# Class 11 NCERT Solutions- Chapter 1 Sets – Miscellaneous Exercise on Chapter 1 | Set 1

**Question 1: Decide, among the following sets, which sets are subsets of one and 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

x^{2}– 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 aboveconditions 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 henceC – B ⊂ C – Ais 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 thatA = (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 thatA ∪ (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 = Cis must for A ∩ B = A ∩ C