Question

Universal set \[U = \left\{ {x:{x^5} - 6{x^4} + 11{x^3} - 6{x^2} = 0} \right\}\] , \[A = \left\{ {x:{x^2} - 5x + 6 = 0} \right\}\] and \[B = \left\{ {x:{x^2} - 3x + 2 = 0} \right\}\]. Then \[(A \cap B)'\] is equal to
A.\[\left\{ {1,3} \right\}\]
B.\[\left\{ {1,2,3} \right\}\]
C.\[\left\{ {0,1,3} \right\}\]
D.\[\left\{ {0,1,2,3} \right\}\]

Answer 1

Hint: The question is related to set theory. The given sets are of set builder form, so first we convert them into roster form. Then by knowing the definition of intersection and complement of a set we are going to simplify the given question. Then we are going to choose the appropriate option.

Complete answer:
Universal set : A universal set is a set which contains all the elements or objects of other sets, including its own elements. It is usually denoted by the symbol ‘U’.
Union of sets: If set A and set B are two sets, then A union B is the set that contains all the elements of set A and set B. It is denoted as \[A \cup B\].
Intersection of sets: If set A and set B are two sets, then A intersection B is the set that contains only the common elements between set A and set B. It is denoted as \[A \cap B\].
Complement of a set: The complement of any set, say A , is the set of all elements in the universal set that are not in set A. It is denoted by \[A'\].
On considering the given question
\[U = \left\{ {x:{x^5} - 6{x^4} + 11{x^3} - 6{x^2} = 0} \right\}\]
This is in the set builder form, we convert it into roster form.
Given condition \[{x^5} - 6{x^4} + 11{x^3} - 6{x^2} = 0\]
On taking \[{x^2}\]common we get ,
\[ \Rightarrow {x^2}({x^3} - 6{x^2} + 11x - 6) = 0\]
On factorization we get ,
\[\begin{array}{*{20}{c}}
  3 \\
  {} \\
  {}
\end{array}\left| \!{\overline {\,

  1\,\,\,\, - 6\,\,\,\,\,11\,\, - 6 \\
  \underline {0\,\,\,\,\,\,\,3\,\,\, - 9\,\,\,\,\,\,6} \\
  1\,\,\, - 3\,\,\,\,\,\,\,\,2\,\,\,\,\,\,0 \\
  \,}} \right. \]
\[ \Rightarrow {x^2}(x - 3)({x^2} - 3x + 2) = 0\]
On simplification
\[ \Rightarrow {x^2}(x - 3)({x^2} - 2x - x + 2) = 0\]
\[ \Rightarrow {x^2}(x - 3)(x(x - 2) - 1(x - 2)) = 0\]
\[ \Rightarrow {x^2}(x - 3)(x - 2)(x - 1) = 0\]
On equating each term to zero we get the values of x are 0, 1, 2, 3
Therefore \[U = \left\{ {0,1,2,3} \right\}\]
Similarly we have \[A = \left\{ {x:{x^2} - 5x + 6 = 0} \right\}\]
given condition \[{x^2} - 5x + 6 = 0\]
On factorization we get ,
\[ \Rightarrow {x^2} - 3x - 2x + 6 = 0\]
\[ \Rightarrow x(x - 3) - 2(x - 3) = 0\]
\[ \Rightarrow (x - 3)(x - 2) = 0\]
On equating each term to zero we get the values of x are 2 and 3
Therefore \[A = \left\{ {2,3} \right\}\]
Similarly we have \[B = \left\{ {x:{x^2} - 3x + 2 = 0} \right\}\]
Given condition \[{x^2} - 3x + 2 = 0\]
On factorization we get ,
\[ \Rightarrow {x^2} - 2x - x + 2 = 0\]
\[ \Rightarrow x(x - 2) - 1(x - 2) = 0\]
\[ \Rightarrow (x - 1)(x - 2) = 0\]
On equating each term to zero we get the values of x are 1 and 2
Therefore \[B = \left\{ {1,2} \right\}\]
Therefore we get \[A \cap B = \left\{ 2 \right\}\]
Hence \[(A \cap B)' = \left\{ {0,1,3} \right\}\]
Therefore option (3) is the correct answer.

Note:
This question can also be solved by Demorgan's law where it is stated as \[\left( {A \cap B} \right)' = A' \cup B'\](for intersection). So first we determine the complement set of both A and B. The set will be \[A' = \{ 0,1\} \] and \[B' = \{ 0,3\} \]. On combining these sets we get \[\left( {A \cap B} \right)' = \{ 0,1,3\} \].