Questions & Answers: "Relation"

A relation from set A to set B is any subset of the Cartesian product A × B. First, we find the number of elements in A × B: n(A × B) = n(A) × n(B) = 3 × 2 = 6. The total number of subsets (and thus total relations) is 2^n, where n is the number of elements in the Cartesian product. So, total relations = 2^6 = 64. The question specifically asks for 'non-empty' relations. Since the empty set (null relation) is a subset of every set, we must subtract that one case. Therefore, 64 - 1 = 63. Option A is incorrect because it includes the empty relation. Option C represents only the number of elements in the Cartesian product, not the number of relations. Option D is simply n(A)+n(B), which is irrelevant to the count of relations.

To find the Range, we need to find all possible values of 'y' such that x is a natural number (N = {1, 2, 3, ...}). From x + 2y = 10, we get x = 10 - 2y. Since x must be a natural number, 10 - 2y > 0, which implies 2y < 10 or y < 5. Also, since y must be a natural number, the possible values for y are {1, 2, 3, 4}. Let's check x for these values: If y=1, x=8 (∈ N); If y=2, x=6 (∈ N); If y=3, x=4 (∈ N); If y=4, x=2 (∈ N). If y=5, x=0 (not a natural number). Thus, the relation R = {(8,1), (6,2), (4,3), (2,4)}. The Range is the set of second elements: {1, 2, 3, 4}. Option A is the Domain of the relation. Option B includes 5, which results in x=0 (not in N). Option D is missing the value 1.

We are given that n(A × B) = 9. Since n(A × B) = n(A) × n(B) and 9 is a perfect square (3 × 3), it implies n(A) = 3 and n(B) = 3 (if A=B). In a Cartesian product A × B, the first element of each ordered pair belongs to set A and the second element belongs to set B. From the pair (-1, 0), we know -1 ∈ A and 0 ∈ B. From the pair (0, 1), we know 0 ∈ A and 1 ∈ B. Since the total number of elements in A must be 3 to satisfy n(A) × n(B) = 9, and we have identified -1 and 0 as elements of A from the given pairs, the elements of A must be {-1, 0, 1} assuming A = B as per standard NCERT textbook problems of this type. Even if A was not equal to B, the first elements of the pairs provided are -1 and 0; however, to reach a product of 9, set A must have 3 elements. The only option containing 3 elements that fits the observations is {-1, 0, 1}. Options A, C, and D only contain 2 elements, which would make n(A × B) either 4 or 6, not 9.

Given x + 2y = 8, where x, y ∈ N. We solve for y: 2y = 8 - x, so y = (8 - x)/2. For y to be a natural number, (8 - x) must be even and positive. If x=2, y=3. If x=4, y=2. If x=6, y=1. If x=8, y=0 (not a natural number). Thus, the values of y are {1, 2, 3}. Option A is the domain. Option C includes 4, which would require x=0 (not a natural number). Option D includes 4 and excludes 1.

Prime numbers less than 10 are {2, 3, 5, 7}. The relation is defined as (x, x^3), so the range consists of the cubes of these primes: 2^3=8, 3^3=27, 5^3=125, and 7^3=343. Option B is the domain. Option C incorrectly includes 1 (1 is not prime). Option D includes 64 (4^3), but 4 is not a prime number.

A relation from A to B is any subset of the Cartesian product A × B. The number of elements in A × B is n(A) × n(B) = 3 × 4 = 12. The number of subsets of a set with n elements is 2^n. Therefore, the number of relations is 2^12. Option A (12) is just the number of elements in the product, not the number of subsets. Option B (2^7) incorrectly adds the number of elements instead of multiplying them. Option D (4^3) is the number of functions from A to B, not relations.

The inverse of a relation R, denoted by R⁻¹, is obtained by swapping the elements of each ordered pair in R. If (a, b) ∈ R, then (b, a) ∈ R⁻¹. For R = {(1, 3), (2, 4)}, the inverse is R⁻¹ = {(3, 1), (4, 2)}. Option A only swaps one pair. Option B is the original relation. Option D swaps the components incorrectly.

The total number of relations from A to B is the total number of subsets of A × B, which is 2^(n(A×B)) = 2^(mn). A 'non-empty' relation means we must exclude the empty set (null relation). Thus, the count is 2^(mn) - 1. Option C includes the empty relation. Option A is just the count of ordered pairs. Option D is related to the count of functions.

In the relation R, y = 3x. Since y must belong to set A (y ≤ 14), we check values of x: if x=1, y=3; if x=2, y=6; if x=3, y=9; if x=4, y=12. If x=5, y=15, which is not in A. Thus, the possible values for x (the domain) are {1, 2, 3, 4}. Option C represents the range, not the domain. Option A includes 5, which results in a value outside the set. Option D includes all elements, but not all elements satisfy the condition within the set constraints.

