Sets

A collection can include different types of elements. Whereas a set is a well-defined collection of distinct elements and it is enclosed in “{}.”

E.g.:

The collection of good teachers is a collection but not a set

Whereas when we take all the teachers of the school, then it is set. Because when we take all the teachers, then it is a definite number, so it called a set. But when we say, good teachers, it changes opinion wise, so we can say it is not definite hence we can call it a collection as it is not well defined.

(i) It is a set. As we are saying all numbers less than 50 it is a definite quantity hence it is a set.

(ii) It is not a set. Because when we say good hockey players, it changes opinion wise and cannot be well defined or is not countable.

(iii) It is a set. As we are saying a collection of all the girls in your class it is a definite quantity hence it is a set.

(iv) It is not a set. Because when we say a collection of all talented writers of India, it changes opinion wise and cannot be well defined or is not countable.

(v) It is not a set. Because when we say a collection of all talented writers of India, it changes opinion wise and cannot be well defined or is not countable.

(vi) It is a set. As we are saying a collection of novels written by Munshi Prem Chand.it is a definite quantity hence it is a set.

(vii) It is a set. As we are saying a collection of all the questions in this chapter. It is a definite quantity hence it is a set.

(viii) It is a set. As we are saying a collection of all months of a year beginning with the letter J. It is a definite quantity hence it is a set.

(ix) It is not a set. Because when we say a collection of most dangerous animals of the world it changes opinion wise and we cannot define dangerousness of an animal.

(x) It is a set as we are saying a collection of prime integers. It is a definite quantity hence it is a set.

First of all it is important that you know what does (ϵ) this symbol means. In terms of mathematics it is used as belongs to () is used as does not belong to.

i.4 ……A

As 4 is there in Set A then we can say 4 belongs to A

∴ 4 ϵ A

ii. -4……A

As -4 is not there in Set A then we can say 4 does not belong to A

∴ 4 A

iii. 12…..A

As 12 is not there in Set A then we can say 12 does not belong to A

∴ 12 A

iv.9 ……A

As 9 is there in Set A then we can say 9 belongs to A

∴ 9 ϵ A

v.0 ……A

As 0 is there in Set A, then we can say 0 belongs to A

∴ 0 ϵ A

vi.-2 ……A

As -2 is not there in Set A then we can say -2 does not belong to A

∴ -2 A

Here, Question is

Now, A'=U∖A

⇒(A' )'=(U∖A)'=U'∖A'

⇒(A' )'=U'∖(U∖A)

⇒(A' )'=ϕ∖U+A

⇒(A' )'=A

A∖B is the set of elements which are in A but not in B

⇒A∖B is the set of elements which are in A and in B'

⇒A∖B is the set of elements of A⋂B'

Let A be the set with n elements.

Each member of A has two possibilities either present or absent.

⇒ Total possible subsets of A are 2×2×2×…n times = 2ⁿ

Here correct answer is D. none of this as A⋂(A⋂B)=(A⋂A)⋂B)

=A⋂B

Or we have to give one condition that then correct answer is A because A⋂(A⋂B)=(A⋂A)⋂B)

=A⋂B

=A (∵ A⊂B⇒A⋂B=A)

A is not correct answer because 4 is not in A.

B is not correct answer because {4} is not in A.

C is not correct answer because 2 is in B but not in A.

So, D is correct answer.

A∖B=A⋂B'

≠(A∖B)⋂(B∖A)

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

A∖B={1,2,3}∖{3,4,5}

={1,2,3}∖{3}

={1,2}

B∖A={3,4,5}∖{1,2,3,4}

={3,4,5}∖{3}

={4,5}

AΔB=(A∖B)⋃(B∖A)

={1,2}⋃{4,5}

={1,2,4,5}

(A∖B)⋃(B∖A)=[A⋂B' ]⋃[B⋂A']

=[A⋃(B⋂A' )]⋂[B'⋃(B⋂A' )]

=[(A⋂B)⋃(A⋂A' )]⋂[(B'⋂B)⋃(B'⋂A' )]

=[(A⋂B)⋃ϕ]⋂[ϕ⋃(B'⋂A' )]

=[(A⋂B)]⋂[(B'⋂A' )

=[(A⋂B)]⋂[(B⋃A)' ]

=(B⋃A)'⋂(A⋂B)

=(A⋃B)∖(A⋂B)

A is true statement because

A∖B is the set of elements which are in A but not in B

⇒A∖B is the set of elements which are in A and in B'

⇒A∖B is the set of elements of A⋂B'

∴A∖B=A⋂B'

B is true statement because

A∖B is the set of elements which are in A but not in B

⇒A∖B is the set of elements which are in A but not in A⋂B'

⇒ A∖B=A∖(A⋂B')

C is false statement because

A∖B=A⋂B' and A∖B^'=A⋂B

∴A⋂B'≠A⋂B⇒A∖B≠B∖A.

(A⋂B)∖(A⋂C)=(A⋂B)∖A⋃(A⋂B)∖C (∵ De Morgen’s law with difference and intersection.)

=ϕ⋃(A⋂B)∖C

=(A⋂B)∖C

=A⋂(B∖C) (∵ intersection with difference is difference with intersection)

Here, A= {x: xϵ R, x>4} and B= {x: xϵ R, x<5}

A⋂B= {x: xϵ R, x>4} and {x: x ϵ R, x<5}

= {xϵ R:x>4, x<5}

={xϵR:4<x<5}

= (4,5)

Here, n(A)=200 , n(B)=300 , n(A⋂B)=100,n(U)=700

n(A⋃B)=n(A)+n(B)-n(A⋂B)

=200+300-100

=400

n(A'⋂B' )=n(U)∖n(A⋃B)

=700-400

=300

In this question we cannot find as we haven’t required values and information about their union or about universal set.

So, correct answer is D

Here, n(A)=5

⇒n(P(A) )=2⁵=32

So,A has total 32 subsets but one of them is A itself.

∴ Proper subsets of A are 32-1=31.

We know x≠x is false for any x.

∴ the set builder form for null set is {x:x≠x}.

Here, A and B are two disjoint sets.

⇒A⋂B=ϕ

⇒n(A⋂B)=0

n(A⋃B)=n(A)+n(B)-n(A⋂B)

=n(A)+n(B)-0

=n(A)+n(B)

If A⋃B=A

Then, any element is in B then it is in A⋃B=A

So, B⊂A

If B⊂A

Then, any element of A⋃B is in A.

So, A⋃B=A

Hence, A⋃B=A iff B⊂A

n(A⋃B) =n(A)+n(B)-n(A⋂B)

∴n(A⋂B) =n(A)+n(B)-n(A⋃B)

∴n(A⋂B) =70+60-110

∴n(A⋂B) =20

A⋂(A⋂B) ^c =A⋂ (A^c⋃B^c )

= (A ⋂ A^c)⋃(A ⋂ B^c)

=ϕ⋃ (A ⋂ B^c)

=A ⋂ B^c

A∖ B is the set of elements which are in A but not in B

⇒A∖B is the set of elements which are in A and in B'

⇒A∖ B is the set of elements of A⋂B'

∴A∖B=A⋂B'

=A⋂B ̅

Let A be the set of population travels by car(in percentage) and

B be the set of population travels by bus(in percentage).

Then A⋂B is the set of population travels by car and bus and

A⋃B is the set of population travels by car or bus.

Then,n(A)=20 , (B)=50 ,n(A⋂B)=10

n(A⋃B)=n(A)+n(B)-n(A⋂B)

=20+50-10

=60

∴ The set of population travels by car or bus is 60%

A⋂B=B which means elements of are in the both sets A and B.

⇒ All the elements of are contained in the intersection of A and B which is equal to B.

⇒B⊂A.

Let U be the set of students which are interviewed by investigator

A be the set of students which take milk

B be the set of students which take coffee

C be the set of students which take tea

Then ,n(U)=100 ,n(A)=12, n(B)=5 ,n(C)=8

Also, (A⋂B⋂C) means students who drink all of three=10

(A⋂B) means students who drink milk and coffee =20-10=10

(B⋂C) means students who drink coffee and tea =20-10=10

(A⋂C) means students who drink milk and tea =25-10=15

A⋃B⋃C is the total of above =12+5+8+10+10+15+10=70

The number of students who didn’t take any drink =100-70=30

Let A and B be the set which contain m and n elements respectively.

