General chapter practice

To determine membership, solve the defining equation rather than guessing from the options. Starting with x + 2 = 5, subtract 2 from both sides: x = 5 − 2 = 3. Since 3 is a natural number, it satisfies both conditions and belongs to Y. Testing the other choices in the equation gives values different from 5, so option B is the only correct answer.

A correct set-builder description must generate every listed element and no extra element. Each member of Z is a multiple of 5, and the inclusive bounds 5 ≤ x ≤ 20 restrict the multiples to 5, 10, 15, and 20 exactly. Option B also includes 1, 2, 4, and 20; option C includes many even numbers; option D gives primes. Therefore option A is correct.

The governing concept is factor identification. A positive factor of 8 divides 8 exactly, leaving remainder zero. Checking the listed values, 8 ÷ 1 = 8, 8 ÷ 2 = 4, 8 ÷ 4 = 2, and 8 ÷ 8 = 1. These are all the positive factors of 8. Multiples of 8 continue as 8, 16, 24, and so on, so option A is correct.

The governing concept is subset notation. A ⊆ B means that every element of A must also belong to B. The elements 1, 2, and 3 all occur in B, so A is a subset of B. Conversely, B is not a subset of A because 4 is missing from A. The sets are also unequal, and 4 ∉ A. Therefore option A is correct.

The governing concept is the cardinality of a set. The notation n(A) denotes the number of distinct elements in set A; it does not denote the largest element, the sum of the elements, or a numerical operation on them. In A = {2, 4, 6}, the distinct members are 2, 4, and 6, so the set contains exactly three elements. Therefore n(A) = 3 and option B is correct. Option A counts too few elements, option C adds an element that is not present, and option D mistakes the largest listed member, 6, for the number of members. If an element were repeated in a set, it would still be counted only once, but no repetition occurs here. Directly counting the displayed distinct members gives the answer.

The governing set principle is that repeated listings do not create new elements. Although five entries are written, the distinct elements of B are only 2, 4, and 6. We can rewrite the set without duplicates as B = {2, 4, 6}. Therefore its cardinality is n(B) = 3. The value 5 counts written entries, not distinct set elements, so option A is correct.

The governing concept is the definition of an empty set: it is a set containing no elements. The symbol ∅ is specially used for this set. In contrast, 0 is a number, while {0} is a one-element set whose only member is 0. Similarly, {1} contains one element, namely 1. Therefore, option C alone represents the empty set; the braces in B and D show that those sets are not empty.

The governing concept is cardinality, written n(C), which means the number of distinct elements in set C. The empty set C = ∅ has no elements at all, so counting its elements gives n(C) = 0. The symbol ∅ in option C denotes the set itself, not the numerical size of that set. Option B would describe a singleton, and option D is incorrect because the cardinality of an empty set is well-defined.

The governing concept is roster notation: a set is written by listing its distinct elements, and repetition does not create a new element. MATHEMATICS contains M, A, T, H, E, I, C, and S. Although M, A, and T occur more than once in the word, each is listed only once in a set. Thus option A is correct; B incorrectly repeats letters, C lists vowels only, and D omits several valid letters.

The governing idea is interpreting a set-builder condition with a strict inequality. Taking natural numbers as 1, 2, 3, and so on, the condition x < 6 produces E = {1, 2, 3, 4, 5}. Since the sign is strictly less than, 6 is excluded rather than included. Therefore 1, 3, and 5 belong to E, while option D, 6, does not.

The governing concept is set-builder notation: the rule must generate every member of F and no other member. Whole numbers are 0, 1, 2, 3, 4, and so on; imposing x < 5 leaves exactly {0, 1, 2, 3, 4}. Under the usual school convention used here, natural numbers begin at 1, so B omits 0. C gives larger integers and D gives only even numbers. Hence A is correct.

The governing concept is classification of sets by size and clarity of definition. The condition “month having 31 days” is objective and identifies exactly seven names: January, March, May, July, August, October, and December. The set therefore has a limited number of clearly specified elements, so it is finite. It is neither empty nor infinite, and the rule is well-defined. Thus option A is correct.

