Sets

Any well-defined collection of objects is called a set. By 'well-defined collection of objects' we mean that given a collection and an object, it should be possible to decide beyond doubt whether the object belongs to the given collection or not.

(i) A collection of popular cinema actors of India:

It is not a well-defined collection, since an actor considered popular by one person may not be considered popular by another.

Hence, this collection is not a set.

(ii) The collection of even natural numbers less than 51:

Given any object, we can decide beyond doubt whether it is an even natural number less than 51 or not. The collection is 2, 4, 6, ....., 50.

Hence, this collection is a set.

(iii) The collection of counting numbers less than 1:

The counting numbers are 1, 2, 3, ..... and we can decide beyond doubt that none of them is less than 1. So the collection is well-defined, though it contains no object at all.

Hence, this collection is a set (it is the empty set).

(iv) Collection of interesting books written by Shakespeare:

It is not a well-defined collection, since a book considered interesting by one person may not be considered interesting by another.

Hence, this collection is not a set.

(v) The collection of novels written by Munshi Prem Chand:

Given any novel, we can decide beyond doubt whether it was written by Munshi Prem Chand or not.

Hence, this collection is a set.

(vi) The collection of 10 most talented students of your school:

It is not a well-defined collection, since a student considered talented by one person may not be considered talented by another. So the 10 students cannot be fixed beyond doubt.

Hence, this collection is not a set.

(vii) Collection of all rivers flowing in India:

Given any river, we can decide beyond doubt whether it flows in India or not.

Hence, this collection is a set.

(viii) Collection of 5 rivers flowing in India:

Many rivers flow in India and it is not stated which 5 of them are to be taken. So, given a river, we cannot decide whether it belongs to this collection or not.

Hence, this collection is not a set.

(ix) Collection of all rational numbers which lie between −1 and 1:

Given any number, we can decide beyond doubt whether it is rational and whether it lies between −1 and 1.

Hence, this collection is a set.

(x) A team of eleven best cricketers of the world:

It is not a well-defined collection, since a cricketer considered best by one person may not be considered best by another.

Hence, this collection is not a set.

(xi) A collection of most dangerous animals of the world:

It is not a well-defined collection, since an animal considered most dangerous by one person may not be considered most dangerous by another.

Hence, this collection is not a set.

Given,

A = {3, 5, 7, 9, 11}

The elements of A are the five numbers 3, 5, 7, 9 and 11.

(i) 3 ∈ A

The element 3 belongs to the set A.

Hence, the statement 3 ∈ A is true.

(ii) 5, 9 ∈ A

Both the elements 5 and 9 belong to the set A.

Hence, the statement 5, 9 ∈ A is true.

(iii) 8 ∉ A

The number 8 is not one of the elements 3, 5, 7, 9, 11 of A, so 8 does not belong to A.

Hence, the statement 8 ∉ A is true.

(iv) 9 ∉ A

The number 9 is one of the elements of A, so 9 belongs to A i.e. 9 ∈ A.

Hence, the statement 9 ∉ A is not true, because 9 ∈ A.

(v) {3} ∈ A

Here {3} is a set having one element 3. The elements of A are the numbers 3, 5, 7, 9, 11 and not the set {3}.

Hence, the statement {3} ∈ A is not true, because {3} is a set and not an element of A.

(vi) {5, 7} ∈ A

Here {5, 7} is a set having two elements 5 and 7. The elements of A are numbers, not sets, so the set {5, 7} is not an element of A.

Hence, the statement {5, 7} ∈ A is not true, because {5, 7} is a set and not an element of A.

The multiples of 6 are 6, 12, 18, 24, 30, 36, 42, 48, 54, ..... Out of these, the ones less than 50 are 6, 12, 18, 24, 30, 36, 42 and 48.

Hence, the required set in roster form is {6, 12, 18, 24, 30, 36, 42, 48}.

The counting numbers less than 10 are 1, 2, 3, 4, 5, 6, 7, 8, 9. Taking each of these as the denominator with numerator 1, we get the fractions $\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}, \frac{1}{8}, \frac{1}{9}$.

Hence, the required set in roster form is $\bm{\bigg\lbrace1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}, \frac{1}{6}, \frac{1}{7}, \frac{1}{8}, \frac{1}{9}\bigg\rbrace}$.

Resolving 84 into prime factors,

84 = 2 × 2 × 3 × 7 = 2² × 3 × 7

∴ The prime factors of 84 are 2, 3 and 7.

Hence, the required set in roster form is {2, 3, 7}.

The integers lying between −4 and 8 are −3, −2, −1, 0, 1, 2, 3, 4, 5, 6, 7. Out of these the odd ones are −3, −1, 1, 3, 5 and 7.

Hence, the required set in roster form is {−3, −1, 1, 3, 5, 7}.

Here x ∈ N and x + 6 < 10

⇒ x < 10 − 6

⇒ x < 4

As x is a natural number less than 4, x = 1, 2, 3.

Hence, the required set in roster form is {1, 2, 3}.

Here x is an integer and x + 6 > 10

⇒ x > 10 − 6

⇒ x > 4

As x is an integer greater than 4, x = 5, 6, 7, .....

Hence, the required set in roster form is {5, 6, 7, .....}.

Here x is an integer and x + 6 < 10

⇒ x < 10 − 6

⇒ x < 4

As x is an integer less than 4, x = 3, 2, 1, 0, −1, −2, .....

Hence, the required set in roster form is {....., −2, −1, 0, 1, 2, 3}.

$\frac{60}{x}$ is a natural number only when x is a positive divisor of 60.

Resolving 60 into prime factors, 60 = 2² × 3 × 5, so the positive divisors of 60 are

1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

Hence, the required set in roster form is {1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60}.

As n ∈ N and n ≤ 6, n = 1, 2, 3, 4, 5, 6.

Putting these values in 3n + 5, we get

⇒ 3(1) + 5 = 8,

⇒ 3(2) + 5 = 11,

⇒ 3(3) + 5 = 14,

⇒ 3(4) + 5 = 17,

⇒ 3(5) + 5 = 20,

⇒ 3(6) + 5 = 23.

Hence, the required set in roster form is {8, 11, 14, 17, 20, 23}.

Here x ∈ N and 4x − 3 ≤ 15

⇒ 4x ≤ 15 + 3

⇒ 4x ≤ 18

⇒ x ≤ $\frac{18}{4}$

⇒ x ≤ 4.5

As x is a natural number not exceeding 4.5, x = 1, 2, 3, 4.

Hence, the required set in roster form is {1, 2, 3, 4}.

Here x ∈ N and x² < 40.

The squares of the natural numbers are 1, 4, 9, 16, 25, 36, 49, ..... Out of these, 1, 4, 9, 16, 25 and 36 are less than 40, and they are the squares of 1, 2, 3, 4, 5 and 6.

Hence, the required set in roster form is {1, 2, 3, 4, 5, 6}.

Here x ∈ Z and x² < 16.

The squares of the integers 0, ±1, ±2, ±3 are 0, 1, 4, 9 and each of these is less than 16, while (±4)² = 16 which is not less than 16.

Hence, the required set in roster form is {−3, −2, −1, 0, 1, 2, 3}.

In our number system every number is written with the help of the ten digits 0, 1, 2, ....., 9.

Hence, the required set in roster form is {0, 1, 2, ....., 9}.

The letters of the word TRIGONOMETRY are T, R, I, G, O, N, O, M, E, T, R, Y. While listing the elements of a set, it is sufficient to list each member only once, so the repeated letters O, T and R are written only once.

Hence, the required set in roster form is {T, R, I, G, O, N, M, E, Y}.

The vowels in the English alphabet are a, e, i, o, u. Out of these, a, e, i and o come before q while u comes after q.

Hence, the required set in roster form is {a, e, i, o}.

The letters of the English alphabet which precede k are a, b, c, d, e, f, g, h, i, j. Out of these, the consonants are b, c, d, f, g, h and j.

Hence, the required set in roster form is {b, c, d, f, g, h, j}.

(i) {1, 3, 5, 7, 9, 11, 13}

Each member of the given set is an odd natural number and the members begin with 1 and end with 13.

Hence, the given set in builder form is {x : x is an odd natural number less than 14}.

(ii) {2, 4, 6, 8, ...}

Each member of the given set is an even natural number and the listing does not end.

Hence, the given set in builder form is {x : x is an even natural number}.

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

The given members are

3 = 3 × 1, 6 = 3 × 2, 9 = 3 × 3, 12 = 3 × 4, 15 = 3 × 5, so each member is of the form 3n where n takes the values 1 to 5.

Hence, the given set in builder form is {x : x = 3n, n ∈ N and n ≤ 5}.

(iv) {2, 4, 8, 16, 32, 64}

The given members are

2 = 2¹, 4 = 2², 8 = 2³, 16 = 2⁴, 32 = 2⁵, 64 = 2⁶, so each member is of the form 2ⁿ where n takes the values 1 to 6.

Hence, the given set in builder form is {x : x = 2ⁿ, n ∈ N and n ≤ 6}.

(v) {5, 25, 125, 625}

The given members are

5 = 5¹, 25 = 5², 125 = 5³, 625 = 5⁴, so each member is of the form 5ⁿ where n takes the values 1 to 4.

Hence, the given set in builder form is {x : x = 5ⁿ, n ∈ N and n ≤ 4}.

(vi) {1, 4, 9, 16, ..., 100}

The given members are

1 = 1², 4 = 2², 9 = 3², 16 = 4², ....., 100 = 10², so each member is of the form n² where n takes the values 1 to 10.

Hence, the given set in builder form is {x : x = n², n ∈ N and n ≤ 10}.

(vii) {1, 4, 9, 16, 25, ...}

The given members are

1 = 1², 4 = 2², 9 = 3², 16 = 4², 25 = 5², ..... and the listing does not end, so each member is of the form n² where n is any natural number.

Hence, the given set in builder form is {x : x = n², n ∈ N}.

(viii) $\bigg\lbrace\frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, ...\bigg\rbrace$

In each member of the given set the denominator is one more than the numerator, and the numerator begins with 1 and goes on increasing without end. So each member is of the form $\frac{n}{n + 1}$ where n is any natural number.

Hence, the given set in builder form is $\bigg\lbrace x : x = \frac{n}{n + 1}, n \in \textbf{N}\bigg\rbrace$.

A set which does not contain any element is called the null set (or the empty set).

(i) Set of even prime numbers:

2 is a prime number and it is even, so the given set contains the element 2 i.e. the set is {2}.

Hence, this set is not a null set.

(ii) Set of odd natural numbers divisible by 2:

Every number divisible by 2 is an even number, so no odd natural number can be divisible by 2. Therefore the given set contains no element.

Hence, this set is a null set.

(iii) Set of all Indian kids 5 metres tall:

No kid in India is 5 metres tall, so the given set contains no element.

Hence, this set is a null set.

(iv) {x : x ∈ N, x < 5 and x > 8}

A natural number x cannot satisfy x < 5 and x > 8 at the same time, since a number less than 5 can never be greater than 8. So the given set contains no element.

Hence, this set is a null set.

(v) {x : x is a point common to any two parallel straight lines}

Two parallel straight lines never intersect, so they have no point in common. Therefore the given set contains no element.

Hence, this set is a null set.

(vi) {x : x is a student of your school presently studying in both classes XI and XII}

At a given time a student can be studying in only one class, so no student can be studying in both classes XI and XII at present. Therefore the given set contains no element.

Hence, this set is a null set.

A set that contains a limited (definite) number of different elements is called a finite set, and a set that contains an unlimited number of different elements is called an infinite set.

(i) The set of days of a week:

There are exactly 7 days in a week, namely Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday. So the number of elements is limited.

Hence, the set of days of a week is a finite set.

(ii) The set of numbers which are multiples of 7:

The multiples of 7 are 7, 14, 21, 28, ..... and this listing never comes to an end.

Hence, the set of numbers which are multiples of 7 is an infinite set.

(iii) The set of animals living on Earth:

At any given moment the number of animals living on Earth, though very large, is a definite number and can be counted.

Hence, the set of animals living on Earth is a finite set.

(iv) The set of consonants in the English alphabet:

The English alphabet has 26 letters out of which 5 are vowels, so the number of consonants is 26 − 5 = 21, which is a definite number.

Hence, the set of consonants in the English alphabet is a finite set.

(v) The set of circles drawn in a plane:

In a plane a circle can be drawn with any point as centre and any positive number as radius, so an unlimited number of circles can be drawn.

Hence, the set of circles drawn in a plane is an infinite set.

(vi) The set of prime numbers which are less than one crore:

The prime numbers less than one crore are 2, 3, 5, 7, 11, ....., and since only those below one crore are taken, they are definite in number.

Hence, the set of prime numbers which are less than one crore is a finite set.

The number of different elements in a finite set A is called the cardinal number of A and is denoted by n(A).

(i) { }

The set { } contains no element at all, so it is the empty set.

Hence, the cardinal number of { } is 0.

(ii) {0}

The set {0} contains exactly one element, namely the number 0.

Hence, the cardinal number of {0} is 1.

(iii) A = {1, 2, 2, 1, 3}

If one or more elements of a set are repeated, the set remains the same. So on writing each element only once,

A = {1, 2, 3}, which has 3 different elements.

Hence, the cardinal number of A is 3.

(iv) The set of all Indians having 8 legs:

No Indian has 8 legs, so this set is the empty set.

Hence, the cardinal number of the set of all Indians having 8 legs is 0.

(v) The set of all letters in the word PRINCIPAL:

The letters of the word PRINCIPAL are P, R, I, N, C, I, P, A, L. Writing each letter only once, the set is {P, R, I, N, C, A, L}, which has 7 different elements.

Hence, the cardinal number of the set of all letters in the word PRINCIPAL is 7.

(vi) The set of all vowels in the word PRINCIPAL:

The vowels occurring in the word PRINCIPAL are I, I and A. Writing each letter only once, the set is {I, A}, which has 2 different elements.

Hence, the cardinal number of the set of all vowels in the word PRINCIPAL is 2.

