Set, Relations and FunctionsPYQ June 22Question 1897 of 217
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If the universal set E={x∣x is a positive integer <25}\displaystyle E = \{x \mid x \text{ is a positive integer } < 25\}, A={2,6,8,14,22}\displaystyle A = \{2, 6, 8, 14, 22\}, B={4,8,10,14}\displaystyle B = \{4, 8, 10, 14\}

Options

A(A∩B)c=Ac∪Bc\displaystyle (A \cap B)^c = A^c \cup B^c
B(A∩B)c=Ac∩Bc\displaystyle (A \cap B)^c = A^c \cap B^c
C(A−B)c=ϕ\displaystyle (A - B)^c = \phi
DNone of these
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Correct Answer

✅ Option a — (A∩B)c=Ac∪Bc\displaystyle (A \cap B)^c = A^c \cup B^c

All Options:

  • A(A∩B)c=Ac∪Bc\displaystyle (A \cap B)^c = A^c \cup B^c
  • B(A∩B)c=Ac∩Bc\displaystyle (A \cap B)^c = A^c \cap B^c
  • C(A−B)c=ϕ\displaystyle (A - B)^c = \phi
  • DNone of these

Detailed Solution & Explanation

We are given:
Universal set E={1,2,3,…,24}\displaystyle E = \{1, 2, 3, \dots, 24\}
Subsets A={2,6,8,14,22}\displaystyle A = \{2, 6, 8, 14, 22\} and B={4,8,10,14}\displaystyle B = \{4, 8, 10, 14\}
Let us evaluate Option A: (A∩B)c=Ac∪Bc\displaystyle (A \cap B)^c = A^c \cup B^c.
By **De Morgan's Laws** in set theory, for any two sets A\displaystyle A and B\displaystyle B within a universal set E\displaystyle E, the complement of their intersection is the union of their complements:
(A∩B)c=Ac∪Bc(A \cap B)^c = A^c \cup B^c
Let us verify this explicitly:
- Intersection: A∩B={8,14}\displaystyle A \cap B = \{8, 14\}
- Complement of intersection: (A∩B)c=E−{8,14}\displaystyle (A \cap B)^c = E - \{8, 14\}
- Complement of A\displaystyle A: Ac=E−{2,6,8,14,22}\displaystyle A^c = E - \{2, 6, 8, 14, 22\}
- Complement of B\displaystyle B: Bc=E−{4,8,10,14}\displaystyle B^c = E - \{4, 8, 10, 14\}
- Union of complements: Ac∪Bc=E−(A∩B)=E−{8,14}\displaystyle A^c \cup B^c = E - (A \cap B) = E - \{8, 14\}
Since both sides are identical, De Morgan's Law holds.
Hence, **Option A** is the correct answer.

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