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Sets and Functions - Venn Diagrams and Operations on Sets

Grade 11ICSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

A Set is a well-defined collection of distinct objects. The Universal Set (UU) represents the totality of all elements under consideration for a specific problem. Visually, UU is depicted as a large rectangle which acts as the boundary for all other sets.

The Union of two sets AA and BB, denoted as ABA \cup B, is the set containing all elements that belong to AA, or BB, or both. In a Venn diagram, this is represented by shading the entire region covered by the two overlapping circles AA and BB.

The Intersection of two sets AA and BB, denoted as ABA \cap B, consists of elements that are common to both sets. Visually, this is the region where the circles for AA and BB overlap. If the circles do not overlap, the sets are disjoint and the intersection is the empty set \emptyset.

The Complement of a set AA, written as AA' or AcA^c, includes all elements in the universal set UU that are not in AA. Visually, this corresponds to the entire area inside the universal rectangle excluding the interior of circle AA.

The Difference of sets ABA - B (also called the relative complement) consists of elements that belong to AA but not to BB. On a Venn diagram, this is shown as the 'crescent' part of circle AA that does not share any area with circle BB.

The Symmetric Difference of sets AA and BB, denoted as AΔBA \Delta B, is the set of elements belonging to either AA or BB but not both. Visually, it is the union of the two circles excluding their shared overlapping intersection region.

Disjoint Sets are sets that have no elements in common, meaning AB=A \cap B = \emptyset. Visually, they are represented as two separate circles within the universal rectangle that do not touch or intersect.

A Subset relationship, ABA \subseteq B, occurs when every element of set AA is also an element of set BB. Visually, circle AA is drawn entirely inside the boundary of circle BB.

📐Formulae

n(AB)=n(A)+n(B)n(AB)n(A \cup B) = n(A) + n(B) - n(A \cap B)

n(ABC)=n(A)+n(B)+n(C)n(AB)n(BC)n(CA)+n(ABC)n(A \cup B \cup C) = n(A) + n(B) + n(C) - n(A \cap B) - n(B \cap C) - n(C \cap A) + n(A \cap B \cap C)

De Morgan's First Law: (AB)=AB(A \cup B)' = A' \cap B'

De Morgan's Second Law: (AB)=AB(A \cap B)' = A' \cup B'

n(AB)=n(A)n(AB)n(A - B) = n(A) - n(A \cap B)

n(AΔB)=n(AB)+n(BA)=n(AB)n(AB)n(A \Delta B) = n(A - B) + n(B - A) = n(A \cup B) - n(A \cap B)

n(A)=n(U)n(A)n(A') = n(U) - n(A)

💡Examples

Problem 1:

In a class of 50 students, 30 study Mathematics, 25 study Physics, and 10 study both subjects. Find the number of students who study: (i) either Mathematics or Physics, and (ii) neither of the two subjects.

Solution:

Let MM be the set of students studying Mathematics and PP be the set of students studying Physics. Given: n(U)=50n(U) = 50, n(M)=30n(M) = 30, n(P)=25n(P) = 25, and n(MP)=10n(M \cap P) = 10. \n(i) To find those studying either subject, we find the union: \nn(MP)=n(M)+n(P)n(MP)n(M \cup P) = n(M) + n(P) - n(M \cap P) \nn(MP)=30+2510=45n(M \cup P) = 30 + 25 - 10 = 45. \n(ii) To find those studying neither, we find the complement of the union: \nn((MP))=n(U)n(MP)n((M \cup P)') = n(U) - n(M \cup P) \nn((MP))=5045=5n((M \cup P)') = 50 - 45 = 5.

Explanation:

We use the Principle of Inclusion-Exclusion to find the number of students in at least one set, then subtract from the total universal set to find those outside both sets.

Problem 2:

If U={x:xN,x10}U = \{x : x \in \mathbb{N}, x \leq 10\}, A={1,3,5,7,9}A = \{1, 3, 5, 7, 9\}, and B={2,3,5,7}B = \{2, 3, 5, 7\}, verify De Morgan's First Law: (AB)=AB(A \cup B)' = A' \cap B'.

Solution:

Step 1: Find ABA \cup B. \nAB={1,2,3,5,7,9}A \cup B = \{1, 2, 3, 5, 7, 9\}. \nStep 2: Find the LHS (AB)(A \cup B)' relative to U={1,2,3,4,5,6,7,8,9,10}U = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}. \n(AB)={4,6,8,10}(A \cup B)' = \{4, 6, 8, 10\}. \nStep 3: Find AA' and BB'. \nA={2,4,6,8,10}A' = \{2, 4, 6, 8, 10\} \nB={1,4,6,8,9,10}B' = \{1, 4, 6, 8, 9, 10\}. \nStep 4: Find the RHS ABA' \cap B'. \nAB={4,6,8,10}A' \cap B' = \{4, 6, 8, 10\}. \nSince LHS = RHS, the law is verified.

Explanation:

This demonstrates De Morgan's Law by calculating the complement of a union and showing it equals the intersection of the individual complements.