General chapter practice

The governing concept is the number of elements in a set. The English alphabet contains exactly 26 letters, so V has a fixed and limited number of elements. A set with a definite limited cardinality is called a finite set. It is not empty because letters exist, not infinite because the number is bounded, and the description clearly defines its members. Therefore, option A is correct.

The governing concept is that a set must be well-defined: membership should be decidable without personal opinion. A list of 40 students identifies a definite collection, so every member can be recognized objectively. The terms beautiful, good, and tall may be judged differently by different people unless a precise standard is supplied. Thus only the collection in option A is clearly a set.

A set must have a clear membership rule. ‘Tasty’ depends on personal taste, so people may disagree about which fruits belong to that collection; it is therefore not well-defined. The states of India, days of a week and natural numbers below 10 can be identified objectively. Hence option C is correct.

First solve the condition x² = 9. The two integer solutions are x = 3 and x = −3. The second condition restricts x to natural numbers; under the standard school convention, −3 is not natural, while 3 is natural. Therefore the set contains only one element, and its roster form is W = {3}. Option B is correct; option A ignores the natural-number restriction.

The governing idea is roster representation: list every element satisfying the stated rule. Integers include negative numbers, zero, and positive numbers. Because both inequalities are inclusive, the endpoints −2 and 2 must be included. Listing all integers from −2 through 2 gives −2, −1, 0, 1, and 2. Hence option A is the complete and correct roster form.

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 notation n(A) represents the cardinality of set A, meaning the number of distinct elements in the set. A contains the three different elements 2, 4, and 6, so n(A) = 3. The number 6 is merely one member of A, not the cardinality. Likewise, 2 and 4 are elements rather than counts. Therefore option B is correct.

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.