A two digit number has a tens digit from 1 to 9 and a units digit from 0 to 9. Taking the pairs of digits whose sum is 8, the numbers are

17, 26, 35, 44, 53, 62, 71, 80.

∴ A = {17, 26, 35, 44, 53, 62, 71, 80}, which has 8 elements.

Hence, the cardinal number of the set A is 8.

$\frac{30}{x}$ is a natural number only when x is a positive divisor of 30.

Resolving 30 into prime factors, 30 = 2 × 3 × 5, so the positive divisors of 30 are

1, 2, 3, 5, 6, 10, 15, 30, which are 8 in number.

Hence, the cardinal number of the required set is 8.

The letters of the word CHANDIGARH are C, H, A, N, D, I, G, A, R, H. Writing each letter only once,

X = {C, H, A, N, D, I, G, R}, which has 8 different elements.

Hence, the cardinal number of the set X is 8.

(i) 37 ∉ {x : x has exactly two positive factors}

The positive factors of 37 are 1 and 37 only, because 37 is a prime number. So 37 has exactly two positive factors and therefore 37 belongs to the given set i.e. 37 ∈ {x : x has exactly two positive factors}.

Hence, the statement 37 ∉ {x : x has exactly two positive factors} is false.

(ii) 35 ∈ {x : x has exactly four positive factors}

Since 35 = 5 × 7, the positive factors of 35 are 1, 5, 7 and 35. So 35 has exactly four positive factors and therefore 35 belongs to the given set.

Hence, the statement 35 ∈ {x : x has exactly four positive factors} is true.

(iii) 496 ∈ {y : the sum of all positive factors of y is 2y}

Resolving 496 into prime factors,

496 = 2⁴ × 31

∴ All the positive factors of 496 are 1, 2, 4, 8, 16, 31, 62, 124, 248 and 496.

The sum of all these positive factors = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 + 496

= 992

= 2 × 496

So the sum of all the positive factors of 496 is twice 496, and therefore 496 belongs to the given set.

Hence, the statement 496 ∈ {y : the sum of all positive factors of y is 2y} is true.

(iv) 3 ∉ {x : x⁴ − 5x³ + 2x² − 112x + 6 = 0}

Putting x = 3 in x⁴ − 5x³ + 2x² − 112x + 6, we get:

⇒ 3⁴ − 5 × 3³ + 2 × 3² − 112 × 3 + 6

⇒ 81 − 135 + 18 − 336 + 6

⇒ −366 ≠ 0

So x = 3 does not satisfy the given equation and therefore 3 does not belong to the given set.

Hence, the statement 3 ∉ {x : x⁴ − 5x³ + 2x² − 112x + 6 = 0} is true.

Here x is an integer and x < 5, so

A = {....., −2, −1, 0, 1, 2, 3, 4}

As the integers go on decreasing without end, the listing never comes to an end.

Hence, the set A is an infinite set.

A whole number divisible by both 4 and 9 must be divisible by their L.C.M.

L.C.M. of 4 and 9 = 36

∴ A = {0, 36, 72, 108, .....}

As the multiples of 36 go on without end, the listing never comes to an end.

Hence, the set A is an infinite set.

2 is the only even prime number, and every even number greater than 2 is divisible by 2 and hence is not prime. So there is no even prime number greater than 2.

∴ P = φ, which is a finite set with n(P) = 0.

Hence, the set P is a finite set and its cardinal number is 0.

Resolving 84 into prime factors,

84 = 2² × 3 × 7

∴ F = {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84}, which has 12 elements.

Hence, the set F is a finite set and its cardinal number is 12.

Taking the two digit numbers whose digits add up to 12, we get

B = {39, 48, 57, 66, 75, 84, 93}, which has 7 elements.

Hence, the set B is a finite set and its cardinal number is 7.

Here x ∈ W and 3x − 7 ≤ 8

⇒ 3x ≤ 15

⇒ x ≤ 5

As x is a whole number not exceeding 5,

C = {0, 1, 2, 3, 4, 5}, which has 6 elements.

Hence, the set C is a finite set and its cardinal number is 6.

Here x = 5n where n ∈ N and x < 20.

Putting n = 1, 2, 3, we get x = 5, 10, 15 and for n = 4 we get x = 20 which is not less than 20.

∴ The set = {5, 10, 15}, which has 3 elements.

Hence, the given set is a finite set and its cardinal number is 3.

Here x = 5n where n is an integer and x < 20.

Since n can be any integer, taking n = 3, 2, 1, 0, −1, −2, ..... we get

the set = {....., −10, −5, 0, 5, 10, 15}

As the negative multiples of 5 go on without end, the listing never comes to an end.

Hence, the given set is an infinite set.

Here n ∈ W and n ≤ 10, so n takes the 11 values 0, 1, 2, ....., 10.

Putting these values in $\frac{n}{n + 1}$, we get

$0, \frac{1}{2}, \frac{2}{3}, \frac{3}{4}, \frac{4}{5}, \frac{5}{6}, \frac{6}{7}, \frac{7}{8}, \frac{8}{9}, \frac{9}{10}, \frac{10}{11}$

All these 11 values are different.

Hence, the given set is a finite set and its cardinal number is 11.

Here n ∈ N and 5 < n < 20, so n takes the 14 values 6, 7, 8, ....., 19.

Putting these values in $\frac{2n}{n + 3}$, we get

$\frac{12}{9}, \frac{14}{10}, \frac{16}{11}, \frac{18}{12}, \frac{20}{13}, \frac{22}{14}, \frac{24}{15}, \frac{26}{16}, \frac{28}{17}, \frac{30}{18}, \frac{32}{19}, \frac{34}{20}, \frac{36}{21}, \frac{38}{22}$

All these 14 values are different.

Hence, the given set is a finite set and its cardinal number is 14.

For A,

⇒ x + 2 = 3

⇒ x = 1

So, A = {1}.

The natural numbers less than 2 give only x = 1, so B = {1}.

Both the sets have exactly the same element 1.

Hence, A = B.

For A, x ∈ N and 3x − 1 < 2

⇒ 3x < 2 + 1

⇒ 3x < 3

⇒ x < 1

There is no natural number less than 1, so A = φ.

For B, x ∈ W and x < 1. The only whole number less than 1 is 0, so B = {0}.

Here A has no element while B has the element 0.

Hence, A ≠ B.

Resolving 36 into prime factors, 36 = 2² × 3², so the prime factors of 36 are 2 and 3.

∴ A = {2, 3} and B = {1, 2, 3, 4, 6, 9, 12}

Here 1 ∈ B but 1 ∉ A, so the two sets do not have exactly the same elements.

Hence, A ≠ B.

For A, x ∈ I and x² ≤ 4. The squares of 0, ±1, ±2 are 0, 1, 4 and each of these is less than or equal to 4, while (±3)² = 9 > 4.

∴ A = {−2, −1, 0, 1, 2}

For B, x² − 3x + 2 = 0

⇒ (x − 1)(x − 2) = 0

⇒ x = 1 or x = 2

∴ B = {1, 2}

Here −2 ∈ A but −2 ∉ B.

Hence, A ≠ B.

Given,

A = {2, 3}

For B, x² + 5x + 6 = 0

⇒ (x + 2)(x + 3) = 0

⇒ x = −2 or x = −3

∴ B = {−2, −3}

The elements of A are positive while those of B are negative.

Hence, A ≠ B.

The letters of the word LOYAL are L, O, Y, A, L, so A = {L, O, Y, A}.

The letters of the word ALLOY are A, L, L, O, Y, so B = {A, L, O, Y}.

Both the sets consist of exactly the same four letters A, L, O and Y, and the order of listing does not matter.

Hence, A = B.

The letters of the word WOLF are W, O, L, F, so A = {W, O, L, F}.

The letters of the word FOLLOW are F, O, L, L, O, W, so B = {F, O, L, W}.

Both the sets consist of exactly the same four letters F, O, L and W.

Hence, A = B.

Given,

A = {10, 20, 30, 40, 50, .....}

The natural numbers which are multiples of 10 are 10, 20, 30, 40, 50, ....., so B = {10, 20, 30, 40, 50, .....}.

Both the sets consist of exactly the same elements.

Hence, A = B.

Given,

A = {1, 2, 3, 4, 5}, B = {2, 3, 4} and C = {2, 4, 5}

∴ n(A) = 5, n(B) = 3 and n(C) = 3

We know that,

A ⊂ B if every member of A is also a member of B, and A ↔ B if n(A) = n(B).

(i) A ⊂ B

1 ∈ A but 1 ∉ B, so every member of A is not a member of B.

Hence, the statement A ⊂ B is false.

(ii) A ⊂ C

1 ∈ A but 1 ∉ C, so every member of A is not a member of C.

Hence, the statement A ⊂ C is false.

(iii) B ⊂ A

Each of the members 2, 3, 4 of B is also a member of A.

Hence, the statement B ⊂ A is true.

(iv) B ⊂ C

3 ∈ B but 3 ∉ C, so every member of B is not a member of C.

Hence, the statement B ⊂ C is false.

(v) C ⊂ A

Each of the members 2, 4, 5 of C is also a member of A.

Hence, the statement C ⊂ A is true.

(vi) C ⊂ B

5 ∈ C but 5 ∉ B, so every member of C is not a member of B.

Hence, the statement C ⊂ B is false.

(vii) B = C

B = {2, 3, 4} and C = {2, 4, 5}. Here 3 ∈ B but 3 ∉ C, so the two sets do not have exactly the same elements.

Hence, the statement B = C is false.

(viii) φ ⊂ B

The empty set has no element, so it is a subset of every set.

Hence, the statement φ ⊂ B is true.

(ix) A ↔ B

n(A) = 5 and n(B) = 3, so n(A) ≠ n(B).

Hence, the statement A ↔ B is false.

(x) B ↔ C

n(B) = 3 and n(C) = 3, so n(B) = n(C).

Hence, the statement B ↔ C is true.

(xi) A ↔ C

n(A) = 5 and n(C) = 3, so n(A) ≠ n(C).

Hence, the statement A ↔ C is false.

(i) 4 ...... {1, 2, 3, 4}

The number 4 is one of the elements of the set {1, 2, 3, 4}, so 4 belongs to it.

Hence, 4 ∈ {1, 2, 3, 4}.

(ii) −5 ...... {2, 3, 4, 5, 6}

The elements of {2, 3, 4, 5, 6} are 2, 3, 4, 5, 6 and −5 is not one of them.

Hence, −5 ∉ {2, 3, 4, 5, 6}.

(iii) {2} ...... {2, 3, 4}

Here {2} is a set whose only member 2 is also a member of {2, 3, 4}, so {2} is a subset of {2, 3, 4}.

Hence, {2} ⊂ {2, 3, 4}.

(iv) {a, b, c} ...... {b, a, c}

The order of listing the elements of a set can be changed without altering the set. Both sets consist of exactly the elements a, b and c.

Hence, {a, b, c} = {b, a, c}.

(v) {a, b, c} ...... {a, b, b, a, c}

If one or more elements of a set are repeated, the set remains the same, so {a, b, b, a, c} = {a, b, c}. Both sets consist of exactly the elements a, b and c.

Hence, {a, b, c} = {a, b, b, a, c}.

(vi) {a, i, u} ...... {a, a, i}

On removing the repetition, {a, a, i} = {a, i}. Every member of {a, i} is also a member of {a, i, u}, and u ∈ {a, i, u} but u ∉ {a, i}. So {a, i, u} contains {a, i}.

Hence, {a, i, u} ⊃ {a, a, i}.

(vii) MUMBAI ... {x : x is a capital city of countries in Asia}:

MUMBAI is a city of India but it is not the capital of India; the capital of India is New Delhi. So MUMBAI is not a member of the set of capital cities of countries in Asia.

Hence, MUMBAI ∉ {x : x is a capital city of countries in Asia}.

(i) {a, b} ⊄ {b, c, a}

The members of {a, b} are a and b, and both a ∈ {b, c, a} and b ∈ {b, c, a}. So every member of {a, b} is a member of {b, c, a} i.e. {a, b} ⊂ {b, c, a}.

Hence, the statement {a, b} ⊄ {b, c, a} is false.

(ii) {1, 2, 3} ⊂ {1, 3, 5}

Here 2 ∈ {1, 2, 3} but 2 ∉ {1, 3, 5}, so every member of {1, 2, 3} is not a member of {1, 3, 5}.

Hence, the statement {1, 2, 3} ⊂ {1, 3, 5} is false.

(iii) {a} ⊂ {a, b, c}

The only member of {a} is a and a ∈ {a, b, c}.

Hence, the statement {a} ⊂ {a, b, c} is true.

(iv) {a} ∈ {a, b, c}

The elements of {a, b, c} are the letters a, b and c, and not the set {a}. So {a} is not an element of {a, b, c}.

Hence, the statement {a} ∈ {a, b, c} is false.

(v) {a, e} ⊂ {x : x is a vowel in the English alphabet}

The set of vowels in the English alphabet is {a, e, i, o, u}. Both a and e are members of this set.

Hence, the statement {a, e} ⊂ {x : x is a vowel in the English alphabet} is true.

(vi) {x : x is an even natural number less than 8} ⊂ {x : x is a natural number which divides 36}

The even natural numbers less than 8 are 2, 4 and 6, so the first set is {2, 4, 6}.

Since 36 = 2² × 3², the natural numbers which divide 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36, so the second set is {1, 2, 3, 4, 6, 9, 12, 18, 36}.

Each of the members 2, 4 and 6 of the first set is also a member of the second set.

Hence, the given statement is true.

(i) 2 ⊂ {1, 2, 3}

The symbol ⊂ is used between two sets, but 2 is a number and not a set.

Hence, the statement 2 ⊂ {1, 2, 3} is false.

(ii) {2} ∈ {1, 2, 3}

The elements of {1, 2, 3} are the numbers 1, 2 and 3, and not the set {2}.

Hence, the statement {2} ∈ {1, 2, 3} is false.

(iii) φ ∈ {1, 2, 3}

The elements of {1, 2, 3} are the numbers 1, 2 and 3, and φ is not one of them.

