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**Hint**: To discuss the required result. We first convert or write all given set in Roster form if any of one is given in set builder form and then after that we see that if all elements of one set is contained in another set then the set will be a sub set.

**:**

__Complete step-by-step answer__To find the required solution of the given problem. We first write given set in Roster form which is as follows:

For option (A) we have:

$ A = \left\{ {x:x \in R\,\,and\,\,satisfy\,\,{x^2} - 8x + 12 = 0} \right\} $

Clearly the above set is not in a Roster form. Therefore, we will solve the quadratic equation to find its roots.

We can find roots of given quadratic by middle term splitting method:

We have,

\[

{x^2} - 8x + 12 = 0 \\

\Rightarrow {x^2} - \left( {6 + 2} \right)x + 12 = 0 \\

\Rightarrow {x^2} - 6x - 2x + 12 = 0 \\

\Rightarrow x\left( {x - 6} \right) - 2\left( {x - 6} \right) = 0 \\

\Rightarrow \left( {x - 6} \right)\left( {x - 2} \right) = 0 \\

x - 6 = 0\,\,or\,\,x - 2 = 0 \\

x = 6\,\,\,or\,\,\,x = 2 \\

\]

Therefore, using above we have

\[{\text{A = \{ 2,6\} }}\]

And all other sets are already given in Roster form.

Therefore, from above we see elements in $ A = \left\{ {2,6} \right\} $ and $ B = \left\{ {2,4,6} \right\} $ .

Since all elements of set A belong to set B.

Therefore, we can say that set ‘A’ is a subset of set ‘B’ or we write it as $ A \subseteq C $ .

Also, from above we see that elements of ‘A’ and ‘C’ are given as: $ A = \left\{ {2,6} \right\} $ and $ C = \left\{ {2,4,6,8.......} \right\} $ .

Clearly all elements of set ‘A’ belong to set ‘C’. Hence, we can say that set ‘A’ is a subset of set ‘C’ and we write it as: $ A \subseteq C $

Also, elements of set ‘B’ and set ‘C’ are given as \[{\text{B = \{ 2,4,6\} }}\]and \[{\text{C = \{ 2,4,6,8}}...{\text{\} }}\].

Clearly we see that all elements of set B belongs to set C. Hence we can say that set ‘B’ is a subset of ‘C’ and we write it as: $ B \subseteq C $

Also, from above we see that elements of set ‘D’ and elements of set ‘A’ are given as $ D = \left\{ 6 \right\} $ and $ A = \left\{ {2,6} \right\} $ .

Clearly we see that all elements of set ‘D’ belong to set ‘A’. Hence, we can say that set ‘D’ is a subset of set ‘A’ and we write it as: $ D \subseteq A $

Also, from above we see that element of set ‘D’ and set ‘B’ are given as $ D = \left\{ 6 \right\} $ and $ B = \left\{ {2,4,6} \right\} $ .

Clearly we see that all elements of set ‘D’ belong to set ‘B’. Hence, we can say that set ‘D’ is a subset of set ‘B’ and we write it as: $ D \subseteq B $

Also, from above we see that elements of set ‘D’ and set ‘C’ are given as $ D = \left\{ 6 \right\} $ and $ C = \left\{ {2,4,6,.....} \right\} $ .

Clearly we see that all elements of set ‘D’ belong to set ‘C’. Hence, we can say that set ‘D’ is a subset of set ‘C’ and we write it as: $ D \subseteq C $

Hence, from above we see that set ‘D’ is a subset of set ‘A’, set ‘B’ and set ‘C’. Also, set ‘A’ is a subset of set ‘B’ and set ‘C’ and set ‘B’ is a subset of set ‘C’.

**Note**: One set is called a subset if all its elements belong to another set, if even only one element doesn't belong to another, then it will not be considered as a subset. Hence, to discuss whether a set is a subset or not. First express given sets in Roster form if sets are given in Builder form as doing problems in builder form students may make mistakes in considering elements of the given set.

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Decide, among the following sets, which sets are subsets of one and another.

$

A = \left\{ {x:x \in R\,\,and\,\,x\,\,satisfy\,\,{x^2} - 8x + 12 = 0} \right\} \\

B = \left\{ {2,4,6} \right\} \\

C = \left\{ {2,4,6,8,......} \right\} \\

D = \left\{ 6 \right\} \\

$

$

A = \left\{ {x:x \in R\,\,and\,\,x\,\,satisfy\,\,{x^2} - 8x + 12 = 0} \right\} \\

B = \left\{ {2,4,6} \right\} \\

C = \left\{ {2,4,6,8,......} \right\} \\

D = \left\{ 6 \right\} \\

$

Class 11 MATHS Miscellaneous (Question - 1) | Sets Class 11 Chapter 1| NCERT | Ratan Kalra Sir

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