Then n(P(A) )=2^m and n(P(B) )=2ⁿ

Also given that,n(P(A) )=n(P(B) )+48

Above equation is only true when and .

Let M,P and C denote the set of students who opting mathematics ,physics and chemistry respectively.

Number of students who opted.

mathematics only =n(M)-n(M⋂P)-n(M⋂C)+n(M⋂P⋂C)

=100-30-28+18

=60

If elements are not repeated then number of elements in A₁⋃A₂⋃A₃⋃….⋃A₃₀=30*5=150

But, each element of S is in 10 of

If elements are not repeated then number of elements in B₁⋃B₂⋃B₃⋃….⋃B_n=3*n

But, each element of S is in of

∴3× n=15× 9

∴n=45

Let A and B be the set which contain m and n elements respectively.

Then n(P(A) )=2^m and n(P(B) )=2ⁿ

Also given that,n(P(A) )=n(P(B) )+112

⇒2^m =2ⁿ +112

Above equation is only true when m=7 and n=4.

A⋂(A⋃B)'=A⋂(A'⋂B' )

=(A⋂A' )⋂(A⋂B')

=ϕ⋂(A⋂B' )

=ϕ

Here, F₁ is the set of all parallelograms

We know that every rectangle, rhombus and square in plane are parallelogram but trapezium is not parallelogram

So, .

Here x:x it is read as x is such that x

So now when we read whole sentence it becomes x is such that x is a letter before e in the English alphabet. Now letters before e are a,b,c,d.

∴ Roster form will be {a,b,c,d}.

First thing analyze the given data. x ϵ N that implies x is a natural number.

x² < 25

⇒ x < ±5

As x belongs to the natural number that means x < 5.

All numbers less than 5 are 1,2,3,4.

∴ Roster form will be {1,2,3,4}.

X is a natural number and is between 10 and 20.

X is such that x is a prime number between 10 and 20.

Prime numbers between 10 and 20 are 11,13,17,19.

∴ Roster form will be {11,13,17,19}.

X is a natural number also x = 2n

∴ Roster form will be {2,4,6,8…..}.

This an infinite set.

Any real number is equal to its value it is neither less nor greater.

According to the question we have to write Roster form of such real numbers which has value less than itself. As there are no such numbers.

∴ Roster form will be ϕ.

This is called as null set.

All numbers which are divisor of 60 are = 1,2,3,4,5,6,10,12,15,20,30,60.

Now numbers which are prime are = 2,3,5.

∴ Roster form will be {2,3,5}.

Numbers which have sum as 8 are = 17,26,35,44,53,62,71,80

⇒ Roster form will be {17,26,35,44,53,62,71,80}.

All letters means no letter should be repeated

Trigonometry = t,r,i,g,o,n,m,e,y

∴ Roster form will be {t,r,i,g,o,n,m,e,y}

All letters means no letter should be repeated

Better = b,e,t,r

∴ Roster form will be {b,e,t,r}.

(i) {x:x ϵ N,x<7}

This is read as x is such that x belongs to natural number and x is less than 7. It satisfies all condition of roster form.

(ii) {x ϵ z: x=1/n+1,n ϵ W}

This is read as x is such that x is an integer greater than or equal to 0. And it’s value is 1/x+1.

(iii) {x N: x = 3n, n W}

(iv) {x:x ϵ N,9<x<16}

(v) {x:x=0}

(vi) {x ϵ N: x=n²,n≤10, n ϵ N}

(vii) {x N: x = 2n, n N}

(viii) {x N: x = 5ⁿ,0<n<6 N}

First of all, x is an integer hence it can be positive and negative also.

X² ≤ 10

X≤ √10

X =±1,±2,±3

A={±1,±2,±3}

_{Substituting the values of n we will get the solutions}

At n=1,

At n=1,

At n=1,

At n=1,

At n=1,

∴

∴ x is an integer between -1/2 and 9/2

So all integers between given values -0.5<x<4.5

0,1,2,3,4

∴ C = {0,1,2,3,4}

All vowels in equation are E,U,A,I,O

∴ D={A,E,I,O,U}

All months of a year do not have 31 days :

February, April, June, September, November.

E: { February, April, June,September,November}

Letters in word M, I, S, P.

F={M,I,S,P}.

When we are dealing with the set builder - roster form match the following it is always better to go in the direction in which we need to convert set builder to roster rather than roster to set builder.

For i. {x:x+5=5, x z}

X is such that x+5 = 5

⇒ x = 5-5

⇒ x = 0

Roster form = {0}

i.e third option

For ii. {x:x is a prime natural number and a divisor of 10}

All natural numbers that are divisor of 10 are = 1,2,5,10

Numbers that are prime are = 2,5

Roster form= {2,5}

i.e. sixth option.

For iii. {x:x is a letter of the word “RAJASTHAN”}

All letters means no letter should be repeated

RAJASTHAN = R,A,J,S,T,H,N

∴ Roster form will be {A,H,J,R,S,T,N}

i.e. fifth option

For iv. {x:x is a natural and divisor of 10}

All natural numbers that are divisor of 10 are = 1,2,5,10.

Roster form= {1,2,5,10}

i.e. Fourth option.

For v. {x:x² – 25 =0}

X²-25 = 0

X =√25

X =±5.

∴ Roster form will be {5,-5}

i.e. second option

for vi. {x:x is a letter of word “APPLE”}

All letters means no letter should be repeated

APPLE = A,P,L,E

∴ Roster form will be {A,E,L,P}

i.e. first option.

Set of all vowels which precede q.

Precede means to come before.

A, E, I, O these are the vowels they come before q.

B = {A,E,I,O}.

Every odd number has an odd cube

Odd numbers can be represented as 2n+1.

{2n+1:n ϵ W} or

{1,3,5,7,……}

Here we can see denominator is square of numerator +1.

We can set builder form as

⇒

Note: Equivalent set are different from equal sets, Equivalent sets are those which have equal number of elements they do not have to be same.

A = {1, 2, 3}

Number of elements = 3

B={t, p, q, r, s}

Number of elements = 5

C={α, β, γ}

Number of elements = 3

D={a, e, I, o, u}

Number of elements = 5

Set A is equivalent with Set C and Set B is equivalent with Set D.

For A

Letters in word reap

A ={R, E, A, P} ={A, E, P, R}

For B

Letters in word paper

B ={P, A, E, R} = {A, E, P, R}

For C

Letters in word rope

C = {R, O, P, E} = {E, O, P, R}.

Set A = Set B

Because every element of set A is present in set B

But Set C is not equal to either of them because all elements are not present.

Note: The Empty set is the set which does not contain any element. Most of the people get confused whether {0} is an empty set or not. It is not because it contains an element 0.

(i) All numbers ending with 0. Except 0 is divisible by 5 and are even. Hence it is not an example of empty set.

(ii) 2 is a prime number and is even, and it is the only prime which is even. So no this not an example of the empty set.

(iii) There is not natural number whose square is 2. So it is an example of empty set.

(iv) Never can a number be simultaneously less than 8 and greater than 12. Hence it is an example of the empty set.

(v) No two parallel lines can never have a common point. Hence it is an example of empty set.

(i) In a plane there can be infinite concentric circles. Hence it is an infinite set.

(ii) There are just 26 letters in English Alphabets and are finite. Hence it is finite set.

(iii) It is an infinite set because, natural numbers greater than 5 is infinite.

(iv) It is a finite set. Natural numbers start from 1 and there are 199 numbers less than 200. Hence it is finite.

(v) It is an infinite set. Because integers less than 5 are infinite.

(vi) It is an infinite set. Because between two real numbers, there are infinite real numbers.

NOTE: A set is said to be equal with another set if all elements of both the sets are equal and same.

A = {1, 2, 3}

B ={x ∈ R:x²–2x+1=0}

x²–2x+1=0

(x–1)² = 0

∴ x=1.

