Last updated at Dec. 8, 2016 by Teachoo

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Ex 1.2,6 (Method 1) From the sets given below, select equal sets: A = {2, 4, 8, 12}, B = {1, 2, 3, 4}, C = {4, 8, 12, 14}, D= {3, 1, 4, 2}, E = {–1, 1}, F = {0, a}, G = {1, –1}, H = {0, 1} 12 is in set A, but on not in set B, D,E,F,G,H ⇒ A ≠ B , A ≠ D , A ≠ E , A ≠ F , A ≠ G , A ≠ H Also, 2 is in set A, but not in set C ∴ A ≠ C 3 is in set B, but not in set C, D, E, F, G, H ⇒ B ≠ C , B ≠ E , B ≠ F , B ≠ G , B ≠ H , We see that each element of set B & D are same, Hence, B = D A = {2, 4, 8, 12}, B = {1, 2, 3, 4}, C = {4, 8, 12, 14}, D= {3, 1, 4, 2}, E = {–1, 1}, F = {0, a}, G = {1, –1}, H = {0, 1} 12 is in set C, but not in set D , E, F, G, H ∴ C ≠ D, C ≠ E, C ≠ F, C ≠ G, C ≠ H 4 is in set D, but not in set E, F, G, H ∴ D ≠ E, D ≠ F, D ≠ G, D ≠ H Now, set E & set G have the same elements, so they are equal ∴ E = G A = {2, 4, 8, 12}, B = {1, 2, 3, 4}, C = {4, 8, 12, 14}, D= {3, 1, 4, 2}, E = {–1, 1}, F = {0, a}, G = {1, –1}, H = {0, 1} –1 is in set E, but not in set F, H ∴ E ≠ F, E ≠ H a is in set F, but not in set G, H ∴ F ≠ G, F ≠ H, –1 is in set H, but not in set H ∴ G ≠ H Hence, among the given sets, B = D and E = G. Ex 1.2,6 (Method 2) From the sets given below, select equal sets: A = {2, 4, 8, 12}, B = {1, 2, 3, 4}, C = {4, 8, 12, 14}, D= {3, 1, 4, 2}, E = {–1, 1}, F = {0, a}, G = {1, –1}, H = {0, 1} 12 ∈ A, but 12 ∉ B , 12 ∉ D , 12 ∉ E , 12 ∉ F , 12 ∉ G , 12 ∉ H , ⇒ A ≠ B , A ≠ D , A ≠ E , A ≠ F , A ≠ G , A ≠ H Also, 2 ∈ A, but 2 ∉ C ∴ A ≠ C 3 ∈ B, 3 ∉ C, 3 ∉ E, 3 ∉ F, 3 ∉ G, 3 ∉ H ⇒ B ≠ C , B ≠ E , B ≠ F , B ≠ G , B ≠ H , We see that each element of set B & D are same, hence, B = D A = {2, 4, 8, 12}, B = {1, 2, 3, 4}, C = {4, 8, 12, 14}, D= {3, 1, 4, 2}, E = {–1, 1}, F = {0, a}, G = {1, –1}, H = {0, 1} 12 ∈ C, but 12 ∉ D, 12 ∉ E, 12 ∉ F, 12 ∉ G, 12 ∉ H ∴ C ≠ D, C ≠ E, C ≠ F, C ≠ G, C ≠ H 4 ∈ D, but 4 ∉ E, 4 ∉ F, 4 ∉ G, 4 ∉ H ∴ D ≠ E, D ≠ F, D ≠ G, D ≠ H Similarly, E ≠ F, E ≠ H F ≠ G, F ≠ H, G ≠ H Also, E = G Hence, among the given sets, B = D and E = G.

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.