Set Concepts

A collection is a set only if it is a well defined collection of objects.

(i) The collection of good students in your school — The word 'good' is not well defined as it varies from person to person. It is not a set.

(ii) The collection of the numbers between 30 and 45 — This is a well defined collection. It is a set.

(iii) The collection of tall people in your colony — The word 'tall' is not well defined as it varies from person to person. It is not a set.

(iv) The collection of interesting books in your school library — The word 'interesting' is not well defined as a book may be interesting to one person and not to another. It is not a set.

(v) The collection of books in the library and are of your interest — Here the books are described with reference to your own interest, so for any given book it can be decided, without doubt, whether it belongs to the collection or not. This is well defined. It is a set.

Hence, (ii) and (v) are sets; whereas (i), (iii) and (iv) are not sets because these collections are not well defined.

(i) Set {4, 5, 8} is same as the set {5, 4, 8} and the set {8, 4, 5}.

The elements of a set can be written in any order, so all three sets have the same elements.

True

(ii) Sets {a, b, m, n} and {a, a, m, b, n, n} are same.

Repeated elements are written only once, so {a, a, m, b, n, n} = {a, b, m, n}.

True

(iii) Set of letters in the word 'suchismita' is {s, u, c, h, i, m, t, a}.

The letters of 'suchismita' are s, u, c, h, i, s, m, i, t, a. Writing each letter only once, we get {s, u, c, h, i, m, t, a}.

True

(iv) Set of letters in the word 'MAHMOOD' is {M, A, H, O, D}.

The letters of 'MAHMOOD' are M, A, H, M, O, O, D. Writing each letter only once, we get {M, A, H, O, D}.

True

(i) Since 6 is an element of A, 6 ∈ A

(ii) Since 10 is not an element of B, 10 ∉ B

(iii) Since 18 is an element of B, 18 ∈ B

(iv) 6 + 3 = 9, which is an element of B, (6 + 3) ∈ B

(v) 15 − 9 = 6, which is not an element of B, (15 − 9) ∉ B

(vi) Since 12 is an element of A, 12 ∈ A

(vii) 6 + 8 = 14, which is not an element of A, (6 + 8) ∉ A

(viii) Since both 6 and 8 are elements of A, 6 and 8 ∈ A

(i) The odd whole numbers between 15 and 27 are 17, 19, 21, 23 and 25.

{17, 19, 21, 23, 25}

(ii) The letters of the word CHITAMBARAM are C, H, I, T, A, M, B, A, R, A, M. Writing each letter only once, we get C, H, I, T, A, M, B and R.

A = {c, h, i, t, a, m, b, r}

(iii) The even numbers from 15 to 26 are 16, 18, 20, 22, 24 and 26.

B = {16, 18, 20, 22, 24, 26}

(iv) The vowels used in the word ARITHMETIC are A, I, E, I. Writing each vowel only once, we get A, I and E.

P = {a, i, e}

(v) The first eight whole numbers are 0, 1, 2, 3, 4, 5, 6 and 7. Their squares are 0, 1, 4, 9, 16, 25, 36 and 49.

S = {0, 1, 4, 9, 16, 25, 36, 49}

(vi) The integers between 7 and 94 which are divisible by 6 are 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84 and 90.

{12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90}

(vii) The composite numbers between 2 and 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16 and 18.

C = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18}

(viii) The prime numbers from 2 to 23 are 2, 3, 5, 7, 11, 13, 17, 19 and 23.

D = {2, 3, 5, 7, 11, 13, 17, 19, 23}

(ix) The natural numbers below 30 which are divisible by 2 are 2, 4, 6, 8, ..., 28 and those divisible by 5 are 5, 10, 15, 20, 25. Taking all such numbers, we get 2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18, 20, 22, 24, 25, 26 and 28.

E = {2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18, 20, 22, 24, 25, 26, 28}

(x) The factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

F = {1, 2, 3, 4, 6, 8, 12, 24}

(xi) Three closed figures in Geometry are a triangle, a circle and a square.

