General chapter practice

The governing concept is listing members that satisfy a set-builder rule. In this question, natural numbers are taken as 1, 2, 3, and so on. The numbers less than 4 are therefore 1, 2, and 3, giving K = {1, 2, 3}. Zero is excluded under this convention, and the strict inequality x < 4 excludes 4. Hence option A contains exactly the required elements.

The governing concept is identifying a set by a common number property. Each listed number—2, 3, 5, 7, and 11—has exactly two positive divisors, 1 and itself, so each is prime; all are also less than 12. These are precisely the prime numbers below 12. Option B would additionally include 1 and 9, C describes a different parity class, and D lists 1, 2, 3, 4, 6, and 12. Therefore A is correct.

The decimal digits are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9. An even digit is divisible by 2 without a remainder. The even digits are therefore 0, 2, 4, 6, and 8. Since these are five distinct elements, the cardinality of E is 5, so option B is correct. Option A misses one even digit, while C and D count too many.

Each member of F can be expressed as a power of 2: 2=2¹, 4=2², 8=2³, and 16=2⁴. Every one of these values is less than 20, and these are exactly the powers of 2 below 20 when positive integral exponents are intended. They are not all prime, they are not the factors of 20, and none is odd. Thus option A is correct.

A singleton set is a set containing exactly one distinct element. In G={5}, the only element listed is 5, so the cardinality of G is 1 and G is a singleton set. It is not empty because it contains 5, and it is not infinite because it has only one element. The label universal set cannot be decided without a stated universal context. Therefore option A is correct.

The set-builder condition requires a natural number x to be strictly greater than 5 and strictly less than 9. Checking the integers between the two endpoints gives 6, 7, and 8. Because both inequality signs are strict, neither 5 nor 9 belongs to H. Therefore the correct roster form is {6,7,8}, which is option B. The other choices include one or both excluded endpoints.

Whole numbers are the non-negative integers: 0, 1, 2, 3, and so on. The condition x≤3 includes every whole number up to and including 3, so the complete set is I={0,1,2,3}. Option A incorrectly excludes 0; option C excludes the allowed endpoint 3; and option D keeps only one of the four qualifying values. Hence option B is correct.

The listed elements 3, 6, 9, and 12 are all multiples of 3. Restricting x to the inclusive interval 3≤x≤12 gives exactly those four multiples: 3, 6, 9, and 12. Thus option A reproduces J without adding or losing an element. The other descriptions produce different sets: factors of 3 are much fewer, and even or prime numbers do not match the roster.

In roster notation, the symbols inside braces are the elements of the set. Since a, b, and d are explicitly listed in L, the statement b∈L is true, so option A is correct. The symbol c is absent, making c∈L false. The statement d∉L is also false because d is present. Finally, L∈b reverses the usual membership relation and is not justified by the given information.

To test membership, compare each proposed number with the elements listed in M. The numbers 2, 5, and 8 occur explicitly, so 2∈M, 5∈M, and 8∈M are true statements. The number 6 does not occur in the roster, so 6∈M is false. Therefore option D is the required answer. The corresponding true non-membership statement would be 6∉M.

A fundamental property of sets is that repetition does not create new elements. The three occurrences of 4 represent one distinct element, and the two occurrences of 6 represent another. Removing repeated copies therefore gives N={4,6}, which has cardinality 2. It is not a five-element set, and deleting either 4 or 6 would change the collection. Hence option A is correct.

The governing concept is the cardinality of a set. In a set, repeated occurrences of the same element are counted only once. Removing repetitions from Y={3,3,5,7,7,7} gives the distinct set {3,5,7}. Thus n(Y)=3, so option A is correct. Option D incorrectly counts all six written occurrences, while B and C do not represent the number of distinct elements.

The governing idea is conversion from set-builder form to roster form. A positive factor of 18 must divide 18 exactly. The factor pairs 1×18, 2×9, and 3×6 produce the complete list {1,2,3,6,9,18}; therefore option A is correct. Option B is mainly a list of even numbers, C contains multiples of 18, and D includes 5, which is not a factor of 18.

The governing concept is counting elements in an inclusive finite interval. Because both inequalities use ≤, the endpoints 2 and 6 are included. Listing the natural numbers gives A={2,3,4,5,6}, so there are 5 elements and option B is correct. Option A results from omitting one endpoint, while C and D do not count the complete listed set.

The governing concept is set-builder notation: the stated rule must generate every element of B and no extra element. The even whole numbers less than 8 are 0, 2, 4, and 6, exactly B, so option A is correct. Option C may exclude 0 and includes odd natural numbers; B describes odd digits, and D describes factors of 6, neither of which matches B.

The governing conditions must both be applied. A negative integer is less than zero, so 0 is excluded. The integers satisfying x≥−3 while remaining negative are −3, −2, and −1. Hence option A is the correct roster form. Options B and C incorrectly include zero, and D adds −4 even though −4<−3, so it fails the inequality.

The governing concept is the distinction between finite and infinite sets. The calendar has a fixed number of months, and exactly four have 30 days: April, June, September, and November. Thus D has four elements and is a finite set, so option A is correct. It is not empty because these months exist, not infinite because the yearly month list is bounded, and not unclear because membership is definite.

A prime number has exactly two positive divisors, 1 and itself. Among the digits 0 through 9, the prime digits are 2, 3, 5, and 7, so option A is the correct roster form. Zero and 1 are not prime; including 1 causes the error in B. Option C lists even composite digits, while D includes 9, which is composite, and omits 2.

The governing concept is that a set contains distinct elements only. The letters in ALGEBRA are A, L, G, E, B, R, and A; the second A is a repetition, not a new set element. Therefore the roster form is {A,L,G,E,B,R}, making option A correct. Option B repeats A, whereas C omits several letters and D omits A and E.

The governing concept is subset notation. G⊆H means every element of G must also belong to H. Since both 1 and 2 are in H, G⊆H is true, so option A is correct. H is not a subset of G because H contains 3, which is absent from G. Therefore option B is false; 3∈G is false, and the sets cannot be equal because their elements differ.