A known property of Cartesian products is that (A × B) ∩ (B × A) = (A ∩ B) × (B ∩ A). If n(A ∩ B) = k, then the number of common elements is k^2. Here k = 3, so 3^2 = 9. Option A is k, not k^2. Option B is 2k. Option D assumes the products are disjoint, which is only true if A and B are disjoint.

The number of elements in the Cartesian product A × B is given by n(A × B) = n(A) × n(B). Here, n(A) = 3 and B = {3, 4, 5} so n(B) = 3. Therefore, n(A × B) = 3 × 3 = 9. Option B incorrectly adds the counts. Option D treats it as 3^3.

We check each x ∈ A against y ∈ B: 2 divides 6 and 10; 3 divides 3 and 6; 4 divides nothing in B; 5 divides 10. Combining these, R = {(2, 6), (2, 10), (3, 3), (3, 6), (5, 10)}. Option B misses (3, 3). Option C is the inverse relation (yRx). Option D contains pairs where x does not divide y.

For x to be in the domain, there must exist an integer y such that x^2 + y^2 = 9. If x=0, y²=9 (y=±3). If x=3, y²=0 (y=0). If x=-3, y²=0 (y=0). If x=2, y² = 9 - 4 = 5. Since √5 is not an integer, x=2 cannot be in the domain of a relation defined over integers Z. Thus, 2 is the correct answer. All other options result in integer values for y.

The relation is defined by y = 3x. Since y must belong to set A (y ≤ 14), the possible values for x are: if x=1, y=3; if x=2, y=6; if x=3, y=9; if x=4, y=12. If x=5, y=15, which is not in A. The domain is the set of all first elements of the ordered pairs in R, which is {1, 2, 3, 4}. {3, 6, 9, 12} is the range, not the domain. {1, 2, 3, 4, 5} is incorrect because 5 has no image in A.

A × B is the set of all ordered pairs (a, b) where a ∈ A and b ∈ B. For A={1,2} and B={3,4}, the pairs are (1,3), (1,4), (2,3), and (2,4). Option 2 is B × A. Option 3 is just a subset of A ∪ B elements paired incorrectly. Option 4 represents sums, which is not the Cartesian product.

The range is the set of all second elements in the relation. For x=0, y=5; x=1, y=6; x=2, y=7; x=3, y=8; x=4, y=9; x=5, y=10. Thus, the range is {5, 6, 7, 8, 9, 10}. Option 1 is the domain. Options 3 and 4 contain incorrect values not generated by the relation's rule.

A relation from A to B is any subset of A × B. The number of elements in A × B is pq. The total number of subsets (relations) is 2^(pq). Since the question asks for 'non-empty' relations, we subtract the empty set (void relation). Therefore, the answer is 2^(pq) - 1. Option 1 includes the empty relation, while options 2 and 4 are incorrect formulas for the number of relations.

The number of elements in the Cartesian product of two sets A and B is given by n(A × B) = n(A) × n(B). Here, n(A) = 3 and B = {3, 4, 5}, so n(B) = 3. Thus, n(A × B) = 3 × 3 = 9. Option 3 is correct. Options 3 and 6 are incorrect counts, and 27 would be the number of elements in A × B × C if n(C) was also 3.

Two ordered pairs are equal if and only if their corresponding first elements are equal and their corresponding second elements are equal. Setting x/3 + 1 = 5/3 gives x/3 = 5/3 - 1 = 2/3, so x = 2. Setting y - 2/3 = 1/3 gives y = 1/3 + 2/3 = 1. Therefore, x=2 and y=1. Other options fail to satisfy both linear equations simultaneously.

By definition, the range of a relation is the set of all second elements of the ordered pairs, which is always a subset of the second set B (the Codomain). The domain is not always equal to A (it is a subset), and the range is not always equal to B (it is a subset). Thus, Option 3 is the only universally true statement.

Given x + 2y = 8 and x, y ∈ N. If y=1, x=6; if y=2, x=4; if y=3, x=2. If y=4, x=0 (not a natural number). Thus, the possible values for x (the domain) are {2, 4, 6}. Option 2 is the range {1, 2, 3}. Option 3 includes 8, which would make y=0 (not natural). Option 4 is too broad.

The number of elements in A is n(A) = 3 and in B is n(B) = 2. The number of elements in A × B is 3 × 2 = 6. The number of relations is the number of subsets of A × B, which is 2^n(A × B) = 2^6 = 64. Option 1 is just the number of pairs. Option 4 is 2^3, which is incorrect.

A relation R on set A is an Identity Relation if every element of A is related to itself only. Here, each element of {1, 2, 3} is paired with itself and nothing else. Universal relation would contain all 9 possible pairs. Void relation would be empty. Inverse relation refers to reversing the pairs of another relation.