Chapter 2 - Relations and Functions — Online MCQ Test
MATHS · CLASS 11 FIRST PUC · Karnataka State Board
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This chapter covers: Relations and Functions.
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Chapter 2 - Relations and Functions — Important Questions & Answers
What is a relation in mathematics?
- A. A subset of the Cartesian product of two sets
- B. A function that maps every element to itself
- C. A set of ordered pairs with no specific rule
- D. An equivalence between two algebraic expressions
Answer: A. A subset of the Cartesian product of two sets
A relation is formally defined as a subset of the Cartesian product A × B, where A and B are two sets.
A relation is formally defined as a subset of the Cartesian product A × B, where A and B are two sets.
If A = {1, 2, 3} and B = {a, b}, how many elements are in A × B?
- A. 3
- B. 5
- C. 6
- D. 2
Answer: C. 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.
If f(x) = x² and g(x) = x + 1, what is (f ∘ g)(2)?
- A. 5
- B. 9
- C. 8
- D. 7
Answer: B. 9
(f ∘ g)(2) = f(g(2)) = f(3) = 3² = 9. First apply g, then apply f to the result.
(f ∘ g)(2) = f(g(2)) = f(3) = 3² = 9. First apply g, then apply f to the result.
A relation that is reflexive, antisymmetric, and transitive is called a _____.
- A. Equivalence relation
- B. Partial order relation
- C. Congruence relation
- D. Total relation
Answer: B. Partial order relation
A partial order (or partial ordering) is a relation that is reflexive, antisymmetric, and transitive.
A partial order (or partial ordering) is a relation that is reflexive, antisymmetric, and transitive.
Consider the function f: Z → Z defined by f(n) = n/2 if n is even, and f(n) = (n+1)/2 if n is odd. Is f surjective?
- A. Yes, f is surjective because every integer has a pre-image
- B. No, because f is not defined for all integers
- C. Yes, f is surjective because it covers all even integers
- D. No, because negative integers cannot be in the range
Answer: A. Yes, f is surjective because every integer has a pre-image
For any integer m in Z, we can find an integer n such that f(n) = m. For example, if m ≥ 0, take n = 2m (even), then f(2m) = m. Thus f is onto (surjective).
For any integer m in Z, we can find an integer n such that f(n) = m. For example, if m ≥ 0, take n = 2m (even), then f(2m) = m. Thus f is onto (surjective).