Class 11 RD Sharma Solutions - Chapter 1 Permutations And Combinations- Exercise 1.5

In this section, we explore Chapter 1 of the Class 11 RD Sharma textbook, which focuses on Sets. Exercise 1.5 is aimed at strengthening students' understanding of set theory, including various operations and properties associated with sets.

Class 11 RD Sharma Solutions - Chapter 1 Sets - Exercise 1.5

This section provides detailed solutions for Exercise 1.5 from Chapter 1 of the Class 11 RD Sharma textbook. These solutions are designed to aid students in mastering the concepts of set theory, ensuring a strong mathematical foundation for future topics.

Question 1. If A and B are two sets such that A ⊂ B, then Find:

(i) A ⋂ B

(ii) A ⋃ B

Solution:

(i) A ∩ B

A ∩ B denotes A intersection B, that is the common elements of both the sets.

Given A ⊂ B, every element of A is contained in B.

∴ A ∩ B = A

(ii) A ⋃ B

A ∪ B denotes A union B, that is it contains elements of either of the set.

Given A ⊂ B, B is having all elements including elements of A.

∴ A ∪ B = B

Question 2. If A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8}, C = {7, 8, 9, 10, 11} and D = {10, 11, 12, 13, 14}. Find:

(i) A ∪ B

(ii) A ∪ C

(iii) B ∪ C

(iv) B ∪ D

(v) A ∪ B ∪ C

(vi) A ∪ B ∪ D

(vii) B ∪ C ∪ D

(viii) A ∩ (B ∪ C)

(ix) (A ∩ B) ∩ (B ∩ C)

(x) (A ∪ D) ∩ (B ∪ C).

Solution:

We know,

 X ∪ Y = {a: a ∈ X or a ∈ Y}

X ∩ Y = {a: a ∈ X and a ∈ Y}

(i) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

A ∪ B = Union of two sets A and B = {x: x ∈ A or x ∈ B}

= {1, 2, 3, 4, 5, 6, 7, 8}

(ii) A = {1, 2, 3, 4, 5}

C = {7, 8, 9, 10, 11}

A ∪ C = Union of two sets A and C =  {x: x ∈ A or x ∈ C}

= {1, 2, 3, 4, 5, 7, 8, 9, 10, 11}

(iii) B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

B ∪ C = {x: x ∈ B or x ∈ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

(iv) B = {4, 5, 6, 7, 8}

D = {10, 11, 12, 13, 14}

B ∪ D = {x: x ∈ B or x ∈ D}

= {4, 5, 6, 7, 8, 10, 11, 12, 13, 14}

(v) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

A ∪ B = {x: x ∈ A or x ∈ B}

= {1, 2, 3, 4, 5, 6, 7, 8}

A ∪ B ∪ C = {x: x ∈ A ∪ B or x ∈ C}

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

(vi) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

D = {10, 11, 12, 13, 14}

A ∪ B = {x: x ∈ A or x ∈ B}

= {1, 2, 3, 4, 5, 6, 7, 8}

A ∪ B ∪ D = {x: x ∈ A ∪ B or x ∈ D}

= {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14}

(vii) B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

D = {10, 11, 12, 13, 14}

B ∪ C = {x: x ∈ B or x ∈ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

B ∪ C ∪ D = {x: x ∈ B ∪ C or x ∈ D}

= {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}

(viii) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

B ∪ C = {x: x ∈ B or x ∈ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

A ∩ B ∪ C = {x: x ∈ A and x ∈ B ∪ C}

= {4, 5}

(ix) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

(A ∩ B) = {x: x ∈ A and x ∈ B}

= {4, 5}

(B ∩ C) = {x: x ∈ B and x ∈ C}

= {7, 8}

(A ∩ B) ∩ (B ∩ C) = {x: x ∈ (A ∩ B) and x ∈ (B ∩ C)}

= ϕ

(x) A = {1, 2, 3, 4, 5}

B = {4, 5, 6, 7, 8}

C = {7, 8, 9, 10, 11}

D = {10, 11, 12, 13, 14}

A ∪ D = {x: x ∈ A or x ∈ D}

= {1, 2, 3, 4, 5, 10, 11, 12, 13, 14}

B ∪ C = {x: x ∈ B or x ∈ C}

= {4, 5, 6, 7, 8, 9, 10, 11}

(A ∪ D) ∩ (B ∪ C) = {x: x ∈ (A ∪ D) and x ∈ (B ∪ C)}

= {4, 5, 10, 11}

Question 3. Let A = {x: x ∈ N}, B = {x: x = 2n, n ∈ N), C = {x: x = 2n – 1, n ∈ N} and, D = {x: x is a prime natural number} Find:

(i) A ∩ B

(ii) A ∩ C

(iii) A ∩ D

(iv) B ∩ C

(v) B ∩ D

(vi) C ∩ D

Solution:

Let us assume, 

A = All natural numbers i.e. {1, 2, 3…..}

B = All even natural numbers i.e. {2, 4, 6, 8…}

C = All odd natural numbers i.e. {1, 3, 5, 7……}

D = All prime natural numbers i.e. {1, 2, 3, 5, 7, 11, …}

(i) A ∩ B

A contains all elements of the set B.

