Set theory practice · Mixed difficulty

Option C (7 and 11) is correct because the minimum value of n(A ∪ B) occurs when A is a subset of B, and the maximum value occurs when A and B are disjoint.

Option D is correct because A contains multiples of 3, B contains multiples of 4, and their intersection contains multiples of both 3 and 4, which are multiples of 12.

Option A ($\left\{(x, y): x^2+y^2 \leq 1 \leq x+y, x, y \in R\right\}$) is correct because it represents an infinite set of real numbers within the specified region.

Option C ($B$) is correct because if $A$ is a subset of $B$, then the union of $A$ and $B$ is equal to $B$.

The correct answer is 45 because the cardinality of the Cartesian product of sets is the product of their individual cardinalities.

Option C (3) is correct because the minimum possible value of $n(B \cap C)$ is equal to $n(A \cap B \cap C)$ when $A \cap B \cap C$ is the entire intersection of B and C.

Option C ($\phi$) is correct because the intersection of a set with the complement of a superset is always the empty set.

Option D ($A \cup B = \{1,2,3,5,9\}$) is false because the union of A and B should include all elements from both sets, which is not accurately represented in this option.

Option D (An equivalence relation) is correct because the relation is reflexive, symmetric, and transitive.

Option B (B) is correct because if $A \cup B = A$, it implies that B is a subset of A, and therefore, the intersection of A and B is B.

Option D ($\phi$) is correct because if the sets are disjoint, their intersection is empty, making the intersection of all three sets also empty, and the intersection of any set with the empty set is empty.

Option A ($B \cap C^{\prime}$) is correct because the given set expression simplifies to $B \cap C^{\prime}$ using set identities.

Option B ($A \cap B \ne \phi$) is correct because the two curves y = e^x and y = e^-x intersect at the point (0, 1), indicating a non-empty intersection.

Option C (100) is correct because the union of nested sets is the largest set, and n(A_99) = 99 + 1 = 100.

Option D ($\{\phi,\{x\},\{y\}\{x, y\}\}$) is correct because it lists all possible subsets of the set A, including the empty set and the set itself.

Option D ($P=Q$) is correct because both sets are equivalent, resulting in the same value for $\tan \theta$.

Option A (300) is correct because it is the number of integers in the given range that are neither multiples of 3 nor multiples of 4.

Option A (11) is correct because the least possible value of n(A Δ B) is achieved when n(A ∩ B) is maximized, given that A ∩ B is not empty.

Option C (300) is correct because it is the number of elements in the complement of the union of A and B, calculated using De Morgan's law and set theory principles.

Option B (${A_1} \cup {A_2} \cup {A_3} = {B_1} \cup {B_2} \cup {B_3}$) is correct because the sets B_i are constructed to be disjoint and their union is equal to the union of the A_i.

Option D ($A \cap B$ contains three elements only) is correct because the graphs of y² = x and y = |x| intersect at three points: (0, 0), (1, 1), and (1, -1).

Option C ($\frac{1}{\log _{\mathrm{xy}} \mathrm{xyz}}+\frac{1}{\log _{\mathrm{yz}} \mathrm{xyz}}+\frac{1}{\log _{\mathrm{zx}} \mathrm{xyz}}=2$) is correct because using the change of base formula and properties of logarithms, the expression simplifies to 2.

Option C (10) is correct because the greatest possible value of $n(A \cap B \cap C)$ is limited by the smallest of $n(A \cap B)$, $n(B \cap C)$, and $n(A \cap C)$, which is 10.

Option A ($A \subset B$) is correct because the region defined by A (a square) is entirely contained within the region defined by B (an ellipse).

Option D is correct because the term "intelligent" is subjective and lacks a clear, objective definition, making the collection not well-defined.

Option D (3) is correct because it represents the number of elements in the complement of the union of sets A and B, calculated by subtracting the number of elements in the union from the total number of elements in the universal set.

Option B (15) is correct because it is the number of people who invest in both National Savings Certificates and shares, calculated using set theory principles.

Option B (8) is correct because the number of subsets containing 2, 3, and 5 is the same as the number of subsets of the remaining elements, which is 2^(6-3) = 2³ = 8.

The correct answer is D (30) because it is derived by equating the total number of elements in the union of sets X_i and Y_i, considering the multiple occurrences of each element.

Option A ($21N$) is correct because the intersection of multiples of 3 and multiples of 7 are multiples of 21.