Chapter 1: Relations and Functions — Online MCQ Test
MATHS · CLASS 12 SECOND PUC · Karnataka State Board
Practice Chapter 1: Relations and Functions with a free chapter-wise online MCQ test.
This chapter covers: Chapter 1: Relations and Functions CBSC.
AI-generated questions from basic to board-exam level, with instant results and explanations.
Chapter 1: Relations and Functions — Important Questions & Answers
A relation R from set A to set B is defined as:
- A. A subset of A × B
- B. A subset of A ∪ B
- C. A subset of A ∩ B
- D. Equal to A × B
Answer: A. A subset of A × B
By definition in NCERT, a relation from set A to set B is any subset of the Cartesian product A × B.
By definition in NCERT, a relation from set A to set B is any subset of the Cartesian product A × B.
If A = {1, 2, 3} and B = {4, 5}, then the number of elements in A × B is:
- A. 5
- B. 6
- C. 8
- D. 10
Answer: B. 6
The Cartesian product A × B has |A| × |B| = 3 × 2 = 6 elements.
The Cartesian product A × B has |A| × |B| = 3 × 2 = 6 elements.
A function f: A → B is defined as a relation where:
- A. Every element of A has at least one image in B
- B. Every element of A has exactly one image in B
- C. Every element of B has a pre-image in A
- D. Some elements of A may not have images in B
Answer: B. Every element of A has exactly one image in B
A function requires that each element in the domain has exactly one corresponding element in the codomain.
A function requires that each element in the domain has exactly one corresponding element in the codomain.
For the relation R = {(1, 2), (2, 3), (1, 3)} on A = {1, 2, 3}, which property does R satisfy?
- A. Reflexive
- B. Symmetric
- C. Transitive
- D. Equivalence
Answer: C. Transitive
R is transitive because (1, 2) ∈ R and (2, 3) ∈ R implies (1, 3) ∈ R. It's not reflexive or symmetric.
R is transitive because (1, 2) ∈ R and (2, 3) ∈ R implies (1, 3) ∈ R. It's not reflexive or symmetric.
If f: A → B and g: B → C are functions, for (g ∘ f)⁻¹ to exist, which condition must hold?
- A. Both f and g must be injective
- B. Both f and g must be surjective
- C. Both f and g must be bijective
- D. f must be injective and g must be surjective
Answer: C. Both f and g must be bijective
For the composition to have an inverse, both functions must be bijective so that (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹ exists.
For the composition to have an inverse, both functions must be bijective so that (g ∘ f)⁻¹ = f⁻¹ ∘ g⁻¹ exists.