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Misc 6 (MCQ) - Chapter 1 Class 12 Relation and Functions
Last updated at Dec. 16, 2024 by Teachoo
Finding number of relations
Chapter 1 Class 12 Relation and Functions
Concept wise
- Relations - Definition
- Empty and Universal Relation
- To prove relation reflexive, transitive, symmetric and equivalent
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Finding number of relations
- Function - Definition
- To prove one-one & onto (injective, surjective, bijective)
- Composite functions
- Composite functions and one-one onto
- Finding Inverse
- Inverse of function: Proof questions
- Binary Operations - Definition
- Whether binary commutative/associative or not
- Binary operations: Identity element
- Binary operations: Inverse
Transcript
Misc 6 Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are reflexive and symmetric but not transitive is (A) 1 (B) 2 (C) 3 (D) 4 Total possible pairs = { (1, 1) , (1, 2), (1, 3), (2, 1) , (2, 2), (2, 3), (3, 1) , (3, 2), (3, 3) } Total possible pairs = { (1, 1) , (1, 2), (1, 3), (2, 1) , (2, 2), (2, 3), (3, 1) , (3, 2), (3, 3) } Reflexive means (a, a) should be in relation . So, (1, 1) , (2, 2) , (3, 3) should be in a relation Symmetric means if (a, b) is in relation, then (b, a) should be in relation . So, since (1, 2) is in relation, (2, 1) should be in relation & since (1, 3) is in relation, (3, 1) should be in relation Transitive means if (a, b) is in relation, & (b, c) is in relation, then (a, c) is in relation We need relation which is not transitive. So, we cannot add any more pair in the relation. If we add (2, 3), we need to add (3, 2) for symmetric, but it would become transitive then Relation R1 = { So, there is only 1 possible relation Correct answer is A.
