Tamil Nadu Board Samacheer Kalvi solutions for Class 9th Mathematics Answers Guide Chapter 1 Set Language Exercise 1.4 [Page 22]
Exercise 1.4 | Q 1. (i) | Page 22
If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find (P ∪ Q) ∪ R
Solution
P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}
P ∪ Q = {1, 2, 5, 7, 9} ∪ {2, 3, 5, 9, 11}
= {1, 2, 3, 5, 7, 9, 11}
(P ∪ Q) ∪ R = {1, 2, 3, 5, 7, 9, 11} ∪ {3, 4, 5, 7, 9}
= {1, 2, 3, 4, 5, 7, 9, 11}
Exercise 1.4 | Q 1. (ii) | Page 22
If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find (P ∩ Q) ∩ S
Solution
P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}
P ∩ Q = {1, 2, 5, 7, 9} ∩ {2, 3, 5, 9, 11}
= {2, 5, 9}
(P ∩ Q) ∩ S = {2, 5, 9} ∩ {2, 3, 4, 5, 8}
= {2, 5}
Exercise 1.4 | Q 1. (iii) | Page 22
If P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8} then find (Q ∩ S) ∩ R
Solution
P = {1, 2, 5, 7, 9}, Q = {2, 3, 5, 9, 11}, R = {3, 4, 5, 7, 9} and S = {2, 3, 4, 5, 8}
Q ∩ S = {2, 3, 5, 9, 11} ∩ {2, 3, 4, 5, 8}
= {2, 3, 5}
(Q ∩ S) ∩ R = {2, 3, 5} ∩ {3, 4, 5, 7, 9}
= {3, 5}
Exercise 1.4 | Q 2 | Page 22
Test for the commutative property of union and intersection of the sets
P = {x : x is a real number between 2 and 7} and
Q = {x : x is an irrational number between 2 and 7}
Solution
Commutative Property of union of sets
(A ∪ B) = (B ∪ A)
Here P = {3, 4, 5, 6}, Q = {√3, √5, √6}
P ∪ Q = {3, 4, 5, 6} ∪ {√3, √5, √6}
= {3, 4, 5, 6, √3, √5, √6} ...(1)
Q ∪ P = {√3, √5, √6} ∪ {3, 4, 5, 6}
= {√3, √5, √6, 3, 4, 5, 6} ...(2)
(1) = (2)
∴ P ∪ Q = Q ∪ P
∴ It is verified that union of sets is commutative.
Commutative Property of intersection of sets (P ∩ Q) = (Q ∩ P)
P ∩ Q = {3, 4, 5, 6} ∩ {√3, √5, √6} = { } ...(1)
Q ∩ P = {√3, √5, √6} ∩ {3, 4, 5, 6} = { } ...(2)
From (1) and (2)
P ∩ Q = Q ∩ P
∴ It is verified that intersection of sets is commutative.
Exercise 1.4 | Q 3 | Page 22
If A = {p, q, r, s}, B = {m, n, q, s, t} and C = {m, n, p, q, s}, then verify the associative property of union of sets
Solution
Associative Property of union of sets
A ∪ (B ∪ C) = (A ∪ B) ∪ C
(B ∪ C) = {m, n, q, s, t} ∪ {m, n, p, q, s}
= {m, n, p, q, s, t}
A ∪ (B ∪ C) = {p, q, r, s} ∪ {m, n, p, q, s, t}
= {m, n, p, q, r, s, t} ...(1)
(A ∪ B) = {p, q, r, s} ∪ {m, n, q, s, t}
= {m, n, p, q, r, s, t}
(A ∪ B) ∪ C = {m, n, p, q, r, s, t} ∪ {m, n, p, q, s}
= {m, n, p, q, r, s, t} ...(2)
From (1) and (2)
It is verified that A ∪ (B ∪ C) = (A ∪ B) ∪ C
Exercise 1.4 | Q 4 | Page 22
Verify the associative property of intersection of sets for A = {-11, √2, √5, 7}, B = {√3, √5, 6, 13} and C = {√2, √3, √5, 9}
Solution
Associative Property of intersection of sets A ∩ (B ∩ C) = (A ∩ B) ∩ C
B ∩ C = {√3, √5, 6, 13} ∩ {√2, √3, √5, 9} = {√5}
A ∩ (B ∩ C) = {-11, √2, √5, 7} ∩ {√5} = {√5} ...(1)
A ∩ B = {-11, √2, √5, 7} ∩ {√3, √5, 6, 13} = {√5}
(A ∩ B) ∩ C = {√5} ∩ {√2, √3, √5, 9} = {√5} ...(2)
From (1) and (2),
It is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C
Exercise 1.4 | Q 5 | Page 22
If A = {x : x = 2n, n ∈ W and n < 4}, B = {x : x = 2n, n ∈ N and n ≤ 4} and C = {0, 1, 2, 5, 6}, then verify the associative property of intersection of sets
Solution
A = {x : x = 2n, n ∈ W, n < 4}
⇒ x = 20 = 1
x = 21 = 2
x = 22 = 4
x = 23 = 8
∴ A = {1, 2, 4, 8}
B = {x : x = 2n, n ∈ N and n ≤ 4}
⇒ x = 2 × 1 = 2
x = 2 × 2 = 4
x = 2 × 3 = 6
x = 2 × 4 = 8
∴ B = {2, 4, 6, 8}
C = {0, 1, 2, 5, 6}
Associative property of intersection of sets
A ∩ (B ∩ C) = (A ∩ B) ∩ C
B ∩ C = {2, 6}
A ∩ (B ∩ C) = {1, 2, 4, 8} ∩ {2, 6}
= {2} ...(1)
A ∩ B = {1, 2, 4, 8} ∩ {2, 4, 6, 8}
= {2, 4, 8}
(A ∩ B) ∩ C = {2, 4, 8} ∩ {0, 1, 2, 5, 6}
= {2} ...(2)
From (1) and (2)
It is verified that A ∩ (B ∩ C) = (A ∩ B) ∩ C