Question 1: Decide, among the following sets, which sets are subsets of one and another:
A = {x : x ∈ R and x satisfy x2– 8x + 12 = 0}, B = {2, 4, 6},  C = {2, 4, 6, 8, . . . }, D = {6}Â
Solution:
At first, simplifying for set AÂ
x2– 8x + 12 = 0Â
(x-6)(x-2) = 0Â
x= 6 or 2Â
Now, A = {2,6}Â
A set X is said to be a subset of a set Y if every element of X is also an element of Y.Â
Therefore, we can write : A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C
Question 2: In each of the following, determine whether the statement is true or false.
If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B, then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
(iv) If A ⊄ B and B ⊄ C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A
Solution:
(i) FalseÂ
Example : Let x=1, A={1,2,3} and B = {{1,2,3},{4,5,6}}Â
Here x ∈ A and A ∈ B but x ∉ BÂ
(ii) FalseÂ
Example : Let A = {1}, B = {1,3} and C = {{1,3},{5,7}}Â
Here A ⊂ B and B ∈ C but A ∉ CÂ
(iii) TrueÂ
Proof : A ⊂ B : Set B contains all the elements in set AÂ
B ⊂ C : Set C contains all the elements in set BÂ
Hence, by transitivity property, set C contains all the elements in set A i.e. A ⊂ CÂ
(iv) FalseÂ
Example : Let A = {1,2}, B = {3,4,5} and C = {1,2,6,7}Â
Here A ⊄ B and B ⊄ C but A ⊂ CÂ
(v) FalseÂ
Example : Let x=1, A = {1,2} and B = {3,4,5}Â
Here x ∈ A and A ⊄ B but x ∉ BÂ
(vi) TrueÂ
Proof : A ⊂ B : Set B contains all the elements in set AÂ
So if x ∉ B then also x ∉ A
Question 3: Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.
Solution:
Given : A ∪ B = A ∪ C …(1)Â
A ∩ B = A ∩ C …(2)Â
A ∪ B = A + B – (A ∩ B) // by principle of inclusion-exclusionÂ
A ∪ C = A + C – (A ∩ C) // by principle of inclusion-exclusionÂ
A + B – (A ∩ B) = A + C – (A ∩ C) // from (1)Â
A + B – (A ∩ B) = A + C – (A ∩ B) // from (2)Â
B = CÂ
Question 4: Show that the following four conditions are equivalent :
(i) A ⊂ B (ii) A – B = φ (iii) A ∪ B = B (iv) A ∩ B = A
Solution:
//Showing (i)=(ii)Â
A ⊂ B means Set B contains all the elements in set A i.e. set A does not have any different element from BÂ
It means A – B = φÂ
//Showing (i)=(iii)Â
All elements of set A are there in Set B so A ∪ B = BÂ
//Showing (i)=(iv)Â
All elements of set A are there in Set B so A ∩ B = AÂ
From above explanation we can say that all above conditions are equivalent.
Question 5: Show that if A ⊂ B, then C – B ⊂ C – A
Solution:
//By taking an exampleÂ
Let A = {1,2} and B = {1,2,3,4,5}Â
and C = {2,5,6,7,8}Â
Set B contains all the elements in set A so A ⊂ BÂ
Now C-B = {6,7,8} //elements present in C but not in BÂ
C-A = {5,6,7,8} //elements present in C but not in AÂ
It is clearly seen that, Set C-A contains all the elements in Set C-B hence C – B ⊂ C – A is proved.
Question 6: Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
Solution:
(A ∩ B) ∪ (A – B)Â
(A ∩ B) ∪ (A ∩ B’) //(A – B) = (A ∩ B’)Â
A ∩ (B ∪ B’)Â
A ∩ U //B ∪ B’ = U where U represents universal setÂ
A //by identity propertyÂ
Hence, it is proved that A = (A ∩ B) ∪ (A – B)Â
A ∪ (B – A)Â
A ∪ (B ∩ A’)Â
//B – A = B ∩ A’Â
(A ∪ B) ∩ (A ∪ A’)Â
//distributive lawÂ
(A ∪ B) ∩ UÂ
//A ∪ A’ = U where U represents universal setÂ
(A ∪ B)Â
Hence, it is proved that A ∪ (B – A) = (A ∪ B)
Question 7: Using properties of sets, show that
(i) A ∪ (A ∩ B) = A (ii) A ∩ (A ∪ B) = A
Solution:
(i) A ∪ (A ∩ B)Â
(A ∪ A) ∩ (A ∪ B) //distributive lawÂ
A ∩ (A ∪ B) //A ∪ A = AÂ
A //by absorption lawÂ
(ii) A ∩ (A ∪ B)Â
(A ∩ A) ∪ (A ∩ B) //distributive lawÂ
A ∪ (A ∩ B) //A ∩ A = AÂ
A //by absorption law
Question 8: Show that A ∩ B = A ∩ C need not imply B = C
Solution:
Let us assume that B ≠CÂ
Take A = {1,2}, B ={2,3} and C = {3,4} // here B ≠CÂ
A ∩ B = {2}Â
A ∩ C = φÂ
Here we can see that A ∩ B ≠A ∩ CÂ
Therefore, our assumption was wrongÂ
Hence, B = C is must for A ∩ B = A ∩ C
Question 9: Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B
Solution:
A = A ∩ (A ∪ X) //absorption lawÂ
A = A ∩ (B ∪ X) //given that A ∩ X = B ∩ XÂ
= (A ∩ B) ∪ (A ∩ X) //distributive lawÂ
= (A ∩ B) ∪ φ //given that A ∩ X = φÂ
A = (A ∩ B) …(1)Â
Repeating above process by taking B = B ∩ (B ∪ X)Â
We get B = (A ∩ B) …(2)Â