G = {triangle, circle, square}

(xii) Here x ∈ W and x < 10, so x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9

H = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

(xiii) Here x ∈ N and 2x − 3 ≤ 17

As, 2x − 3 ≤ 17

⇒ 2x ≤ 17 + 3

⇒ 2x ≤ 20

⇒ x ≤ 10

So x = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

J = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(xiv) Here x is an integer and − 3 < x < 5, so x = −2, −1, 0, 1, 2, 3, 4

K = {−2, −1, 0, 1, 2, 3, 4}

(i) {3, 6, 9, 12, 15}

These are natural numbers divisible by 3 which are less than 18.

{x : x is a natural number divisible by 3; x < 18}

(ii) {2, 3, 5, 7, 11, 13, ...}

These are the prime numbers.

{x : x is a prime number}

(iii) {1, 4, 9, 16, 25, 36}

These are perfect squares of natural numbers up to 6, i.e. 1², 2², 3², 4², 5², 6².

{x : x is a perfect square natural number and x ≤ 36}

(iv) {0, 2, 4, 6, 8, 10, 12, ...}

These are whole numbers divisible by 2.

{x : x is a whole number divisible by 2}

(v) {Monday, Tuesday, Wednesday}

These are the first three days of the week.

{x : x is one of the first three days of the week}

(vi) {23, 25, 27, 29, ...}

These are odd natural numbers greater than or equal to 23.

{x : x is an odd natural number; x ≥ 23}

(vii) $\Bigg\lbrace\frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}, \frac{1}{8}\Bigg\rbrace$

Each element is of the form $\frac{1}{n}$ where n takes the natural number values from 3 to 8.

{x : x = $\frac{1}{n}$, where n is a natural number; 3 ≤ n ≤ 8}

(viii) {42, 49, 56, 63, 70, 77}

These are natural numbers divisible by 7 lying between 42 to 77 inclusive.

{x : x is a natural number divisible by 7; 42 ≤ x ≤ 77}

A = {x : x is a multiple of 2 and is less than 25}

The multiples of 2 less than 25 are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22 and 24.

A = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}

B = {x : x is a square of a natural number and is less than 25}

The squares of natural numbers less than 25 are 1² = 1, 2² = 4, 3² = 9 and 4² = 16 (since 5² = 25 is not less than 25).

B = {1, 4, 9, 16}

C = {x : x is a multiple of 3 and is less than 25}

The multiples of 3 less than 25 are 3, 6, 9, 12, 15, 18, 21 and 24.

C = {3, 6, 9, 12, 15, 18, 21, 24}

D = {x : x is a prime number less than 25}

The prime numbers less than 25 are 2, 3, 5, 7, 11, 13, 17, 19 and 23.

D = {2, 3, 5, 7, 11, 13, 17, 19, 23}

(i) A leap year has 366 days, so set A has 366 elements.

n(A) = 366

(ii) The numbers on the face of a clock are 1, 2, 3, ..., 12, so B has 12 elements.

n(B) = 12

(iii) C = {x : x ∈ N and x ≤ 7} = {1, 2, 3, 4, 5, 6, 7}, so C has 7 elements.

n(C) = 7

(iv) The letters of the word PANIPAT are P, A, N, I, P, A, T. Writing each letter only once, we get P, A, N, I and T, i.e. D = {P, A, N, I, T}.

n(D) = 5

(v) The prime numbers between 5 and 15 are 7, 11 and 13, i.e. E = {7, 11, 13}.

n(E) = 3

(vi) F = {x : x ∈ Z and −2 < x ≤ 5} = {−1, 0, 1, 2, 3, 4, 5}, so F has 7 elements.

n(F) = 7

(vii) The perfect square numbers x ∈ N with x ≤ 30 are 1, 4, 9, 16 and 25, i.e. G = {1, 4, 9, 16, 25}.

n(G) = 5

(i) {natural numbers more than 100} = {101, 102, 103, ...}, which never ends. Infinite set

