General chapter practice

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.

A number x is a square root of 16 when x² = 16. Over the integers, both −4 and 4 satisfy this equation because (−4)² = 16 and 4² = 16. However, the set specifically requires x ∈ N, and the given natural numbers are positive, so −4 is excluded. Therefore P = {4}. Option C confuses the square with its root, while 2 does not satisfy x² = 16.

Use the defining condition x² < 10 and remember that W contains only non-negative integers. We have 0² = 0, 1² = 1, 2² = 4 and 3² = 9, all less than 10. But 4² = 16, so 4 is excluded. Negative numbers cannot be included because they are not whole numbers. Hence the roster form is Q = {0, 1, 2, 3}, so option A is correct.

The listed elements are consecutive squares: 1 = 1², 4 = 2², 9 = 3², 16 = 4² and 25 = 5². In option A, n takes every natural-number value from 1 through 5, so x = n² produces exactly these five elements and no others. Option B stops before n = 5 and omits 25; option C gives even multiples, while option D gives 2, 3, 4, 5 and 6.

For an integer x, the condition |x| < 3 means that x is fewer than three units from zero. It is equivalent to −3 < x < 3. The integers strictly between these endpoints are −2, −1, 0, 1 and 2, so V = {−2, −1, 0, 1, 2}. The endpoints −3 and 3 are excluded because the inequality is strict, and zero is included because |0| = 0.

The smallest two-digit number is 10. The question asks for natural numbers strictly less than 10, and the stated convention is N = {1, 2, 3, ...}. Therefore the members are 1, 2, 3, 4, 5, 6, 7, 8 and 9. Zero is excluded because it is not in the given N, while 10 is excluded because the condition says less than 10. Thus option A is correct.

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.

First list the positive factors of 18: 1, 2, 3, 6, 9 and 18. Then apply the second condition, namely that the factor must be prime. Only 2 and 3 have exactly two positive divisors. The number 1 is neither prime nor composite, while 6, 9 and 18 are composite. Therefore E = {2, 3}, making option A correct.

The inequalities are strict, so the endpoints 10 and 20 are not included. The natural numbers between them are 11, 12, 13, 14, 15, 16, 17, 18 and 19. Selecting only the even values leaves 12, 14, 16 and 18. Thus H has four elements, written n(H) = 4. Counting 10 or 20 would incorrectly treat < as ≤, and counting every integer would ignore the evenness condition.

The positive multiples of 6 below 40 are found by calculating 6 × 1 through 6 × 6: 6, 12, 18, 24, 30 and 36. Therefore 18, 24 and 36 satisfy both conditions and belong to K. Although 42 is also divisible by 6, it is greater than 40, so it fails the phrase “less than 40.” Hence 42, option D, does not belong to K.

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 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 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 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.