from (1) and (2)Â
A = B
Question 10: Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ
Solution:
(A ∩ B) should be non-empty means atleast one element is common in between themÂ
… same for (B ∩ C) & (A ∩ C)Â
A ∩ B ∩ C = φ means there should not be any element common in all the three sets A, B and CÂ
Let A = {1,2} B = {2,3} and C = {1,3}Â
A ∩ B = {2}Â
B ∩ C = {3}Â
A ∩ C = {1}Â
and A ∩ B ∩ C = φ
Deleted Questions from NCERT
Assume that P (A) = P (B). Show that A = B
Solution:
P(X) represents power set of set XÂ
To prove A = B we have to prove that A ⊂ B and B ⊂ AÂ
(eg : if A = {1,2} then P(A) = {φ, {1}, {2}, {1,2}})Â
Power set of any set contains all the possible subsets of it.Â
A ∈ P(A)Â
as P(A)=P(B) so A ∈ P(B)Â
If A is P(B) then clearly A is subset of B.Â
A ⊂ B …(1)Â
Repeating above process for B ∈ P(B) we getÂ
B ⊂ A …(2)Â
From above equations,Â
A = B
Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer
Solution:
It is FalseÂ
Let A = {1,2} and B = {2,3}Â
A ∪ B = {1,2,3}Â
P(A) = {φ, {1}, {2}, {1,2}}Â
P(B) = {φ, {2}, {3}, {2,3}}Â
P(A) ∪ P(B) = {φ, {1}, {2}, {3}, {1,2}, {2,3}} …(1)Â
P(A ∪ B) = {φ, {1}, {2}, {3}, {1,2}, {2,3}, {1,3}, {1,2,3}} …(2)Â
P(A) ∪ P(B) ≠P(A ∪ B) //from (1) and (2)
In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Solution:
There are total 600 studentsÂ
Let A and B represents sets of students taking tea and coffee respectivelyÂ
n(A) = 150Â
n(B) = 225Â
Students taking both tea and coffee = n(A ∩ B) = 100Â
Students taking either tea or coffee = n(A ∪ B) = ?Â
By principle of inclusion-exclusion,Â
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)Â
= 150 + 225 – 100Â
= 275Â
Now,Â
Number of students who neither take tea nor coffee = total students – Number of Students who either take tea or coffeeÂ
= 600 – 275Â
= 325Â
There are 325 students who neither take tea nor coffeeÂ
In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Solution:
Let A and B represents sets of students who knows Hindi and English respectivelyÂ
n(A) = 100Â
n(B) = 50Â
Number of students who know both languages = n(A ∩ B) = 25Â
It is given that each student knows either Hindi or EnglishÂ
Hence, Number of Students in a group = n(A ∪ B)Â
By principle of inclusion-exclusion,Â
n(A ∪ B) = n(A) + n(B) – n(A ∩ B)Â
= 100 + 50 – 25Â
= 125Â
Total students in the group are 125
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:Â
(i) the number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper
Solution:
(i) Total number of people in survey = 60Â
Let H, I, T represents sets of people reading newspaper H, I and T respectivelyÂ
n(H) = 25Â
n(I) = 26Â
n(T) = 26Â
People reading both H and I = n(H ∩ I) = 9Â
People reading both H and T = n(H ∩ T) = 11Â
People reading both T and I = n(T ∩ I) = 8Â
People reading all the three newspapers = n(H ∩ I ∩ T) = 3Â
(i) Number of people who read at least one of the newspapers is given by n(H ∪ I ∪ T)Â
By principle of inclusion-exclusion,Â
n(H ∪ I ∪ T) = n(H) + n(I) + n(T) – n(H ∩ I) – n(H ∩ T) – n(T ∩ I) + n(H ∩ I ∩ T)Â
= 25 + 26 + 26 – 9 – 11 – 8 + 3Â
= 52Â
Number of people who read at least one of the newspapers is 52Â
(ii) Take a look at following Venn diagram
Number of people who read exactly one newspaper are represented by green color in above diagramÂ
= n(H ∪ I ∪ T) – n(H ∩ I) – n(H ∩ T) – n(T ∩ I) + (2 x n(H ∩ I ∩ T))Â
= 52 – 9 – 11 – 8 + (2 x 3)Â
= 24 + 6Â
= 30Â
Number of people who read exactly one newspaper are 30
Â
Â
In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
Solution:
Let A, B, C represents sets of people who liked product A, B and C respectivelyÂ
n(A) = 21Â
n(B) = 26Â
n(C) = 29Â
People who liked product A and B both = n(A ∩ B) = 14Â
People who liked product A and C both = n(A ∩ C) = 12Â
People who liked product B and C both = n(B ∩ C) = 14Â
People who liked all the three products = n(A ∩ B ∩ C) = 8Â
Take a look at following Venn diagram
Number of people who only like product C are represented by green colourÂ
= n(C) – n(A ∩ C) – n(B ∩ C) + n(A ∩ B ∩ C)Â
= 29 – 12 – 14 + 8Â
= 11Â
Number of people who only like product C are 11
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