B = {1}

C= {1, 2, 2, 3}

In sets we do not repeat elements hence C can be written as {1, 2, 3}

D = {x ∈R : x³ – 6x²+11x – 6 = 0}.

For x = 1

= (1)³–6(1)²+11(1)–6

= 1–6+11–6

= 0

For x =2

= (2)³–6(2)²+11(2)–6

= 8–24+22–6

= 0

For x =3

= (3)³–6(3)²+11(3)–6

= 27–54+33–6

= 0

As cubic equation has three roots at max so the roots are 1, 2, 3

∴ D = {1, 2, 3}

Hence Set A, C and D are equal.

A = {2, 3}

B = {–2, –3}

As A and B do not have exactly same elements hence they are not equal.

Every letter in WOLF

A ={W, O, L, F} = {F, L, O, W}

Every letter in FOLLOW

B = {F, O, L, W} = {F, L, O, W}

As A and B have same number of elements which are exactly same hence they are equal sets.

We first of all need to manipulate some of the sets

D = {3, 1, 2, 4} = {1, 2, 3, 4, }

F = {8, 4, 12} = {4, 8, 12}

Equivalent sets:

i. A, E, H (all of them have exactly two elements in them)

ii. B, D, G (all of them have exactly four elements in them)

iii. C, F (all of them have exactly three elements in them)

Equal sets :

i. B, D (all of them have exactly the same elements, so they are equal)

ii. C, F (all of them have exactly the same elements, so they are equal)

A = x is a natural number. And x is less than 3

So all natural numbers less than 3 constitute set A.

{1, 2}

A ={1, 2}

B = {1, 2}

C = {1, 3}

D = x is a natural number. And x is less than 5 and is odd.

So all odd natural numbers less than 5 constitute set D.

{1, 3}

D = {1, 3}

E ={1, 2, 1, 1}

We don’t repeat same elements in a set.

∴ E = {1, 2}

F ={1, 1, 3}

We don’t repeat same elements in a set.

∴ F = {1, 3}

∴ A = {1, 2}

B = {1, 2}

C = {1, 3}

D = {1, 3}

E = {1, 2}

F = {1, 3}

Now, we can see clearly that set A, B, E are equal and set C, D, F are equal.

For “CATRACTR”

Letters in word are

{C, A, T, R} ={A, C, R, T}

For “TRACT”

Letters in word are

{T, R, A, C} = {A, C, R, T}

As we see letters need to spell cataract is equal to set of letters need to spell tract.

Hence Proved.

(i) False

No, it is not necessary for any two set A and B to be either A B or B A.

Let A = {1,2} and B ={a,b}

Here neither A B nor B A.

(ii) False

A = {1,2,3} It is subset of infinite set N and is finite.

(iii) True

Even if we think logically then also smaller part of something finite can never be infinite.

(iv) False

Null set or empty set does not have a proper subset.

(v) False

We do not repeat elements in a set, so the given set becomes {a,b} which is a finite set.

(vi) True

In both of the number of the set of elements are same.

(vii) False

In A = {1}

The subsets are ϕ and {1} which are finite.

(i) True

1 belongs to the given set as it is present in it.

(ii) False

{a} ⊂ {b,c,a}

This is the write form to write it or a ϵ {b,c,a}

(iii) False

{a} ⊂ {b,c,a}

This is the write form to write it or a ϵ {b,c,a}

(iv) True

We do not repeat same elements in a given set.

∴ The given set can be written as {a,b}

(v) False

X+8 = 8

i.e. x = 0

{0} It is not a null set

A = x² – 8x + 12=0

= (x–6)(x–2) =0

= x=2 or x=6

A = {2,6}

B ={2,4,6}

C= {2,4,6,8}

D ={6}

So we can say

D⊂A⊂B⊂C

(i) True

A rational number is represented by the form p/q where p and q are integers and (q not equal to 0) keeping q=1 we can place any number as p. Which then will be an integer.

(ii) True

All crows are birds, so they are contained in the set of all birds.

(iii) False

Every square can be a rectangle, but every rectangle cannot be a square.

(iv) False

Every square can be a rectangle, but every rectangle cannot be a square.

(v) False

P = {a}

B = {{a}}

But {a} =P

B ={P}

Hence they are not equal.

(vi) True

A = For “LITTLE”

A = {L,I,T,E} = {E,I,L,T}

B = For “TITLE”

B = {T,I,L,E} = {E,I,L,T}

(i) In this a isn’t subset of given set but belongs to the given set.

∴ The correct form would be

a ϵ {a,b,c}

(ii) In this {a} is subset of {a,b,c}

∴ The correct form would be

{a} ⊂ {a,b,c}

(iii) In this a is not the element of the set.

∴ The correct form would be

{a} ϵ {{a},b}

(iv) In this {a} is not athe subset of given set

∴ The correct form would be

{a} ϵ {{a},b}

(v) {b,c} is not a subset of given set. But it belongs to the given set.

∴ The correct form would be

{b,c} ϵ {a,{b,c}}

(vi) {a,b} is not a subset of given set

∴ The correct form would be

{a,b} {a,{b,c}}

(vii) ϕ does not belong to given set but it is subset

∴ The correct form would be

ϕ ⊂ {a,b}

(viii) True ϕ is subset of every set

(ix) X+3=3

X=0

{0}

It is not ϕ

∴ The correct form would be

{x:x + 3 = 3}= {0}

(i) False

{c,d} is not a subset of A but it belong to A.

{c,d} ϵ A

(ii) True

{c,d} ϵ A

(iii) True

{c,d} is a subset of A.

(iv) It is true that a belong to A.

(v) False

a is not a subset of A but it belongs to A

(vi) True

(vii) False

{a,b,e} ⊂ A this is the correct form.

(viii) False

{a,b,c} is not a subset of A

(ix) False

ϕ is a subset of A.

ϕ ⊂ A.

(x) False

(i) False

1 is not an element of A.

(ii) True

{1,2,3} ϵ A. this is correct.

(iii) True.

{6, 7, 8} ϵ A.

(iv) False

2 is not an element of A, and ϕ is also not an element of A. Hence, false.

(v) False

2 alone is not an element of A. Hence, false.

(vi) True

{1, 2, 3} is the set belonging to A. {2, {1}} does not belong to A.

(vii) True

{{2}}, {1}} does not belongs to A. Hence, true.

(viii) False

Φ does not belong to A.

(i) True

Φ is a member of set A. Hence, true.

(ii) True

{ Φ} is a member of set A. Hence, true.

(iii) False

1 alone is not a member of A. Hence, false.

(iv) False

We can see that 2 is a member of set A, {2, Φ} is not. Hence, false

(v) True

2 is a member of set A. Hence, true.

(vi) True

{1} is not a member of set A.

(vii) True

Neither {2} and nor {1} is a member of set A. Hence, true.

(viii) True

All three are members of set A. Hence, true.

(ix) False

{{ϕ}} is not a member of set A. Hence, false.

(i) Subsets of given set are {a}, ϕ.

(ii) Subsets of given set are {0},{1},{0,1},ϕ.

(iii) Subsets of given set are {a},{b},{c},{a,b},{b,c},{a,c},{a,b,c},ϕ.

(iv) Subsets of given set are {1},{{1}},{1,{1}},ϕ.

(v) Subsets of given set are{ϕ},ϕ.

(i) {1},{2},ϕ

(ii) {1},{2},{3},{1,2},{2,3},{1,3},ϕ.

(iii) ϕ ,{1}.

A total number of subsets for any given ser is 2ⁿ.

In every set there is only one improper set i.e. the set itself.

∴ proper subsets would be 2ⁿ–1.

Let A⊆ ϕ,

If A is a subset of an empty set, then A is the empty set.

∴ A = ϕ

Now let A = ϕ,

This means A is an empty set.

As we know that every set is a subset of itself.

∴ A ⊆ ϕ

Thus, we have,

A⊆ ϕ ⇔ A=ϕ

Hence, Proved.

We have A B, B C and C A \

∴ A B C A.

Now, A is a subset of B and B is a subset of C, so

A is a subset of C, i.e., A C

Also, C A

Hence C= A.