Hence, the statement φ ∈ {1, 2, 3} is false.

(iv) 0 ∈ φ

The empty set φ contains no element at all, so 0 cannot belong to φ.

Hence, the statement 0 ∈ φ is false.

(v) {1, 2} ⊂ {1, {2, 3}}

The elements of {1, {2, 3}} are the number 1 and the set {2, 3}. Now 2 ∈ {1, 2} but 2 ∉ {1, {2, 3}}, so every member of {1, 2} is not a member of {1, {2, 3}}.

Hence, the statement {1, 2} ⊂ {1, {2, 3}} is false.

(vi) φ ⊂ {φ}

The empty set has no element, so it is a subset of every set, and {φ} is a set.

Hence, the statement φ ⊂ {φ} is true.

Given,

A = {1, 2, {3, 4}, 5}

The elements of A are the numbers 1, 2, 5 and the set {3, 4}. Note carefully that 3 ∉ A and 4 ∉ A.

(i) {3, 4} ⊂ A

The members of {3, 4} are 3 and 4, and 3 ∉ A.

Hence, the statement {3, 4} ⊂ A is not true, because 3 is not an element of A.

(ii) {3, 4} ∈ A

The set {3, 4} is itself one of the elements of A.

Hence, the statement {3, 4} ∈ A is true.

(iii) {{3, 4}} ⊂ A

The only member of {{3, 4}} is the set {3, 4}, and {3, 4} ∈ A.

Hence, the statement {{3, 4}} ⊂ A is true.

(iv) {1, 2, 5} ⊂ A

Each of the members 1, 2 and 5 of {1, 2, 5} is also an element of A.

Hence, the statement {1, 2, 5} ⊂ A is true.

(v) {1, 2, 5} ∈ A

The elements of A are 1, 2, {3, 4} and 5; the set {1, 2, 5} is not one of them.

Hence, the statement {1, 2, 5} ∈ A is not true, because {1, 2, 5} is not an element of A.

(vi) {1, 2, 3} ⊂ A

The members of {1, 2, 3} are 1, 2 and 3, and 3 ∉ A.

Hence, the statement {1, 2, 3} ⊂ A is not true, because 3 is not an element of A.

(vii) 1 ∈ A

The number 1 is one of the elements of A.

Hence, the statement 1 ∈ A is true.

(viii) 1 ⊂ A

The symbol ⊂ is used between two sets, but 1 is a number and not a set. The correct statement here is 1 ∈ A.

Hence, the statement 1 ⊂ A is not true, because 1 is an element and not a set.

(ix) φ ∈ A

The elements of A are 1, 2, {3, 4} and 5, and φ is not one of them.

Hence, the statement φ ∈ A is not true, because φ is not an element of A.

(x) φ ⊂ A

The empty set has no element, so it is a subset of every set.

Hence, the statement φ ⊂ A is true.

(xi) {φ} ⊂ A

The only member of {φ} is φ, and φ ∉ A as shown in part (ix).

Hence, the statement {φ} ⊂ A is not true, because φ is not an element of A.

Given,

ξ = {1, 2, 3, ....., 10}

Resolving 66 into prime factors,

66 = 2 × 3 × 11

∴ The prime factors of 66 are 2, 3 and 11.

It is understood that A is a subset of the universal set ξ, so the members of A are to be taken only from ξ. As ξ = {1, 2, 3, ....., 10}, the prime factors 2 and 3 belong to ξ but 11 does not belong to ξ.

Hence, the set A is {2, 3}.

If A is a set with n(A) = m, then the number of subsets of A is 2^m.

(i) {a}

Here n({a}) = 1, so the number of subsets = 2¹ = 2.

Hence, the subsets of {a} are φ and {a}.

(ii) {a, b}

Here n({a, b}) = 2, so the number of subsets = 2² = 4.

Hence, the subsets of {a, b} are φ, {a}, {b} and {a, b}.

(iii) φ

Here n(φ) = 0, so the number of subsets = 2⁰ = 1. The only subset of the empty set is the empty set itself.

Hence, the only subset of φ is φ.

(iv) {φ, 1}

The set {φ, 1} has two elements, namely φ and 1, so the number of subsets = 2² = 4.

Hence, the subsets of {φ, 1} are φ, {φ}, {1} and {φ, 1}.

Given,

A = {−1, 0, 2}

∴ n(A) = 3

The set formed by all the subsets of a given set A is called the power set of A and is denoted by P(A).

Number of elements in P(A) = 2^n(A) = 2³ = 8

Writing down all the subsets of A, we have

The subset with no element : φ

The subsets with one element : {−1}, {0}, {2}

The subsets with two elements : {−1, 0}, {−1, 2}, {0, 2}

The subset with three elements : {−1, 0, 2}

Hence, the power set of A is P(A) = {φ, {−1}, {0}, {2}, {−1, 0}, {−1, 2}, {0, 2}, {−1, 0, 2}}.

Given,

A = {−3, −1, 0, 1, 4}

∴ n(A) = 5

If A is a set with n(A) = m, then the number of subsets of A is 2^m and the number of proper subsets of A is 2^m − 1, because the set A itself is a subset of A but not a proper subset of A.

∴ Number of proper subsets of A = 2⁵ − 1

= 32 − 1

= 31.

Hence, the number of proper subsets of A is 31.

Given,

ξ = {1, 2, 3, ....., 40}

Resolving 42 into prime factors,

42 = 2 × 3 × 7

∴ The factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.

It is understood that A is a subset of the universal set ξ, so the members of A are to be taken only from ξ = {1, 2, 3, ....., 40}. Out of the above factors, 42 does not belong to ξ, so it cannot be taken.

∴ A = {1, 2, 3, 6, 7, 14, 21}, which has 7 elements.

Hence, n(A) = 7.

In our number system every number is written with the help of the ten digits 0, 1, 2, ....., 9.

∴ ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

It is understood that A is a subset of ξ, so the members of A are to be taken only from ξ.

The multiples of 3 lying in ξ are 0, 3, 6 and 9, since 0 = 3 × 0, 3 = 3 × 1, 6 = 3 × 2 and 9 = 3 × 3.

∴ A = {0, 3, 6, 9}, which has 4 elements.

Hence, n(A) = 4.

We use the following forms of intervals:

(a, b) = {x : x ∈ R, a < x < b}, [a, b] = {x : x ∈ R, a ≤ x ≤ b},

(a, b] = {x : x ∈ R, a < x ≤ b}, [a, b) = {x : x ∈ R, a ≤ x < b},

(a, ∞) = {x : x ∈ R, x > a}, [a, ∞) = {x : x ∈ R, x ≥ a},

(−∞, a) = {x : x ∈ R, x < a}, (−∞, a] = {x : x ∈ R, x ≤ a}.

(i) {x : x ∈ R, −3 < x ≤ 5}

Here −3 is not included but 5 is included, so the interval is open on the left and closed on the right.

Hence, the given set as an interval is (−3, 5].

(ii) {x : x ∈ R, −5 < x < −1}

Here neither −5 nor −1 is included, so the interval is open at both ends.

Hence, the given set as an interval is (−5, −1).

(iii) {x : x ∈ R, 2 ≤ x ≤ 7}

Here both 2 and 7 are included, so the interval is closed at both ends.

Hence, the given set as an interval is [2, 7].

(iv) {x : x ∈ R, 0 ≤ x < 3}

Here 0 is included but 3 is not included, so the interval is closed on the left and open on the right.

Hence, the given set as an interval is [0, 3).

(v) {x : x ∈ R, x ≤ 5}

Here x can be any real number up to and including 5, so the interval extends indefinitely to the left and is closed on the right.

Hence, the given set as an interval is (−∞, 5].

(vi) {x : x ∈ R, x < −3}

Here x can be any real number less than −3 but −3 is not included.

Hence, the given set as an interval is (−∞, −3).

(vii) {x : x ∈ R, x ≥ −2}

Here x can be any real number from −2 onwards, −2 being included.

Hence, the given set as an interval is [−2, ∞).

(i) (−2, 0]

The interval (−2, 0] is open on the left and closed on the right, so −2 is excluded and 0 is included.

Hence, the given interval in set builder form is {x : x ∈ R, −2 < x ≤ 0}.

(ii) (2, 7)

The interval (2, 7) is open at both ends, so both 2 and 7 are excluded.

Hence, the given interval in set builder form is {x : x ∈ R, 2 < x < 7}.

(iii) [−5, −2]

The interval [−5, −2] is closed at both ends, so both −5 and −2 are included.

Hence, the given interval in set builder form is {x : x ∈ R, −5 ≤ x ≤ −2}.

(iv) [−9, 4)

The interval [−9, 4) is closed on the left and open on the right, so −9 is included and 4 is excluded.

Hence, the given interval in set builder form is {x : x ∈ R, −9 ≤ x < 4}.

(v) (3, ∞)

The interval (3, ∞) consists of all real numbers greater than 3, and 3 is excluded.

Hence, the given interval in set builder form is {x : x ∈ R, x > 3}.

(vi) (−∞, −1]

The interval (−∞, −1] consists of all real numbers less than or equal to −1.

Hence, the given interval in set builder form is {x : x ∈ R, x ≤ −1}.

(vii) (−∞, 4)

The interval (−∞, 4) consists of all real numbers less than 4, and 4 is excluded.

Hence, the given interval in set builder form is {x : x ∈ R, x < 4}.

Given,

ξ = {1, 2, 3, ....., 12}

It is understood that A and B are subsets of ξ, so their members are to be taken only from ξ.

For the set A,

⇒ 2x + 3 ≤ 18

⇒ 2x ≤ 18 − 3

⇒ 2x ≤ 15

⇒ x ≤ $\frac{15}{2}$

⇒ x ≤ 7.5

The members of ξ which are not greater than 7.5 are 1, 2, 3, 4, 5, 6, 7.

∴ A = {1, 2, 3, 4, 5, 6, 7}

For the set B,

x² ≤ 40

The squares of 1, 2, 3, 4, 5, 6 are 1, 4, 9, 16, 25, 36 and each of these is not greater than 40, while 7² = 49 > 40.

∴ B = {1, 2, 3, 4, 5, 6}

Hence, A = {1, 2, 3, 4, 5, 6, 7} and B = {1, 2, 3, 4, 5, 6}.

Given,

ξ = {0, 1, 2, 3, ....., 50}

It is understood that A, B and C are subsets of ξ, so their members are to be taken only from ξ.

(i) For the set A, x = 6n where n ∈ W.

Putting n = 0, 1, 2, ....., we get 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, ..... Out of these, only those up to 50 belong to ξ.

∴ A = {0, 6, 12, 18, 24, 30, 36, 42, 48}

For the set B, x = 7n where n ∈ W.

Putting n = 0, 1, 2, ....., we get 0, 7, 14, 21, 28, 35, 42, 49, 56, ..... Out of these, only those up to 50 belong to ξ.

∴ B = {0, 7, 14, 21, 28, 35, 42, 49}

For the set C, resolving 72 into prime factors, 72 = 2³ × 3², so the factors of 72 are

1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72.

Out of these, 72 does not belong to ξ, so it cannot be taken.

∴ C = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}

Hence, A = {0, 6, 12, 18, 24, 30, 36, 42, 48}, B = {0, 7, 14, 21, 28, 35, 42, 49} and C = {1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36}.

(ii) Counting the different elements of each set,

A has 9 elements, B has 8 elements and C has 11 elements.

Hence, n(A) = 9, n(B) = 8 and n(C) = 11.

(i) The letters of the word ACCUMULATOR are A, C, C, U, M, U, L, A, T, O, R. While listing the elements of a set, each member is written only once, so the repeated letters C, U and A are written only once.

Hence, A in roster form is {A, C, U, M, L, T, O, R}.

(ii) The vowels in the English alphabet are A, E, I, O, U. Out of the elements of A, the vowels are A, U and O.

∴ The set of vowels in A is V = {A, U, O}, which has 3 elements.

Hence, the cardinal number of the set of vowels in A is 3.

(iii) Let V = {A, U, O} be the set of vowels in A, so n(V) = 3.

Number of elements in P(V) = 2^n(V) = 2³ = 8

Writing down all the subsets of V, we have

The subset with no element : φ

The subsets with one element : {A}, {U}, {O}

The subsets with two elements : {A, U}, {A, O}, {U, O}

The subset with three elements : {A, U, O}

Hence, the power set of the set of vowels in A is P(V) = {φ, {A}, {U}, {O}, {A, U}, {A, O}, {U, O}, {A, U, O}}.

Let A = {0, 1, 2, 3}, so n(A) = 4.

Number of subsets of A = 2⁴ = 16

Number of proper subsets of A = 2⁴ − 1 = 16 − 1 = 15

(The set A itself is a subset of A but not a proper subset of A, so it is left out.)

Writing down all the subsets of A other than A itself, we have

The subset with no element : φ

The subsets with one element : {0}, {1}, {2}, {3}

The subsets with two elements : {0, 1}, {0, 2}, {0, 3}, {1, 2}, {1, 3}, {2, 3}

The subsets with three elements : {0, 1, 2}, {0, 1, 3}, {0, 2, 3}, {1, 2, 3}

Hence, the proper subsets of {0, 1, 2, 3} are φ, {0}, {1}, {2}, {3}, {0, 1}, {0, 2}, {0, 3}, {1, 2}, {1, 3}, {2, 3}, {0, 1, 2}, {0, 1, 3}, {0, 2, 3} and {1, 2, 3}.

If A is a set with n(A) = m, then the number of subsets of A is 2^m and the number of proper subsets of A is 2^m − 1.

When A has five elements, n(A) = 5.

∴ Number of subsets of A = 2⁵ = 32

When A has six elements, n(A) = 6.

∴ Number of proper subsets of A = 2⁶ − 1 = 64 − 1 = 63

Hence, A will have 32 subsets when it has five elements, and 63 proper subsets when it has six elements.

Let the two sets be A and B, then n(A) = m and n(B) = k.

As the first set has more subsets than the second set, m > k.

Now, number of subsets of A = 2^m and number of subsets of B = 2^k.

