As shown in the figure, in the rectangular coordinate system, point a is a point on the image of inverse scale function Y1 = KX, the positive half axis of ab ⊥ X axis is at point B, and C is the midpoint of OB; the image of primary function y2 = ax + B passes through two points a and C, and the Y axis is at point d (0, - 2), if s △ AOD = 4 (1) (2) observe the image, please point out the value range of X on the right side of y-axis when Y1 > Y2

As shown in the figure, in the rectangular coordinate system, point a is a point on the image of inverse scale function Y1 = KX, the positive half axis of ab ⊥ X axis is at point B, and C is the midpoint of OB; the image of primary function y2 = ax + B passes through two points a and C, and the Y axis is at point d (0, - 2), if s △ AOD = 4 (1) (2) observe the image, please point out the value range of X on the right side of y-axis when Y1 > Y2

(1) Make AE ⊥ Y axis at e, ∵ s △ AOD = 4, OD = 2 ≌ 12od · AE = 4 ≌ AE = 4 (1 point) ≁ ab ⊥ ob, C is the midpoint of ob, ≁ doc = ∠ ABC = 90 °, OC = BC, ≌ OCD = ∠ BCA ≌ RT ≌ RT ≌ ABC ≌ AB = od = 2 ≁ a (4,2) (2 points) substitute a (4,2) into Y1 = KX, get k = 8, the analytic formula of inverse ratio function is: Y1 = 8x, (3 points) substitute a (4,2) and D (0, - 2) into y2 = a X + B, we get 4A + B = 2B = − 2. The solution is a = 1b = − 2. The analytic formula of the first-order function is: y2 = X-2; (4 points) (2) on the right side of the y-axis, when Y1 > Y2, 0 < x < 4. (6 points)