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}
We know,
8 ∈ A, 8 ∉ B, 8 ∉ D, 8 ∉ E, 8 ∉ F, 8 ∉ G, 8 ∉ H
A ≠ B, A ≠ D, A ≠ E, A ≠ F, A ≠ G, A ≠ H
It can be re-written as
2 ∈ A, 2 ∉ C
So, A ≠ C
3 ∈ B, 3 ∉ C, 3 ∉ E, 3 ∉ F, 3 ∉ G, 3 ∉ H
B ≠ C, B ≠ E, B ≠ F, B ≠ G, B ≠ H
It can be re-written as
12 ∈ C, 12 ∉ D, 12 ∉ E, 12 ∉ F, 12 ∉ G, 12 ∉ H
So, C ≠ D, C ≠ E, C ≠ F, C ≠ G, C ≠ H
4 ∈ D, 4 ∉ E, 4 ∉ F, 4 ∉ G, 4 ∉ H
Therefore, D ≠ E, D ≠ F, D ≠ G, D ≠ H
Here, E ≠ F, E ≠ G, E ≠ H
F ≠ G, F ≠ H, G ≠ H
Order in which the elements of a set are listed is not significant.
B = D and E = G
Therefore, among the given sets, B = D and E = G.