4-3-1 \\
& \Rightarrow 0 \\
\end{align} $
So, we get the first value of the set X as 0.
Similarly, by substituting the value of n as 2 in X as:
$ \begin{align}
& {{4}^{2}}-3\left( 2 \right)-1=16-6-1 \\
& \Rightarrow 9 \\
\end{align} $
So, we get the second value of the set X as 9.
Similarly, by substituting the value of n as 3 in X as:
$ \begin{align}
& {{4}^{3}}-3\left( 3 \right)-1=64-9-1 \\
& \Rightarrow 54 \\
\end{align} $
So, we get the third value of the set X as 54.
Then, we get the set X as {0, 9, 54, ....}.
Now, by substituting the value of n as 1 in Y as:
$ \begin{align}
& 9\left( 1-1 \right)=9\left( 0 \right) \\
& \Rightarrow 0 \\
\end{align} $
So, we get the first value of the set Y as 0.
Similarly, by substituting the value of n as 2 in Y as:
$ \begin{align}
& 9\left( 2-1 \right)=9\left( 1 \right) \\
& \Rightarrow 9 \\
\end{align} $
So, we get the second value of the set Y as 9.
Similarly, by substituting the value of n as 3 in Y as:
$ \begin{align}
& 9\left( 3-1 \right)=9\left( 2 \right) \\
& \Rightarrow 18 \\
\end{align} $
So, we get the third value of the set Y as 18.
Similarly, by substituting the value of n as 4 in Y as:
$ \begin{align}
& 9\left( 4-1 \right)=9\left( 3 \right) \\
& \Rightarrow 27 \\
\end{align} $
So, we get the fourth value of the set Y as 27.
Similarly, by substituting the value of n as 5 in Y as:
$ \begin{align}
& 9\left( 5-1 \right)=9\left( 4 \right) \\
& \Rightarrow 36 \\
\end{align} $
So, we get the fifth value of the set Y as 36.
Similarly, by substituting the value of n as 6 in Y as:
$ \begin{align}
& 9\left( 6-1 \right)=9\left( 5 \right) \\
& \Rightarrow 45 \\
\end{align} $
So, we get the sixth value of the set Y as 45.
Similarly, by substituting the value of n as 7 in Y as:
$ \begin{align}
& 9\left( 7-1 \right)=9\left( 6 \right) \\
& \Rightarrow 54 \\
\end{align} $
So, we get the seventh value of the set Y as 54.
Then, we get the set Y as {0, 9, 18, 27, 36, 45, 54, ....}.
So, we can clearly observe from the two sets values that X is a subset of Y where the subset means that X has elements which are present in Y compulsorily.
Now, we need to find the value of union of X and Y which means all the values which are commonly taken as once and remaining values are also considered.
Then, we get $ X\subset Y $ because of it, we get:
$ X\cup Y=Y $
So, the correct answer is “Option (d)”.
Note: Now, to solve these types of the questions we need to know some of the basic rules of the sets which is a series of values defined by some relation. So, the basic rule is that if X is the subset of Y( $ X\subset Y $ ), then $ X\cup Y=Y $ is the result.
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