(ii) A = {x : x is an integer between 1 and 2}. There is no integer between 1 and 2. Null set (φ)

(iii) B = {x : x ∈ W; x is less than 100} = {0, 1, 2, ..., 99}, which has a limited number of elements. Finite set

(iv) Set of mountains in the world has an unlimited number of elements. Infinite set

(v) {multiples of 8} = {8, 16, 24, 32, ...}, which never ends. Infinite set

(vi) {even numbers not divisible by 2}. Every even number is divisible by 2, so there is no such number. Null set (φ)

(vii) {squares of natural numbers} = {1, 4, 9, 16, ...}, which never ends. Infinite set

(viii) {coins used in India} has a limited number of elements. Finite set

(ix) C = {x | x is a prime number between 7 and 10}. The numbers between 7 and 10 are 8 and 9 which are not prime, so there is no prime number between 7 and 10. Null set (φ)

(x) Planets of the solar system are limited in number. Finite set

Two sets are disjoint if they have no element in common.

(i) {0, 1, 2, 6, 8} and {odd numbers less than 10} = {1, 3, 5, 7, 9}. The element 1 is common to both sets. Not disjoint.

(ii) {birds} and {trees}. A bird is not a tree, so they have no element in common. Disjoint.

(iii) {x : x is a fan of cricket} and {x : x is a fan of football}. A person can be a fan of both cricket and football, so the sets can have common elements. Not disjoint.

(iv) A = {natural numbers less than 10} = {1, 2, 3, ..., 9} and B = {x : x is a multiple of 5} = {5, 10, 15, ...}. The element 5 is common to both sets. Not disjoint.

(v) {people living in Calcutta} and {people living in West Bengal}. Calcutta is a city in West Bengal, so the people living in Calcutta also live in West Bengal. Not disjoint.

Hence, only the pair in (ii) is disjoint.

Two sets are equal if they have exactly the same elements, and equivalent if they have the same number of elements.

(i) A = {first four natural numbers} = {1, 2, 3, 4} and B = {first four whole numbers} = {0, 1, 2, 3}.

n(A) = 4 = n(B), but the elements are different.

Equivalent

(ii) A = {letters of FOLLOW} = {F, O, L, W} and B = {letters of WOLF} = {W, O, L, F}.

Both sets have exactly the same elements.

Equal

(iii) E = {even natural numbers less than 10} = {2, 4, 6, 8} and O = {odd natural numbers less than 9} = {1, 3, 5, 7}.

n(E) = 4 = n(O), but the elements are different.

Equivalent

(iv) A = {days of the week starting with letter S} = {Saturday, Sunday} and B = {days of the week starting with letter T} = {Tuesday, Thursday}.

n(A) = 2 = n(B), but the elements are different.

Equivalent

(v) M = {multiples of 2 and 3 between 10 and 20} = {multiples of 6 between 10 and 20} = {12, 18} and N = {multiples of 2 and 5 between 10 and 20} = {multiples of 10 between 10 and 20} = { }.

n(M) = 2 and n(N) = 0, so the sets are neither equal nor equivalent.

None

(vi) 70 = 2 × 5 × 7, so P = {prime numbers which divide 70 exactly} = {2, 5, 7}.

105 = 3 × 5 × 7, so Q = {prime numbers which divide 105 exactly} = {3, 5, 7}.

n(P) = 3 = n(Q), but the elements are different.

Equivalent

(vii) A = {0², 1², 2², 3², 4²} = {0, 1, 4, 9, 16} and B = {16, 9, 4, 1, 0} = {0, 1, 4, 9, 16}.

Both sets have exactly the same elements.

Equal

(viii) E = {8, 10, 12, 14, 16} and F = {even natural numbers between 6 and 18} = {8, 10, 12, 14, 16}.

Both sets have exactly the same elements.

Equal

(ix) A = {letters of SUPERSTITION} = {S, U, P, E, R, T, I, O, N}, so n(A) = 9.