Empty set has zero element.

∴ Power set of ϕ has 2⁰ = 1 element.

(i) Set of all triangles.

(ii) Set of all triangles.

A ∩ B means A intersection B. Common elements of A and B come in this group.

Given A ⊂ B

i.e. B is having all elements that is present in A.

∴ A ∩ B = A.

A ∪ B means A union B. All elements of A and B come in this set.

Given A ⊂ B

i.e. B is having all elements including elements of A.

∴ A ∪ B = B.

Note: In general X∪Y = {a: aϵX or aϵY}

X∩Y = {a:aϵX and aϵY}.

i. A ∪ B = {x: x ϵ A or x ϵ B}

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

ii. A ∪ C = {x: x ϵ A or x ϵ C}

= {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}

iii. B ∪ C = {x: x ϵ B or x ϵ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

iv. B ∪ D = {x: x ϵ B or x ϵ D}

= {4, 5, 6, 7, 8, 10, 11, 12, 13, 14}

v. A ∪ B = {x: x ϵ A or x ϵ B}

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

A ∪ B ∪ C = {x: x ϵ A ∪ B or x ϵ C}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

vi. A ∪ B = {x: x ϵ A or x ϵ B}

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

A ∪ B ∪ D = {x: x ϵ A ∪ B or x ϵ D}

= {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14}

vii. B ∪ C = {x: x ϵ B or x ϵ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

B ∪ C ∪ D = {x: x ϵ B ∪ C or x ϵ D}

= {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

viii. B ∪ C = {x: x ϵ B or x ϵ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

A ∩ B ∪ C = {x:x ϵ A and x ϵ B ∪ C}.

= {4, 5}

ix. (A ∩ B) = {x:x ϵ A and x ϵ B}.

= {4, 5}

(B ∩ C) = {x:x ϵ B and x ϵ C}

= {7, 8}

(A ∩ B) ∩ (B ∩ C) = {x:x ϵ (A ∩ B) and x ϵ (B ∩ C)}.

= ϕ

x. A ∪ D = {x: x ϵ A or x ϵ D}

= {1, 2, 3, 4, 5, 10, 11, 12, 13, 14}.

B ∪ C = {x: x ϵ B or x ϵ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

(A ∪ D) ∩ (B ∪ C) = {x:x ϵ (A ∪ D) and x ϵ (B ∪ C)}.

= {4, 5, 10, 11}.

A = All natural numbers i.e. {1, 2, 3…..}

B = All even natural numbers i.e. {2, 4, 6, 8…}

C = All odd natural numbers i.e. {1, 3, 5, 7……}

D = All prime natural numbers i.e. {1, 2, 3, 5, 7, 11, …}

i. A ∩ B

A contains all elements of B.

∴ B ⊂ A

∴ A ∩ B = B

ii. A ∩ C

A contains all elements of C.

∴ C ⊂ A

∴ A ∩ C = C

iii. A ∩ D

A contains all elements of D.

∴ D ⊂ A

∴ A ∩ D = D

iv. B ∩ C

B ∩ C = ϕ

There is no natural number which is both even and odd at same time.

v. B ∩ D

B ∩ D = 2

2 is the only natural number which is even and a prime number.

vi. C ∩ D

C ∩ D = {1, 3, 5, 7…}

Every prime number is odd except 2.

A – B is defined as {xϵA : x∉B}

i. A – B is defined as {x ϵ A : x ∉ B}

A = {3, 6, 12, 15, 18, 21}

B = {4, 8, 12, 16, 20}

A – B = {3, 6, 15, 18, 21}

ii. A – C is defined as {x ϵ A : x ∉ C}

A = {3, 6, 12, 15, 18, 21}

C = {2, 4, 6, 8, 10, 12, 14, 16}

A – C = {3, 15, 18, 21}

iii. A – D is defined as {x ϵ A : x ∉ D}

A = {3, 6, 12, 15, 18, 21}

D = {5, 10, 15, 20}.

A – D = {3, 6, 12, 18, 21}

iv. B – A is defined as {x ϵ B : x ∉ A}

A = {3, 6, 12, 15, 18, 21}

B = {4, 8, 12, 16, 20}

B – A = {4, 8, 16, 20}

v. C – A is defined as {x ϵ C : x ∉ A}

A = {3, 6, 12, 15, 18, 21}

C = {2, 4, 6, 8, 10, 12, 14, 16}

C – A = {2, 4, 8, 10, 14, 16}

vi. D – A is defined as {x ϵ D : x ∉ A}

A = {3, 6, 12, 15, 18, 21}

D = {5, 10, 15, 20}.

D – A = {5, 10, 20}.

vii. B – C is defined as {x ϵ B : x ∉ C}

B = {4, 8, 12, 16, 20}

C = {2, 4, 6, 8, 10, 12, 14, 16}

B – C = {4, 8, 20}

viii. B – D is defined as {x ϵ B : x ∉ D}

B = {4, 8, 12, 16, 20}

D = {5, 10, 15, 20}

B – D = {4, 8, 12, 16}

A’ means Complement of A with respect to universal set U.

So, A’ = U – A

U – A is defined as {x ϵ U : x ∉ A}

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

A = {1, 2, 3, 4}

A’ = {5, 6, 7, 8, 9}

B’ means Complement of B with respect to universal set U.

So, B’ = U – B

U – B is defined as {x ϵ U : x ∉ B}

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

B = {2, 4, 6, 8}

B’ = {1, 3, 5, 7, 9}

(A ∩ C) = {x:x ϵ A and x ϵ C}.

= {3, 4}

(A∩C)’ means Complement of (A∩C) with respect to universal set U.

So, (A∩C)’ = U – (A∩C)

U – (A∩C)’ is defined as {x ϵ U : x ∉ (A∩C)’}

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

(A∩C)’ = {3, 4}

U – (A∩C)’ = {1, 2, 5, 6, 7, 8, 9}

A ∪ B = {x: x ϵ A or x ϵ B}

= {1, 2, 3, 4, 6, 8}

(A∪B)’ means Complement of (A∪B) with respect to universal set U.

So, (A∪B)’ = U – (A∪B)’

U – ( A∪B)’ is defined as {x ϵ U : x ∉ (A∪B)’}

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

(A∪B)’ = {1, 2, 3, 4, 6, 8}

U – ( A∪B)’ = {5, 9}

A’ = U – A

((A’)’ = (U – A)’

= U – (U – A)

= A

A = {1, 2, 3, 4}

B – C is defined as {x ϵ B : x ∉ C}

B = {2, 4, 6, 8}

C = {3, 4, 5, 6}

B – C = {2, 8}

Now, (B – C)’ = U – (B – C)

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

B – C = {2, 8}

(B – C)’ = {1, 3, 4, 5, 6, 7, 9}.

A ∪ B = {x: x ϵ A or x ϵ B}

= {2, 3, 4, 5, 6, 7, 8}

(A∪B)’ means Complement of (A∪B) with respect to universal set U.

So, (A∪B)’ = U – (A∪B)’

U – ( A∪B)’ is defined as {x ϵ U : x ∉ (A∪B)’}

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

(A∪B)’ = {2, 3, 4, 5, 6, 7, 8}

U – ( A∪B)’ = {1, 9}

Now

A’ means Complement of A with respect to universal set U.

So, A’ = U – A

U – A is defined as {x ϵ U : x ∉ A}

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

A = {2, 4, 6, 8}

A’ = {1, 3, 5, 7, 9}

B’ means Complement of B with respect to universal set U.

So, B’ = U – B

U – B is defined as {x ϵ U : x ∉ B}

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

B = {2, 3, 5, 7}.

B’ = {1, 4, 6, 8, 9}

A’ ∩ B’ = = {x:x ϵ A’ and x ϵ C’}.

= {1, 9}

Hence verified.

(A ∩ B) = {x:x ϵ A and x ϵ B}.

= {2}

(A∩B)’ means Complement of (A∩B) with respect to universal set U.