∴ B ⊂ A = {2, 4, 6, 8…}

∴ A ∩ B = B

(ii) A ∩ C

A contains all elements of the set C.

∴ C ⊂ A = {1, 3, 5…}

∴ A ∩ C = C

(iii) A ∩ D

A contains all elements of the set D.

∴ D ⊂ A = {2, 3, 5, 7..}

∴ A ∩ D = D

(iv) B ∩ C

B ∩ C = ϕ

There cannot be any natural number which is both even and odd at same time.

(v) B ∩ D

B ∩ D = 2

{2} is the only natural number possible which is even and a prime number.

(vi) C ∩ D

C ∩ D = {1, 3, 5, 7…}

= D – {2}

Therefore, Every prime number is odd except {2}.

Question 4. Let A = {3,6,12,15,18,21}, B = {4,8,12,16,20}, C = {2,4,6,8,10,12,14,16} and D = {5,10,15,20}

Find:

(i) A-B 

(ii) A-C

(iii) A-D

(iv) B-A

(v) C-A

(vi) D-A

(vii) B-C

(viii) B-D

Solution: 

For any two sets A and B, A-B is the set of elements belonging in A and not in B.

that is, A-B = {x : x ∈ A and x ∉ B}

(i) A-B = {x : x ∈ A and x ∉ B} = {3,6,15,18,21}

(ii) A-C = {x : x ∈ A and x ∉ C} = {3,15,18,21}

(iii) A-D = {x : x ∈ A and x ∉ D} = {3,6,12,18,21}

(iv) B-A = {x : x ∈ B and x ∉ A} = {4,8,16,20}

(v) C-A = {x : x ∈ C and x ∉ A} = {2,3,8,10,14,16}

(vi) D-A = {x : x ∈ D and x ∉ A} = {5,10,20}

(vii) B-C = {x : x ∈ B and x ∉ C} = {20}

(viii) B-D = {x : x ∈ B and x ∉ D} = {4,8,12,16}

Question 5. Let U = {1,2,3,4,5,6,7,8,9}, A = {1,2,3,4}, B = {2,4,6,8} and C = {3,4,5,6}. Find:

(i) A'

(ii) B'

(iii) (A ∩ C)'

(iv) (A U B)'

(v) (A')'

(vi) (B-C)'

Solution:

(i) A' ={x : x ∈ U and x ∉ A}

= {5,6,7,8,9}

(ii) B' ={x : x ∈ U and x ∉ B}

 = {1,3,5,7,9}

(iii) (A ∩ C)' = A' U C' = {x : x ∈ U and x ∉ C and x ∉ A} 

= {1,2,5,6,7,8,9}

(iv) (A U B)' = {x : x ∈ U and x ∉ A ∩ B} 

= {5,7,9}

(vi) (A')' = {x : x ∈ A} because, complement of complement cancels with each other

= {1,2,3,4}

(vi) (B-C)' = {x : x ∈ U and x ∉ B-C}

={1,3,4,5,6,7,9}

Question 6. Let U = {1,2,3,4,5,6,7,8,9}, A = {2,4,6,8}, B = {2,3,5,7}. Verify that:

(i) (A U B)' = A' ∩ B'

(ii) (A ∩ B)' = A' U B'

Solution:

(i) We have, LHS = (A U B)'

Computing (A U B) = {2,3,4,5,6,7,8}

Now (A U B)' = U - (A U B) 

= {x : x ∈ U and x ∉ A U B}

= {1,9}

RHS = A' ∩ B'

Now, A' = {1,3,5,7,9}

B' = {1,4,6,8,9}

A' ∩ B' = {x : x ∈ A' and x ∈ B'}

={1,9}

Therefore, LHS = RHS

(ii) We have, LHS = (A ∩ B)'

Computing (A ∩ B) = {2}

Now, (A ∩ B)' = U - (A ∩ B)

={x : x ∈ U and x ∉ A ∩ B}

={1,3,4,5,6,7,8,9}

RHS = A' U B'

Computing A' = {1,3,5,7,9}

B' = {1,4,6,8,9}

Now A' U B' = {1,3,4,5,7,8,9}

Therefore, LHS= RHS.

Summary

Exercise 1.5 covers advanced set operations, including intersection, union, and complement. Students learn to find the intersection, union, and complement of two sets, and understand the properties of these operations.

• Class 11 RD Sharma Solutions - Chapter 1 Sets - Exercise 1.4
• Class 11 RD Sharma Solutions - Chapter 1 Sets - Exercise 1.6