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

• Last Updated : 11 Feb, 2021

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

(i) A â‹‚ B

(ii) A â‹ƒ B

Solution:

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

Given A âŠ‚ B, every element of A is contained in B.

âˆ´ A âˆ© B = A

(ii) A â‹ƒ B

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

Given A âŠ‚ B, B is having all elements including elements of A.

âˆ´ A âˆª 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 âˆª B

(ii) A âˆª C

(iii) B âˆª C

(iv) B âˆª D

(v) A âˆª B âˆª C

(vi) A âˆª B âˆª D

(vii) B âˆª C âˆª D

(viii) A âˆ© (B âˆª C)

(x) (A âˆª D) âˆ© (B âˆª C).

Solution:

We know,

X âˆª Y = {a: a âˆˆ X or a âˆˆ Y}

X âˆ© Y = {a: a âˆˆ X and a âˆˆ Y}

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

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

A âˆª B = Union of two sets A and B = {x: x âˆˆ A or x âˆˆ B}

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

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

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

A âˆª C = Union of two sets A and C =  {x: x âˆˆ A or x âˆˆ 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 âˆª C = {x: x âˆˆ B or x âˆˆ C}

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

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

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

B âˆª D = {x: x âˆˆ B or x âˆˆ 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 âˆª B = {x: x âˆˆ A or x âˆˆ B}

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

A âˆª B âˆª C = {x: x âˆˆ A âˆª B or x âˆˆ 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 âˆª B = {x: x âˆˆ A or x âˆˆ B}

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

A âˆª B âˆª D = {x: x âˆˆ A âˆª B or x âˆˆ 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 âˆª C = {x: x âˆˆ B or x âˆˆ C}

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

B âˆª C âˆª D = {x: x âˆˆ B âˆª C or x âˆˆ 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 âˆª C = {x: x âˆˆ B or x âˆˆ C}

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

A âˆ© B âˆª C = {x: x âˆˆ A and x âˆˆ B âˆª C}

= {4, 5}

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

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

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

(A âˆ© B) = {x: x âˆˆ A and x âˆˆ B}

= {4, 5}

(B âˆ© C) = {x: x âˆˆ B and x âˆˆ C}

= {7, 8}

= Ï•

(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 âˆª D = {x: x âˆˆ A or x âˆˆ D}

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

B âˆª C = {x: x âˆˆ B or x âˆˆ C}

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

(A âˆª D) âˆ© (B âˆª C) = {x: x âˆˆ (A âˆª D) and x âˆˆ (B âˆª C)}

= {4, 5, 10, 11}

### Question 3. Let A = {x: x âˆˆ N}, B = {x: x = 2n, n âˆˆ N), C = {x: x = 2n â€“ 1, n âˆˆ N} and, D = {x: x is a prime natural number} Find:

Solution:

Let us assume,

A = All natural numbers i.e. {1, 2, 3â€¦..}

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

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

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

A contains all elements of the set B.

âˆ´ B âŠ‚ A = {2, 4, 6, 8â€¦}

âˆ´ A âˆ© B = B

A contains all elements of the set C.

âˆ´ C âŠ‚ A = {1, 3, 5â€¦}

âˆ´ A âˆ© C = C

A contains all elements of the set D.

âˆ´ D âŠ‚ A = {2, 3, 5, 7..}

âˆ´ A âˆ© D = D

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

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

C âˆ© D = {1, 3, 5, 7â€¦}

= D â€“ {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 âˆˆ A and x âˆ‰ B}

(i) A-B = {x : x âˆˆ A and x âˆ‰ B} = {3,6,15,18,21}

(ii) A-C = {x : x âˆˆ A and x âˆ‰ C} = {3,15,18,21}

(iii) A-D = {x : x âˆˆ A and x âˆ‰ D} = {3,6,12,18,21}

(iv) B-A = {x : x âˆˆ B and x âˆ‰ A} = {4,8,16,20}

(v) C-A = {x : x âˆˆ C and x âˆ‰ A} = {2,3,8,10,14,16}

(vi) D-A = {x : x âˆˆ D and x âˆ‰ A} = {5,10,20}

(vii) B-C = {x : x âˆˆ B and x âˆ‰ C} = {20}

(viii) B-D = {x : x âˆˆ B and x âˆ‰ 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’

(iv) (A U B)’

(v) (A’)’

(vi) (B-C)’

Solution:

(i) A’ ={x : x âˆˆ U and x âˆ‰ A}

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

(ii) B’ ={x : x âˆˆ U and x âˆ‰ B}

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

(iii) (A âˆ© C)’ = A’ U C’ = {x : x âˆˆ U and x âˆ‰ C and x âˆ‰ A}

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

(iv) (A U B)’ = {x : x âˆˆ U and x âˆ‰ A âˆ© B}

= {5,7,9}

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

= {1,2,3,4}

(vi) (B-C)’ = {x : x âˆˆ U and x âˆ‰ 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’ âˆ© B’

(ii) (A âˆ© 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 âˆˆ U and x âˆ‰ A U B}

= {1,9}

Now, A’ = {1,3,5,7,9}

B’ = {1,4,6,8,9}

A’ âˆ© B’ = {x : x âˆˆ A’ and x âˆˆ B’}

={1,9}

Therefore, LHS = RHS

(ii) We have, LHS = (A âˆ© B)’

Computing (A âˆ© B) = {2}

={x : x âˆˆ U and x âˆ‰ A âˆ© 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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