
Decide among the following sets, which sets are subset of one and another
\[
A = \{ x:x \in R\,and\,x\,satisfy\,{x^2} - 8x + 12 = 0\} \\
B = \{ 2,4,6\} \\
C = \{ 2,4,6,8...\} \\
D = \{ 6\} \\
\]
Answer
593.4k+ views
Hint:
Let us first find all the elements of all the given sets and then using the definition of subset, as if the first set’s elements lie completely in another set’s element then the first set is a subset of another set. So finding the element of each set so we can easily compare which set is a subset of another set.
Complete step by step solution:
Let the given sets in set builder form and rosters form be
\[
A = \{ x:x \in R\,and\,x\,satisfy\,{x^2} - 8x + 12 = 0\} \\
B = \{ 2,4,6\} \\
C = \{ 2,4,6,8...\} \\
D = \{ 6\} \\
\]
First, calculating the elements of set A which are roots for the given equation of \[{x^2} - 8x + 12 = 0\] .
So, finding the root for above quadratic using factorization method as
\[ \Rightarrow {x^2} - 6x - 2x + ( - 6)( - 2) = 0\]
Now, taking common from the above terms as,
\[ \Rightarrow x(x - 6) - 2(x - 6) = 0\]
Again taking common from above terms as
\[ \Rightarrow (x - 6)(x - 2) = 0\]
Hence, the value of x is \[x = 2,6\]
Hence, set A can be given as \[A = \{ 2,6\} \]
The given set B has three element as \[B = \{ 2,4,6\} \]
The third set has all the even terms in it as \[C = \{ 2,4,6,8...\} \]
While set D is singleton set given as \[D = \{ 6\} \]
Now, we have to see that which set has all its element lying in the another set thus we have to find all the possible subsets as,
For set A,
\[A \subset B,A \subset C\]
For set B,
\[B \subset C\]
Set C is not subset of any set,
And for set D,
\[D \subset A,D \subset B,D \subset C\]
Hence, above all the possible subsets are shown.
Note:
Roster or tabular form: In roster form, all the elements of a set are listed, the elements are being separated by commas and are enclosed within braces \[\{ \,\} \].
Set-builder form: In the set builder form, all the elements of the set must possess a single property to become the member of that set.
In mathematics, a set A is a subset of a set B if all elements of A are also elements of B; B is then a superset of A. It is possible for A and B to be equal; if they are unequal, then A is a proper subset of B.
Let us first find all the elements of all the given sets and then using the definition of subset, as if the first set’s elements lie completely in another set’s element then the first set is a subset of another set. So finding the element of each set so we can easily compare which set is a subset of another set.
Complete step by step solution:
Let the given sets in set builder form and rosters form be
\[
A = \{ x:x \in R\,and\,x\,satisfy\,{x^2} - 8x + 12 = 0\} \\
B = \{ 2,4,6\} \\
C = \{ 2,4,6,8...\} \\
D = \{ 6\} \\
\]
First, calculating the elements of set A which are roots for the given equation of \[{x^2} - 8x + 12 = 0\] .
So, finding the root for above quadratic using factorization method as
\[ \Rightarrow {x^2} - 6x - 2x + ( - 6)( - 2) = 0\]
Now, taking common from the above terms as,
\[ \Rightarrow x(x - 6) - 2(x - 6) = 0\]
Again taking common from above terms as
\[ \Rightarrow (x - 6)(x - 2) = 0\]
Hence, the value of x is \[x = 2,6\]
Hence, set A can be given as \[A = \{ 2,6\} \]
The given set B has three element as \[B = \{ 2,4,6\} \]
The third set has all the even terms in it as \[C = \{ 2,4,6,8...\} \]
While set D is singleton set given as \[D = \{ 6\} \]
Now, we have to see that which set has all its element lying in the another set thus we have to find all the possible subsets as,
For set A,
\[A \subset B,A \subset C\]
For set B,
\[B \subset C\]
Set C is not subset of any set,
And for set D,
\[D \subset A,D \subset B,D \subset C\]
Hence, above all the possible subsets are shown.
Note:
Roster or tabular form: In roster form, all the elements of a set are listed, the elements are being separated by commas and are enclosed within braces \[\{ \,\} \].
Set-builder form: In the set builder form, all the elements of the set must possess a single property to become the member of that set.
In mathematics, a set A is a subset of a set B if all elements of A are also elements of B; B is then a superset of A. It is possible for A and B to be equal; if they are unequal, then A is a proper subset of B.
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