# Class 11 Maths solutions of important questions of ‘Sets’

Class 11 Maths solutions of important questions of chapter 1-Sets are going to help every maths student of class 11 in boosting their preparation for the class 11 exam of CBSE board. The solutions of important questions of chapter 1-Sets will clear your basics of maths which is required to study for gaining higher maths study. All questions are solved by an expert teacher of maths as per the CBSE norms. All important questions of chapter 1- sets of class 11 CBSE are solved in a step by step method.

**Class 11 Maths solutions of important questions of chapter 1-Sets**

**Study the basics of the Sets, see the video**

**Q1.If A = { 1,2,3,4,5,6}, B = {2,4,6, 8} then find A – B.**

Ans.We are given the sets A ={ 1,2,3,4,5,6}, B = {2,4,6, 8}

A – B = { 1,2,3,4,5,6}- {2,4,6, 8} = {1, 3, 5}

**Q2. Let A and B be two sets containing 3 and 6 elements respectively. Find the maximum and the minimum number of elements in A ∪ B.**

Ans.There may be the case when atleast 3 elements are common between both sets

Let a set A = {a, b, c} and B = {a, b, c, d, e, f}

∴ A ∪ B = {a, b, c, d, e, f} implies that the minimum number of elements in A ∪ B are = 6

There may be the case when there are no any elements are common between both sets

lf A = {a, b, c}, B = { d, e, f, g, h, i}

A ∪ B = {a, b, c, d, e, f,g,h,i} implies that the maximum number of elements in A ∪ B are = 9

**Q3.If A = {(x,y) : x² + y²= 25 where x, y ∈ W } write a set of all possible ordered pair .**

Ans. We are given the set A = {(x,y) : x² + y²= 25 where x, y ∈ W }

All possible ordered pair of set A are following

For x = 0,y =5, x=3,y=4,for x =4, y =3,for x=5,y =0

A = {(0,5),(3,4),(4,3),(5,0)}

**Q4.If A = {1,2,3}, B = {4, 5, 6} and C ={5} verify that A ∪ ( B ∩ C) = (A ∪ B) ∩ A ∪ C.**

We are given the sets A = {1,2,3}, B = {4, 5, 6} ,C = {5}

B ∩ C = {5}

LHS

A ∪ ( B ∩ C) = {1,2,3,5}

(A ∪ B) and A ∪ C

A ∪ B = {1,2,3,4,5,6} and A ∪ C = {1,2,3,5}

RHS

(A ∪ B) ∩ A ∪ C = {1,2,3,5}

Therefore

A ∪ ( B ∩ C) = (A ∪ B) ∩ A ∪ C, Hence proved

**Q5.From the adjoining Venn diagram, write the value of the following.**

(**a) A ‘**

**(b) B’**

**(c) (A ∩ B)’**

Ans. From the venn diagram ,we have

U = {1,2,3,4………15}

A = {7,9.11}. B ={11,12,13,14}

A’ = U – A = {1,2,3,4………15} – {7,9.11} = {1,2,3,4,5,6,8,10,12,13,14,15}

B’ = U – B = {1,2,3,4………15} – {11,12,13,14}= {1,2,3,4,5,6,7,8,9,10,,15}

We have,(A ∩ B) = 11

(A ∩ B)’ = U – (A ∩ B) = {1,2,3,4………15} – {11} = (1,2,3,4,5,6,7,8,9,10,12,13,14,15}

**Q6. If P(A) = P(B) show that A = B.**

Ans. P(A} and P(B) implies that both are power sets of A and B respectively

Every set is an element of its power set , so A ∈ P(A)

Since, we are given that

P(A) = P(B)

Therefore, A ∈ P(B)

Indicates that every element of A belongs to the set B

So, A ⊂ B……(i)

Similarly B ∈ P(B)

Since, we are given that

P(A) = P(B)

Indicates that every element of B belongs to the set A

So, B ⊂ A……(ii)

From (i) and (ii), we get

A = B, Hence proved

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**Q7. Let A and B be sets ; if A∩X = B∩X = ∅ and A∪X = B∪X for some set X.Show that A=B.**

Ans. We are given that A∩X = B∩X = ∅ and A∪X = B∪X

To prove A=B

Proof. A∪X = B∪X (given)

Multiplying both sides by A∩

A∩ (A∪X) =A∩ (B∪X)

Using distributive property

(A∩ A) ∪ (A ∩ X) = (A∩ B )∪ (A∩ X)

A∩Φ = (A∩ B) ∪ Φ

A = (A∩ B) …………(i)

A∪X = B∪X

Multiplying both sides by B∩

B∩(A∪X) = B∩(B∪X)

(B∩ A) ∪ (B ∩ X) = (B∩ B )∪ (B∩ X)

(B∩ A) ∪φ = B ∪ φ

B = (B∩ A)

B = (A∩ B) …………(ii)

From (i) and (ii)

