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.

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 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.

A number x is a square root of 16 when x² = 16. Over the integers, both −4 and 4 satisfy this equation because (−4)² = 16 and 4² = 16. However, the set specifically requires x ∈ N, and the given natural numbers are positive, so −4 is excluded. Therefore P = {4}. Option C confuses the square with its root, while 2 does not satisfy x² = 16.

Use the defining condition x² < 10 and remember that W contains only non-negative integers. We have 0² = 0, 1² = 1, 2² = 4 and 3² = 9, all less than 10. But 4² = 16, so 4 is excluded. Negative numbers cannot be included because they are not whole numbers. Hence the roster form is Q = {0, 1, 2, 3}, so option A is correct.

The listed elements are consecutive squares: 1 = 1², 4 = 2², 9 = 3², 16 = 4² and 25 = 5². In option A, n takes every natural-number value from 1 through 5, so x = n² produces exactly these five elements and no others. Option B stops before n = 5 and omits 25; option C gives even multiples, while option D gives 2, 3, 4, 5 and 6.

For an integer x, the condition |x| < 3 means that x is fewer than three units from zero. It is equivalent to −3 < x < 3. The integers strictly between these endpoints are −2, −1, 0, 1 and 2, so V = {−2, −1, 0, 1, 2}. The endpoints −3 and 3 are excluded because the inequality is strict, and zero is included because |0| = 0.

The smallest two-digit number is 10. The question asks for natural numbers strictly less than 10, and the stated convention is N = {1, 2, 3, ...}. Therefore the members are 1, 2, 3, 4, 5, 6, 7, 8 and 9. Zero is excluded because it is not in the given N, while 10 is excluded because the condition says less than 10. Thus option A is correct.

First list the positive factors of 18: 1, 2, 3, 6, 9 and 18. Then apply the second condition, namely that the factor must be prime. Only 2 and 3 have exactly two positive divisors. The number 1 is neither prime nor composite, while 6, 9 and 18 are composite. Therefore E = {2, 3}, making option A correct.

The inequalities are strict, so the endpoints 10 and 20 are not included. The natural numbers between them are 11, 12, 13, 14, 15, 16, 17, 18 and 19. Selecting only the even values leaves 12, 14, 16 and 18. Thus H has four elements, written n(H) = 4. Counting 10 or 20 would incorrectly treat < as ≤, and counting every integer would ignore the evenness condition.

The positive multiples of 6 below 40 are found by calculating 6 × 1 through 6 × 6: 6, 12, 18, 24, 30 and 36. Therefore 18, 24 and 36 satisfy both conditions and belong to K. Although 42 is also divisible by 6, it is greater than 40, so it fails the phrase “less than 40.” Hence 42, option D, does not belong to K.