Relations

Correct answer: A. A relation from A to B

Explanation: A relation from set \(A\) to set \(B\) is defined as any subset of \(A\times B\), so option A is correct. Not every relation is a function: a function must assign exactly one image in \(B\) to each element of \(A\). Option D is incorrect because an identity relation is generally defined on the same set. Exam tip: remember ‘any subset’ for a relation and ‘exactly one image for every element’ for a function.

Correct answer: A. (p\in A) and (q\in B)

Explanation: In a Cartesian product, the first component comes from the first set and the second from the second set. Therefore (p\in A) and (q\in B).

Correct answer: A. Symmetric but not reflexive

Explanation: If (a+b) is odd, then (b+a) is also odd, so it is symmetric. But (a+a) is even, so it is not reflexive.

Correct answer: A. Reflexive and transitive

Explanation: It is reflexive because ((1,1)) and ((2,2)) are present. The pair ((1,2)) causes no violation of transitivity, so it is also transitive.

Correct answer: A. It is symmetric but not reflexive

Explanation: Both ((1,2)) and ((2,1)) are present, so it is symmetric. But ((1,1)) and ((2,2)) are absent, so it is not reflexive.

Correct answer: A. Because \((1,3)\) is absent

Explanation: For transitivity, whenever \((a,b)\in R\) and \((b,c)\in R\), the pair \((a,c)\) must also belong to R. Here, \((1,2)\in R\) and \((2,3)\in R\), so \((1,3)\) must be in R; however, it is absent. Therefore, R is not transitive. Exam tip: diagonal pairs such as \((1,1),(2,2),(3,3)\) do not compensate for a missing required pair.

Correct answer: A. (5)

Explanation: All elements in the given set are non-negative, so (a^2=b^2) gives (a=b). Therefore only (5) diagonal pairs exist.

Correct answer: A. \((-2,2)\)

Explanation: For the pair \((-2,2)\), \((-2)^2=2^2=4\), so it satisfies the condition \(a^2=b^2\) and hence \((-2,2)\in R\). In the other options, the squares of the two numbers are unequal. Exam tip: when a relation involves squares, check numbers with opposite signs as well.

Correct answer: A. (R) is reflexive and transitive

Explanation: Every (a) divides itself and (a\mid b,\ b\mid c) imply (a\mid c). Hence it is reflexive and transitive.

Correct answer: A. Reflexive, symmetric, transitive

Explanation: The definition of equivalence relation includes reflexive, symmetric, and transitive properties. In exams, check all three separately.

Correct answer: A. \((4,2)\notin R\)

Explanation: For the ordered pair \((4,2)\), \(a+b=4+2=6\). The relation \(R\) contains only those ordered pairs for which \(a+b\le 5\), but \(6\le 5\) is false. Therefore, \((4,2)\notin R\). Option B incorrectly treats the inequality condition as satisfied. Exam tip: To test whether an ordered pair belongs to a relation, substitute its two components directly into the defining condition.

Correct answer: A. Symmetric but not reflexive

Explanation: Direct answer: option A, symmetric but not reflexive. The rule is R = {(a,b): |a-b|=1} on A={1,2,3,4,5}. Symmetric means that whenever (a,b) belongs, (b,a) also belongs. Since |a-b|=|b-a|, this always happens. For example, (1,2) is present and (2,1) is also present. Reflexive means every (a,a) must belong. But |a-a|=0, not 1, so no pair such as (1,1) belongs. Therefore it is not reflexive. C, “transitive,” is false: (1,2) and (2,3) are present, but (1,3) is absent because |1-3|=2. D, equivalence, is false because an equivalence relation must be reflexive, symmetric and transitive; this relation fails reflexivity and transitivity. B is false because it claims reflexivity. Memory cue: distance exactly 1 gives two-way pairs, but never self-pairs.

Correct answer: A. ((2,2))

Explanation: To be reflexive, both ((1,1)) and ((2,2)) are needed. ((1,1)) is already present, so ((2,2)) must be added.