B = {letters of JURISDICTION} = {J, U, R, I, S, D, C, T, O, N}, so n(B) = 10.

Since n(A) ≠ n(B), the sets are neither equal nor equivalent.

None

(i) The set of triangles having three equal sides. An equilateral triangle has three equal sides, so such triangles exist. Not an empty set.

(ii) The set of lions in your class. There are no lions in a class. Empty set.

(iii) {x : x + 3 = 2 and x ∈ N}

x + 3 = 2

⇒ x = 2 − 3

⇒ x = −1

Since −1 is not a natural number, there is no such x. Empty set.

(iv) P = {x : 3x = 0}

⇒ 3x = 0

⇒ x = 0

So P = {0}, which has one element. Not an empty set.

Hence, the sets in (ii) and (iii) are the empty sets.

(i) All examples of the empty set are equal. There is one and only one empty set, so all its examples are equal. True

(ii) All examples of the empty set are equivalent. Each empty set has cardinal number 0, so they are equivalent. True

(iii) If two sets have the same cardinal number, they are only equivalent, not necessarily equal. False

(iv) If n(A) = n(B), then A and B are equivalent sets. By definition of equivalent sets, this is correct. True

(v) If B = {x : x + 4 = 4}, then B is not the empty set. Here x + 4 = 4 ⇒ x = 0, so B = {0}, which is not empty. False

(vi) The set of all points in a line is not a finite set. A line contains an unlimited number of points, so it is an infinite set. False

(vii) The set of letters in your Mathematics book is not an infinite set. The letters used are from the English alphabet, which are limited (26), so it is a finite set. False

(viii) If M = {1, 2, 4, 6} and N = {x : x is a factor of 12}, here N = {1, 2, 3, 4, 6, 12} ≠ M. False

(ix) The set of whole numbers greater than 50 is an infinite set. It is {51, 52, 53, ...}, which never ends. True

(x) If A and B are two different infinite sets, then n(A) = n(B). Two infinite sets are always equivalent. True

φ — This is the symbol for the empty set, so it represents the null set.

{0} — This is a set with one element 0, so it is not the null set.

0 — This is a number, not a set.

{ } — This is a pair of braces with no element in it, so it represents the null set.

{φ} — This is a set with one element φ, so it is not the null set.

Hence, φ and { } represent the null set.

(i) If each element of set P is also an element of set Q, then P is said to be subset of Q and Q is said to be superset of P.

(ii) Every set is a subset of itself.

(iii) The empty set is a subset of every set.

(iv) If A is proper subset of B, then n(A) is less than n(B).

(i) B = {5, 8}. Both 5 and 8 are elements of A. B is a subset of A.

(ii) C = {0}. 0 is not an element of A. C is not a subset of A.

(iii) D = {7, 9, 10}. 10 is not an element of A. D is not a subset of A.

(iv) E = { }. The empty set is a subset of every set. E is a subset of A.

(v) F = {8, 7, 9, 5}. All elements 8, 7, 9 and 5 are in A. F is a subset of A.

Hence, (i), (iv) and (v) are subsets of A.

A set is a proper subset of P if all its elements are in P and it is not equal to P.

(i) A = {3, 4}. Both 3 and 4 are in P, and A ≠ P. A is a proper subset of P.

(ii) B = { }. The empty set is a proper subset of every non-empty set. B is a proper subset of P.

(iii) C = {23, 45}. 23 and 45 are not in P. C is not a proper subset of P.

(iv) D = {6, 5, 4}. 6 is not in P. D is not a proper subset of P.

(v) E = {0}. 0 is not in P. E is not a proper subset of P.

Hence, A and B are proper subsets of P.

A = {even numbers less than 12} = {2, 4, 6, 8, 10}, B = {2, 4}, C = {1, 2, 3}, D = {2, 6}, E = {4}.