So, (A∩B)’ = U – (A∩B)

U – (A∩B)’ is defined as {x ϵ U : x ∉ (A∩B)’}

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

(A∩B)’ = {2}

U – (A∩B)’ = {1, 3, 4, 5, 6, 7, 8, 9}

A’ means Complement of A with respect to universal set U.

So, A’ = U – A

U – A is defined as {x ϵ U : x ∉ A}

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

A = {2, 4, 6, 8}

A’ = {1, 3, 5, 7, 9}

B’ means Complement of B with respect to universal set U.

So, B’ = U – B

U – B is defined as {x ϵ U : x ∉ B}

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

B = {2, 3, 5, 7}.

B’ = {1, 4, 6, 8, 9}

A’ ∪ B’ = {x: x ϵ A or x ϵ B}

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

Hence verified.

A ∪ {1, 2} = {1, 2, 3, 5, 9}

Elements of A and {1, 2} together give us the result

So smallest se of A can be

A = {1, 2, 3, 5, 9} – {1, 2}

A = {3, 5, 9}.

L.H.S =

(B ∩ C) = {x:x ϵ B and x ϵ C}

= {5, 6}

A ∪ (B ∩ C) = {x: x ϵ A or x ϵ (B ∩ C) }

= {1, 2, 4, 5, 6}.

R.H.S =

(A ∪ B) = {x: x ϵ A or x ϵ B }

= {1, 2, 4, 5, 6}.

(A ∪ C) = {x: x ϵ A or x ϵ C }

= {1, 2, 4, 5, 6, 7}.

(A ∪ B) ∩ (A ∪ C) = {x:x ϵ (A ∪ B) and x ϵ (A ∪ C) }.

= {1, 2, 4, 5, 6}.

Hence Verified.

L.H.S.

(B ∪ C) = {x: x ϵ B or x ϵ C }

= {2, 3, 4, 5, 6, 7}.

(A ∩ (B ∪ C)) = {x:x ϵ A and x ϵ (B ∪ C)}

= {2, 4, 5}

R.H.S:

(A ∩ B) = {x:x ϵ A and x ϵ B}

= {2, 5}

(A ∩ C) = {x:x ϵ A and x ϵ C}

= {4, 5}

(A ∩ B) ∪ (A ∩ C) = {x:x ϵ (A∩B) and x ϵ (A∩C) }.

= {2, 4, 5}.

Hence verified.

B–C is defined as {x ϵ B : x ∉ C}

B = {2, 3, 5, 6}

C = {4, 5, 6, 7}

B–C = {2, 3}

(A ∩ (B – C)) = {x:x ϵ A and x ϵ (B – C)}

= {2}

R.H.S:

(A ∩ B) = {x:x ϵ A and x ϵ B}

= {2, 5}

(A ∩ C) = {x:x ϵ A and x ϵ C}

= {4, 5}

(A ∩ B) – (A ∩ C) is defined as {x ϵ (A ∩ B) : x ∉ (A ∩ C)}

= {2}.

Hence Verified.

L.H.S.

(B ∪ C) = {x: x ϵ B or x ϵ C }

= {2, 3, 4, 5, 6, 7}.

A–(B ∪ C) is defined as {x ϵ A : x ∉ (B ∪ C)}

A = {1, 2, 4, 5}

(B ∪ C) = {2, 3, 4, 5, 6, 7}.

A–(B ∪ C) = {1}

R.H.S

(A – B) =

A–B is defined as {x ϵ A : x ∉ B}

A = {1, 2, 4, 5}

B = {2, 3, 5, 6}

A–B = {1, 4}

(A–C) =

A–C is defined as {x ϵ A : x ∉ C}

A = {1, 2, 4, 5}

C = {4, 5, 6, 7}

A–C = {1, 2}

(A – B) ∩ (A–C) = {x:x ϵ (A – B) and x ϵ (A – C) }.

= {1, 2}

Hence verified.

(B ∩ C) = {x:x ϵ B and x ϵ C}

= {5, 6}

(A – (B ∩ C)) =

A–(B ∩ C) is defined as {x ϵ A : x ∉(B ∩ C)}

A = {1, 2, 4, 5}

(B ∩ C) = {5, 6}

(A – (B ∩ C)) = {1, 2, 4}

R.H.S:

(A – B) =

A–B is defined as {x ϵ A : x ∉ B}

A = {1, 2, 4, 5}

B = {2, 3, 5, 6}

A–B = {1, 4}

(A–C) =

A–C is defined as {x ϵ A : x ∉ C}

A = {1, 2, 4, 5}

C = {4, 5, 6, 7}

A–C = {1, 2}

(A – B) ∪ (A–C) = {x:x ϵ (A – B) OR x ϵ (A – C) }.

= {1, 2, 4}

Hence verified.

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

Now,

B ∪ C = {5, 6}

A ∩ (B ∪ C) = {5}

Similarly finding out R.H.S we get,

A ∩ B = {2, 5}

A ∩ C = {4, 5}

(A ∩ B) ∩ (A ∩ C) = {5}

L.H.S = R.H.S

Hence, Verified.

A ∪ B = {x: x ϵ A or x ϵ B }

= {2, 3, 5, 7, 9 }

(A∪B)’ means Complement of (A∪B) with respect to universal set U.

So, (A∪B)’ = U– (A∪B)’

U–( A∪B)’ is defined as {x ϵ U : x ∉ (A∪B)’}

U = {2, 3, 5, 7, 9}

(A∪B)’ = {2, 3, 5, 7, 9 }

U–( A∪B)’ = ϕ

Now

A’ means Complement of A with respect to universal set U.

So, A’ = U–A

U–A is defined as {x ϵ U : x ∉ A}

U = {2, 3, 5, 7, 9}

A = {3, 7}

A’ = {2, 5, 9}

B’ means Complement of B with respect to universal set U.

So, B’ = U–B

U–B is defined as {x ϵ U : x ∉ B}

U = {2, 3, 5, 7, 9}

B = {2, 5, 7, 9}.

B’ = {3}

A’ ∩ B’ = = {x:x ϵ A’ and x ϵ C’ }.

= ϕ.

Hence verified.

(A ∩ B}’ = A’ ∪ B.’

(A ∩ B) = {x:x ϵ A and x ϵ B }.

= {7}

(A∩B)’ means Complement of (A∩B) with respect to universal set U.

So, (A∩B)’ = U–(A∩B)

U–(A∩B)’ is defined as {x ϵ U : x ∉ (A∩B)’}

U = {2, 3, 5, 7, 9}

(A∩B)’ = {7}

U–(A∩B)’ = {2, 3, 5, 9}

A’ means Complement of A with respect to universal set U.

So, A’ = U–A

U–A is defined as {x ϵ U : x ∉ A}

U = {2, 3, 5, 7, 9}

A = {3, 7}

A’ = {2, 5, 9}

B’ means Complement of B with respect to universal set U.

So, B’ = U–B

U–B is defined as {x ϵ U : x ∉ B}

U = {2, 3, 5, 7, 9}

B = {2, 5, 7, 9}.

B’ = {3}

A’ ∪ B’ = {x: x ϵ A or x ϵ B }

= {2, 3, 5, 9}

Hence verified.

Let an element be p such that it belongs to B

∴ p ϵ B

⇒ p ϵ B ∪ A

⇒B ⊂ A ∪ B

Let an element be p such that it belongs to B

∴ p ϵ A ∩ B

⇒ p ϵ A and p ϵ B

⇒A ∩ B ⊂ A

Let

p ϵ A ⊂ B.

⇒ x ϵ B

Let and p ϵ A ∩ B

⬄ x ϵ A and x ϵ B

∴ (A ∩ B) = A.

To show that the following four statements are equivalent, we need to show that 1⇒2, 2⇒3, 3⇒ 4, 4⇒1

We first show that 1⇒ 2

Now A–B = {xϵ A: x ∉ B} .As A ⊂ B,

∴ Each element of A is an element of B,

∴ A–B = ϕ

Hence we proved 1⇒ 2.

We need to show that 2 ⇒ 3

So assume that A–B = ϕ

To show: A∪B = B.

∵ A – B = ϕ

∴ Every element of A is an element of B.