A = B, Hence proved

**Q8.If A ={1,2,3,4,5},then write the proper subsets of A.**

Ans. The number of elements in the given sets A ={1,2,3,4,5} are =5

The number of proper subsets of any set are = 2^{n} – 1

^{ Where n = number of elements = 5}

^{ The number of proper subsets of any set are = }2^{5} – 1 =32 – 1 = 31

**Q9.Write the following sets in the Roster form**

**(i) A={x : x ∈ R, 2x+11 =15}**

**(ii)B={x |x² =x, x ∈ R}**

**(iii)C={x = x is a positive factor of the prime number p}**

Ans.(i)We have, A={x : x ∈ R, 2x+11 =15}

2x+1= 15 ⇒x= 2

∴ A = {2}

(ii)We have,B={x |x² =x, x ∈ R}

∴ x² = x ⇒ x²-x= 0 ⇒x(x-1)= 0 ⇒x=0,1

∴ B ={0,1}

(iii)We have, C={x = x is a positive factor of the prime number p}

Sice positive factors of a prime u∪mer are 1 ad the number itself,we have

C={1,p}

**Q10.For all sets A,B and C show that (A – B) ∩****(A – C) = A – (B ∪ C).**

Ans. Considering that

x ∈ (A – B)∩(A – C)

⇒ x ∈ (A – B) and x ∈ (A – C)

⇒ (x ∈ A and x∉ B) and (x ∈ A and x∉ C)

⇒ (x ∈ A ) and (x ∉B and x∉ C )

⇒(x ∈ A ) x∉ (B ∪C)∈

⇒x ∈ A – (B ∪C)

⇒(A – B) ∩(A – C)⊂A – (B ∪ C)…….(i)

Now,Considering that

y ∈A – (B ∪ C)

⇒ y ∈A and y ∉ (B ∪ C)

⇒y ∈A and (y ∉B and y ∉ C)

⇒(y ∈A and y ∉B) and (y ∈A and y ∉ C)

⇒y ∈ (A – B) and y ∈ (A – C)

⇒y ∈ (A – B) ∩ y ∈ (A – C)

⇒A – (B ∪ C) ⊂(A – B) ∩(A – C)………(ii)

From (i) and (ii)

(A – B) ∩(A – C)= A – (B ∪ C), Hence proved

**Q11.Let A,B and C be the sets such that A∪B = A∪C and A ∩B = A ∩C,show that B = C.**

Ans. According to question, A ∪ B = A ∪ C and A ∩ B = A ∩ C

To show, B = C

Let us assume, x ∈ B So, x ∈ A ∪ B

x ∈ A ∪ C

Hence, x ∈ A or x ∈ C

when x ∈ A, then x ∈ B

∴ x ∈ A ∩ B

As, A ∩ B = A ∩ C

So, x ∈ A ∩ C

∴ x ∈ A or x ∈ C

x ∈ C

∴ B ⊂ C

similarly, it can be shown that C ⊂ B

Hence, B = C

**Q12.Show that for any sets A and B**

**A = (A∩B)∪(A-B) **

Ans.We have to prove

A = (A∩B)∪(A-B)

Taking RHS and solving it

(A∩B)∪(A-B)

Using the property

A -B = A -(A∩B)

A-B = A∩B’

= (A∩B) ∪ (A∩B’)

Applying distributive property

A∩(B∪C) = (A∩B) ∪(A∩C)

Replacing C = B’ in LHS

A-B = A∩(B∪B’)

= A∩ (U) [since B∪B’ = U]

= A [since A∩ U = A]

= RHS, Hence proved

**Q13. Write the following sets in the roster form:**

**(i) A = {x : x ∈ R, 2x + 15 = 15}**

**(ii) B = {x : x² = x, x ∈ R}**

Ans(i) It is given to us that set

A = {x : x ∈ R, 2x + 11 = 15}

2x + 11 = 15

2x = 15 – 11

2x = 4

x = 2

Therefore in roster form it is written as A = {2}

(ii) It is given to us that

B = {x : x² =x, x ∈ R}

x² = x

x² – x = 0

x(x – 1) = 0

x = 0, x = 1

Therefore in roster form it is written as

B = {0, 1}

**Q14. If A and B are subsets of the universal set U, then show that **

**(i) A ⊂ A ∪ B**

Ans. Let’s prove that

A ⊂ A ∪ B

Let x ∈ A or x ∈ B

If x ∈ A then x ∈ A ∪ B

Hence A ⊂ A ∪ B

**Q15. A,B and C are subsets of universal set if A = {2,4,6,8,12,20}, B ={3,6.9.12.15},C ={5,10,15,20} and U is the set of all whole numbers,draw a venn diagram showing the relation of U,A,B and C.**

Ans.

**Q16.If A and B are two sets such that A ⊂B then find:**

**(i) A ∩ B (ii) A ∪ B**

Ans**.(i) **A ∩ B implies all the elements which are available in A and B

Let A = {1,2,3}, B = {1,2,3,4,5}⇒ A ⊂B

A ∩ B = {1,2.3} = A

∴A ∩ B = A

A ⊂B implies that all the elements of A are in B

(ii) A ∪ B implies all the elements which are available in A or B

Let A = {1,2,3}, B = {1,2,3,4,5}⇒ A ⊂B

A ∪ B = {1,2,3,4,5} = B

∴ A ∪ B = B

**Q17.Let A ={{2,3,4},{5,6},{6,7,8}}.Determine which of the following is true or false.**

**(i) 4 ∈ A (ii) {2,3,4} ⊂ A (iii) {6,7,8}∈ A (iv) {5,6} ∈ A (v) Φ∈ A (vi) Φ ⊂ A**

Ans. (i) False**, **since 4 is not an element of the set A, it is the part of the element {2,3,4}

(ii) False, {2,3,4} is an element of A,therefore it is written as {2,3,4}∈ A

(iii) True, since {6,7,8} is an element of A

(iv) True, since {5,6} is an element of A

(v) False,Φ is not an element of any set,it is a subset of every set

(vi) True, since Φ is a subset of every set

**Class 11 Maths solutions of important questions of chapter 1-Sets**

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