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• RD Sharma Class 11 Solutions for Maths

# Class 11 RD Sharma Solutions – Chapter 1 Sets – Exercise 1.5

### 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}.

### 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.

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