CBSE Class 11 Maths Sets NCERT Exercise 1.4 Answers

(Q1 - Q5)

Q1: Find the union of each of the following pairs of sets:
(i) X = {1,3,5} Y = {1,2,3}

(ii) A = {a,e,i,o,u}, B = {a,b,c}

(iii) A = {x : x is a natural number and multiple of 3}
B = {x : x is a natural number less than 6}

(iv) A = {x : x is a natural number and 1 < x ≤ 6 }
B = {x : x is a natural number and 6 < x < 10 }

(v) A = {1, 2, 3}, B = φ

(i) X = {1,3,5} Y = {1,2,3}
X ∪ Y = {1,2,3,5}

(ii)  A = {a,e,i,o,u}, B = {a,b,c}
A ∪ B = {a,b,c,e,i,o,u}

(iii)  A = {x : x is a natural number and multiple of 3} = {3,6,9,...}
B = {x : x is a natural number less than 6}  = {1,2,3,4,5,6}
A ∪ B = {1,2,4,5,3,6,9,12..}
A ∪ B =  {x: x = 1,2,4,5 or a multiple of 3 }

(iv) A = {x : x is a natural number and 1 < x ≤ 6 }  = {2,3,4,5,6}
B = {x : x is a natural number and 6 < x < 10 } = {7,8,9}

A ∪ B = {2,3,4,5,6,7,8,9}
∴ A ∪ B = {x : x ∈ N and 1 < x <10}

(v) A = {1, 2, 3}, B = φ
A ∪ B = {1, 2, 3}

Q2: Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?

Answer: Given, A = { a, b }, B = {a, b, c}

Yes, A ⊂ B

A ∪ B = {a, b, c} = B

Q3: If A and B are two sets such that A ⊂ B, then what is A ∪ B ?

Answer: If A and B are two sets such that A ⊂ B, then A∪B = B.

Q4: If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 } and D = { 7, 8, 9, 10 }; 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

A = {1, 2, 3, 4},
B = {3, 4, 5, 6},
C = {5, 6, 7, 8 }
D = { 7, 8, 9, 10 }

(i) A ∪ B  = {1,2,3,4,5,6}
(ii) A ∪ C = {1,2,3,4,5,6,7,8}
(iii) B ∪ C = {3,4,5,6,7,8}
(iv) B ∪ D = {3,4,5,6,7,8,9,10}
(v) A ∪ B ∪ C  = {1,2,3,4,5,6,7,8}
(vi) A ∪ B ∪ D = {1,2,3,4,5,6,7,8,9,10}
(vii) B ∪ C ∪ D = {3,4,5,6,7,8,9,10}

Q5: Find the intersection of each pair of sets

(i) X = {1,3,5} Y = {1,2,3}

(ii) A = {a,e,i,o,u}, B = {a,b,c}

(iii) A = {x : x is a natural number and multiple of 3}
B = {x : x is a natural number less than 6}

(iv) A = {x : x is a natural number and 1 < x ≤ 6 }
B = {x : x is a natural number and 6 < x < 10 }

(v) A = {1, 2, 3}, B = φ

(i) X = {1,3,5} Y = {1,2,3}
X ∩ Y = {1,3}

(ii) A = {a,e,i,o,u}, B = {a,b,c}
A ∩ B = {a}

(iii)  A = {x : x is a natural number and multiple of 3} = {3,6,9,...}
B = {x : x is a natural number less than 6}  = {1,2,3,4,5,6}
A ∩ B = {3}

(iv) A = {x : x is a natural number and 1 < x ≤ 6 }  = {2,3,4,5,6}
B = {x : x is a natural number and 6 < x < 10 } = {7,8,9}
A ∩ B = {φ}

(v) A = {1, 2, 3}, B = φ
A ∩ B = {φ}