Learn Exercise 1.4 with Free Lessons & Tips

(i)X-Y={a,c}

Elements present in X but not in Y

(ii)Y-X={f,g}

Elements present in Y but not in X

(iii)X∩Y={b,d}

elements common to X  and Y

Q is the subset of R.

set of real numbers other than rational numbers are irrational numbers.

R-Q=irrational numbers

(i) False

As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}

⇒ {2, 3, 4, 5} ∩ {3, 6} = {3}

(ii) False

As a ∈ {a, e, i, o, u}, a ∈ {a, b, c, d}

⇒ {a, e, i, o, u } ∩ {a, b, c, d} = {a}

(iii) True

As {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ

(iv) True

As {2, 6, 10} ∩ {3, 7, 11} = 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 …}

As 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

A∪ B = {2, 3, 4, 5, 6, 7, 8, 9}

∴ A∪ B = {x: x ∈ N and 1 x

(v) A = {1, 2, 3}, B = Φ

A∪ B = {1, 2, 3}

Here, A = {a, b} and B = {a, b, c}

Yes, A ⊂ B.

A∪ B = {a, b, c} = B

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

1.{1,2,3,4,5,6}

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

3.{3,4,5,6,7,8}

4.{3,4,5,6,7,8,9,10}

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

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

7.{3,4,5,6,7,8,9,10}

(i) A ∩ B = {7, 9, 11}

(ii) B ∩ C = {11, 13}

(iii) A ∩ C ∩ D = { A ∩ C} ∩ D = {11} ∩ {15, 17} = Φ

(iv) A ∩ C = {11}

(v) B ∩ D = Φ

(vi) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

= {7, 9, 11} ∪ {11} = {7, 9, 11}

(vii) A ∩ D = Φ

(viii) A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)

= {7, 9, 11} ∪Φ = {7, 9, 11}

(ix) (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}

(x) (A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17) ∩ {7, 9, 11, 13, 15}

= {7, 9, 11, 15}

A = {x: x is a natural number} = {1, 2, 3, 4, 5 …}

B ={x: x is an even natural number} = {2, 4, 6, 8 …}

C = {x: x is an odd natural number} = {1, 3, 5, 7, 9 …}

D = {x: x is a prime number} = {2, 3, 5, 7 …}

(i) A ∩B = {x: x is a even natural number} = B

(ii) A ∩ C = {x: x is an odd natural number} = C

(iii) A ∩ D = {x: x is a prime number} = D

(iv) B ∩ C = Φ

(v) B ∩ D = {2}

(vi) C ∩ D = {x: x is odd prime number}

(i) {1, 2, 3, 4}

{x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}

Now, {1, 2, 3, 4} ∩ {4, 5, 6} = {4}

Therefore, this pair of sets is not disjoint.

(ii) {a, e, i, o, u} ∩ (c, d, e, f} = {e}

Therefore, {a, e, i, o, u} and (c, d, e, f} are not disjoint.

(iii) {x: x is an even integer} ∩ {x: x is an odd integer} = Φ

Therefore, this pair of sets is disjoint.