Relations

Correct answer: C. (4)

(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

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

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

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

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

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

Elements with the same parity are related, so reflexive, symmetric, and transitive properties all hold. Hence (R) is an equivalence relation.

Correct answer: B. \(\{1,2,3\}\)

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. \(\{4,5,6\}\)

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)

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

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)

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})

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

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})

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

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)

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

(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

(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)

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)})

For reflexivity, all diagonal pairs are compulsory. The smallest relation keeps only those required pairs.

Correct answer: B. (3)

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}

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

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))

(\operatorname{lcm}(2,3)=6), so ((2,3)\in R). In the other options, the (\operatorname{lcm}) is not (6).