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

Question

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

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Katherine 3 weeks 2021-09-09T16:16:34+00:00 1 Answer 0 views 0

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    2021-09-09T16:18:24+00:00

    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) }

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