Given,

Number of subsets of the first set is 112 more than the number of subsets of the second set.

⇒ 2^m − 2^k = 112

⇒ 2^k(2^{m − k} − 1) = 112

⇒ 2^k(2^{m − k} − 1) = 16 × 7

⇒ 2^k(2^{m − k} − 1) = 2⁴ × 7

As 2^{m − k} − 1 is an odd number and 2^k is a power of 2, comparing both sides we get

2^k = 2⁴ and 2^{m − k} − 1 = 7

From,

⇒ 2^k = 2⁴

⇒ k = 4

From,

⇒ 2^{m − k} − 1 = 7

⇒ 2^{m − k} = 8

⇒ 2^{m − k} = 2³

⇒ m − k = 3

⇒ m − 4 = 3

⇒ m = 3 + 4

⇒ m = 7

Hence, m = 7 and k = 4.

Given,

A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8}

(i) A ∪ B

A ∪ B is the set of all the elements which belong to A or B or both. Collecting all such elements and writing each of them only once,

A ∪ B = {1, 2, 3, 4, 5, 6, 8}

Hence, A ∪ B = {1, 2, 3, 4, 5, 6, 8}.

(ii) A ∩ B

A ∩ B is the set of all the elements which belong to both A and B. The elements common to A and B are 2, 4 and 6.

∴ A ∩ B = {2, 4, 6}

Hence, A ∩ B = {2, 4, 6}.

(iii) A − B

A − B is the set of all the elements which belong to A but do not belong to B. Removing 2, 4, 6 from A, the remaining elements are 1, 3 and 5.

∴ A − B = {1, 3, 5}

Hence, A − B = {1, 3, 5}.

(iv) B − A

B − A is the set of all the elements which belong to B but do not belong to A. Removing 2, 4, 6 from B, the remaining element is 8.

∴ B − A = {8}

Hence, B − A = {8}.

Given,

A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 5, 6, 8, 9, 10}

The symmetric difference of A and B is the set of all elements which belong to A or B but do not belong to A ∩ B, and

A Δ B = (A − B) ∪ (B − A)

Now, the elements of A which do not belong to B are 1 and 3.

∴ A − B = {1, 3}

The elements of B which do not belong to A are 8, 9 and 10.

∴ B − A = {8, 9, 10}

∴ A Δ B = (A − B) ∪ (B − A) = {1, 3} ∪ {8, 9, 10} = {1, 3, 8, 9, 10}

Hence, A Δ B = {1, 3, 8, 9, 10}.

For the set A, x is a natural number with 1 < x ≤ 6, so x = 2, 3, 4, 5, 6.

∴ A = {2, 3, 4, 5, 6}

For the set B, x is a natural number with 6 < x < 10, so x = 7, 8, 9.

∴ B = {7, 8, 9}

(i) A ∪ B

Collecting all the elements which belong to A or B or both,

A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}

Hence, A ∪ B = {2, 3, 4, 5, 6, 7, 8, 9}.

(ii) A ∩ B

There is no element which belongs to both A and B, so A and B are disjoint sets.

∴ A ∩ B = φ

Hence, A ∩ B = φ.

(iii) A − B

As no element of A belongs to B, no element of A has to be removed.

∴ A − B = {2, 3, 4, 5, 6} = A

Hence, A − B = A = {2, 3, 4, 5, 6}.

(iv) B − A

As no element of B belongs to A, no element of B has to be removed.

∴ B − A = {7, 8, 9} = B

Hence, B − A = B = {7, 8, 9}.

Two sets A and B are called disjoint if A ∩ B = φ i.e. if they have no element in common.

(i) {a, e, i, o, u} and {c, d, e, f}

Let A = {a, e, i, o, u} and B = {c, d, e, f}.

The letter e belongs to both the sets, so A ∩ B = {e} ≠ φ.

Hence, this pair of sets is not disjoint.

(ii) {2, 6, 10, 14, 18} and {3, 7, 11, 15}

Let A = {2, 6, 10, 14, 18} and B = {3, 7, 11, 15}.

No element of A is a member of B, so A ∩ B = φ.

Hence, this pair of sets is disjoint.

(iii) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 6}

Let A = {1, 2, 3, 4} and B = {x : x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}.

The number 4 belongs to both the sets, so A ∩ B = {4} ≠ φ.

Hence, this pair of sets is not disjoint.

(iv) {x : x is an even integer} and {x : x is an odd integer}

Let A = {x : x is an even integer} and B = {x : x is an odd integer}.

An integer is either even or odd but never both, so no integer can belong to both A and B.

∴ A ∩ B = φ

Hence, this pair of sets is disjoint.

Given,

ξ = {1, 2, 3, 4, 5, 6, 7, 8}, A = {1, 2, 3, 5, 6} and B = {2, 3, 4, 7, 8}

The complement of a set is obtained by removing its elements from the universal set ξ.

∴ A' = ξ − A = {4, 7, 8} and B' = ξ − B = {1, 5, 6}

(i) A ∪ B

Collecting all the elements which belong to A or B or both,

A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8} = ξ

Hence, A ∪ B = ξ = {1, 2, 3, 4, 5, 6, 7, 8}.

(ii) A ∩ B

The elements common to A and B are 2 and 3.

∴ A ∩ B = {2, 3}

Hence, A ∩ B = {2, 3}.

(iii) A − B

Removing from A the elements 2 and 3 which also belong to B,

A − B = {1, 5, 6}

Hence, A − B = {1, 5, 6}.

(iv) B − A

Removing from B the elements 2 and 3 which also belong to A,

B − A = {4, 7, 8}

Hence, B − A = {4, 7, 8}.

(v) A ∩ B'

Here B' = {1, 5, 6} and A = {1, 2, 3, 5, 6}.

The elements common to A and B' are 1, 5 and 6.

∴ A ∩ B' = {1, 5, 6}

Hence, A ∩ B' = {1, 5, 6}.

(vi) (A ∪ B)'

From part (i), A ∪ B = ξ, so removing all the elements of ξ from ξ leaves nothing.

∴ (A ∪ B)' = ξ − ξ = φ

Hence, (A ∪ B)' = φ.

(vii) A' ∩ B'

Here A' = {4, 7, 8} and B' = {1, 5, 6}.

There is no element common to A' and B'.

∴ A' ∩ B' = φ

Hence, A' ∩ B' = φ.

(viii) A' ∪ B

Here A' = {4, 7, 8} and B = {2, 3, 4, 7, 8}.

Collecting all the elements which belong to A' or B or both,

A' ∪ B = {2, 3, 4, 7, 8}

Hence, A' ∪ B = {2, 3, 4, 7, 8}.

(ix) (A − B)'

From part (iii), A − B = {1, 5, 6}.

Removing the elements 1, 5, 6 from ξ,

(A − B)' = {2, 3, 4, 7, 8}

Hence, (A − B)' = {2, 3, 4, 7, 8}.

Given,

A ⊂ B, so every member of A is also a member of B.

(i) A ∪ B

A ∪ B is the set of all the elements which belong to A or B or both. Since every element of A already belongs to B, no new element is contributed by A.

∴ A ∪ B = B

Hence, A ∪ B = B.

(ii) A ∩ B

A ∩ B is the set of all the elements which belong to both A and B. Since every element of A also belongs to B, all the elements of A are common to A and B, and no element outside A can be common.

∴ A ∩ B = A

Hence, A ∩ B = A.

Here ξ = N. It is understood that A, B, C, D, E, F, G and H are all subsets of N, and the complement of a set consists of those natural numbers which do not belong to the set.

(i) A = {x : x is an odd natural number}

A natural number which is not odd is even.

Hence, A' = {x : x is an even natural number}.

(ii) B = {x : x is an even natural number}

A natural number which is not even is odd.

Hence, B' = {x : x is an odd natural number}.

(iii) C = {x : x is a multiple of 5}

Hence, C' = {x : x ∈ N and x is not a multiple of 5}.

(iv) D = {x : x is divisible by 2 and 3}

A natural number divisible by both 2 and 3 must be divisible by their L.C.M., which is 6.

∴ D = {x : x ∈ N and x is divisible by 6}

Hence, D' = {x : x ∈ N and x is not divisible by 6}.

(v) E = {x : x is a perfect cube}

Hence, E' = {x : x ∈ N and x is not a perfect cube}.

(vi) F = {x : x + 5 = 8}

Given,

x + 5 = 8 ⇒ x = 3, so F = {3}.

The complement of F consists of all natural numbers other than 3.

Hence, F' = {x : x ∈ N and x ≠ 3}.

(vii) G = {x : x ≥ 7}

Given,

G = {x : x ≥ 7} = {7, 8, 9, 10, .....}

The natural numbers which are not greater than or equal to 7 are 1, 2, 3, 4, 5, 6.

Hence, G' = {1, 2, 3, 4, 5, 6}.

(viii) H = {x : 3x − 1 < 14}

Given,

3x − 1 < 14

⇒ 3x < 15

⇒ x < 5

∴ H = {1, 2, 3, 4}

The natural numbers which are not less than 5 are 5, 6, 7, .....

Hence, H' = {x : x ∈ N and x ≥ 5}.

From the given Venn diagram,

ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {1, 2, 4, 7} (the elements written inside the circle A)

B = {0, 1, 3, 7, 8} (the elements written inside the circle B)

(i) A ∪ B

Collecting all the elements which belong to A or B or both,

A ∪ B = {0, 1, 2, 3, 4, 7, 8}

Hence, A ∪ B = {0, 1, 2, 3, 4, 7, 8}.

(ii) A ∩ B

The elements lying in the common (overlapping) region of the two circles are 1 and 7.

∴ A ∩ B = {1, 7}

Hence, A ∩ B = {1, 7}.

(iii) A − B

Removing from A the elements 1 and 7 which also belong to B,

A − B = {2, 4}

Hence, A − B = {2, 4}.

(iv) (A ∩ B)'

From part (ii), A ∩ B = {1, 7}.

Removing the elements 1 and 7 from ξ,

(A ∩ B)' = {0, 2, 3, 4, 5, 6, 8, 9}

Hence, (A ∩ B)' = {0, 2, 3, 4, 5, 6, 8, 9}.

From the given Venn diagram,

ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {0, 2, 3, 4, 8} (the elements written inside the circle A)

B = {0, 3, 5, 6, 8} (the elements written inside the circle B)

(i) A'

A' consists of those elements of ξ which do not belong to A. Removing 0, 2, 3, 4, 8 from ξ,

A' = {1, 5, 6, 7, 9}

Hence, A' = {1, 5, 6, 7, 9}.

(ii) B'

B' consists of those elements of ξ which do not belong to B. Removing 0, 3, 5, 6, 8 from ξ,

B' = {1, 2, 4, 7, 9}

Hence, B' = {1, 2, 4, 7, 9}.

(iii) (A ∩ B)'

The elements lying in the common region of the two circles are 0, 3 and 8.

∴ A ∩ B = {0, 3, 8}

Removing 0, 3 and 8 from ξ,

(A ∩ B)' = {1, 2, 4, 5, 6, 7, 9}

Hence, (A ∩ B)' = {1, 2, 4, 5, 6, 7, 9}.

Given,

ξ = {x : x ∈ N and x < 10} = {1, 2, 3, 4, 5, 6, 7, 8, 9}

It is understood that A and B are subsets of ξ, so their members are to be taken only from ξ.

For the set A, x = 2y + 1 where y ∈ N.

Putting y = 1, 2, 3, 4, we get x = 3, 5, 7, 9 and for y = 5 we get x = 11 which does not belong to ξ.

∴ A = {3, 5, 7, 9}

For the set B, x = 3y − 1 where y ∈ N.

Putting y = 1, 2, 3, we get x = 2, 5, 8 and for y = 4 we get x = 11 which does not belong to ξ.

∴ B = {2, 5, 8}

Now, A' consists of those elements of ξ which do not belong to A.

∴ A' = ξ − A = {1, 2, 4, 6, 8}

∴ A' ∪ B = {1, 2, 4, 6, 8} ∪ {2, 5, 8} = {1, 2, 4, 5, 6, 8}

Hence, A' ∪ B = {1, 2, 4, 5, 6, 8}.

Given,

ξ = {1, 2, 3, ....., 10}

It is understood that A and B are subsets of ξ, so their members are to be taken only from ξ.

The prime numbers lying in ξ are 2, 3, 5 and 7.

∴ A = {2, 3, 5, 7}

The even integers lying in ξ are 2, 4, 6, 8 and 10.

∴ B = {2, 4, 6, 8, 10}

(i) A − B

Removing from A the element 2 which also belongs to B,

A − B = {3, 5, 7}

Hence, A − B = {3, 5, 7}.

(ii) A ∩ B'

B' consists of those elements of ξ which do not belong to B.

∴ B' = ξ − B = {1, 3, 5, 7, 9}

The elements common to A = {2, 3, 5, 7} and B' = {1, 3, 5, 7, 9} are 3, 5 and 7.

∴ A ∩ B' = {3, 5, 7}

Hence, A ∩ B' = {3, 5, 7}.

In our decimal system every number is written with the help of the ten digits 0, 1, 2, ....., 9.

∴ ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

It is understood that A and B are subsets of ξ, so their members are to be taken only from ξ.

The prime numbers lying in ξ are 2, 3, 5 and 7.

∴ A = {2, 3, 5, 7}

For the set B, x² < 25. The squares of 0, 1, 2, 3, 4 are 0, 1, 4, 9, 16 and each of these is less than 25, while 5² = 25 which is not less than 25.

∴ B = {0, 1, 2, 3, 4}

(i) B − A

Removing from B the elements 2 and 3 which also belong to A,

B − A = {0, 1, 4}

Hence, B − A = {0, 1, 4}.

(ii) A ∪ B

Collecting all the elements which belong to A or B or both,

A ∪ B = {0, 1, 2, 3, 4, 5, 7}

Hence, A ∪ B = {0, 1, 2, 3, 4, 5, 7}.

(iii) (A ∪ B)'

Removing the elements of A ∪ B from ξ, the elements left are 6, 8 and 9.

∴ (A ∪ B)' = {6, 8, 9}

Hence, (A ∪ B)' = {6, 8, 9}.

