General chapter practice

The governing concept is the empty set: it has no element satisfying its defining condition. For every natural number x, x² is non-negative, so x² = -1 has no natural-number solution. Therefore C contains no element and is empty. In contrast, A contains 4, B contains -4 and 4, and D contains 0 and 1. Hence option C is the only correct answer.

The governing idea is the cardinality of a finite set, found by counting its distinct members. The positive multiples of 11 below 100 are 11×1 through 11×9: 11, 22, 33, 44, 55, 66, 77, 88 and 99. The next multiple is 110, which is not less than 100. Thus L has 9 elements, so option B is correct.

The governing concept is filtering a set by all stated conditions. Two-digit natural numbers below 20 are 10 through 19. Among these, 11, 13, 17 and 19 have exactly two positive factors, so they are prime. Number 10 is composite, 12 is composite, and 2, 3, 5 and 7 are only one-digit primes. Therefore option A gives exactly M.

The governing concept is membership in a set of whole numbers. Whole numbers are 0, 1, 2, 3, 4, and so on. Applying x≤4 gives N={0,1,2,3,4}. Hence 0 and 4 belong to N, whereas 5 does not. Therefore 0∈N, making option D correct. The important distinction is that zero is included in the whole-number system.

The governing concept is roster form: list every integer satisfying all conditions, without adding endpoints excluded by strict inequalities. From -5<x<5, the possible integers are -4 through 4. Selecting the odd ones gives -3, -1, 1 and 3. Thus option A is correct. Option B wrongly includes -5 and 5, C lists even integers, and D ignores the odd-number condition.

The governing concept is testing set membership by checking every defining condition. The natural multiples of 7 not exceeding 35 are 7, 14, 21, 28 and 35. Among the choices, 28 is divisible by 7 and satisfies 28≤35. Number 30 is not divisible by 7, while 36 and 42 exceed the upper limit. Therefore option A belongs to R.

The governing concept is cardinality: n(T) is the number of distinct elements in T. A positive factor divides 21 exactly. Since 21=1×21=3×7, its positive factors are 1, 3, 7 and 21. There are four distinct factors, so n(T)=4 and option C is correct. The factors 1 and 21 must not be omitted, and negative factors are excluded by “positive.”

The governing idea is roster form: list every element satisfying the stated condition exactly once. Whole numbers are W={0,1,2,3,...}. The values less than 3 are 0, 1, and 2. The word “or” adds 5 as another member, so R={0,1,2,5}. Option A omits 0, option C incorrectly includes 3, and option D omits the values below 3.

The governing concept is cardinality, or the number of distinct members of a set. The natural factors of 16 are 1, 2, 4, 8, and 16. The condition x=10 contributes one additional element because 10 is not already a factor of 16. Thus S={1,2,4,8,10,16}, which has six distinct elements, so option C is correct. No element is counted twice.

The governing concept is intersection of an integer interval with the set of even integers. First list the integers satisfying -6≤x<0: -6,-5,-4,-3,-2,-1. Among them, the even integers are -6, -4, and -2. The endpoint 0 is excluded because the inequality is strict, and negative integers can still be even when divisible by 2. Therefore option A is correct.

The governing inequality is “not greater than 4,” which means x≤4. With the convention used here, N={1,2,3,...}; therefore the natural numbers satisfying the condition are 1, 2, 3, and 4. The equality sign includes the boundary value 4, while values 5 and above are excluded. Option B would be correct only under a convention that includes 0 in N, but the question’s expected convention is positive natural numbers.

The governing idea is intersection: a number must be both a factor of 100 and a two-digit natural number. The positive factors of 100 are 1,2,4,5,10,20,25,50,100. Two-digit numbers range from 10 through 99, so the qualifying factors are 10,20,25, and 50. The one-digit factors and 100 are excluded, while 40 is not a factor of 100. Thus option A is correct.

The governing concept is equality of sets: two sets are equal when they contain exactly the same elements, regardless of order. Solving x²=4 over the integers gives x=2 or x=-2, since 2²=4 and (-2)²=4. Therefore X={-2,2}, which has precisely the same members as Y={-2,2}. Thus X=Y, so option A is correct; X is not just {2}, and Y contains no 4.

