Relations

Correct answer: A. ((1,3))

Explanation: Since ((1,2)\in R) and ((2,3)\in R), transitivity requires ((1,3)). It is missing, so (R) is not transitive.

Correct answer: B. \(\{2,3,4\}\)

Explanation: The relation requires \(a+b=6\) with \(a,b\in A\). The valid ordered pairs are \((2,4),(3,3),(4,2)\). Therefore, the set of their first components is \(\{2,3,4\}\), so option B is correct. Option C incorrectly omits 4, while 1 cannot be included because it would require \(b=5\), which is not in \(A\). Exam tip: The domain of a relation is the set of all first components of its ordered pairs.

Correct answer: B. 2

Explanation: Check the ordered pairs formed from elements of \(A\) against \(a^2+b^2=25\). Since \(3^2+4^2=9+16=25\) and \(4^2+3^2=16+9=25\), the pairs \((3,4)\) and \((4,3)\) belong to \(R\). No other pair from \(A\times A\) satisfies the equation. Hence, \(R\) contains \(2\) ordered pairs. Exam tip: In a relation, \((a,b)\) and \((b,a)\) are counted separately because order matters, unless the two entries are equal.

Correct answer: B. Symmetric but not reflexive

Explanation: If (a\ne b), then (b\ne a), so it is symmetric. But ((a,a)) never occurs, so it is not reflexive.

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

Explanation: The equivalence class of \(a\) is \([a]_R=\{x\in A:(a,x)\in R\}\). Here, the pairs \((1,1)\) and \((1,2)\) belong to \(R\), whereas \((1,3)\) does not. Therefore, \([1]_R=\{1,2\}\). Option D is incorrect because \(3\) is not related to \(1\). Exam tip: To find \([a]_R\), identify the ordered pairs whose first component is \(a\), then collect their second components.

Correct answer: A. \(\{2,5\}\)

Explanation: The class \([2]\) contains all elements of \(A\) that are congruent to \(2\) modulo \(3\). Both \(2\) and \(5\) leave remainder \(2\) on division by \(3\), so \([2]=\{2,5\}\). Option B, \(\{1,4\}\), is actually the class \([1]\), since both elements leave remainder \(1\). Exam tip: To find an equivalence class, collect all elements having the same remainder modulo the given number.

Correct answer: A. ({1,3},{2,4})

Explanation: An even sum means both numbers have the same parity. Thus the classes are odd ({1,3}) and even ({2,4}).

Correct answer: A. \((5,2)\)

Explanation: By the definition of a symmetric relation, if \((a,b)\in R\), then \((b,a)\in R\) must also hold. Here, \(a=2\) and \(b=5\), so \((5,2)\in R\) is necessary. The presence of \((2,2)\) or \((5,5)\) is related to reflexivity, not symmetry. Exam tip: To test symmetry, interchange the two elements of every ordered pair and check whether the reversed pair is also present.

Correct answer: C. \((1,6)\)

Explanation: Transitivity states that if \((a,b)\in R\) and \((b,c)\in R\), then \((a,c)\in R\). Substituting \(a=1\), \(b=4\), and \(c=6\) gives \((1,6)\in R\), so option C is correct. Options A and B reverse the direction of the given ordered pairs, while option D is not required by transitivity. Exam tip: in a transitive chain, match the second element of the first pair with the first element of the second pair, then join the remaining outer elements.

Correct answer: B. {3,4}

Explanation: The only ordered pairs from A satisfying a+b=7 are (3,4) and (4,3). Therefore, the range—the set of second components of the ordered pairs—is {3,4}. Option A incorrectly includes every element of A; for instance, pairing 2 would require 5, but 5∉A. Exam tip: the range of a relation is formed from the second coordinates of its ordered pairs.

Correct answer: A. Transitive but not reflexive

Explanation: The required ((1,3)) from ((1,2)) and ((2,3)) is present, so the key transitivity condition holds. Diagonal pairs are absent, so it is not reflexive.

Correct answer: C. ((3,3)\in R)

Explanation: Symmetry lets us infer ((4,3)\notin R) from ((3,4)\notin R). But the presence of diagonal pairs is not guaranteed by symmetry.

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

Explanation: For ((1,1)), (1+1=2) is prime, so it is in (R). However, all listed pairs are actually in (R), so this is a trap question.

Correct answer: D. (3,4)

Explanation: For an ordered pair \((a,b)\) to belong to \(R\), the sum \(a+b\) must be prime. In option D, \(3+4=7\), and 7 is prime; hence \((3,4)\in R\). The sums for the other options are 4, 8, and 6, none of which is prime. Exam tip: calculate the sum of the two entries and then test whether the sum is prime.

Correct answer: A. Symmetric

Explanation: If (a+b=6), then (b+a=6), so the relation is symmetric. But all ((a,a)) are not present, so it is not reflexive.

Correct answer: A. (I_A)

Explanation: This relation sends each (a) to (5-a), and applying it twice returns (a) to itself. Therefore (R\circ R=I_A).

Correct answer: B. (2^{10})

Explanation: Reflexivity fixes the (5) diagonal pairs, and symmetry lets the remaining pairs be chosen in unordered pairs. Therefore the number is (2^{\frac{5(5-1)}{2}}=2^{10}).

Correct answer: A. Reflexive and transitive but not symmetric

Explanation: For every (a\in A), (a\mid a), and if (a\mid b), (b\mid c), then (a\mid c). But (2\mid4) while (4\nmid2), so it is not symmetric.

Correct answer: A. Equivalence relation

Explanation: (a+a) is even, (a+b) even implies (b+a) even, and same parity gives transitivity. In exams, identify the even and odd equivalence classes.

Correct answer: A. (2^{n^2})

Explanation: (A\times A) has (n^2) ordered pairs, and a relation can be any subset of it. Hence the number of relations is (2^{n^2}).

Correct answer: A. (2^{mn})

Explanation: (A\times B) has (mn) ordered pairs. Each pair may be included or excluded, so there are (2^{mn}) possible relations.

Correct answer: A. (2^6)

Explanation: A reflexive relation must contain ((1,1),(2,2),(3,3)). The remaining (9-3=6) pairs are optional, so the answer is (2^6).

Correct answer: A. (2^{12})

Explanation: There are (4^2=16) ordered pairs, and (4) diagonal pairs are compulsory. Thus (12) pairs are optional, giving (2^{12}).

Correct answer: A. (2^{10})

Explanation: For a symmetric relation, the number of independent choices is (\frac{4(4+1)}{2}=10). Hence the number of relations is (2^{10}).

Correct answer: A. (2^{10})

Explanation: Reflexivity fixes the (5) diagonal pairs. Symmetry leaves only (\frac{5(5-1)}{2}=10) off-diagonal blocks free, so the answer is (2^{10}).