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Class 11 NCERT Solutions- Chapter 1 Sets – Miscellaneous Exercise on Chapter 1 | Set 1
• Last Updated : 05 Apr, 2021

### 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}

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)

### (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

### Chapter 1 Sets –  Miscellaneous Exercise on Chapter 1 | Set 2

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