(i) B ⊂ A. The elements 2 and 4 of B are in A and B ≠ A. True

(ii) C ⊆ A. The elements 1 and 3 of C are not in A. False

(iii) D ⊂ C. The element 6 of D is not in C. False

(iv) D ⊄ A. The elements 2 and 6 of D are both in A, so D ⊂ A. Hence D ⊄ A is incorrect. False

(v) E ⊇ B. This means B ⊆ E. The element 2 of B is not in E = {4}. False

(vi) A ⊇ B ⊇ E. Here B ⊆ A (2, 4 are in A) and E ⊆ B (4 is in B). So A ⊇ B ⊇ E holds. True

Hence, statements (i) and (vi) are true.

A = {a, c}, B = {p, q, r}.

C = Set of digits used to form the number 1351. The digits are 1, 3, 5, 1. Writing each digit only once, C = {1, 3, 5}.

Subsets of A : { }, {a}, {c}, {a, c}

Subsets of B : { }, {p}, {q}, {r}, {p, q}, {p, r}, {q, r}, {p, q, r}

Subsets of C : { }, {1}, {3}, {5}, {1, 3}, {3, 5}, {1, 5}, {1, 3, 5}

If a set has n elements, then the number of its subsets is 2ⁿ and the number of its proper subsets is 2ⁿ − 1.

(i) A = {p, q, r} has 3 elements, so the number of subsets = 2³ = 8

(ii) B = {5, 4, 6, 8} has 4 elements, so the number of proper subsets = 2⁴ − 1 = 16 − 1 = 15

(iii) C = {0} has 1 element, so the number of subsets = 2¹ = 2

(iv) M = {x : x ∈ N and x < 3} = {1, 2} has 2 elements, so the number of proper subsets = 2² − 1 = 4 − 1 = 3

The universal set is {4, 5, 6, 7, 8, 9, 10, 11, 12, 13}.

(i) The even numbers in the universal set are 4, 6, 8, 10 and 12.

A = {4, 6, 8, 10, 12}

(ii) The odd numbers greater than 8 in the universal set are 9, 11 and 13.

B = {9, 11, 13}

(iii) The prime numbers in the universal set are 5, 7, 11 and 13.

C = {5, 7, 11, 13}

(iv) The even numbers less than 10 in the universal set are 4, 6 and 8.

D = {4, 6, 8}

A = {4, 5, 6, 7, 8} and B = {6, 8, 10, 12}.

(i) A ∪ B = {all elements of A and of B} = {4, 5, 6, 7, 8, 10, 12}

(ii) A ∩ B = {elements common to A and B} = {6, 8}

(iii) A − B = {elements of A which are not in B} = {4, 5, 7}

(iv) B − A = {elements of B which are not in A} = {10, 12}

A = {3, 5, 7, 9, 11} and B = {4, 7, 10}.

(i) A has 5 elements. n(A) = 5

(ii) B has 3 elements. n(B) = 3

(iii) A ∪ B = {3, 4, 5, 7, 9, 10, 11}

A ∪ B = {3, 4, 5, 7, 9, 10, 11} and n(A ∪ B) = 7

(iv) A ∩ B = {elements common to A and B} = {7}

A ∩ B = {7} and n(A ∩ B) = 1

A = {2, 4, 6, 8} and B = {3, 6, 9, 12}.

(i) A ∩ B = {elements common to A and B} = {6}

A ∩ B = {6} and n(A ∩ B) = 1

(ii) A − B = {elements of A which are not in B} = {2, 4, 8}

A − B = {2, 4, 8} and n(A − B) = 3

(iii) B has 4 elements. n(B) = 4

P = {x : x is a factor of 12} = {1, 2, 3, 4, 6, 12}.

Q = {x : x is a factor of 16} = {1, 2, 4, 8, 16}.

(i) P has 6 elements. n(P) = 6

(ii) Q has 5 elements. n(Q) = 5

(iii) Q − P = {elements of Q which are not in P} = {8, 16}

Q − P = {8, 16} and n(Q − P) = 2

M = {x : x is a natural number between 0 and 8} = {1, 2, 3, 4, 5, 6, 7}.

