Class 11 NCERT Solutions Chapter 1 Sets Miscellaneous Exercise on Chapter 1 | Set 1 (original) (raw)
Last Updated : 23 Jul, 2025
In Class 11 Mathematics, Chapter 1 focuses on "Sets" a fundamental topic in modern mathematics. The Sets form the foundation for the various branches of mathematics including algebra, probability, and statistics. The Miscellaneous Exercise of Chapter 1 provides a diverse set of problems designed to test and reinforce understanding of set theory concepts. This exercise is crucial for mastering the basics of the sets and their operations which are pivotal for solving more complex mathematical problems.
Miscellaneous Exercise
The Miscellaneous Exercise of Chapter 1 in the NCERT Class 11 textbook covers the various problems related to sets. This exercise aims to:
- Reinforce the fundamental concepts of the set theory.
- Provide practice with the different types of problems including the operations on sets, Venn diagrams, and set relations.
- Enhance problem-solving skills and ensure a thorough understanding of the topic.
**Question 1: Decide, among the following sets, which sets are subsets of one another:
**A = {x : x ∈ R and x satisfy x 2 – 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8, . . . }, D = {6}
**Solution:
At first, simplifying for set A
x2– 8x + 12 = 0
(x-6)(x-2) = 0
x= 6 or 2
Now, A = {2,6}
A set X is said to be a subset of a set Y if every element of X is also an element of Y.
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) False**
Example : Let x=1, A={1,2,3} and B = {{1,2,3},{4,5,6}}
Here x ∈ A and A ∈ B but x ∉ B
****(ii) False**
Example : Let A = {1}, B = {1,3} and C = {{1,3},{5,7}}
Here A ⊂ B and B ∈ C but A ∉ C
****(iii) True**
Proof : A ⊂ B : Set B contains all the elements in set A
B ⊂ C : Set C contains all the elements in set B
Hence, by transitivity property, set C contains all the elements in set A i.e. A ⊂ C
****(iv) False**
Example :Let A = {1,2}, B = {3,4,5} and C = {1,2,6,7}
Here A ⊄ B and B ⊄ C but A ⊂ C
****(v) False**
Example : Let x=1, A = {1,2} and B = {3,4,5}
Here x ∈ A and A ⊄ B but x ∉ B
****(vi) True**
Proof : A ⊂ B : Set B contains all the elements in set A
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)
A ∩ B = A ∩ C ...(2)
A ∪ B = A + B - (A ∩ B) // by principle of inclusion-exclusion
A ∪ C = A + C - (A ∩ C) // by principle of inclusion-exclusion
A + B - (A ∩ B) = A + C - (A ∩ C) // from (1)
A + B - (A ∩ B) = A + C - (A ∩ B) // from (2)
**B = C
**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)
A ⊂ B means Set B contains all the elements in set A i.e. set A does not have any different element from B
It means A - B = φ
//Showing (i)=(iii)
All elements of set A are there in Set B so A ∪ B = B
//Showing (i)=(iv)
All elements of set A are there in Set B so A ∩ B = A
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 example
Let A = {1,2} and B = {1,2,3,4,5}
and C = {2,5,6,7,8}
Set B contains all the elements in set A so A ⊂ B
Now C-B = {6,7,8} //elements present in C but not in B
C-A = {5,6,7,8} //elements present in C but not in A
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: Assume that P (A) = P (B). Show that A = B
**Solution:
P(X) represents power set of set X
To prove A = B we have to prove that A ⊂ B and B ⊂ A
(eg : if A = {1,2} then P(A) = {φ, {1}, {2}, {1,2}})
Power set of any set contains all the possible subsets of it.
A ∈ P(A)
as P(A)=P(B) so A ∈ P(B)
If A is P(B) then clearly A is subset of B.
A ⊂ B ...(1)
Repeating above process for B ∈ P(B) we get
B ⊂ A ...(2)
From above equations,
**A = B
**Question 7: Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer
**Solution:
**It is False
Let A = {1,2} and B = {2,3}
A ∪ B = {1,2,3}
P(A) = {φ, {1}, {2}, {1,2}}
P(B) = {φ, {2}, {3}, {2,3}}
P(A) ∪ P(B) = {φ, {1}, {2}, {3}, {1,2}, {2,3}} ...(1)
P(A ∪ B) = {φ, {1}, {2}, {3}, {1,2}, {2,3}, {1,3}, {1,2,3}} ...(2)
**P(A) ∪ P(B) ≠ P(A ∪ B) //from (1) and (2)
**Question 8: Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
**Solution:
(A ∩ B) ∪ (A – B)
(A ∩ B) ∪ (A ∩ B’) //(A – B) = (A ∩ B’)
A ∩ (B ∪ B’)
A ∩ U //B ∪ B’ = U where U represents universal set
A //by identity property
Hence, it is proved that **A = (A ∩ B) ∪ (A – B)A ∪ (B - A)
A ∪ (B ∩ A’)
//B – A = B ∩ A’
(A ∪ B) ∩ (A ∪ A’)
//distributive law
(A ∪ B) ∩ U
//A ∪ A’ = U where U represents universal set
(A ∪ B)
Hence, it is proved that **A ∪ (B – A) = (A ∪ B)
**Question 9: Using properties of sets, show that
****(i) A ∪ (A ∩ B) = A (ii) A ∩ (A ∪ B) = A**
**Solution:
****(i)** A ∪ (A ∩ B)
(A ∪ A) ∩ (A ∪ B) //distributive law
A ∩ (A ∪ B) //A ∪ A = A
A //by absorption law****(ii)** A ∩ (A ∪ B)
(A ∩ A) ∪ (A ∩ B) //distributive law
A ∪ (A ∩ B) //A ∩ A = A
A //by absorption law
**Question 10: Show that A ∩ B = A ∩ C need not imply B = C
**Solution:
Let us assume that B ≠ C
Take A = {1,2}, B ={2,3} and C = {3,4} // here B ≠ C
A ∩ B = {2}
A ∩ C = φ
Here we can see that A ∩ B ≠ A ∩ C
Therefore, our assumption was wrong
Hence, **B = C is must for A ∩ B = A ∩ C