The governing concept is membership by divisibility. A natural number is a factor of 12 when it divides 12 exactly, leaving remainder zero. The positive divisors are found from the factor pairs 1×12, 2×6, and 3×4, giving H = {1, 2, 3, 4, 6, 12}. Option B is mostly a list of even numbers, C contains multiples of 12, and D contains odd numbers, so only A is complete and correct.

The governing concept is combining an integer classification with a strict inequality. Negative integers are less than zero, and the integers greater than −4 are −3, −2, −1, 0, 1, and so on; restricting them to negative integers leaves only −3, −2, and −1. The strict sign x > −4 excludes −4, while 0 is not negative. Therefore option A is the correct roster form.

The governing idea is set-builder description: a condition must produce every member of the given set and no extra member. The prime numbers less than 10 are exactly 2, 3, 5, and 7, so option A matches T. Odd numbers would also include 1 and 9; even numbers would include 2, 4, 6, and 8; divisors of 10 are 1, 2, 5, and 10. Therefore A is unambiguous.

The governing concept is interpreting strict and inclusive inequalities in a set definition. The condition 2<x excludes 2, while x≤6 includes 6. Because x is natural, the resulting set is U={3,4,5,6}. Thus 3, 4, and 6 are members, but 2 is not. Option D is therefore the only element outside U; confusing < with ≤ would lead to a wrong choice.

A factor of 6 is a natural number that divides 6 exactly, leaving remainder zero. The factor pairs are 1×6 and 2×3, so the positive natural factors are 1, 2, 3, and 6. Zero cannot be a factor because division by zero is undefined, and 12 is not a divisor of 6. Therefore option A is the complete roster form of V.

The governing process is substitution in a set-builder rule. Whole numbers less than 4 are n=0,1,2,3. Substituting into x=2n+1 gives x=1,3,5,7 respectively. Hence C={1,3,5,7}, which is option A. Option B incorrectly starts with n=1; option C ignores the rule, and option D includes 0, which the formula never produces for these values.

The governing concept is counting integers in a half-open interval. Since -3<x, the integer -3 is excluded; since x≤2, the endpoint 2 is included. Listing the permitted integers gives -2,-1,0,1,2. There are five distinct values, so the cardinality of D is 5. Therefore option B is correct; counts of 4, 6, or 7 result from mishandling an endpoint or adding an outside integer.

A roster form lists every element explicitly. The positive multiples of 6 below 30 are obtained as 6×1=6, 6×2=12, 6×3=18, and 6×4=24. The next multiple, 30, is excluded because the condition says less than 30, and 0 is not positive. Thus E={6,12,18,24}, so option A is correct.

Using nonnegative integer exponents, the relevant powers are 10⁰=1, 10¹=10, and 10²=100. The condition x<100 includes 1 and 10 but excludes 100, so G={1,10}. Zero is not the value of 10⁰; it is the exponent in that expression. Therefore option A is correct, while the other choices include the boundary or an invalid value.

The governing concept is membership under inclusive bounds. Because 1≤x≤4 includes both endpoints, the set is H={1,2,3,4}. Consequently, 1∈H, 3∈H, and 4∈H are true statements. The number 5 lies beyond the upper limit, so 5∈H is false. Hence option C is the required answer; treating the question as asking for a true statement would reverse the task.

The governing concept is identifying positive divisors and writing them in roster form. The factor pairs of 9 are 1×9 and 3×3, so its positive factors are 1, 3, and 9. Six is not a factor because 9÷6 is not an integer, and zero cannot be a factor because division by zero is undefined. Therefore option A gives the complete set.

The governing idea is converting strict inequalities into a roster form. Greater than 2 means x>2, so 2 is excluded; less than 9 means x<9, so 9 is also excluded. The natural numbers between these boundaries are 3,4,5,6,7,8. Thus option B contains exactly the required members. The other options include one or both excluded endpoints.

A constant term is a term that contains no variable. In 7a - 4, the term 7a contains the variable a, whereas -4 contains no variable. Therefore, -4 is the constant term. The option 7 is only the coefficient of a in 7a; it is not a separate term in the expression. Exam tip: To identify the constant term, choose the term that has no variable.