N = {x : x is a natural number from 5 to 10} = {5, 6, 7, 8, 9, 10}.

(i) M − N = {elements of M which are not in N} = {1, 2, 3, 4}

M − N = {1, 2, 3, 4} and n(M − N) = 4

(ii) N − M = {elements of N which are not in M} = {8, 9, 10}

N − M = {8, 9, 10} and n(N − M) = 3

A = {x : x is a natural number divisible by 2 and x < 16} = {2, 4, 6, 8, 10, 12, 14}.

B = {x : x is a whole number divisible by 3 and x < 18} = {0, 3, 6, 9, 12, 15}.

(i) A has 7 elements. n(A) = 7

(ii) B has 6 elements. n(B) = 6

(iii) A ∩ B = {elements common to A and B} = {6, 12}

A ∩ B = {6, 12} and n(A ∩ B) = 2

(iv) A − B = {elements of A which are not in B} = {2, 4, 8, 10, 14}

n(A − B) = 5

Given n(A) = 75, n(B) = 65 and n(A ∩ B) = 45.

(i) n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

= 75 + 65 − 45

= 140 − 45

n(A ∪ B) = 95

(ii) n(A − B) = n(A) − n(A ∩ B)

= 75 − 45

n(A − B) = 30

(iii) n(B − A) = n(B) − n(A ∩ B)

= 65 − 45

n(B − A) = 20

Given n(A) = 45, n(B) = 38 and n(A ∪ B) = 70.

(i) n(A ∩ B) = n(A) + n(B) − n(A ∪ B)

= 45 + 38 − 70

= 83 − 70

n(A ∩ B) = 13

(ii) n(A − B) = n(A) − n(A ∩ B)

= 45 − 13

n(A − B) = 32

(iii) n(B − A) = n(B) − n(A ∩ B)

= 38 − 13

n(B − A) = 25

Given n(A) = 30, n(B) = 27 and n(A ∪ B) = 45.

(i) n(A ∩ B) = n(A) + n(B) − n(A ∪ B)

= 30 + 27 − 45

= 57 − 45

n(A ∩ B) = 12

(ii) n(A − B) = n(A) − n(A ∩ B)

= 30 − 12

n(A − B) = 18

Given n(A) = 31, n(B) = 20 and n(A ∩ B) = 6.

(i) n(A − B) = n(A) − n(A ∩ B)

= 31 − 6

n(A − B) = 25

(ii) n(B − A) = n(B) − n(A ∩ B)

= 20 − 6

n(B − A) = 14

(iii) n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

= 31 + 20 − 6

= 51 − 6

n(A ∪ B) = 45

The elements 0, 3, 6, 9, 12, ... are multiples of 3 starting from 0. So x = 3n where n must include 0, i.e. n ∈ W.

When n ∈ W, x = 3n gives 0, 3, 6, 9, 12, ...

A = {x : x = 3n and n ∈ W}

Hence, Option 4 is the correct option.

Here n ∈ N and n < 5, so n = 1, 2, 3, 4.

When n = 1, x = 1² − 1 = 0

When n = 2, x = 2² − 1 = 3

When n = 3, x = 3² − 1 = 8

When n = 4, x = 4² − 1 = 15

So the set is {0, 3, 8, 15}.

Hence, Option 3 is the correct option.

Since n(A) = n(B) = 8, the two sets have the same number of elements, so they are equivalent. They need not have the same elements, so they need not be equal.

A and B are equivalent sets

Hence, Option 2 is the correct option.

Here n ∈ W and n < 3, so n = 0, 1, 2.

When n = 0, x = 0³ = 0

When n = 1, x = 1³ = 1

When n = 2, x = 2³ = 8

So A = {0, 1, 8}, which has 3 elements.

Number of subsets = 2³.

Hence, Option 2 is the correct option.

