General chapter practice

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.

Test the rule 2n² for the first five natural numbers. For n=1,2,3,4,5, the values are 2(1²)=2, 2(2²)=8, 2(3²)=18, 2(4²)=32, and 2(5²)=50. These exactly reproduce every element of I, so option A is correct. The ordinary squares give 1,4,9,16,25; the first even numbers are different, and the multiples of 2 less than 50 form a much larger set.

The governing idea is to list the natural numbers satisfying both conditions. Taking N as {1,2,3,...}, the numbers not exceeding 10 are 1 through 10. The positive factors of 10 are 1, 2, 5 and 10. Removing these from the list leaves 3, 4, 6, 7, 8 and 9. Therefore option A is correct. Option B lists the excluded factors, while C and D either include a factor or incorrectly include 0.

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.

A number divisible by both 4 and 5 must be a multiple of their least common multiple. Since lcm(4,5)=20, every such positive natural number is at least 20; the first one is 20 itself. However, the condition requires x<20, so 20 is excluded and no natural number satisfies both conditions. Therefore N is the empty set, making option A correct. Options B, C and D fail the divisibility-and-bound test.

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.