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Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 1 Set Language Ex 1.5
Question 1.
Using the adjacent Venn diagram, find the following sets:
(i) A – B
(ii) B – C
(iii) A’∪B’
(iv) A’∩B’
(v) (B∪C)’
(vi) A – (B∪C)
(vii) A – (B∩C)
Solution:
From the diagram we get
U = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8},
A= {-2,-1, 3, 4, 6}, B = {-2,-1, 5, 7, 8}
C = {-3, -2, 0, 3, 8}
A’ = U – A = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8} – {-2, -1, 3, 4, 6}
= {-3, 0, 1, 2, 5, 7, 8}
B’ = U – B = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8} – {-2, -1, 5, 7, 8}
= {-3, 0, 1, 2, 3, 4, 6}
B∪C = {-2, -1, 5, 7, 8} ∪ {-3, -2, 0, 3, 8} = {-3, -2, -1, 0, 3, 5, 7, 8}
B∩C = {-2, -1, 5, 7, 8} ∩ {-3, -2, 0, 3, 8} = {-2, 8}
(i) A – B = {3, 4, 6}
(ii) B – C = {-1, 5, 7}
(iii) A’∪B’= {-3, 0, 1, 2, 5, 7, 8} ∪ {-3, 0, 1, 2, 3, 4, 6}
= {-3, 0, 1, 2, 3, 4, 5, 6, 7, 8}
(iv) A’∩B’ = {-3, 0, 1, 2, 5, 7, 8} ∩ {-3, 0, 1, 2, 3, 4, 6}
= {-3, 0, 1, 2}
(v) (B∪C)’ = U – (B∪C)= {-3,-2,-1,0, 1,2, 3,4, 5, 6, 7, 8} – {-3, -2, -1, 0, 3, 5, 7, 8}
= {1, 2, 4, 6}
(vi) A – (B∪C) = {-2, -1, 3, 4, 6} – {-3, -2, -1, 0, 3, 5, 7, 8} = {4, 6}
(vii) A – (B∩C) = {-2,-1, 3, 4, 6} – {-2, 8} = {-1, 3, 4, 6}
Question 2.
If K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h} then find the following:
(i) K∪(L∩M)
(ii) K∩(L∪M)
(iii) (K∪L) ∩ (K∪M)
(iv) (K∩L) ∪ (K∩M)
and verify distributive laws.
Solution:
K = {a, b, d, e, f}, L = {b, c, d, g} and M = {a, b, c, d, h}
(i) K∪(L∩M)
(L∩M) = {b, c, d, g} ∩ [a, b, c, d, h}
= {b, c, d}
K∪(L∩M) = {a, b, d, e, f} ∪ {b, c, d}
= {a, b, c, d, e, f}
(ii) K∩(L∪M)
(L∪M) = {b, c, d, g} ∪ {a, b, c, d, h}
= {a, b, c, d, g, h}
K∩(L∪M) = {a, b, d, e, f} ∩ {a, b, c, d, g, h}
= {a, b, d }
(iii) (K∪L) ∩ (K∪M)
(K∪L) = {a, b, d, e, f} ∪ {b, c, d, g}
= {a, b, c, d, e, f, g}
(K∪M) = {a, b, d, e, f} ∪ {a, b, c, d, h}
= {a, b, c, d, e, f, h}
(K∪L) ∩ (K∪M) = {a, b, c, d, e, f, g} ∩ {a, b, c, d, e, f, h}
= {a, b, c, d, e, f}
(iv) (K∩L) ∪ (K∩M)
(K∩L) = {a, b, d, e, f) ∩ {b, c, d, g}
= {b, d}
(K∩M) = {a, b, d, e, f} ∩ {a, b, c, d, h}
= {a, b, d}
(K∩L) ∪ (K∩M) = {b, d} ∪ [a, b, d}
= {a, b, d}
From (ii) & (iv) we get, K∩(L∪M) = (K∩L) ∪ (K∩M)
From (i) & (iii) we get, K∪(L∩M) = (K∪L) ∩ (K∪M)
Question 3.
For A = {x : x ∈ Z, -2 < x ≤ 4}, B = {x : x ∈ W, x ≤ 5}, C = {-4, -1, 0, 2, 3, 4}
verify A∪(B∩C) = (A∪B) ∩ (A∪C).
Solution:
A = {-1, 0, 1, 2, 3, 4}, B = {0, 1, 2, 3, 4, 5} and C = {-4, -1, 0, 2, 3, 4}
B∩C = {0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 2, 3, 4}
= {0, 2, 3, 4}
A∪(B∩C) = {-1, 0, 1, 2, 3, 4} ∪ {0, 2, 3, 4}
= {-1, 0, 1, 2, 3, 4} ……..(1)
A∪B = {-1, 0, 1, 2, 3, 4} ∪ {0, 1, 2, 3, 4, 5}
= {-1, 0, 1, 2, 3, 4, 5}
A∪C = {-1, 0, 1, 2, 3, 4} ∪ {-4, -1, 0, 2, 3, 4}
= {-4, -1, 0, 1, 2, 3, 4}
(A∪B) ∩ (A∪C) = {-1, 0, 1, 2, 3, 4, 5} ∩ {-4, -1, 0, 1, 2, 3, 4}
= {-1, 0, 1, 2, 3, 4} ……..(2)
From (1) and (2) we get A∪(B∩C) = (A∪B) ∩ (A∪C).
Question 4.
Verify A∪(B∩C) = (A∪B) ∩ (A∪C) using Venn diagrams.
Solution:
From (ii) and (v) we get A∪(B∩C) = (A∪B) ∩ (A∪C).
Question 5.
If A = {b, c, e, g, h}, B = {a, c, d, g, f}, and C = {a, d, e, g, h}, then show that A – (B∩C) = (A – B) ∪ (A – C).