Given,

A = (0, 3] = {x : x ∈ R, 0 < x ≤ 3} and B = [2, 6] = {x : x ∈ R, 2 ≤ x ≤ 6}

The sets A and B on the real axis are represented below:

(i) A ∪ B

A ∪ B is the set of all real numbers which belong to A or B or both. The two intervals overlap between 2 and 3, so together they cover all real numbers x with 0 < x ≤ 6, where 0 is excluded and 6 is included.

∴ A ∪ B = (0, 6]

Hence, A ∪ B = (0, 6].

(ii) A ∩ B

A ∩ B is the set of all real numbers which belong to both A and B. A real number lies in both the intervals only when 2 ≤ x ≤ 3, both 2 and 3 being included.

∴ A ∩ B = [2, 3]

Hence, A ∩ B = [2, 3].

(iii) A − B

A − B is the set of all real numbers which belong to A but do not belong to B. Removing from (0, 3] the part [2, 3], we are left with 0 < x < 2, where 0 is excluded and 2 is also excluded because 2 ∈ B.

∴ A − B = (0, 2)

Hence, A − B = (0, 2).

(iv) B − A

B − A is the set of all real numbers which belong to B but do not belong to A. Removing from [2, 6] the part [2, 3], we are left with 3 < x ≤ 6, where 3 is excluded because 3 ∈ A and 6 is included.

∴ B − A = (3, 6]

Hence, B − A = (3, 6].

Given,

A = [−1, 1) = {x : x ∈ R, −1 ≤ x < 1} and B = [0, 4] = {x : x ∈ R, 0 ≤ x ≤ 4}

Here the universal set is R, the set of all real numbers.

The sets A and B on the real axis are represented below:

(i) A'

A' consists of all real numbers which do not belong to A, i.e. all x < −1 together with all x ≥ 1. Note that −1 ∈ A so −1 is excluded from A', while 1 ∉ A so 1 is included in A'.

∴ A' = (−∞, −1) ∪ [1, ∞)

Hence, A' = (−∞, −1) ∪ [1, ∞).

(ii) B'

B' consists of all real numbers which do not belong to B, i.e. all x < 0 together with all x > 4. Note that both 0 and 4 belong to B, so both are excluded from B'.

∴ B' = (−∞, 0) ∪ (4, ∞)

Hence, B' = (−∞, 0) ∪ (4, ∞).

(iii) A ∪ B

The two intervals overlap between 0 and 1, so together they cover all real numbers x with −1 ≤ x ≤ 4, both −1 and 4 being included.

∴ A ∪ B = [−1, 4]

Hence, A ∪ B = [−1, 4].

(iv) A ∩ B

A real number lies in both the intervals only when 0 ≤ x < 1, where 0 is included and 1 is excluded because 1 ∉ A.

∴ A ∩ B = [0, 1)

Hence, A ∩ B = [0, 1).

(v) A − B

Removing from [−1, 1) the part [0, 1), we are left with −1 ≤ x < 0, where −1 is included and 0 is excluded because 0 ∈ B.

∴ A − B = [−1, 0)

Hence, A − B = [−1, 0).

(vi) B − A

Removing from [0, 4] the part [0, 1), we are left with 1 ≤ x ≤ 4, where 1 is included because 1 ∉ A, and 4 is included.

∴ B − A = [1, 4]

Hence, B − A = [1, 4].

The letters of the word JAIPUR are J, A, I, P, U, R and the letters of the word JODHPUR are J, O, D, H, P, U, R.

∴ A = {J, A, I, P, U, R} and B = {J, O, D, H, P, U, R}

(i) A ∪ B

Collecting all the letters which belong to A or B or both, and writing each of them only once,

A ∪ B = {J, A, I, P, U, R, O, D, H}

Hence, A ∪ B = {J, A, I, P, U, R, O, D, H}.

(ii) A ∩ B

The letters common to A and B are J, P, U and R.

∴ A ∩ B = {J, P, U, R}

Hence, A ∩ B = {J, P, U, R}.

(iii) A − B

Removing from A the letters J, P, U, R which also belong to B,

A − B = {A, I}

Hence, A − B = {A, I}.

(iv) B − A

Removing from B the letters J, P, U, R which also belong to A,

B − A = {O, D, H}

Hence, B − A = {O, D, H}.

Verification

Counting the elements of the sets found above,

n(A) = 6, n(B) = 7, n(A ∪ B) = 9, n(A ∩ B) = 4, n(A − B) = 2

(a) n(A) + n(B) − n(A ∩ B) = 6 + 7 − 4 = 9 = n(A ∪ B)

Hence, the result n(A ∪ B) = n(A) + n(B) − n(A ∩ B) is verified.

(b) n(A) − n(A ∩ B) = 6 − 4 = 2 = n(A − B)

and n(A ∪ B) − n(B) = 9 − 7 = 2 = n(A − B)

Hence, the result n(A − B) = n(A) − n(A ∩ B) = n(A ∪ B) − n(B) is verified.

Given,

ξ = {0, 1, 2, 3, 4, 5, 6, 7}, A = {2, 5, 7}, B = {0, 2, 3, 7} and C = {0, 3, 4, 6}

(i) (A ∪ B)'

Collecting all the elements which belong to A or B or both,

A ∪ B = {0, 2, 3, 5, 7}

Removing these elements from ξ, the elements left are 1, 4 and 6.

∴ (A ∪ B)' = {1, 4, 6}

Hence, (A ∪ B)' = {1, 4, 6}.

(ii) A − C

A − C is the set of all the elements which belong to A but do not belong to C. None of the elements 2, 5, 7 of A belongs to C, so no element has to be removed.

∴ A − C = {2, 5, 7}

Hence, A − C = {2, 5, 7}.

(iii) A ∩ (B ∪ C)

First, collecting all the elements which belong to B or C or both,

B ∪ C = {0, 2, 3, 4, 6, 7}

Now, the elements common to A = {2, 5, 7} and B ∪ C = {0, 2, 3, 4, 6, 7} are 2 and 7.

∴ A ∩ (B ∪ C) = {2, 7}

Hence, A ∩ (B ∪ C) = {2, 7}.

For the set A, x ∈ N and 1 ≤ x < 4, so x = 1, 2, 3.

Putting these values in 2x, we get 2, 4, 6.

∴ A = {2, 4, 6}

For the set B, x ∈ N and 2 ≤ x < 5, so x = 2, 3, 4.

Putting these values in x + 2, we get 4, 5, 6.

∴ B = {4, 5, 6}

For the set C, x ∈ N and 3 < x < 6, so x = 4, 5.

∴ C = {4, 5}

(i) A ∩ B

The elements common to A and B are 4 and 6.

∴ A ∩ B = {4, 6}

Hence, A ∩ B = {4, 6}.

(ii) A ∪ B

Collecting all the elements which belong to A or B or both,

A ∪ B = {2, 4, 5, 6}

Hence, A ∪ B = {2, 4, 5, 6}.

(iii) (A ∪ B) ∩ C

The elements common to A ∪ B = {2, 4, 5, 6} and C = {4, 5} are 4 and 5.

∴ (A ∪ B) ∩ C = {4, 5}

Hence, (A ∪ B) ∩ C = {4, 5}.

Verification

Counting the elements of the sets found above,

n(A) = 3, n(B) = 3, n(A ∪ B) = 4, n(A ∩ B) = 2

∴ n(A) + n(B) − n(A ∩ B) = 3 + 3 − 2 = 4 = n(A ∪ B)

Hence, the result n(A ∪ B) = n(A) + n(B) − n(A ∩ B) is verified.

In our decimal system every number is written with the help of the ten digits 0, 1, 2, ....., 9.

∴ ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

The prime numbers lying in ξ are 2, 3, 5 and 7.

∴ A = {2, 3, 5, 7}

The perfect squares lying in ξ are 0 = 0², 1 = 1², 4 = 2² and 9 = 3².

∴ B = {0, 1, 4, 9}

Taking complements with respect to ξ,

A' = ξ − A = {0, 1, 4, 6, 8, 9} and B' = ξ − B = {2, 3, 5, 6, 7, 8}

(i) A ∪ A' = ξ

A ∪ A' = {2, 3, 5, 7} ∪ {0, 1, 4, 6, 8, 9} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} = ξ

Hence, the result A ∪ A' = ξ is verified.

(ii) A ∩ A' = φ

No element belongs to both A and A', because A' is formed by leaving out exactly the elements of A.

∴ A ∩ A' = φ

Hence, the result A ∩ A' = φ is verified.

(iii) A − B = A ∩ B'

None of the elements 2, 3, 5, 7 of A belongs to B, so

A − B = {2, 3, 5, 7}

Also, the elements common to A = {2, 3, 5, 7} and B' = {2, 3, 5, 6, 7, 8} are 2, 3, 5 and 7.

∴ A ∩ B' = {2, 3, 5, 7}

∴ A − B = A ∩ B'

Hence, the result A − B = A ∩ B' is verified.

(iv) (A ∩ B)' = A' ∪ B'

There is no element common to A and B, so A ∩ B = φ.

∴ (A ∩ B)' = φ' = ξ = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Also, A' ∪ B' = {0, 1, 4, 6, 8, 9} ∪ {2, 3, 5, 6, 7, 8} = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

∴ (A ∩ B)' = A' ∪ B'

Hence, the result (A ∩ B)' = A' ∪ B' is verified.

(v) (A ∪ B)' = A' ∩ B'

A ∪ B = {2, 3, 5, 7} ∪ {0, 1, 4, 9} = {0, 1, 2, 3, 4, 5, 7, 9}

Removing these elements from ξ, the elements left are 6 and 8.

∴ (A ∪ B)' = {6, 8}

Also, the elements common to A' = {0, 1, 4, 6, 8, 9} and B' = {2, 3, 5, 6, 7, 8} are 6 and 8.

∴ A' ∩ B' = {6, 8}

∴ (A ∪ B)' = A' ∩ B'

Hence, the result (A ∪ B)' = A' ∩ B' is verified.

(vi) (A')' = A

Here A' = {0, 1, 4, 6, 8, 9}.

Removing the elements of A' from ξ, the elements left are 2, 3, 5 and 7.

∴ (A')' = {2, 3, 5, 7} = A

Hence, the result (A')' = A is verified.

Given,

n(A) = 28, n(B) = 32 and n(A ∪ B) = 50

We know that,

⇒ n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

Substituting values we get :

⇒ 50 = 28 + 32 − n(A ∩ B)

⇒ 50 = 60 − n(A ∩ B)

⇒ n(A ∩ B) = 60 − 50 = 10

Hence, n(A ∩ B) = 10.

Given,

n(X) = 21, n(Y) = 32 and n(X ∩ Y) = 11

We know that,

n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y)

⇒ n(X ∪ Y) = 21 + 32 − 11

⇒ n(X ∪ Y) = 53 − 11 = 42

Hence, X ∪ Y has 42 elements.

Given,

n(A) = 20, n(B) = 18 and n(A ∩ B) = 5

(i) n(A ∪ B)

We know that,

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

⇒ n(A ∪ B) = 20 + 18 − 5 = 33

Hence, n(A ∪ B) = 33.

(ii) n(A − B)

The set A − B is obtained from A by removing those elements of A which also belong to B, i.e. the elements of A ∩ B.

∴ n(A − B) = n(A) − n(A ∩ B) = 20 − 5 = 15

Hence, n(A − B) = 15.

(iii) n(B − A)

Similarly, the set B − A is obtained from B by removing the elements of A ∩ B.

∴ n(B − A) = n(B) − n(A ∩ B) = 18 − 5 = 13

Hence, n(B − A) = 13.

Given,

n(A ∪ B) = 18, n(A − B) = 5 and n(B − A) = 3

The sets A − B, A ∩ B and B − A are mutually disjoint and together they make up A ∪ B.

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

⇒ 18 = 5 + 3 + n(A ∩ B)

⇒ 18 = 8 + n(A ∩ B)

⇒ n(A ∩ B) = 18 − 8 = 10

Hence, n(A ∩ B) = 10.

Given,

n(A − B) = 5, n(B − A) = 3 and n(A ∩ B) = 10

(i) n(A ∪ B)

The sets A − B, A ∩ B and B − A are mutually disjoint and together they make up A ∪ B.

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

⇒ n(A ∪ B) = 5 + 3 + 10 = 18

Hence, n(A ∪ B) = 18.

(ii) n(A)

The set A is made up of the two disjoint parts A − B and A ∩ B.

∴ n(A) = n(A − B) + n(A ∩ B) = 5 + 10 = 15

Hence, n(A) = 15.

(iii) n(B)

The set B is made up of the two disjoint parts B − A and A ∩ B.

∴ n(B) = n(B − A) + n(A ∩ B) = 3 + 10 = 13

Hence, n(B) = 13.

Given,

n(ξ) = 50, n(A) = 28 and n(B) = 30

It is understood that every set under consideration is a subset of the universal set ξ.

∴ A ∪ B ⊂ ξ

⇒ n(A ∪ B) ≤ n(ξ)

⇒ n(A ∪ B) ≤ 50

So n(A ∪ B) can never exceed 50, and the value 50 is actually possible.

[Taking n(A ∪ B) = 50 gives n(A ∩ B) = n(A) + n(B) − n(A ∪ B) = 28 + 30 − 50 = 8, which is a permissible value since 8 ≤ 28 and 8 ≤ 30.]

Hence, the greatest value of n(A ∪ B) is 50.

Given,

n(ξ) = 48 and n(A' ∪ B') = 36

By De Morgan's law,

A' ∪ B' = (A ∩ B)'

∴ n((A ∩ B)') = 36

We know that,

n(X') = n(ξ) − n(X)

∴ n((A ∩ B)') = n(ξ) − n(A ∩ B)

⇒ 36 = 48 − n(A ∩ B)

⇒ n(A ∩ B) = 48 − 36 = 12

Hence, n(A ∩ B) = 12.