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

Explanation: The domain of a relation is the set of first components of its ordered pairs. The first components here are 1, 2, and 1; removing the repetition gives the domain \(\{1,2\}\). Option B contains the second components and therefore represents the range. Exam tip: for the domain, list the first element of every ordered pair and remove duplicates.

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

Explanation: The range of a relation consists of all distinct second components of its ordered pairs. Here, the second components are 2, 3, and 3, so the range is \(\{2,3\}\). Options A and C use incorrect combinations of components, while option D also includes the first component 1. Exam tip: To find the range, list the second component of every ordered pair and remove repetitions.

Correct answer: A. {x,y}

Explanation: For a relation R from A to B, the codomain is the entire target set B, whether or not every element of B is actually used. Hence, the codomain is {x,y}. The set {1,3} is the domain, while the range is also {x,y} here because both x and y occur as second components. Exam tip: identify the codomain from the stated target set B, but find the range from the actual second components of the ordered pairs.

Correct answer: A. The range is a subset of the codomain

Explanation: The range of a relation consists of the actual output values, whereas the codomain is the set in which those outputs are allowed to lie. Therefore, the range is always a subset of the codomain. Here, \({p,q}\subset {p,q,r}\); the element \(r\) belongs to the codomain but not to the range, so the two sets are not equal. Exam tip: range is always contained in the codomain, but it need not be the whole codomain.

Correct answer: C. (10)

Explanation: A relation from A to B is any subset of the Cartesian product \(A\times B\). The largest possible relation is obtained when every ordered pair in that product is included. The number of ordered pairs in a Cartesian product is found by multiplying the number of elements in the first set by the number in the second set.

Here \(|A|=2\) and \(|B|=5\), so \(|A\times B|=2\times5=10\). A relation cannot contain more than all ten pairs because it is a subset of the product. It can contain fewer pairs, including none, but the maximum is reached by taking \(R=A\times B\). Therefore, option C, 10 ordered pairs, is correct; 25 would incorrectly square or multiply the wrong quantities.

Correct answer: B. 3

Explanation: Using the condition b = a + 3, a = 1 gives (1, 4), a = 2 gives (2, 5), and a = 3 gives (3, 6). Thus, R = {(1, 4), (2, 5), (3, 6)} contains 3 ordered pairs. The value 6 is the number of pairs in the Cartesian product A × B, not in this restricted relation. Exam tip: Apply the defining condition to each element before counting the ordered pairs.

Correct answer: A. \((1,4)\)

Explanation: For \((1,4)\), we have \(a=1\) and \(b=4\), so \(ab=1\times4=4\). Hence, \((1,4)\) belongs to \(R\). The products for the other options are \(6,3\), and \(8\), respectively, so they do not satisfy the relation. Exam tip: To test whether an ordered pair belongs to a relation, verify both the allowed elements and the defining condition.

Correct answer: A. Universal and equivalence

Explanation: It contains all pairs of (A\times A), so it is universal. It is also reflexive, symmetric, and transitive.

Correct answer: A. ((5,3)\in R) cannot occur

Explanation: In an antisymmetric relation, for distinct (a) and (b), both ((a,b)) and ((b,a)) cannot occur together. Since (3\ne5), ((5,3)) cannot occur.

Correct answer: A. It is reflexive and antisymmetric

Explanation: All diagonal pairs are present, so it is reflexive. Since ((2,1)) is absent, ((1,2)) does not break antisymmetry.

Correct answer: A. Reflexive, antisymmetric, and transitive

Explanation: Because (a\ge a) is true, and (a\ge b,\ b\ge c) imply (a\ge c). Also (a\ge b) and (b\ge a) imply (a=b).

Correct answer: C. (64)

Explanation: Since (n(A\times B)=3\times 2=6), the number of relations is (2^6=64). In exams, first count the elements of (A\times B).