2. When a < 0, the image of f(x) is a parabola with a downward opening.
① When the parabola is separated from the X axis, the problem holds. At this time, discriminant = 4+24a < 0, and A.
When a
(2) When the parabola intersects with the X-axis, in order for the problem to be established, it is necessary to satisfy both:
The discriminant formula ≧0, f (- 1) < 0, f (1) < 0.
From the discriminant ≧0, we get: a≧ 1/6, and A < 0, ∴- 1/6 ≦ A < 0.
From f (- 1) < 0, we get: 2a-2-3 < 0, ∴ 2a < 5, ∴ a < 5/2.
From f (1) < 0, we get: 2a+2-3 < 0, ∴ 2a < 1, ∴ a < 1/2.
Synthesize:-1/6 ≦ A < 0, A < 5/2, A < 1/2, and get:-1/6 ≦ A < 0.
To sum up ① ②, we get: a < 0.
-
3. When a > 0, the image of f(x) is a parabola with an upward opening. For the problem to be established, the following two conditions must be met:
Discriminant > 0, f (- 1) < 0, f (1) < 0.
From the discriminant of > 0, we get: a >- 1/6, and a > 0, ∴ a > 0.
From f (- 1) < 0, we get: 2a-2-3 < 0, ∴ 2a < 5, ∴ a < 5/2.
From f (1) < 0, we get: 2a+2-3 < 0, ∴ 2a < 1, ∴ a < 1/2.
Comprehensive: a > 0, a < 5/2, a < 1/2, 0 < a < 1/2.
-
To sum up, it is concluded that the value range of A that meets the conditions is (-∞, 0)∩(0, 1/2).