A = {letters of JANTAR} = {J, A, N, T, R}

B = {letters of AJANTA} = {A, J, N, T}

Every element of B is in A, and B ≠ A (since R ∈ A but R ∉ B). So B is a proper subset of A.

B ⊂ A

Hence, Option 3 is the correct option.

We need integers x with x² < 16, i.e. −4 < x < 4.

The integers satisfying this are −3, −2, −1, 0, 1, 2 and 3 (note that (±4)² = 16, which is not less than 16).

So, the roster form is {−3, −2, −1, 0, 1, 2, 3}.

Hence, Option 3 is the correct option.

A − B = {elements of A which are not in B}.

Since A and B have no common element, all elements of A remain.

So A − B = {7, 8, 9, 10}.

Hence, Option 2 is the correct option.

S = {x : x ∈ Z and −2 < x ≤ 2} = {−1, 0, 1, 2}

A = {−1, 0, 1}

A′ = S − A = {elements of S which are not in A} = {2}

Hence, Option 2 is the correct option.

If a set has n elements, the number of its proper subsets is 2ⁿ − 1.

Here n = 5, so the number of proper subsets = 2⁵ − 1.

Hence, Option 2 is the correct option.

A proper subset of A must contain only elements of A and must not be equal to A.

{6, 7, 8} contains 8 ∉ A; {3, 4} contains 3 ∉ A; {8} contains 8 ∉ A. So these are not subsets of A.

The empty set { } is a proper subset of every non-empty set.

Hence, Option 4 is the correct option.

A ∪ B = {all elements of A and of B} = {0, 1, 2, 3, 6, 7, 8, 9}.

Hence, Option 2 is the correct option.

B ∩ A = {elements common to A and B}. Since A and B have no common element, B ∩ A = { }.

Hence, Option 3 is the correct option.

For any two sets A and B, the correct cardinal property is

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

which can be rearranged as n(A ∪ B) + n(A ∩ B) = n(A) + n(B).

Hence, Option 4 is the correct option.

The number of proper subsets of a set with n elements is 2ⁿ − 1.

Given 2ⁿ − 1 = 2⁴ − 1, so n = 4.

Hence, Option 4 is the correct option.

If every element of A is in B (A ⊆ B) and every element of B is in C (B ⊆ C), then every element of A is in C.

So A ⊆ C.

Hence, Option 4 is the correct option.

Statement 1 : A = {x, y, z, w}, so n(A) = 4. B = {x | x ≤ 4, x ∈ N} = {1, 2, 3, 4}, so n(B) = 4. Since n(A) = n(B), the sets are equivalent. Statement 1 is true.

Statement 2 : Equal sets have the same elements, so they are always equivalent. However, equivalent sets only have the same number of elements and may or may not be equal. Statement 2 is true.

Hence, Option 1 is the correct option.

The smallest composite number is 4, so there is no composite number less than 4. Hence A = { }, which is an empty set.

Thus, assertion (A) is true.

{0} is a set with one element 0, so it is not an empty set.

Thus, reason (R) is true.

Hence, Option 3 is the correct option.

A = {1, 3, 5} and B = {5, 1, 3} = {1, 3, 5}. Since both sets have the same elements, A ⊆ B and B ⊆ A.

Thus, assertion (A) is true.

If B is a subset of A, then each element of B is present in A, not each element of A present in B.

Thus, reason (R) is false.

Hence, Option 1 is the correct option.

No number is both prime and composite, so sets A and B have no element in common, which means they are disjoint sets. Hence the statement that they are not disjoint is false.

Thus, assertion (A) is false.

Two non-empty sets are said to be disjoint if they do not have any element in common, which is the correct definition.

Thus, reason (R) is true.

Hence, Option 2 is the correct option.

A = {a, b, c, d} and B = {a, b, c, d, e, f}. Every element of A is in B and A ≠ B, since B has the extra elements e and f. So A is a proper subset of B.

Thus, assertion (A) is true.

No set is a proper subset of itself, which is a correct statement.

Thus, reason (R) is true.

Hence, Option 3 is the correct option.