The governing concept is selecting elements that satisfy both restrictions. Under the convention N={1,2,3,...}, one-digit natural numbers are 1 through 9. The positive multiples of 4 in that range are 4 and 8 only. Zero is not included under this convention, 12 has two digits, and 1 is not divisible by 4. Hence Y={4,8}, making option A correct.

The governing concept is translating inequalities into roster form. The strict inequality 4<x excludes 4, while x≤10 includes the upper endpoint 10. Listing the natural numbers that satisfy both conditions gives 5,6,7,8,9,10. Option B incorrectly includes 4, and option C incorrectly omits 10. Therefore the correct roster form is option A.

Use the inequality-solving principle first: 2x−1≤7 gives 2x≤8, and division by the positive number 2 gives x≤4. Since x belongs to N, the possible values are I={1,2,3,4} under the usual school convention. This set has four members, so it is finite. It is not empty, not a singleton, and not infinite. Hence option C is correct.

The governing concepts are whole numbers, divisibility and a strict upper bound. Whole numbers include 0, and 0 is divisible by 15 because 0=15×0. The multiples of 15 that are less than 60 are 0, 15, 30 and 45. The value 60 is not allowed because the condition says “less than 60,” not “less than or equal to 60.” Therefore option B is correct.

Translate the verbal conditions into inequalities. “x is not greater than 7” means x≤7, while the second condition is x>2. Combining them gives 2<x≤7. The natural numbers in this interval are 3, 4, 5, 6 and 7, so option A is correct. The value 2 fails x>2, and values 8 and above fail x≤7. Option C wrongly omits 7.

The relevant number-theory fact is that 17 is prime. A prime number has exactly two positive factors: 1 and the number itself. Thus the set described is L={1,17}, which contains exactly two elements. It is therefore finite, but it is neither empty nor a singleton. It cannot be infinite because a fixed positive integer has only finitely many positive factors. Hence option C is correct.

Solve the defining equation over the integers: x²=1. Taking square roots gives x=1 or x=−1, and both values are integers. Verification confirms that 1²=1 and (−1)²=1. Therefore the roster form is M={−1,1}, so option B is correct. Option A omits the negative solution, option C includes 0 even though 0²=0, and option D incorrectly treats the solution set as empty.

The two inequalities impose both a lower and an upper bound equal to 8. A number that satisfies 8≤x and x≤8 must be exactly x=8. Since 8 is a natural number, it belongs to the domain and is valid. Thus P={8}, a singleton set, so option B is correct. The empty set has no member, while options C and D include numbers that fail at least one of the two bounds.

The governing ideas are the definition of whole numbers and the inclusive meaning of “or.” Whole numbers begin at 0, so the values less than 4 are 0, 1, 2 and 3. The separate condition x=6 adds 6 to the set. Combining these distinct values gives R={0,1,2,3,6}, so option B is correct. Option A omits 0, option C wrongly includes 4, and D omits all values below 4.

The governing concept is the roster form of a set and the meaning of a positive divisor. A positive divisor of 12 must be a positive integer that divides 12 exactly, leaving remainder zero. The factor pairs of 12 are 1×12, 2×6, and 3×4. Therefore, collecting one number from each pair and including every distinct factor gives 1, 2, 3, 4, 6, and 12. Hence option A is correct. Option B omits 1 and 12, so it is incomplete. Option C incorrectly includes 8, because 12÷8 is not an integer. Option D includes 0, but division by zero is undefined and zero cannot be a divisor. The listed elements in A are all positive and each divides 12 exactly.

A singleton set is a set containing exactly one distinct element. Using the standard school convention N={1,2,3,...}, examine each condition. In option A, the natural number must be greater than 2 and less than 4, so the only possible value is x=3. Thus the set is {3}, which has exactly one element, making A correct. Option B allows 2, 3, and 4, so it has three elements. Option C allows x=1, but if a convention included zero in N it could contain 0 and 1; in either convention it is not safely a singleton under the stated school convention. Option D allows 1 and 2 because 1²<9 and 2²<9, so it has two elements. Therefore only A satisfies the definition.

A divisor of 24 is a natural number that divides 24 exactly. The positive divisors are 1, 2, 3, 4, 6, 8, 12, and 24. The condition that x must be even removes the odd divisors 1 and 3, leaving 2, 4, 6, 8, 12, and 24. Hence option A gives the complete roster form; C omits 8, while B includes odd divisors.