Step 1: Analyze set A

The inequalities defining set A are:

|α - 1| ≤ 4 => -4 ≤ α - 1 ≤ 4 => -3 ≤ α ≤ 5

|β - 5| ≤ 6 => -6 ≤ β - 5 ≤ 6 => -1 ≤ β ≤ 11

So, A represents a rectangle in the αβ-plane with vertices (-3, -1), (5, -1), (5, 11), and (-3, 11).

Step 2: Analyze set B

The inequality defining set B is:

16(α - 2)² + 9(β - 6)² ≤ 144

Divide by 144:

(α - 2)² / 9 + (β - 6)² / 16 ≤ 1

This represents an ellipse centered at (2, 6) with semi-major axis a = 4 (along the β-axis) and semi-minor axis b = 3 (along the α-axis).

Step 3: Compare the sets A and B

The rectangle A has α ranging from -3 to 5 and β ranging from -1 to 11.

The ellipse B is centered at (2, 6). The α values range from 2 - 3 = -1 to 2 + 3 = 5. The β values range from 6 - 4 = 2 to 6 + 4 = 10.

Since -1 ≥ -3 and 5 ≤ 5 for α, and 2 ≥ -1 and 10 ≤ 11 for β, the ellipse B is contained within the rectangle A. Therefore, B ⊂ A.

Step 4: Verify the options

(A) A ⊂ B is false because we found B ⊂ A.

(B) B ⊂ A is true.

(C) neither A ⊂ B nor B ⊂ A is false.

(D) A ∪ B = {(x, y) : -4 ≤ x ≤ 4, -1 ≤ y ≤ 11} is false. The correct union would be the rectangle A, which is {(α, β) : -3 ≤ α ≤ 5, -1 ≤ β ≤ 11}.