Let the universal set \mathcal{E} = \{x \in \mathbb{Z} : 1 \le x \le 25\}. Two subsets A a… — JAMB UTME

Let the universal set $\mathcal{E} = \{x \in \mathbb{Z} : 1 \le x \le 25\}$ . Two subsets $A$ and $B$ of $\mathcal{E}$ are defined as $A = \{x \in \mathcal{E} : x \text{ is a perfect square}\}$ and $B = \{x \in \mathcal{E} : x \text{ is an odd number}\}$ . What is the cardinality of $(A \cup B)'$ ?

10Cevap
B
12
C
15
D
7

Cevap

The correct answer is 10 because the universal set has 25 elements. The union

A \cup B

contains all 13 odd numbers in the range along with the 2 even perfect squares (

4

and

16

), giving

n(A \cup B) = 15

. Subtracting this from the universal set size gives

25 - 15 = 10

Adım Adım Çözüm

Identify the elements of the universal set and subsets A and B

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

with

n(\mathcal{E}) = 25

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

B = \{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25\}

Explicitly listing elements clarifies overlapping regions.

Find the union set

A \cup B

and its cardinality

A \cup B = \{1, 3, 4, 5, 7, 9, 11, 13, 15, 16, 17, 19, 21, 23, 25\}

, so

n(A \cup B) = 15

The union includes all odd numbers from 1 to 25 plus the even perfect squares (4 and 16).

Calculate the cardinality of the complement

(A \cup B)'

n((A \cup B)') = n(\mathcal{E}) - n(A \cup B) = 25 - 15 = 10

The complement set contains all elements in

\mathcal{E}

that are neither odd nor perfect squares, which are the 10 non-square even numbers:

\{2, 6, 8, 10, 12, 14, 18, 20, 22, 24\}

Anahtar Kavram

Set Complements and De Morgan's Laws / Set Operations

Tahmini Süre:1m 30s

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