General chapter practice

The listed elements 3, 6, 9, and 12 are all multiples of 3. Restricting x to the inclusive interval 3≤x≤12 gives exactly those four multiples: 3, 6, 9, and 12. Thus option A reproduces J without adding or losing an element. The other descriptions produce different sets: factors of 3 are much fewer, and even or prime numbers do not match the roster.

The governing concept is set-builder notation: the stated rule must generate every element of B and no extra element. The even whole numbers less than 8 are 0, 2, 4, and 6, exactly B, so option A is correct. Option C may exclude 0 and includes odd natural numbers; B describes odd digits, and D describes factors of 6, neither of which matches B.

The governing concept is subset notation. G⊆H means every element of G must also belong to H. Since both 1 and 2 are in H, G⊆H is true, so option A is correct. H is not a subset of G because H contains 3, which is absent from G. Therefore option B is false; 3∈G is false, and the sets cannot be equal because their elements differ.

The governing concept is the subset condition: every element of a proposed subset must belong to the original set. Both 2 and 4 are elements of I={2,3,4}, so {2,4} is a subset and option A is correct. Option B contains 5, C contains 1, and D contains 5; none of those outside elements belongs to I, so those options are not subsets.

The elements of B are obtained by multiplying 3 by 0, 1, 2, 3 and 4: 3(0) = 0, 3(1) = 3, 3(2) = 6, 3(3) = 9 and 3(4) = 12. Since zero is needed, the parameter must come from W, the whole numbers. Option B uses N, which excludes zero under the stated convention; C gives even numbers and D omits 0. Hence A is correct.

Solve the defining equation before applying the set restriction. Factorisation gives x² − 5x + 6 = (x − 2)(x − 3) = 0, so x = 2 or x = 3. Both roots are natural numbers, hence both belong to D and D = {2, 3}. The negative values are not roots of this factorisation, 1 and 6 do not make the expression zero, and 0 is not a solution either.

The governing concept is set-builder notation, which describes a common rule and the allowed values of a variable. The elements are successive powers 2¹, 2², 2³, 2⁴ and 2⁵, so x=2ⁿ with n∈N and 1≤n≤5 gives exactly P. Option B gives multiples of 2, C gives squares, and D includes 2⁰=1 while omitting 32. Hence A is correct.

The governing concept is converting a set-builder condition into roster form. Solve 2x+1<10: subtracting 1 gives 2x<9, and dividing by 2 gives x<4.5. Under the stated convention N={1,2,3,...}, the allowed values are 1,2,3,4. Therefore A={1,2,3,4}, so option A is correct. Zero is excluded here, and 5 fails the inequality.

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.

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.

For an integer x, the inequality |x|<3 means that x is less than 3 units from zero. Equivalently, −3<x<3. The integers in this open interval are −2, −1, 0, 1, and 2, so option B is correct. The endpoints −3 and 3 are excluded, while option C wrongly omits zero and option D omits negative integers.

Solve the defining equation by factoring: x²−5x+6=(x−2)(x−3). A product is zero when at least one factor is zero, so x−2=0 or x−3=0. Hence x=2 or x=3; both are integers and satisfy the original equation. Therefore E={2,3}, making option A correct. The other options contain incorrect or extra roots.