Relations

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

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. \(\{2,3,4\}\)

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

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

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. \(\{1,2\}\)

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. \(\{2,5\}\)

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})

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

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

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. \((1,6)\)

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}

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

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)

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))

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)

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

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)

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})

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: B. \(R\) is reflexive and transitive but not symmetric.

For any \(a\in A\), we have \(a\mid a\), so \(R\) is reflexive. If \(a\mid b\) and \(b\mid c\), then \(a\mid c\), so \(R\) is transitive. However, symmetry fails: for example, \(2\mid 4\) is true, so \((2,4)\in R\), but \(4\mid 2\) is false, so \((4,2)\notin R\). Hence \(R\) is reflexive and transitive but not symmetric.

Correct answer: A. Equivalence relation

(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})

(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})

(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)

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})

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})

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})

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}).