So A ⊂ B and therefore A∪B = B

So (2) ⇒ (3) is true.

We need to show that 3 ⇒ 4

Assume that A ∪ B = B

To show: A ∩ B = A.

∵A ∪ B = B

∴ A⊂ B and so A ∩ B = A.

So (3) ⇒ (4) is true.

Finally, now we need to show 4 ⇒ 1.

So, assume A ∩ B = A.

To show: A ⊂ B

∵A ∩ B = A, therefore A⊂B, and so (4) ⇒ (1) is true.

Let A = {1, 2}

B = {2, 3}

C = {2, 4}

Then,

A ∩ B = {2}

A ∩ C = {2}

Hence, A ∩ B = A ∩ C, but clearly B is not equal to C.

Given: A ⊂ B

To show: C–B ⊂ C–A

Let x ϵ C– B

⇒ x ϵ C and x ∉ B [by definition C–B]

⇒ x ϵ C and x ∉ A

⇒ x ϵ C–A

Thus x ϵ C–B ⇒ x ϵ C–A. This is true for all x ϵ C–B.

C – B ⊂ C – A.

A ∪ (A ∩ B) [∵ union is distributive over intersection]

(A ∪ A) ∩ (A ∪ B)[∵ A ∪ A = A]

= A ∩ (A ∪ B)

= A.

= (A ∩ A) ∪ (A ∩ B) [∵intersection is distributive over union]

= A ∪ (A ∩ B) [∵ A ∩ A = A]

= A.

To find sets A, B and C such that A ∩ B = ϕ , A ∩ C = ϕ and A ∩ B ∩ C = ϕ

Take A = {1, 2, 3}

B = {2, 4, 6}

C = {3, 4, 7}

Then,

A ∩ B = {2}

∴ A ∩ B ≠ ϕ

A ∩ C = {3}

∴ A ∩ C ≠ ϕ

C ∩ B = {4}

∴ C ∩ B ≠ ϕ

But A, B and C do not have no elements in common,

∴ A ∩ B ∩ C = ϕ.

Let x be any element of Set A

And y be any element of Set B

Now x≠y A ∩ B = ϕ

This means no element of B should be in A.

Thus, x is an element of A and an element of B’

As x ϵ B’

A B’

Hence, Proved.

Let x ϵ A and y ϵ B

A – B = The set of values of A that are not in B.

A ∩ B = The set containing common values of A and B

B – A = The set of values of B that are not in A.

Two sets X and Y are called disjoint if,

X ∩ Y = ϕ

(A – B) ∩ (A ∩ B) = ((A – B) ∩ A) ∪ ((A – B) ∩B)

(A – B) ∩ (A ∩ B) = ϕ ∪ ϕ

(A – B) ∩ (A ∩ B) = ϕ

Similarly,

(B – A) ∩ (A ∩ B) = ((B – A) ∩ A) ∪ ((B – A) ∩B)

(B – A) ∩ (A ∩ B) = ϕ

Hence, the three sets are pair wise disjoint.

We need to show (A ∪ B) ∩ (A ∪ B’) = A.

Now,

(A ∪ B) ∩ (A ∪ B’) = ((A ∪ B) ∩ A) ∩ B’.

= (A ∩ A) ∪ (B ∩ A)) ∩B’

= A ∩ B’

= A.

To show: A⊂B

⇒ x ∉ A

⇒ x ϵ A and A ⊂ U

⇒ x ϵ A

⇒ x ϵ (A’ ∪ B) [∵ U = A’∪ B]

⇒ x ϵ A’ or x ϵ B

But, x ∉ A’,

∴ x ϵ B

Thus, x ϵ A ⇒ x ϵ B

This is true for all x ϵ A

∴ A ⊂ B

We have B’⊂ A’

To Show: A ⊂ B

Let, x ϵ A

⇒ x∉ A’ [∵ A ∩ A’ = ϕ ]

⇒ x ∉ B’ [ ∵ B’ ⊂ A’ ]

⇒ x ϵ B [∵ B ∩ B’ = ϕ]

Thus, x ϵ A ⇒ x ϵ B

This is true for all x ϵ A

∴ A ⊂ B.

It is a False statement.

Let, A = {3} and B = {4}

Then,

P(A) = {ϕ , {3}}

And P(B) = {ϕ , {4}}

∴ P(A) ∪ P (B) = {ϕ , {1}, {2}}

Now,

A ∪ B = {1, 2}

And P(A ∪ B) = {ϕ , {1}, {2}, {1, 2}}

Hence, P(A) ∪ P(B) ≠ P (A ∪ B)

We know that (A ∩ B) ⊂ A and (A – B) ⊂ A

⇒(A ∩ B) ∩ (A – B)⊂ A….(1)

Let and x ϵ (A ∩ B) ∩ (A – B)

⇒ x ϵ (A ∩ B) and x ϵ (A–B)

⇒ x ϵ A and x ϵ B and x ϵ A and x ∉ B

⇒ x ϵ A and x ϵ A [∵ x ϵ B and x ∉ B are not possible simultaneously]

→ x ϵ A

∴ (A ∩ B) ∩ (A – B)⊂ A…(2)

From (1) and (2), we get

A = (A ∩ B) ∩ (A – B)

Let x ϵ A ∪ (B – A)

⇒ x ϵ A or x ϵ (B–A)

⇒ X ϵ A or x ϵ B or x ∉ A

⇒ x ϵ A or x ϵ B

⇒ x ϵ (A ∪ B)

∴ A ∪ (B – A) ⊂ (A ∪ B)…….(1)

Let and x ϵ (A ∪ B)

⇒ x ϵ A or x ϵ B

⇒ x ϵ A or x ϵ B and x ∉ A

⇒ x ϵ A or x ϵ B–A

⇒ x ϵ A ∪ (B–A)

∴ (A ∪ B) ⊂A ∪ (B – A)…….(2)

From (1) and(2), we get

A ∪ (B – A) = A ∪ B.

To show, A’ – B’ = B – A

We need to show

A’ – B’ ⊆ B – A

B – A ⊆ A’ – B’

Let, x ϵ A’ – B.’

⇒ x ϵ A’ and x ∉ B.’

⇒ x ∉ A and x ϵ B

⇒ x ϵ B – A

It is true for all x x ϵ A’ – B’

∴ A’ – B’ = B – A

Hence Proved.

Expanding

(A ∩ A’) ∪ (A∩ B)

(A ∩ A’) =ϕ

⇒ ϕ ∪ (A∩ B)

(A ∪ ϕ =A)

⇒ (A∩ B)

For any sets A and B we have De morgans law

(A ∪B)’ =A’ ∩B’ , (A ∩ B)’ = A’ ∪ B’

=A – (A–B)

= A ∩ (A–B)’

= A∩(A∩B’)’

= A∩(A’∪ B’)’)

= A ∩ (A’∪B)

= (A ∩ A’) ∪ (A ∩ B)

= ϕ ∪ (A ∩ B)

= (A ∩ B)

= A∩ (A ∪ B’)

= A∩( A’∩ B’) [By De–morgan’s law]

= (A ∩ A’) ∩ B’ [∴ A ∩ A’ = ϕ ]

= ϕ ∩ B’

= ϕ

=RHS

= A Δ (A ∩ B) [∵ E Δ F =(E–F) ∪ (F–E) ]

= (A–( A ∩ B)) ∪ (A ∩B –A) [∵ E – F = E ∩ F’]

= (A ∩ (A ∩ B)’) ∪ (A∩B∩A’)

= (A ∩ (A’∪B’)) ∪ (A∩A’∩B)

= ϕ ∪ (A ∩ B’) ∪ ϕ

= A ∩ B’ [∵A ∩ B’ = A–B]

= A–B

=LHS

∴ LHS=RHS Proved.

We have, ACB.

To show: C – B ⊂ C – A

Let, x ϵ C–B

⇒ x ϵ C and x∉ B

⇒ x ϵ C and x∉ A

⇒ x ϵ C – A

Thus, x ϵ C–B ⇒ x ϵ C – A

This is true for all x ϵ C–B

∴ C – B ⊂ C – A

= (A–B) ∪ (B–B)

= (A–B) ∪ ϕ

= A–B

= (A–A) ∩(A–B)

=ϕ ∩ (A – B)

= A – B.

