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SubjectsMathematics
संबंध
In this Class 11 Mathematics topic, students learn how a relation connects elements of one set with elements of another, building the foundation for the chapter Relations and Functions. The topic introduces ordered pairs, Cartesian products, and the representation of relations as subsets of a Cartesian product. Students also examine the domain, codomain, and range of a relation and learn to interpret relations through rosters, diagrams, and set notation, preparing them to distinguish relations from functions.
Correct answer: C. (4)
Explanation: (R={(1,4),(2,3),(3,2),(4,1)}), so the count is (4). In exams, remember that order matters in ordered pairs.
Correct answer: B. Equivalence relation
Explanation: This is the equality relation. Since a=a, it is reflexive. If aRb, then a=b, so b=a and bRa; hence it is symmetric. Also, aRb and bRc imply a=c, so it is transitive. Therefore, R is an equivalence relation. Exam tip: check reflexive, symmetric and transitive properties together.
Correct answer: A. Reflexive but not symmetric
Explanation: All ((a,a)) pairs are present, so (R) is reflexive. Since ((1,2)) is present but ((2,1)) is not, it is not symmetric.
Correct answer: B. Reflexive and transitive
Explanation: For every (a\in A), (a\le a), so it is reflexive. If (a\le b) and (b\le c), then (a\le c), so it is transitive.
Correct answer: A. Symmetric but not reflexive
Explanation: Every reverse pair is present, so (R) is symmetric. But ((1,1),(2,2),(3,3)) are missing, so it is not reflexive.
Correct answer: C. (R) is an empty relation
Explanation: There is no ordered pair in (R), so it is an empty relation. On non-empty (A), it is not reflexive.
Correct answer: B. Equivalence relation
Explanation: Elements with the same parity are related, so reflexive, symmetric, and transitive properties all hold. Hence (R) is an equivalence relation.
Correct answer: B. \(\{1,2,3\}\)
Explanation: The domain of a relation is the set of all first components of its ordered pairs. Here the first components are 1, 2, 2 and 3; removing the repeated 2 gives \(\{1,2,3\}\). Option A is the range, while options C and D do not correctly include all first components. Exam tip: use the first coordinate for the domain and the second coordinate for the range.
Correct answer: B. \(\{4,5,6\}\)
Explanation: The range of a relation is the set of the second components of its ordered pairs. Here the second components are 4, 4, 5, and 6; since repeated elements are written only once in a set, the range is \(\{4,5,6\}\). Option A is the set of first components, or the domain. Exam tip: in an ordered pair \((a,b)\), select the second component \(b\) to find the range.
Correct answer: D. (4,2)
Explanation: An ordered pair \((a,b)\) belongs to \(R\) only when \(a\) divides \(b\), meaning that \(b/a\) is an integer. We have \(1\mid4\), \(2\mid4\), and \(3\mid3\), but \(4\nmid2\); hence, \((4,2)\notin R\). Exam tip: in a divisibility relation, the first component is the divisor and the second component is the dividend.
Correct answer: A. Reflexive and transitive but not symmetric
Explanation: Every (a\mid a), so it is reflexive, and divisibility is transitive. But (1\mid2) while (2\nmid1), so it is not symmetric.
Correct answer: C. (2^9)
Explanation: Direct answer: Option C, 2^9. A relation on A is any subset of the Cartesian product A×A. Since A has 3 elements, the ordered pairs are 3×3=9: each first element can be paired with each of the three second elements. For every one of these 9 pairs, we have two independent choices: include it in the relation or leave it out. Therefore the number of subsets, and hence relations, is 2×2×... nine times = 2^9=512. Option A, 2^3, counts choices for only three objects and ignores the ordered-pair construction. Option B, 2^6, uses the wrong number of possible pairs. Option C is correct. Option D, 3^2, reverses the exponent idea and is not the number of subsets. No extra condition such as reflexivity or symmetry is imposed, so all subsets count. Memory cue: an n-element set has n^2 ordered pairs and therefore 2^(n^2) relations.
Correct answer: A. (2^{12})
Explanation: In a reflexive relation, (4) diagonal pairs are fixed. The remaining (16-4=12) pairs are free, so the number is (2^{12}).
Correct answer: B. It is symmetric
Explanation: If (|a-b|=1), then (|b-a|=1), so it is symmetric. But ((1,2)) and ((2,3)) hold while ((1,3)) does not, so it is not transitive.
Correct answer: B. ({(x,y):y-x=3})
Explanation: In the inverse, order is reversed, so the original condition (a-b=3) becomes (y-x=3). Do not forget the direction when variables are swapped.
Correct answer: B. Symmetric but not reflexive
Explanation: If (a+b) is odd, then (b+a) is also odd, so it is symmetric. But (a+a=2a) is even, so it is not reflexive.
Correct answer: B. (10)
Explanation: For (5) elements, the number of pairs with (a<b) is (\binom{5}{2}=10). In such questions, count each unordered pair in the correct ordered direction.
Correct answer: B. Irreflexive and transitive
Explanation: (a<a) is never true, so it is irreflexive. If (a<b) and (b<c), then (a<c), so it is transitive.
Correct answer: A. It is an equivalence relation
Explanation: (x^2=x^2), (x^2=y^2\Rightarrow y^2=x^2), and equality is transitive. Hence it is an equivalence relation.
Correct answer: B. (2, 2)
Explanation: The identity relation on A is I_A = {(a, a) : a ∈ A}; hence, both components of every ordered pair must be equal. Therefore, (2, 2) belongs to I_A. The pairs (1, 2), (3, 4), and (4, 1) are not identity pairs because their components differ. Exam tip: In an identity relation, every element is related only to itself.
Correct answer: B. ({(1,1),(2,2),(3,3)})
Explanation: For reflexivity, all diagonal pairs are compulsory. The smallest relation keeps only those required pairs.
Correct answer: B. (3)
Explanation: All three domain elements must appear in some pair, so at least (3) pairs are needed. The (2) range elements can be covered within these (3) pairs.
Correct answer: B. {x, y}
Explanation: The range of a relation is the set of all second components of its ordered pairs. The second components in R are x, x, and y; removing the repetition gives {x, y}. The set {1, 2, 3} is the domain, while {x} and {y} omit one of the attained values. Exam tip: To find the range, collect the second element from every ordered pair and write repeated elements only once.
Correct answer: C. 10
Explanation: For each value of \(a\), count the values of \(b\) satisfying \(a+b\le 5\). When \(a=1\), there are 4 choices for \(b\); for \(a=2,3,4\), there are 3, 2, and 1 choices, respectively. Thus, \(|R|=4+3+2+1=10\), so option C is correct. Exam tip: For a relation defined by an inequality, count the admissible second components for each possible first component and then add them.
Correct answer: A. ((2,3))
Explanation: (\operatorname{lcm}(2,3)=6), so ((2,3)\in R). In the other options, the (\operatorname{lcm}) is not (6).
