Let the universal set be \mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 25\}. If A = \{x \… — JAMB UTME

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 25\}$ . If $A = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\}$ and $B = \{x \in \mathcal{U} : x \text{ is a perfect square}\}$ , what is the cardinality of $(A \cup B)'$ ?

13Cevap
B
12
C
14
D
17

Cevap

The total number of elements in the universal set is 25. Set A has 8 elements (multiples of 3) and set B has 5 elements (perfect squares). The element 9 belongs to both sets. Therefore, the union of A and B contains 8 + 5 - 1 = 12 elements. Subtracting this from the universal set size gives 25 - 12 = 13 elements in the complement.

Adım Adım Çözüm

Determine the elements of the universal set and its cardinality.

\mathcal{U} = \{1, 2, 3, \dots, 25\}

, so

n(\mathcal{U}) = 25$.

The universal set contains all integers from 1 to 25 inclusive.

List the elements of set A and set B.

A = \{3, 6, 9, 12, 15, 18, 21, 24\}

(so

n(A) = 8

) and

B = \{1, 4, 9, 16, 25\}

(so

n(B) = 5

Set A contains multiples of 3 within the domain, and set B contains perfect squares within the domain.

Find the intersection

A \cap B

and calculate the cardinality of the union

A \cup B

A \cap B = \{9\}

, so

n(A \cap B) = 1

. Thus,

n(A \cup B) = n(A) + n(B) - n(A \cap B) = 8 + 5 - 1 = 12

The principle of inclusion-exclusion avoids double-counting the common element 9.

Calculate the cardinality of the complement

(A \cup B)'

n((A \cup B)') = n(\mathcal{U}) - n(A \cup B) = 25 - 12 = 13

The complement consists of all elements in the universal set that do not belong to the union of A and B.

Anahtar Kavram

Complement of Set Union and Inclusion-Exclusion Principle

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