Solution:
A = {b, c, e, g, h} ; B = {a, c, d, g, f}; C = {a, d, e, g, h}
B∩C = {a, c, d, g, i} ∩ {a, d, e, g, h}
= {a, d, g}
A – (B∩C) = {b, c, e, g, h} – {a, d, g}
= {b, c, e, h}…….(1)
A – B = {b, c, e, g, h} – {a, c, d, g, i}
= {b, e, h}
A – C = {b, c, e, g, h} – {a, d, e, g, h}
= {b, c}
(A – B) ∪ (A – C) = {b, e, h} ∪ {b, c}
= {b, c, e, h)……..(2)
From (1) and (2) we get A – (B∩C) = (A – B) ∪ (A – C)
Question 6.
If A= {x : x = 6n, n∈W and n < 6}, B = {x : x = 2n, n∈N and 2 < n ≤ 9} and
C = {x : x = 3n, n∈N and 4 ≤ n < 10}, then show that A – (B∩C) = (A – B) ∪ (A – C)
Solution:
A = {0, 6, 12, 18, 24, 30}; B = {6, 8, 10, 12, 14, 16, 18}; C = {12, 15, 18, 21, 24, 27}
B∩C = {6, 8, 10, 12, 14, 16, 18} ∩ {12, 15, 18, 21, 24, 27}
= {12, 18}
A – (B∩C) = {0, 6, 12, 18, 24, 30} – {12, 18}
= {0, 6, 24, 30}………(1)
A – B = {0, 6, 12, 18, 24, 30} – {6, 8, 10, 12, 14, 16, 18}
= {0, 24, 30}
A – C = {0, 6, 12, 18, 24, 30} – {12, 15, 18, 21, 24, 27}
= {0, 6, 30}
(A – B) ∪ (A – C) = {0, 24, 30} ∪ {0, 6, 30}
= {0, 6, 24, 30}……..(2)
From (1) and (2) we get A – (B∩C) = (A – B) ∪ (A – C).
Question 7.
If A = {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6} and C = {-1, 2, 5, 6, 7}, then show that
A – (B∪C) = (A – B) ∩ (A – C).
Solution:
A= {-2, 0, 1, 3, 5}, B = {-1, 0, 2, 5, 6}, C = {-1, 2, 5, 6, 7}
B∪C = {-1, 0, 2, 5, 6} ∪ {-1, 2, 5, 6, 7}
= {-1, 0, 2, 5, 6, 7}
A – (B∪C) = {-2, 0, 1, 3, 5} – {-1, 0, 2, 5, 6, 7}
= {-2, 1, 3} ………(1)
A – B = {-2, 0, 1, 3, 5} – {-1, 0, 2, 5, 6}
= {-2, 1, 3}
A – C = {-2, 0, 1, 3, 5}- {-1, 2, 5, 6, 7}
= {-2, 0, 1, 3}
(A- B) ∩ (A- C) = {-2, 1, 3} ∩ {-2, 0, 1, 3}
= {-2, 1, 3} ….(2)
From (1) and (2) we get A – (B∪C) = (A – B) ∩ (A – C).
Question 8.
IF A = {y : y = \(\frac{a + 1}{2}\), a ∈ W and a ≤ 5}, B = {y : y = \(\frac{2n – 1}{2}\), n ∈ W and n < 5} and C = {-1, \(-\frac{1}{2}\), 1, \(\frac{3}{2}\), 2} then show that A – (B∪C) = (A – B) ∩ (A – C).
Solution:
From (1) and (2) we get A – (B∪C) = (A – B) ∩ (A – C).
Question 9.
Verify A- (B∩C) = (A – B) ∪ (A – C) using Venn diagrams.
Solution:
From (ii) and (v) we get A- (B∩C) = (A – B) ∪ (A – C).
Question 10.
If U = {4, 7, 8, 10, 11, 12, 15, 16} , A = {7, 8, 11, 12} and B = {4, 8, 12, 15}, then verify De Morgan’s Laws for complementation.
U= {4, 7, 8, 10, 11, 12, 15, 16} , A = {7, 8, 11, 12} and B = {4, 8, 12, 15}
(i) (A∪B)’ = A’∩B’
(ii) (A∩B)’ = A’∪B’
Solution:
(i) A∪B = {7, 8, 11, 12} ∪ {4, 8, 12, 15}
= {4, 7, 8, 11, 12, 15}
(A∪B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 7, 8, 11, 12, 15}
= {10,16} ………(1)
A’ = {4, 7, 8, 10, 11, 12, 15, 16} – {7, 8, 11, 12}
= {4, 10, 15, 16}
B’ = {4, 7, 8, 10, 11, 12, 15, 16} – {4, 8, 12, 15}
= {7, 10, 11, 16}
A’∩B’ = {4, 10, 15, 16} ∩ {7, 10, 11, 16}
= {10,16} ………(2)
From (1) and (2) we get (A∪B)’ = A’∩B’
(ii) A∩B = {7, 8, 11, 12} ∩ {4, 8, 12, 15}
= {8, 12}
(A∩B)’ = {4, 7, 8, 10, 11, 12, 15, 16} – {8, 12}
= {4, 7, 10, 11, 15, 16} ………(1)
A’ = {4, 10, 15, 16}
B’ = {7, 10, 11, 16}
A’∪B’ = {4, 10, 15, 16} ∪ {7, 10, 11, 16}
= {4, 7, 10, 11, 15, 16} ………(2)
From (1) and (2) we get (A∩B)’ = A’∪B’
Question 11.
Verify (A∩B)’ = A∪B’ using Venn diagrams.
Solution:
From (ii) and (i) we get (A∩B)’ = A’∪B’