Class 12 Maths Chapter 1: Relations and Functions MCQs with answers
10 multiple-choice questions on Relations and Functions, each with its answer and the reason, written from the NCERT chapter. For the CBSE board and JEE Main, which are set from the same NCERT chapter. Open a question's solution to check your working; then take the free test to find your next chapter.
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Answer: (b) 64
A relation from A to B is any subset of A × B. A × B has 2 × 3 = 6 elements, so it has 26 = 64 subsets. 6 counts the pairs, not the subsets; 32 and 8 are 25 and 23.
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Answer: (b) R = {(a, b) : a + b = 11}
The largest sum of two elements of A is 5 + 5 = 10, so no pair has a + b = 11 and R = φ. (a) holds for every pair, so it is the universal relation; (c) contains (5, 1); (d) contains (1, 5) and (5, 1).
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Answer: (b) the universal relation
The squares of the elements are 1, 4, 9, 16, so the largest value of |a2 − b2| is 16 − 1 = 15 < 16. Every pair of A × A is in R, so R = A × A, the universal relation (which is also reflexive and symmetric, so (c) and (d) are false).
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Answer: (b) (3, 3)
R is symmetric but not reflexive, since (3, 3) ∉ R; it is not transitive either, since (3, 1), (1, 3) ∈ R but (3, 3) ∉ R. Adding (3, 3) mends both, and R ∪ {(3, 3)} is reflexive, symmetric and transitive (classes {1, 3} and {2}). Any other pair leaves (3, 3) missing, so the relation is still not reflexive.
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Answer: (b) (a, b) ∈ R and (b, c) ∈ R imply (a, c) ∈ R, for all a, b, c ∈ A
This is the definition: a related to b and b related to c must give a related to c. (a) is symmetry, (d) is reflexivity, and (c) reverses the order of the conclusion.
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Answer: (c) 9
P(X) = {φ, {1}, {2}, {1, 2}}. Pairs with A ⊂ B: φ with all four sets (4), {1} with {1} and {1, 2} (2), {2} with {2} and {1, 2} (2), {1, 2} with itself (1): 4 + 2 + 2 + 1 = 9. Equivalently, each of 1 and 2 is in neither set, in B only, or in both, giving 32 = 9. 16 counts all pairs; 5 leaves out the four pairs (A, A).
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Answer: (c) 64
A × A has 9 pairs. A reflexive relation must contain the 3 pairs (a, a); each of the other 9 − 3 = 6 pairs may be in or out, giving 26 = 64. 512 = 29 counts all relations.
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Answer: (c) 8
The three pairs (1, 1), (2, 2), (3, 3) must be present. Symmetry makes the other six pairs come in three couples {(1, 2), (2, 1)}, {(1, 3), (3, 1)}, {(2, 3), (3, 2)}, each wholly in or wholly out: 23 = 8. 64 counts reflexive relations without the symmetry condition.
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Answer: (c) 2
The classes are the six remainder classes on division by 6. Since 20 = 6 × 3 + 2, the remainders 1 and 2 occur four times ({1, 7, 13, 19}, {2, 8, 14, 20}) and the remainders 3, 4, 5, 0 three times. So 2 classes have 4 elements. 6 is the number of classes; 3 is the size of the others.
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Answer: (c) {1, 2}, {3, 4, 5}, {6}
Equivalence classes are mutually disjoint and their union is A. (c) satisfies both. (b) and (d) are not disjoint (3 is in two sets in (b); 2, 4, 6 repeat in (d)), and (a) leaves out 3.