Given,

n(ξ) = 40, n(A) = 22, n(A ∩ B) = 8 and n((A ∪ B)') = 6

We know that,

n((A ∪ B)') = n(ξ) − n(A ∪ B)

⇒ 6 = 40 − n(A ∪ B)

⇒ n(A ∪ B) = 40 − 6 = 34

Also we know that

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

⇒ 34 = 22 + n(B) − 8

⇒ 34 = 14 + n(B)

⇒ n(B) = 34 − 14 = 20

Hence, n(B) = 20.

Given,

n(ξ) = 50, n(A) = 20, n(B) = 26 and n((A ∪ B)') = 6

We know that,

n((A ∪ B)') = n(ξ) − n(A ∪ B)

⇒ 6 = 50 − n(A ∪ B)

⇒ n(A ∪ B) = 50 − 6 = 44

(i) n(A ∩ B)

We know that,

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

⇒ 44 = 20 + 26 − n(A ∩ B)

⇒ 44 = 46 − n(A ∩ B)

⇒ n(A ∩ B) = 46 − 44 = 2

Hence, n(A ∩ B) = 2.

(ii) n(A − B)

The set A − B is obtained from A by removing the elements of A ∩ B.

∴ n(A − B) = n(A) − n(A ∩ B) = 20 − 2 = 18

Hence, n(A − B) = 18.

(i) A ∪ A = A

A ∪ A = {x : x ∈ A or x ∈ A} = {x : x ∈ A} = A. This is the idempotent law for union.

Hence, the statement A ∪ A = A is true.

(ii) A ∩ A = φ

A ∩ A = {x : x ∈ A and x ∈ A} = {x : x ∈ A} = A, and not φ. By the idempotent law, A ∩ A = A.

Hence, the statement A ∩ A = φ is false.

(iii) A ∪ A' = ξ

A ∪ A' = {x : x ∈ A or x ∉ A}. Every element of the universal set either belongs to A or does not belong to A, so this set is ξ.

Hence, the statement A ∪ A' = ξ is true.

(iv) A ∩ A' = φ

For all x ∈ A ∩ A' we must have x ∈ A and x ∉ A, which is impossible. So A ∩ A' contains no element of ξ.

Hence, the statement A ∩ A' = φ is true.

(v) A ⊂ B ⇒ A ∪ B = B

If A ⊂ B, then every element of A is also an element of B, so A contributes no new element to the union.

∴ A ∪ B = B

Hence, the statement A ⊂ B ⇒ A ∪ B = B is true.

(vi) A ⊂ B ⇒ A ∩ B = A

If A ⊂ B, then every element of A is also an element of B, so all the elements of A are common to A and B.

∴ A ∩ B = A

Hence, the statement A ⊂ B ⇒ A ∩ B = A is true.

(vii) A ∩ φ = A

A ∩ φ = {x : x ∈ A and x ∈ φ}. Since φ has no element, no x can satisfy x ∈ φ, so A ∩ φ = φ and not A. This is the boundedness law.

Hence, the statement A ∩ φ = A is false.

(viii) A ∪ φ = A

A ∪ φ = {x : x ∈ A or x ∈ φ} = {x : x ∈ A} = A, since φ contributes no element. This is the identity law.

Hence, the statement A ∪ φ = A is true.

(i) φ' ∩ A = ...

By the law of complementation,

φ' = ξ.

∴ φ' ∩ A = ξ ∩ A = A (by the identity law)

Hence, φ' ∩ A = A.

(ii) ξ' ∩ A = ...

By the law of complementation,

ξ' = φ.

∴ ξ' ∩ A = φ ∩ A = φ (by the boundedness law)

Hence, ξ' ∩ A = φ.

No, the given statement is not true in general.

To justify it, we give an example. Let A = {1, 2} and B = {2, 3}, then

P(A) = {φ, {1}, {2}, {1, 2}} and P(B) = {φ, {2}, {3}, {2, 3}}

∴ P(A) ∪ P(B) = {φ, {1}, {2}, {1, 2}, {3}, {2, 3}}

Now A ∪ B = {1, 2, 3}

∴ P(A ∪ B) = {φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

Here {1, 3} ∈ P(A ∪ B) but {1, 3} ∉ P(A) ∪ P(B), because {1, 3} is neither a subset of A nor a subset of B.

∴ P(A) ∪ P(B) ≠ P(A ∪ B)

Hence, it is not true that P(A) ∪ P(B) = P(A ∪ B).

First we simplify (A − B) ∪ B.

(A − B) ∪ B = (A ∩ B') ∪ B   (∵ X − Y = X ∩ Y')

= (A ∪ B) ∩ (B' ∪ B)   (Distributive law)

= (A ∪ B) ∩ ξ   (∵ B ∪ B' = ξ)

= A ∪ B   (∵ X ∩ ξ = X)

∴ (A − B) ∪ B = A ∪ B ...........(1)

First, let (A − B) ∪ B = A, we shall prove that B ⊂ A.

From (1), A ∪ B = A.

But we know that A ∪ B = A if and only if B ⊂ A.

∴ B ⊂ A

Conversely, let B ⊂ A, we shall prove that (A − B) ∪ B = A.

As B ⊂ A, every element of B is also an element of A, so A ∪ B = A.

∴ From (1), (A − B) ∪ B = A ∪ B = A

Hence, proved that (A − B) ∪ B = A if and only if B ⊂ A.

Let X be any arbitrary element of P(A) ∪ P(B), then

X ∈ P(A) ∪ P(B)

⇒ X ∈ P(A) or X ∈ P(B)

⇒ X ⊂ A or X ⊂ B   (by the definition of a power set)

Now, we know that A ⊂ A ∪ B and B ⊂ A ∪ B.

If X ⊂ A, then X ⊂ A ⊂ A ∪ B, so X ⊂ A ∪ B.

If X ⊂ B, then X ⊂ B ⊂ A ∪ B, so X ⊂ A ∪ B.

Thus in either case, X ⊂ A ∪ B

⇒ X ∈ P(A ∪ B)   (by the definition of a power set)

So every element of P(A) ∪ P(B) is also an element of P(A ∪ B).

Hence, proved that P(A) ∪ P(B) ⊂ P(A ∪ B).

To show that the four conditions are equivalent, it is enough to prove the cycle of implications

(i) ⇒ (ii) ⇒ (iii) ⇒ (iv) ⇒ (i)

(i) ⇒ (ii)

Let A ⊂ B. Then every element of A is also an element of B, so there is no element which belongs to A but does not belong to B.

∴ A − B = φ

(ii) ⇒ (iii)

Let A − B = φ i.e. A ∩ B' = φ.

We know that,

A = A ∩ ξ   (Identity law)

⇒ A = A ∩ (B ∪ B')   (∵ B ∪ B' = ξ)

⇒ A = (A ∩ B) ∪ (A ∩ B')   (Distributive law)

⇒ A = (A ∩ B) ∪ φ   (∵ A ∩ B' = φ, given)

⇒ A = A ∩ B

∴ A ∩ B = A

(iii) ⇒ (iv)

Let A ∩ B = A.

Then A ∪ B = (A ∩ B) ∪ B   (putting A = A ∩ B)

= B   (Absorption law, since B ∪ (B ∩ A) = B)

∴ A ∪ B = B

(iv) ⇒ (i)

Let A ∪ B = B.

We know that,

A ⊂ A ∪ B   (Boundedness law)

⇒ A ⊂ B   (∵ A ∪ B = B, given)

∴ A ⊂ B

Hence, proved that the four conditions A ⊂ B, A − B = φ, A ∩ B = A and A ∪ B = B are equivalent.

(i) (A − B) ∪ B = A ∪ B

(A − B) ∪ B = (A ∩ B') ∪ B   (∵ X − Y = X ∩ Y')

= (A ∪ B) ∩ (B' ∪ B)   (Distributive law)

= (A ∪ B) ∩ ξ   (∵ B ∪ B' = ξ)

= A ∪ B   (∵ X ∩ ξ = X)

Hence, proved that (A − B) ∪ B = A ∪ B.

(ii) (A − B) ∩ B = φ

(A − B) ∩ B = (A ∩ B') ∩ B   (∵ X − Y = X ∩ Y')

= A ∩ (B' ∩ B)   (Associative law)

= A ∩ φ   (∵ B ∩ B' = φ)

= φ   (∵ X ∩ φ = φ)

Hence, proved that (A − B) ∩ B = φ.

(i) (A ∪ B) − B = A − B

(A ∪ B) − B = (A ∪ B) ∩ B'   (∵ X − Y = X ∩ Y')

= (A ∩ B') ∪ (B ∩ B')   (Distributive law)

= (A ∩ B') ∪ φ   (∵ B ∩ B' = φ)

= A ∩ B'   (∵ X ∪ φ = X)

= A − B   (∵ X ∩ Y' = X − Y)

Hence, proved that (A ∪ B) − B = A − B.

(ii) A − (A ∩ B) = A − B

A − (A ∩ B) = A ∩ (A ∩ B)'   (∵ X − Y = X ∩ Y')

= A ∩ (A' ∪ B')   (De Morgan law)

= (A ∩ A') ∪ (A ∩ B')   (Distributive law)

= φ ∪ (A ∩ B')   (∵ A ∩ A' = φ)

= A ∩ B'     (∵ φ ∪ X = X)

= A − B   (∵ X ∩ Y' = X − Y)

Hence, proved that A − (A ∩ B) = A − B.

(i) A − (A − B) = A ∩ B

Solving L.H.S.,

A − (A − B) = A − (A ∩ B')   (∵ X − Y = X ∩ Y')

= A ∩ (A ∩ B')'   (∵ X − Y = X ∩ Y')

= A ∩ (A' ∪ (B')')   (De Morgan law)

= A ∩ (A' ∪ B)   (∵ (B')' = B)

= (A ∩ A') ∪ (A ∩ B)   (Distributive law)

= φ ∪ (A ∩ B)   (∵ A ∩ A' = φ)

= A ∩ B   (∵ φ ∪ X = X)

Hence, proved that A − (A − B) = A ∩ B.

(ii) (A − B) ∪ (A − C) = A − (B ∩ C)

Solving L.H.S.,

(A − B) ∪ (A − C) = (A ∩ B') ∪ (A ∩ C')   (∵ X − Y = X ∩ Y')

= A ∩ (B' ∪ C')   (Distributive law)

= A ∩ (B ∩ C)'   (De Morgan law)

= A − (B ∩ C)   (∵ X ∩ Y' = X − Y)

Hence, proved that (A − B) ∪ (A − C) = A − (B ∩ C).

(iii) (A − B) ∩ (C − B) = (A ∩ C) − B

Solving L.H.S.,

(A − B) ∩ (C − B) = (A ∩ B') ∩ (C ∩ B')   (∵ X − Y = X ∩ Y')

= (A ∩ C) ∩ (B' ∩ B')   (Commutative and Associative laws)

= (A ∩ C) ∩ B'   (∵ B' ∩ B' = B', Idempotent law)

= (A ∩ C) − B   (∵ X ∩ Y' = X − Y)

Hence, proved that (A − B) ∩ (C − B) = (A ∩ C) − B.

(iv) A − (B − C) = (A − B) ∪ (A ∩ C)

Solving L.H.S.,

A − (B − C) = A − (B ∩ C')   (∵ X − Y = X ∩ Y')

= A ∩ (B ∩ C')'   (∵ X − Y = X ∩ Y')

= A ∩ (B' ∪ (C')')   (De Morgan law)

= A ∩ (B' ∪ C)   (∵ (C')' = C)

= (A ∩ B') ∪ (A ∩ C)   (Distributive law)

= (A − B) ∪ (A ∩ C)   (∵ X ∩ Y' = X − Y)

Hence, proved that A − (B − C) = (A − B) ∪ (A ∩ C).

Taking the left hand side,

L.H.S. = (A − B) ∪ (B − A)

= (A ∩ B') ∪ (B ∩ A')   (∵ X − Y = X ∩ Y')

= [(A ∩ B') ∪ B] ∩ [(A ∩ B') ∪ A']   (Distributive law)

= [(A ∪ B) ∩ (B' ∪ B)] ∩ [(A ∪ A') ∩ (B' ∪ A')]   (Distributive law)

= [(A ∪ B) ∩ ξ] ∩ [ξ ∩ (A' ∪ B')]   (∵ B ∪ B' = ξ and A ∪ A' = ξ)

= (A ∪ B) ∩ (A' ∪ B')   (∵ X ∩ ξ = X)

= (A ∪ B) ∩ (A ∩ B)'   (De Morgan law)

= (A ∪ B) − (A ∩ B)   (∵ X ∩ Y' = X − Y)

= R.H.S.

Hence, proved that (A − B) ∪ (B − A) = (A ∪ B) − (A ∩ B).

Any well-defined collection of objects is called a set.

Given any integer, we can decide beyond doubt whether it is even or not, so option 1 is a set.

The months of a year beginning with the letter J are January, June and July, which are fixed beyond doubt, so option 2 is a set.

Given any number, we can decide beyond doubt whether it is prime and less than 20, so option 4 is a set.

But a writer considered most talented by one person may not be considered most talented by another, so the collection of most talented writers of India is not a well-defined collection of objects.

Hence, Option 3 is the correct option.

Given,

A = {1, 2, 3, 4, 5}

The element 0 does not belong to A, while 3 belongs to A.

∴ The statements 0 ∉ A and 3 ∈ A are true.

{3} represents a set whose only member 3 belongs to A, so {3} is a subset of {1, 2, 3, 4, 5}.

∴ The statement {3} ⊂ A is true.

The elements of A are the numbers 1, 2, 3, 4, 5 and not the set {3}, so set {3} is not an element of A.

∴ The statement {3} ∈ A is not true.

Hence, Option 3 is the correct option.

A set which does not contain any element is called the null set.

For option 1,

2x − 1 = 3

⇒ 2x = 3 + 1

⇒ 2x = 4

⇒ x = 2 ∈ N

So the set is {2}, which is not a null set.

For option 2, the squares 1, 4, 9, 16 of 1, 2, 3, 4 are less than 20, so the set is {1, 2, 3, 4}, which is not a null set.

