Relations practice · Mixed difficulty

Option D (Equivalence) is correct because the relation R, where two colleges are related if they are affiliated to the same university, is reflexive, symmetric, and transitive, thus it is an equivalence relation.

Option B (reflexive) is correct because a relation R is reflexive if (a, a) belongs to R for every a in A.

Option A is correct because the relation R is not reflexive, as a square cannot share a common side with itself.

Option B ($\left\{(1,1),(2,2),(3,3)\right\}$) is correct because it represents the composition of the relation $R$ with its inverse $R^{-1}$, where $y = 3x$.

Option D (Reflexive, transitive but not symmetric) is correct because n is always a factor of itself, if n is a factor of m and m is a factor of p, then n is a factor of p, but if n is a factor of m, m is not necessarily a factor of n.

Option C (7) is correct because 7 ordered pairs must be added to R to make it reflexive, symmetric, and transitive, thus forming an equivalence relation.

Option A is correct. Relation S is an equivalence relation because it satisfies reflexivity, symmetry, and transitivity, while R fails to be symmetric.

Option A (Reflexive) is correct. For any real number x, x - x + $\sqrt{2}$ = $\sqrt{2}$, which is irrational, thus xRx.

Option C (Transitive) is correct because for any (a, b) and (b, c) in R, (a, c) is also in R.

Only I is true because if A is a subset of B, then the Cartesian product of A with any set C will also be a subset of the Cartesian product of B with C.

Option B ((3,1)) is correct because adding (3,1) makes the relation symmetric, and since it's already reflexive and transitive, it becomes an equivalence relation.

Option B (symmetric) is correct because if line $l_1$ is perpendicular to $l_2$, then $l_2$ is perpendicular to $l_1$.

Option C ($p q$) is correct because the number of elements in the Cartesian product of two sets is the product of the number of elements in each set.

Option B (${99^2}$) is correct because the number of common elements in $A \times B$ and $B \times A$ is the square of the number of common elements in A and B.

Option C (50) is correct because it is the sum of the squares of the number of elements in set A and set B, calculated using the given information about the Cartesian product.

Option B (equivalence relation) is correct because the relation satisfies reflexive, symmetric, and transitive properties, thus making it an equivalence relation.

Option B (symmetric only) is correct because if (x, y) is in R, then (y, x) is also in R, but the relation is not reflexive or transitive.

Option $2^{mn}-1$ is correct, as the total number of relations from A to B is $2^{mn}$, and subtracting 1 excludes the empty relation.

Option A ($\{(1,0),(1,1),(2,0),(2,1)\}$) is correct because it represents the Cartesian product of A and B, where each element of A is paired with each element of B.

Option A is correct. The relation R is reflexive because x=y is a solution, and it's transitive because if x=3y and y=3z then x=9z, but x=y or x=3y, therefore transitive.

Option B (Neither ${R_1}$ nor ${R_2}$ is an equivalence relation) is correct because neither relation is transitive, and therefore neither can be an equivalence relation.

Option D (2) is correct because it is the number of elements in set C, calculated using the cardinality of the Cartesian product of sets A, B, and C.

Option D (None of the above) is correct because all the given relations $R_1$, $R_2$, and $R_3$ are not equivalence relations on A.

Option C ($A \cap B = \phi$) is correct because the hyperbola $y = \frac{1}{x}$ and the line $y = -x$ do not intersect, meaning their intersection is the empty set.

Option D is correct because all the given statements about the relation R are true.

Option A (Symmetric) is correct because if line $l_1$ is perpendicular to $l_2$, then $l_2$ is perpendicular to $l_1$.

The correct answer is C. The relation $R$ is not symmetric because $(2,3) \in R$ but $(3,2) \notin R$.

Option C ({ (3,3),(3,5),(5,3),(5,5) }) is correct because it represents the composition of the relation R with its inverse.

Option B (Anti-symmetric) is correct because the subset relation is anti-symmetric: if A is a subset of B and B is a subset of A, then A must equal B.

Option D ($p+q=8$) is correct because the order of the differential equation is 2 and the degree is 6 after removing the fractional power, so their sum is 8.