Question

. Let A = {x : x^2 - 5x + 6 = 0}, B={2 ,4}, C = {4, 5}. Write A X (B ∩ C)

Answer 1

Answer :

A×(B∩C) = { (2,4) , (3,4) }

Note :

★ Set : A well defined collection of distinct objects is called a set .

★ Cardinal number / Cardinality : The number of elements/members/objects in a finite set is called cardinal no. / Cardinality .

→ Cardinal no. of a finite set A is denoted by n(A) .

★ Method of representing a set :

a). Roster / Tabular / Listed form

b). Set Builder form

★ Roster form :

→ All elements are listed .

→ Elements are separated by commas .

→ Elements are enclosed within braces { } .

→ The order of writing elements doesn't matter .

→ The elements are not repeated

★ Set builder form :

→ The common properties of elements are written .

→ The elements is described using symbols like x , y , z (mostly x) .

→ Whole description of the elements are enclosed within braces { } .

★ Union of two sets : The union of two sets A and B is the set of all those elements which are either in A or in B or in both .

→ This set is denoted by A U B .

★ Intersection of two sets : The intersection of two sets A and B is the set of all those elements which are in common in both A and B .

→ This set is denoted by A ∩ B .

★ Difference of sets : The difference of two sets A and B in the order ( also called relative complement of B in A ) is the set of all those elements of A which are not the elements of B .

→ It is denoted by (A - B) .

★ Cartesian Product : If A and B are any two non empty sets then the set of all ordered pairs (a,b) such that a € A and b € B is called the cartesian product of sets a with set B and it is denoted by A×B .

→ A×B = { (a,b) : a € A and b € B } .

Solution :

→ Given : A = { x : x² - 5x + 6 = 0 }

B = { 2 , 4 }

C = { 4 , 5 }

→ To find : A×(B∩C)

Firstly ,

Let's find the elements of set A .

We have ;

=> x² - 5x + 6 = 0

=> x² - 2x - 3x + 6 = 0

=> x(x - 2) - 3(x - 2) = 0

=> (x - 2)(x - 3) = 0

=> x = 2 , 3

Hence , A = { 2 , 3 }

Now ,

Let's find (B∩C) .

=> B∩C = { 2 , 4 } ∩ { 4 , 5 }

=> B∩C = { 4 }

Now ,

=> A×(B∩C) = { 2 , 3 } × { 4 }

=> A×(B∩C) = { (2,4) , (3,4) }

Hence ,

A×(B∩C) = { (2,4) , (3,4) }