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Question:

Two sets A and B are as under: A = {(a,b) ∈ R × R: |a| < 1 and |b| < 1}; B = {(a,b) ∈ R × R: 4(a)² + 9(b)² ≤ 36}. Then.

A∩B = φ (an empty set)

Neither A⊆B nor B⊆A

B⊆A

A⊆B

Solution:

A = {(a,b) ∈ R × R: |a| < 1 and |b| < 1}
Now, |a| < 1 ⇒ a ∈ (-1,1) and |b| < 1 ⇒ b ∈ (-1,1)
∴ A = {(a,b): a ∈ (-1,1), b ∈ (-1,1)}
Now, we have B = {(a,b) ∈ R × R: 4(a)² + 9(b)² ≤ 36}
Considering conditions from set A, we can clearly see that they satisfy conditions in set B as maximum value of 4(a)² + 9(b)² = 4(1)² + 9(1)² = 4 + 9 = 13 ≤ 36
Apart from these values of a and b, we have other values satisfying the equation and many more such values.
Hence, A ⊆ B.