Let x ϵ A–(A–B) ⬄ x ϵ A and x ∉ (A–B)

⬄ x ϵ A and x ∉ (A ∩ B)

⬄ x ϵ A∩(A ∩ B)

⬄ x ϵ (A ∩ B)

∴ A–(A–B) = (A ∩ B)

Let x ϵ A ∪ (B –A) ⇒ x ϵ A or x ϵ (B – A)

⇒ x ϵ A or x ϵ B and x ∉ A

⇒ x ϵ B

⇒ x ϵ (A ∪ B) [∵ B ⊂ (A ∪ B)]

This is true for all x ϵ A ∪ (B–A)

∴ A∪(B–A)⊂(A∪B)……(1)

Conversely,

Let x ϵ (A ∪ B) ⇒ x ϵ A or x ϵ B

⇒ x ϵ A or x ϵ (B–A) [∵ B ⊂ (A ∪ B)]

⇒ x ϵ A ∪ (B–A)

∴ (A∪B)⊂ A∪(B–A)……(2)

From 1 and 2 we get…

A ∪ (B – A) = A ∪ B

Let x ϵ A

Then either x ϵ (A–B) or x ϵ (A ∩ B)

⇒ x ϵ (A–B) ∪ (A ∩ B)

∴ A ⊂ (A – B) ∪ (A ∩ B)….(1)

Consverly,

Let x ϵ (A–B) ∪ (A ∩ B)

⇒ x ϵ (A–B) or x ϵ (A ∩ B)

⇒ x ϵ A and x ∉ B or x ϵ B

⇒ x ϵ A

∴ (A–B) ∪ (A ∩ B) ⊂ A……….(2)

∴ From (1) and (2), We get

(A–B) ∪ (A ∩ B) =A

Given:

n (A ∪ B) = 50

n(A) = 28

n(B) = 32

To Find:

n (A ∩ B) =?

We know,

n (A ∪ B) = n(A) + n(B) – n (A ∩ B)

Substituting the values we get

50 = 28+32 – n (A ∩ B)

50 = 60 – n (A ∩ B)

–10 = – n (A ∩ B)

∴ n (A ∩ B) = 10.

Given:

n(P) = 40

n (P ∪ Q) = 60

n (P ∩ Q) =10

n(Q) = ?

We know,

n (P ∪ Q) = n(P) + n(Q) – n (P ∩ Q)

Substituting the values we get

60 = 40+n(Q)–10

60 = 30+ n(Q)

n(Q) =30

∴ Q =has 30 elements.

Given :

20 teachers teach physics or math.

4 teachers teach physics and math.

12 teach maths

Let us denote teachers who teach physics as n(P) and for Maths n(M).

Now ,

20 teachers teach physics or math = n(P ∪ M) = 20

4 teachers teach physics and math = n (P ∩ M) = 4

12 teach maths = n(M) =12

We know,

n (P ∪ M) = n(M) + n(P) – n (P ∩ M)

Substituting the values we get

20 = 12+n(P)–4

20 = 8+ n(P)

n(P) =12

∴ There are 12 physics teachers.

A total number of people = 70.

Number of people liking Coffee = n(C) = 37.

Number of people liking Tea = n(T) = 52

Total number = n (C ∪ T) = 70

Person who likes both would be n(C ∩ T)

We know,

n (C ∪ T) = n(C) + n(T) – n (C ∩ T)

Substituting the values we get

70 = 37+52– n (C ∩ T)

70 = 89 – n (C ∩ T)

n (C ∩ T) =19.

∴ There are 19 persons who like both coffee and tea.

We know,

n (A ∪ B) = n(A) + n(B) – n (A ∩ B)

Substituting the values we get

42 = 20+ n (B) – 4

42 = 16+ n (B)

n (B) =26

n (A – B) = n (A ∪ B)– n (B)

= 42–26

= 16.

n (B – A) = n (B) – n (A ∩ B)

= 26 –4

= 22.

Let us denote the people who like oranges by n(O) = 76

Let us denote the people who like oranges by n(B) = 62

Total number of people will be those who like oranges or banana = n (O ∪ B) =100

People who like both oranges and banana = n (O ∩ B)

We know,

n (O ∪ B) = n(O) + n(B) – n (O ∩ B)

Substituting the values we get

100 = 76+62 – n (O ∩ B)

100= 138 – n (O ∩ B)

n (O ∩ B) = 38

People who like both oranges and banana = 38%.

Let , Total number of people be n(P) = 950

People who can speak English n(E) = 460

People who can speak Hindi n(H) = 750

i. How many can speak both Hindi and English.

People who can speak both Hindi and English = n (H ∩ E)

We know,

n (P) = n(E) + n(H) – n (H ∩ E)

Substituting the values we get

950 = 460+750 – n (H ∩ E)

950= 1210 – n (H ∩ E)

n (H ∩ E)=260.

Number of people who can speak both English and Hindi are 260.

ii. How many can speak Hindi only.

We can see that H is disjoint union of n(H–E) and n (H ∩ E).

(If A and B are disjoint then n (A ∪ B) = n(A) + n(B))

∴ H = n(H–E) ∪ n (H ∩ E).

⇒ n(H) = n(H–E) + n (H ∩ E).

⇒ 750 = n (H – E)+ 260

⇒ n(H–E) = 490.

Only, 490 people speak Hindi.

iii. how many can speak English only.

We can see that E is disjoint union of n(E–H) and n (H ∩ E).

(If A and B are disjoint then n (A ∪ B) = n(A) + n(B))

∴ E = n(E–H) ∪ n (H ∩ E).

⇒ n(E) = n(E–H) + n (H ∩ E).

⇒ 460 = n (H – E)+ 260

⇒ n(H–E) = 200.

Only, 200 people speak English.

Let total number of people n(P) = 50

A number of people who drink Tea n(T) = 30.

A number of people who drink coffee n(C).

n(T–C) = 14

i. how may drink tea and coffee both.

We can see that T is disjoint union of n(T–C) and n (T ∩ C).

(If A and B are disjoint then n (A ∪ B) = n(A) + n(B))

∴ T = n(T–C) ∪ n (T ∩ C).

⇒ n(T) = n(T–C) + n (T ∩ C).

⇒ 30 = 14 + n (T ∩ C).

⇒ n(T ∩ C) = 16.

16 People drink both coffee and tea.

ii. how many drink coffee but not tea.

We know

n (P) = n(T) + n(C) – n (T ∩ C)

Substituting the values we get

50 = 30+n(C) – 16

n(C) = 36.

We can see that T is disjoint union of n(C–T) and n (T ∩ C).

(If A and B are disjoint then n (A ∪ B) = n(A) + n(B))

∴ C = n(C–T) ∪ n (T ∩ C).

⇒ n(C) = n(C–T) + n (T ∩ C).

⇒ 36 = n(C–T) + 16.

⇒ n(C–T) = 20.

20 People drink coffee but not tea.

Total number of People n(P) = 60.

n(H) =25.

n(T) = 26.

n(I) = 26.

n (H ∩ I) = 9

n (H ∩ T) = 11

n (T ∩ I) = 8

n (H ∩T ∩ I) = 3

The people who read at least one newspaper would be n(H or I or T) = n(H∪I∪T)

We know,

n(H∪I∪T) = n(H)+n(I)+n(T) – n (H ∩ I)– n (H ∩ T)– n (T ∩ I)+ n (H ∩T ∩ I)

n(H∪I∪T) = 25+26+26–9–11–8+3

n(H∪I∪T) = 52.

There are 52 people who read at least one newspaper.

No venn diagram provided.

A total number of People n(P) = ?.

People who play Basketball n(B) =21.

People who play Football n(F) = 29.

People who play Hockey n(H) = 26.