For option 4,

3x + 7 = 1

⇒ 3x = 1 − 7

⇒ 3x = −6

⇒ x = −2 ∈ I

So the set is {−2}, which is not a null set.

For option 3, 2 is the only even prime number and every even number greater than 2 is divisible by 2, so it cannot be prime. Therefore this set contains no element.

Hence, Option 3 is the correct option.

A set that contains a limited (definite) number of different elements is called a finite set.

For option 1, the set is {2, 4, 6, 8, .....}, which never comes to an end, so it is an infinite set.

For option 2, the prime numbers are 2, 3, 5, 7, 11, ....., which are infinite in number, so it is an infinite set.

For option 4, the set is {....., 4, 5, 6, 7} and the integers go on decreasing without end, so it is an infinite set.

For option 3, since 128 = 2⁷, the factors of 128 are 1, 2, 4, 8, 16, 32, 64 and 128, which are 8 in number, a definite number.

Hence, Option 3 is the correct option.

A set that contains all the elements under consideration in a given problem is called the universal set. So every one of the sets A, B and C must be a subset of it.

In option 1, the element 9 ∈ A but 9 does not belong to it, so A is not its subset.

In option 2, the element 0 ∈ B but 0 does not belong to it, so B is not its subset.

In option 3, the element 0 ∈ B but 0 does not belong to it, so B is not its subset.

In option 4, the set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Every element of A, every element of B and every element of C belongs to it, so A ⊂ it, B ⊂ it and C ⊂ it.

Hence, Option 4 is the correct option.

If A is a set with n(A) = m, then the number of subsets of A is 2^m.

The set A itself is a subset of A but it is not a proper subset of A, so it has to be left out.

∴ Number of proper subsets of A = 2^m − 1

Here the set contains 4 elements, so m = 4.

∴ Number of proper subsets = 2⁴ − 1 = 16 − 1 = 15

Hence, Option 4 is the correct option.

We know that N ⊂ W ⊂ I ⊂ Q ⊂ R, so N ⊂ R, N ⊂ Q and Q ⊂ R are all correct.

Here T denotes the set of irrational numbers i.e. T = {x : x ∈ R and x ∉ Q}.

Every natural number n can be written as $\frac{n}{1}$, so every natural number is a rational number and hence no natural number is irrational.

∴ N ⊄ T

Hence, Option 4 is the correct option.

Given,

A = [1, 5] = {x : x ∈ R, 1 ≤ x ≤ 5}

and B = [3, 9] = {x : x ∈ R, 3 ≤ x ≤ 9}

A − B is the set of all real numbers which belong to A but do not belong to B.

The part of A which also lies in B is 3 ≤ x ≤ 5. Removing this part from [1, 5], we are left with 1 ≤ x < 3, where 1 is included and 3 is excluded because 3 ∈ B.

∴ A − B = [1, 3)

Hence, Option 4 is the correct option.

Given,

n(A − B) = 10, n(B − A) = 23 and n(A ∪ B) = 50

The sets A − B, A ∩ B and B − A are mutually disjoint and together they make up A ∪ B.

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

⇒ 50 = 10 + 23 + n(A ∩ B)

⇒ 50 = 33 + n(A ∩ B)

⇒ n(A ∩ B) = 50 − 33 = 17

Hence, Option 2 is the correct option.

Given,

A ⊂ B, so every element of A is also an element of B.

As A contributes no new element to the union, A ∪ B = B, so option 1 is correct.

As all the elements of A are common to A and B, A ∩ B = A, so option 2 is correct.

As there is no element of A which does not belong to B, A − B = φ, so option 3 is correct.

But B may contain elements which do not belong to A, so B − A need not be φ.

For example, taking A = {1, 2} and B = {1, 2, 3}, we have A ⊂ B and B − A = {3} ≠ φ.

Hence, Option 4 is the correct option.

Let the two sets be A and B, then n(A) = m and n(B) = n, where m > n.

Number of subsets of A = 2^m and number of subsets of B = 2ⁿ

Given,

The total number of subsets of the first set is 192 more than the total number of subsets of the second set.

⇒ 2^m − 2ⁿ = 192

⇒ 2ⁿ(2^{m − n} − 1) = 192

⇒ 2ⁿ(2^{m − n} − 1) = 64 × 3 = 2⁶ × 3

As 2^{m − n} − 1 is an odd number and 2ⁿ is a power of 2, comparing both sides we get

2ⁿ = 2⁶ and 2^{m − n} − 1 = 3

⇒ n = 6 and 2^{m − n} = 4 = 2²

⇒ n = 6 and m − n = 2

⇒ n = 6 and m = 2 + 6 = 8

Hence, Option 2 is the correct option.

By the absorption law,

A ∩ (A ∪ B) = (A ∪ φ) ∩ (A ∪ B)   (∵ A ∪ φ = A)

= A ∪ (φ ∩ B)   (Distributive law)

= A ∪ φ   (∵ φ ∩ B = φ)

= A   (∵ A ∪ φ = A)

Hence, Option 1 is the correct option.

Given,

A = {0, 1, 2} and B = {2, 3, 4}

We know that,

A Δ B = (A − B) ∪ (B − A)

The elements of A which do not belong to B are 0 and 1.

∴ A − B = {0, 1}

The elements of B which do not belong to A are 3 and 4.

∴ B − A = {3, 4}

∴ A Δ B = {0, 1} ∪ {3, 4} = {0, 1, 3, 4}

Hence, Option 3 is the correct option.

By definition, the symmetric difference of two sets A and B is the set consisting of all elements which belong to A or B but do not belong to A ∩ B.

∴ A Δ B = {x : x ∈ A or x ∈ B but x ∉ A ∩ B}

= (A ∪ B) − (A ∩ B)

Option 1 is not correct, because (A − B) ∪ A = A by the absorption law, since A − B ⊂ A.

Option 2 is not correct, because (B − A) ∪ B = B by the absorption law, since B − A ⊂ B.

Option 4 is not correct, because A ∩ B ⊂ A ∪ B and so (A ∪ B) ∩ (A ∩ B) = A ∩ B.

Hence, Option 3 is the correct option.

We know that,

A Δ B = (A ∪ B) − (A ∩ B) = (A − B) ∪ (B − A)

So options 2 and 3 are both equal to A Δ B.

For option 4, we have

(A ∪ B) − B = (A ∪ B) ∩ B' = (A ∩ B') ∪ (B ∩ B') = (A ∩ B') ∪ φ = A ∩ B' = A − B

and similarly (A ∪ B) − A = B − A.

∴ ((A ∪ B) − B) ∪ ((A ∪ B) − A) = (A − B) ∪ (B − A) = A Δ B, so option 4 is also equal to A Δ B.

For option 1, we know that the sets A − B and B − A are disjoint, since A − B contains only elements which do not belong to B while every element of B − A belongs to B.

∴ (A − B) ∩ (B − A) = φ, which is not equal to A Δ B in general.

Hence, Option 1 is the correct option.

X ∩ (X ∪ Y)' = X ∩ (X' ∩ Y')   (De Morgan law)

= (X ∩ X') ∩ Y'   (Associative law)

= φ ∩ Y'   (∵ X ∩ X' = φ)

= φ   (∵ φ ∩ Z = φ)

Hence, Option 3 is the correct option.

First we simplify (A' ∪ B') − A.

(A' ∪ B') − A = (A' ∪ B') ∩ A'   (∵ X − Y = X ∩ Y')

= (A' ∩ A') ∪ (B' ∩ A')   (Distributive law)

= A' ∪ (A' ∩ B')   (∵ A' ∩ A' = A', Idempotent law)

= A'   (Absorption law)

∴ ((A' ∪ B') − A)' = (A')' = A   (Involution law)

Hence, Option 1 is the correct option.

First we simplify (B' − A),

B' − A = B' ∩ A'   (∵ X − Y = X ∩ Y')

∴ B' ∪ (B' − A) = B' ∪ (B' ∩ A') = B'   (Absorption law)

∴ [B' ∪ (B' − A)]' = (B')' = B   (Involution law)

Hence, Option 2 is the correct option.

Solving,

(A − B) ∩ (C − B) = (A ∩ B') ∩ (C ∩ B')   (∵ X − Y = X ∩ Y')

= (A ∩ C) ∩ (B' ∩ B')   (Commutative and Associative laws)

= (A ∩ C) ∩ B'   (∵ B' ∩ B' = B', Idempotent law)

= (A ∩ C) − B   (∵ X ∩ Y' = X − Y)

Hence, Option 3 is the correct option.

Given,

A and B are disjoint sets, so A ∩ B = φ.

First we simplify (A ∪ B) ∩ B'.

(A ∪ B) ∩ B' = (A ∩ B') ∪ (B ∩ B')   (Distributive law)

= (A ∩ B') ∪ φ   (∵ B ∩ B' = φ)

= A ∩ B'   (∵ X ∪ φ = X)

= A − B   (∵ X ∩ Y' = X − Y)

As A ∩ B = φ, no element of A belongs to B, so on removing the elements of B from A nothing is removed.

∴ A − B = A

∴ A' ∪ ((A ∪ B) ∩ B') = A' ∪ A = ξ   (∵ A ∪ A' = ξ)

Hence, Option 2 is the correct option.

The triangle and the circle are contained in the square, so every point inside the triangle and every point inside the circle is also a point inside the square.

∴ T ⊂ S and C ⊂ S

As T and C contribute no new point to the union,

S ∪ T ∪ C = S

Option 1 is not correct, because the triangle and the circle intersect each other and their common region lies inside the square, so S ∩ T ∩ C ≠ φ.

Option 2 is not correct, because S ⊄ C.

Option 4 is not correct, because S ∪ T = S while S ∩ C = C, and S ≠ C.

Hence, Option 3 is the correct option.

A ∩ B is the set of all those ordered pairs (x, y) which satisfy both the equations, so we solve them simultaneously.

$y = \frac{1}{x}$ and $y = −x$

$\Rightarrow -x = \frac{1}{x}$

⇒ $-x^2$ = 1

⇒ $x^2$ = −1

The square of a real number can never be negative, so there is no real value of x satisfying this equation.

∴ There is no ordered pair common to A and B.

∴ A ∩ B = φ

Hence, Option 3 is the correct option.

Given,

n(ξ) = 50, n(A) = 38 and n(B) = 30

It is understood that every set under consideration is a subset of ξ.

∴ A ∪ B ⊂ ξ

⇒ n(A ∪ B) ≤ n(ξ)

⇒ n(A ∪ B) ≤ 50

We know that,

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

⇒ n(A) + n(B) − n(A ∩ B) ≤ 50

⇒ 38 + 30 − n(A ∩ B) ≤ 50

⇒ 68 − 50 ≤ n(A ∩ B)

⇒ n(A ∩ B) ≥ 18

Hence, Option 4 is the correct option.

Given,

ξ = {1, 2, 3, ....., 40}

It is understood that A and B are subsets of ξ, so their members are to be taken only from ξ.

A number divisible by both 2 and 3 must be divisible by their L.C.M., which is 6.

The multiples of 6 lying in ξ are 6, 12, 18, 24, 30 and 36.

∴ A = {6, 12, 18, 24, 30, 36} and so n(A) = 6

For the set B, x = n² where n ∈ N. The squares lying in ξ are 1, 4, 9, 16, 25 and 36, since 7² = 49 > 40.

∴ B = {1, 4, 9, 16, 25, 36} and so n(B) = 6

∴ n(A) − n(B) = 6 − 6 = 0

Hence, Option 1 is the correct option.

Given,

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

Statement I: A ∩ B is the set of all the elements which belong to both A and B. The only element common to A and B is 6.

∴ A ∩ B = {6}

∴ Statement I is true.

Statement II: A − B is the set of all the elements which belong to A but do not belong to B. Removing 6 from A, the remaining elements are 3 and 9.

∴ A − B = {3, 9}

But the given Statement II says A − B = {3, 9, 2, 4, 8}, which is actually A Δ B and not A − B.

∴ Statement II is false.

Hence, Option 3 is the correct option.

The squares of the integers 0, ±1, ±2, ±3, ±4, ±5 and ±6 are

0, 1, 4, 9, 16, 25 and 36

and each one of these is less than 49, while (±7)² = 49 which is not less than 49.

∴ Reason (R) is true.

As x ∈ I and x² < 49, the only integers x which satisfy the condition are the ones listed above i.e. 0, ±1, ±2, ±3, ±4, ±5 and ±6.

∴ A = {−6, −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, 6}

∴ Assertion (A) is true.

The roster form of A is obtained exactly by picking out those integers whose squares are less than 49, which is what Reason (R) states.

Hence, Assertion (A) and Reason (R) are both true and Reason (R) is the correct explanation of Assertion (A).

We know that the number of subsets of a set containing n elements is 2ⁿ.

∴ Reason (R) is true.

Given,

A = {1, 2, 3}, so n(A) = 3.

The power set P(A) is the set formed by all the subsets of A, so the cardinal number of P(A) is the number of subsets of A.

∴ n(P(A)) = 2^n(A) = 2³ = 8, and not 9.

∴ Assertion (A) is false.

Hence, Assertion (A) is false, but Reason (R) is true.

Given,

A ⊂ B, thus every member of A is also a member of B.

∴ Reason (R) is true.

So, the elements common to A and B (i.e. A ∩ B) will be equal to the members of A.

∴ A ∩ B = A

∴ Assertion (A) is true.

Hence, Assertion (A) and Reason (R) are both true and Reason (R) is the correct explanation of Assertion (A).

Assertion (A): A − B = A − (A ∩ B)

Solving R.H.S.,

A − (A ∩ B) = A ∩ (A ∩ B)'   (∵ X − Y = X ∩ Y')

= A ∩ (A' ∪ B')   (De Morgan law)

= (A ∩ A') ∪ (A ∩ B')   (Distributive law)

= φ ∪ (A ∩ B')   (∵ A ∩ A' = φ)

= A ∩ B' = A − B

∴ Assertion (A) is true.

