请你解答一下图片中的第七题---Question 7: (江西·高考真题) 当 a > 0, b > 0 时, 不等式 -b < 1/x < a 的解是 ( ) A. x < -1/b 或 x > 1/a B. -1/a < x < 1/b C. x < -1/a 或 x > 1/b D. -1/b < x < 0 或 0 < x < 1/a Handwritten Annotations for Question 7: ∵a>0, b>0 ∴ -b<0, a>0. -b < 1/x < a -b < 1/x 且 1/x < a ∵a>0, b>0 1/x > -b => 1/x + b > 0 => (1+bx)/x > 0 If x > 0, 1+bx > 0 => bx > -1 => x > -1/b. Combined with x > 0, gives x > 0. If x < 0, 1+bx < 0 => bx < -1 => x > -1/b. Combined with x < 0, gives -1/b < x < 0. So, -b < 1/x solution is x > 0 or -1/b < x < 0 which simplifies to x < -1/b or x > 0 (My previous analysis was incorrect for the negative case). Let's re-evaluate the handwritten annotations. Handwritten says: -b < 1/x => 1/x > -b If x > 0, 1 > -bx => bx > -1. Since b > 0, x > -1/b. Combined with x > 0, gives x > 0. (This seems wrong based on the subsequent steps in the annotation). Let's follow the annotation steps carefully. -b < 1/x: If x > 0: -bx < 1 => bx > -1 => x > -1/b. Since x > 0, this is x > 0. If x < 0: -bx > 1 => bx < -1. Since b > 0, x < -1/b. Combined with x < 0, this is x < -1/b. So, -b < 1/x solution is x < -1/b or x > 0. This matches the first part of the handwritten annotation conclusion below. 1/x < a: If x > 0: 1 < ax => x > 1/a. Since a > 0, 1/a > 0. Combined with x > 0, this is x > 1/a. If x < 0: 1 > ax => x < 1/a. Since x < 0, this is x < 0. So, 1/x < a solution is x > 1/a or x < 0. This matches the second part of the handwritten annotation conclusion below. Handwritten annotation conclusion steps: ∴ -b < 1/x => x < -1/b 或 x > 0 ∴ 1/x < a => x > 1/a 或 x < 0 To solve -b < 1/x < a, we need the intersection of these two solution sets. Intersection of (x < -1/b or x > 0) and (x > 1/a or x < 0). Let S1 = (-∞, -1/b) U (0, +∞) Let S2 = (-∞, 0) U (1/a, +∞) S1 ∩ S2 = [(-∞, -1/b) U (0, +∞)] ∩ [(-∞, 0) U (1/a, +∞)] = [(-∞, -1/b) ∩ (-∞, 0)] U [(-∞, -1/b) ∩ (1/a, +∞)] U [(0, +∞) ∩ (-∞, 0)] U [(0, +∞) ∩ (1/a, +∞)] Since -1/b < 0 and 1/a > 0: = (-∞, -1/b) U ∅ U ∅ U (1/a, +∞) = (-∞, -1/b) U (1/a, +∞) This corresponds to x < -1/b or x > 1/a. This matches Option A. The handwritten annotation's final check mark is on A. Question 8: (2023·全国·高考真题) 已知集合 M = {-2, -1, 0, 1, 2}, 集合 N = {x|x^2 - x - 6 >= 0}, 则 M ∩ N = ( ) A. {-2, -1, 0, 1} B. {0, 1, 2} C. {-2} D. {2} Handwritten Annotations for Question 8: x^2 - x - 6 >= 0 (x-3)(x+2) >= 0 x <= -2 或 x >= 3 N = (-∞, -2] U [3, +∞) M = {-2, -1, 0, 1, 2} M ∩ N = {-2} (Implicitly derived from M and N)