People who play Basketball and Hockey n(B ∩ H) = 14

People who play Football and Hockey n(H ∩ F) = 15

People who play Basketball and Football n(B ∩ F) = 12

People who play all games. n(H ∩B ∩ F) = 8

Total number of people would be n(H or B or F) = n(H∪B∪F)

We know,

n(H∪B∪F) = n(H)+n(B)+n(F) – n (H ∩ B)– n (H ∩ F)– n (B ∩ F)+ n (H ∩B ∩ F)

n(H∪B∪F) = 26+21+29–14–15–12+8

n(H∪B∪F) = 43.

Hence, there are 43 members in all.

Let

Total number of People n(P) = 1000

People who speak Hindi n(H) = 750

People who speak Bengali n(B) = 400

We have P = n(H∪B)

⇒ n(P) = n(H)+n(B)– n (H ∩B)

= 1000 = 750+400 – n (H ∩B)

⇒n (H ∩B) = 150.

∴ 150 People can speak both languages.

H = (H–B) ∪ (H ∩B) (union is disjoint)

∴ n(H) = n(H–B) +n(H ∩B)

750 = n(H–B)+150

n(H–B) = 600

∴ 600 people speak Hindi.

B = (B–H) ∪ (H ∩B) (union is disjoint)

∴ n(B) = n(B–H) +n(H ∩B)

400 = n(B–H)+150

n(B–H) = 250

∴ 250 people speak Bengali.

Let ,

Total number of People n(P) = 500.

People who watch Basketball n(B) =115.

People who watch Football n(F) = 285.

People who watch Hockey n(H) = 195.

People who watch Basketball and Hockey n(B ∩ H) = 50

People who watch Football and Hockey n(H ∩ F) = 70

People who watch Basketball and Football n(B ∩ F) = 45

People who do not watch any games. n(H∪B∪F)= 50

Now,

n(H∪B∪F)’ = n(P) – n(H∪B∪F)

50 = 500–( n(H)+n(B)+n(F) – n (H ∩ B)– n (H ∩ F)– n (B ∩ F)+ n (H ∩B ∩ F))

50 = 500–(285+195+115–70–50–45 +n (H ∩B ∩ F))

50 = 500–430 + n (H ∩B ∩ F))

n (H ∩B ∩ F) = 70–50

n (H ∩B ∩ F)) = 20

∴ 20 People watch all three games.

Number of people who only watch football

= 285–(50+20+25)

= 285–95

= 190.

Number of people who only watch Hockey

= 195–(50+20+30)

= 195–100

= 95.

Number of people who only watch Basketball

= 115–(25+20+30)

= 115–75

= 40.

Number of people who watch exactly one of the three games

As the sets are pairwise disjoint we can write

= number of people who watch either football only or hockey only or Basketball only

=190+95+40

=325

∴ 325 people watch exactly one of the three games.

Total number of People n(P) = 100.

People who Read Magazine A n(A) =28.

People who Read Magazine B n(B) = 30.

People who Read Magazine C n(C) = 42.

People who Read Magazine A and B n(A ∩ B) = 8

People who Read Magazine B and C n(B ∩ C) = 10

People who v Read Magazine A and C n(A ∩ C) = 5

n(A∩B∩C)= 3

Now,

n(A∪B∪C) = n(A)+n(B)+n(C) – n (A ∩ B)– n (B ∩ C)– n (A ∩ C)+ n (A ∩B ∩ C)

= 28+30+42–8–10–5+3

= 100–20

= 80

People who read none of the three magazines

= n(A∪B∪C)’

= n(P) – n(A∪B∪C)

= 100 –80

= 20.

∴ People who read no magazine = 20

n(C) = 42 –(7+3+2)

= 42–12.

= 30.

Let us denote,

Total number of students by n(P)

Students studying English n(E)

Students Studying Hindi n(H)

Students studying Sanskrit n(S)

According to the question,

n(P) =100, n(E–H) = 23, n(E ∩ S) =8, n(E) = 26, n(S) = 48, n(H ∩ S) =8, n(E∪H∪S)’=24

Number of students studying English only = 18

Now,

n(E∪H∪S)’=24

n(P) – n(E∪H∪S) = 24

100 – 24 = n(E∪H∪S)

n(E∪H∪S) = 76

n(E∪H∪S) = n(E)+ n(H)+ n(S)– n(E ∩ S)– n(E ∩ H)– n(H ∩ S)+ (E∩ H ∩ S)

76 = 26+n(H)+48–3–8–8+3

n(H) = 76–58

n(H) = 18

18 Students are studying Hindi.

From q1 we have n(E ∩ H) = 3

∴ 3 students study both hindi and English.

Let n(P₁) be a number of people liking product P₁.

Let n(P₂) be a number of people liking product P₂.

Let n(P₃) be a number of people liking product P₃.

Then, According to the question:

n(P₁) = 21, n(P₂) = 26, n(P₃) = 29, n(P₁∩ P₂) = 14

n(P₁∩ P₃) = 12, n(P₂∩ P₃) = 14, n(P₁∩ P₂ ∩ P₃) = 8.

∴ Number of people liking product P₃ only:

= 29–(4+8+6)

= 29– 18

=11

Let A be the set with n elements.

Then power set of A contains all the subsets of A.

Each member of A has two possibilities either present or absent.

⇒ Total possible subsets of A are 2×2×2×…n times= 2ⁿ

∴ number of elements in power set of A are 2ⁿ.

Power set contains all the subsets of a set.

Null set has no member in it and each set is a subset of itself (i.e. null set is the only subset of null set)

∴ number of elements in power set of null set is 1.

Here, A={X:XϵN,X is a multiple of 3} and

B={X:XϵN,X is a multiple of 5}.

A⋂B={X:XϵN,X is a multiple of 3} and{X:XϵN,X is a multiple of 5}

={X:XϵN,X is a multiple of 3 and 5}

={X:XϵN,X is a multiple of 3*5=15}

={X:XϵN,X is a multiple of 15}

Here, n(A)=3 and n(B)=6

Now, n(A⋃B)=n(A)+n(B)-n(A⋂B)

=3+6-n(A⋂B)

=9-n(A⋂B)

So,n(A⋃B) is minimum whenever n(A⋂B) is maximum and it is possible only when A⊂B

Now,A⊂B then max(n(A⋂B))=n(A)=3.

∴ min(n(A⋃B) )=9-3=6

Here,={XϵC:X²=1}

={1 ,-1}

B={xϵC:X⁴=1}

={1,-1,i,-i}

Now, A∖B=A∖(A⋂B)

={1,-1}∖{1,-1}

=ϕ

And B∖A=B∖(A⋂B)

={1,-1,i,-i}∖{1,-1}

={i,-i}

Here, A⊂B ⇒ B'⊂A'

∴ A'⋂B'=B’

B'∖A'=B'∖(A'⋂B')

=B'∖B'

=ϕ

Here, n(A)=4 and n(B)=7

Now, n(A⋃B) =n(A)+n(B)-n(A⋂B)

=4+7-n(A⋂B)

=11-n(A⋂B)

So, n(A⋃B) is maximum whenever n(A⋂B) is minimum and it is possible only when A⋂B=ϕ

Now, A⋂B=ϕ then min(n(A⋂B)) =0.

∴ min(n(A⋃B))=11-0=11

Here, and

B={(x,y):y=-x, xϵR

and }

and

={(x, y):-x² =1,xϵR}

={(x, y):x² =-1,xϵR}

=ϕ (∵ there is no such that )

Here, and

A⋂B={(x,y):y=e^x ,xϵR} and {(x,y):y=e^-x ,xϵR}

={(x,y):y= e^x and y= e^-x xϵR}

={(x,y): e^x = e^-x xϵR}

={(x,y):e^2x =1,xϵR}

={(0,1)} (∵ )

n(A⋃B)=n(A)+n(B)-n(A⋂B)

∴n(A⋂B)=n(A)+n(B)-n(A⋃B)

∴n(A⋂B)=20+25-40=5

∴n(A⋂B)=5

Here, n(A)=115 , n(B)=326 , n(A∖B)=47

n(A∖B)=n(A)-n(A⋂B)

∴ n(A⋂B)=n(A)-n(A∖B)

=115-47

=68

Now, n(A⋃B)=n(A)+n(B)-n(A⋂B)

=115+326-68

=373