Reason (R): A − B = (A ∪ B) − B

Solving R.H.S.,

(A ∪ B) − B = (A ∪ B) ∩ B'   (∵ X − Y = X ∩ Y')

= (A ∩ B') ∪ (B ∩ B')   (Distributive law)

= (A ∩ B') ∪ φ   (∵ B ∩ B' = φ)

= A ∩ B' = A − B

∴ Reason (R) is true.

Both are correct results about A − B, but Reason (R) removes B from the bigger set A ∪ B, whereas Assertion (A) removes A ∩ B from A itself. So Reason (R) is not used anywhere in proving Assertion (A).

Hence, Assertion (A) and Reason (R) are both true, but Reason (R) is not the correct explanation of Assertion (A).

Given,

A ⊂ B, thus every element of A is also an element of B.

So, there is no element which belongs to A but does not belong to B.

∴ A − B = φ

∴ Assertion (A) is true.

We know that,

n(A ∪ B) = n(A) + n(B) − n(A ∩ B).

If A and B are disjoint sets, then A ∩ B = φ so that n(A ∩ B) = 0.

∴ n(A ∪ B) = n(A) + n(B)

∴ Reason (R) is true.

Reason (R) is a result about the cardinal number of the union of two disjoint sets, while Assertion (A) is about the difference of two sets one of which is contained in the other. So Reason (R) is not used in proving Assertion (A).

Hence, Assertion (A) and Reason (R) are both true, but Reason (R) is not the correct explanation of Assertion (A).

For the set A,

x² − 8x + 12 = 0

⇒ x² − 6x − 2x + 12 = 0

⇒ x(x − 6) − 2(x − 6) = 0

⇒ (x − 2)(x − 6) = 0

⇒ x − 2 = 0 or x − 6 = 0

⇒ x = 2 or x = 6

∴ A = {2, 6}

Given,

B = {2, 4, 6}, C = {2, 4, 6, 8, 10, .....} and D = {6}

Now we compare the sets one by one.

Both the members 2 and 6 of A belong to B, so A ⊂ B.

Both the members 2 and 6 of A are even natural numbers, so A ⊂ C.

Each of the members 2, 4 and 6 of B is an even natural number, so B ⊂ C.

The only member 6 of D belongs to A, to B and to C, so D ⊂ A, D ⊂ B and D ⊂ C.

Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B and D ⊂ C.

(i) The letters of the word TAMILNADU are T, A, M, I, L, N, A, D, U. While listing the elements of a set, each member is written only once, so the repeated letter A is written only once.

∴ ξ = {T, A, M, I, L, N, D, U}

The vowels in the English alphabet are A, E, I, O, U. Out of the elements of ξ, the vowels are A, I and U.

∴ X = {A, I, U}

Hence, ξ = {T, A, M, I, L, N, D, U} and X = {A, I, U}.

(ii) Counting the different elements, ξ has 8 elements and X has 3 elements.

Hence, n(ξ) = 8 and n(X) = 3.

(iii) Here n(X) = 3.

Number of subsets of X = 2³ = 8

Number of proper subsets of X = 2³ − 1 = 8 − 1 = 7

(The set X itself is a subset of X but not a proper subset of X, so it is left out.)

Hence, the proper subsets of X are φ, {A}, {I}, {U}, {A, I}, {A, U} and {I, U}.

(iv) The power set P(X) is the set formed by all the subsets of X, so the cardinal number of P(X) is the number of subsets of X.

∴ n(P(X)) = 2^n(X) = 2³ = 8

Hence, the cardinal number of the power set of X is 8.

The letters of the word POOR are P, O, O, R. While listing the elements of a set, each member is written only once, so the repeated letter O is written only once.

∴ A = {P, O, R} and n(A) = 3

Number of elements in P(A) = 2^n(A) = 2³ = 8

Writing down all the subsets of A, we have

The subset with no element : φ

The subsets with one element : {P}, {O}, {R}

The subsets with two elements : {P, O}, {P, R}, {O, R}

The subset with three elements : {P, O, R}

Hence, the power set of A is P(A) = {φ, {P}, {O}, {R}, {P, O}, {P, R}, {O, R}, {P, O, R}}.

(i) {−1, 0, 1}

Let A = {−1, 0, 1}, so n(A) = 3.

Number of elements in P(A) = 2³ = 8

Writing down all the subsets of A, we have

The subset with no element : φ

The subsets with one element : {−1}, {0}, {1}

The subsets with two elements : {−1, 0}, {0, 1}, {−1, 1}

The subset with three elements : {−1, 0, 1}

Hence, P(A) = {φ, {−1}, {0}, {1}, {−1, 0}, {0, 1}, {−1, 1}, {−1, 0, 1}}.

(ii) {0, 1, {0, 1}}

Let B = {0, 1, {0, 1}}.

The elements of B are the numbers 0 and 1 and the set {0, 1}, so n(B) = 3.

Number of elements in P(B) = 2³ = 8

Writing down all the subsets of B, we have

The subset with no element : φ

The subsets with one element : {0}, {1}, {{0, 1}}

The subsets with two elements : {0, 1}, {0, {0, 1}}, {1, {0, 1}}

The subset with three elements : {0, 1, {0, 1}}

Hence, P(B) = {φ, {0}, {1}, {{0, 1}}, {0, 1}, {0, {0, 1}}, {1, {0, 1}}, {0, 1, {0, 1}}}.

Given,

A = {2, 3, 5, 7, 8}, B = {1, 5, 9} and A' = {1, 4, 6, 9}

The universal set consists of the elements of A together with the elements of A'.

∴ ξ = A ∪ A' = {2, 3, 5, 7, 8} ∪ {1, 4, 6, 9} = {1, 2, 3, 4, 5, 6, 7, 8, 9}

∴ B' = ξ − B = {2, 3, 4, 6, 7, 8}

(i) (A ∪ B)' = A' ∩ B'

Collecting all the elements which belong to A or B or both,

A ∪ B = {1, 2, 3, 5, 7, 8, 9}

Removing these elements from ξ, the elements left are 4 and 6.

∴ (A ∪ B)' = {4, 6}

Now, the elements common to A' = {1, 4, 6, 9} and B' = {2, 3, 4, 6, 7, 8} are 4 and 6.

∴ A' ∩ B' = {4, 6}

∴ (A ∪ B)' = A' ∩ B'

Hence, the result (A ∪ B)' = A' ∩ B' is verified.

(ii) B − A = A' ∩ B

B − A is the set of all the elements which belong to B but do not belong to A. Removing 5 from B,

B − A = {1, 9}

Now, the elements common to A' = {1, 4, 6, 9} and B = {1, 5, 9} are 1 and 9.

∴ A' ∩ B = {1, 9}

∴ B − A = A' ∩ B

Hence, the result B − A = A' ∩ B is verified.

No, the given statement is not true in general.

To justify it, we give an example. Let

A = {1, 2}, B = {2, 3} and C = {1, 3}

Taking the left hand side,

B − C = {2, 3} − {1, 3} = {2}

∴ A − (B − C) = {1, 2} − {2} = {1}

Taking the right hand side,

A − B = {1, 2} − {2, 3} = {1}

∴ (A − B) − C = {1} − {1, 3} = φ

Here A − (B − C) = {1} and (A − B) − C = φ, and {1} ≠ φ.

∴ A − (B − C) ≠ (A − B) − C

Hence, the statement A − (B − C) = (A − B) − C is not true for all sets A, B and C.

Given,

n(ξ) = 30, n(A') = 15, n(B) = 5 and n(A ∩ B) = 3

(i) n(A)

We know that,

n(A') = n(ξ) − n(A)

⇒ 15 = 30 − n(A)

⇒ n(A) = 30 − 15 = 15

Hence, n(A) = 15.

(ii) n(A ∪ B)

We know that,

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

⇒ n(A ∪ B) = 15 + 5 − 3 = 17

Hence, n(A ∪ B) = 17.

(iii) n(A − B)

The set A − B is obtained from A by removing the elements of A ∩ B.

∴ n(A − B) = n(A) − n(A ∩ B) = 15 − 3 = 12

Hence, n(A − B) = 12.

Given,

n(ξ) = 40, n((A ∪ B)') = 12, n(A − B) = 10 and n(B − A) = 14

We know that,

n((A ∪ B)') = n(ξ) − n(A ∪ B)

⇒ 12 = 40 − n(A ∪ B)

⇒ n(A ∪ B) = 40 − 12 = 28

The sets A − B, A ∩ B and B − A are mutually disjoint and together they make up A ∪ B.

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

⇒ 28 = 10 + 14 + n(A ∩ B)

⇒ 28 = 24 + n(A ∩ B)

⇒ n(A ∩ B) = 28 − 24 = 4

(i) n(A)

The set A is made up of the two disjoint parts A − B and A ∩ B.

∴ n(A) = n(A − B) + n(A ∩ B) = 10 + 4 = 14

Hence, n(A) = 14.

(ii) n(B)

The set B is made up of the two disjoint parts B − A and A ∩ B.

∴ n(B) = n(B − A) + n(A ∩ B) = 14 + 4 = 18

Hence, n(B) = 18.

(iii) n(A ∩ B)

From the working above, n(A ∩ B) = 4.

Hence, n(A ∩ B) = 4.

Given,

n(A ∪ B) = 18, n(A' ∩ B) = 3 and n(A ∩ B') = 5

We know that,

A ∩ B' = A − B and A' ∩ B = B − A

∴ n(A − B) = 5 and n(B − A) = 3

The sets A − B, A ∩ B and B − A are mutually disjoint and together they make up A ∪ B.

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

⇒ 18 = 5 + 3 + n(A ∩ B)

⇒ 18 = 8 + n(A ∩ B)

⇒ n(A ∩ B) = 18 − 8 = 10

Hence, n(A ∩ B) = 10.

Given,

n(A ∪ B) = 21, n(A' ∩ B') = 9 and n(A ∩ B) = 7

By De Morgan's law,

A' ∩ B' = (A ∪ B)'

∴ n((A ∪ B)') = 9

We know that,

n((A ∪ B)') = n(ξ) − n(A ∪ B)

⇒ 9 = n(ξ) − 21

⇒ n(ξ) = 21 + 9 = 30

Now, n((A ∩ B)') = n(ξ) − n(A ∩ B)

⇒ n((A ∩ B)') = 30 − 7 = 23

Hence, n((A ∩ B)') = 23.

Given,

n(ξ) = 50, n(A) = 3x, n(B) = 2x and n(A ∩ B) = x = n((A ∪ B)')

(i) the value of x:

We know that,

n(A ∪ B) = n(A) + n(B) − n(A ∩ B)

⇒ n(A ∪ B) = 3x + 2x − x = 4x ...........(1)

Also we know that

n((A ∪ B)') = n(ξ) − n(A ∪ B)

⇒ x = 50 − n(A ∪ B)

⇒ n(A ∪ B) = 50 − x ...........(2)

From (1) and (2),

4x = 50 − x

⇒ 4x + x = 50

⇒ 5x = 50

⇒ x = 10

Hence, the value of x is 10.

(ii) n(A − B)

From part (i), x = 10.

∴ n(A) = 3x = 3 × 10 = 30 and n(A ∩ B) = x = 10

The set A − B is obtained from A by removing the elements of A ∩ B.

∴ n(A − B) = n(A) − n(A ∩ B) = 30 − 10 = 20

Hence, n(A − B) = 20.

Given,

n(ξ) = 15, A ⊂ B, n(A) = 8 and n(B) = 12

As A ⊂ B, the circle representing A lies completely inside the circle representing B, and both lie inside the rectangle representing ξ.

From the Venn diagram,

Number of elements in A = 8,

Number of elements in B but not in A = n(B) − n(A) = 12 − 8 = 4,

Number of elements outside B = n(ξ) − n(B) = 15 − 12 = 3.

(i) n(A')

We know that,

n(A') = n(ξ) − n(A) = 15 − 8 = 7

Hence, n(A') = 7.

(ii) n(B')

We know that,

n(B') = n(ξ) − n(B) = 15 − 12 = 3

Hence, n(B') = 3.

(iii) n(A ∩ B')

We know that,

A ∩ B' = A − B.

As A ⊂ B, every element of A is also an element of B, so there is no element which belongs to A but not to B.

∴ A ∩ B' = φ and so n(A ∩ B') = 0

Hence, n(A ∩ B') = 0.

(iv) n(A' ∩ B)

We know that,

A' ∩ B = B − A.

As A ⊂ B, we have

n(B − A) = n(B) − n(A) = 12 − 8 = 4

Hence, n(A' ∩ B) = 4.

Given,

n(A) = 10 and n(B) = 7

(i) The least value of n(A ∩ B):

The set A ∩ B has the smallest number of elements when A and B have no element in common, i.e. when they are disjoint. In that case A ∩ B = φ.

∴ The least value of n(A ∩ B) = 0

Hence, the least value of n(A ∩ B) is 0 and it occurs when A ∩ B = φ.

(ii) The greatest value of n(A ∩ B):

Since A ∩ B ⊂ B, we have

⇒ n(A ∩ B) ≤ n(B)

⇒ n(A ∩ B) ≤ 7

The value 7 is attained when every element of B is also an element of A, i.e. when B ⊂ A, for then A ∩ B = B.

∴ The greatest value of n(A ∩ B) = 7

Hence, the greatest value of n(A ∩ B) is 7 and it occurs when B ⊂ A.

(iii) The greatest value of n(A ∪ B):

We know that,

n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 10 + 7 − n(A ∩ B) = 17 − n(A ∩ B)

So n(A ∪ B) is greatest when n(A ∩ B) is least, and from part (i) the least value of n(A ∩ B) is 0.

∴ The greatest value of n(A ∪ B) = 17 − 0 = 17

Hence, the greatest value of n(A ∪ B) is 17 and it occurs when A ∩ B = φ.

(iv) The least value of n(A ∪ B):

From part (iii), n(A ∪ B) = 17 − n(A ∩ B).

So n(A ∪ B) is least when n(A ∩ B) is greatest, and from part (ii) the greatest value of n(A ∩ B) is 7.

∴ The least value of n(A ∪ B) = 17 − 7 = 10

Hence, the least value of n(A ∪ B) is 10 and it occurs when B ⊂ A.