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

Here Class 11 maths solutions of important questions are going to help every maths student of class 11 in boosting their preparation of SA-1 and SA-2 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.

**NCERT solutins class 11 maths -Conic Section**

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**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

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

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**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 umer are 1 ad the number itself,we have

C={1,p}

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

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