HMMT 二月 1998 · 代数 · 第 8 题

8. Problem: Find the set of solutions for x in the inequality > when x 6 = − 2 , x 6 = . x +2 2 x +9 2 9 9 Solution: There are 3 possible cases of x : 1) − < x , 2) ≤ x ≤ − 2, 3) − 2 < x . For the cases (1) and 2 2 (3), x + 2 and 2 x + 9 are both positive or negative, so the following operation can be carried out without changing the inequality sign: x + 1 3 x + 4

x + 2 2 x + 9 2 2 ⇒ 2 x + 11 x + 9 > 3 x + 10 x + 8 2 ⇒ 0 > x − x − 1 √ √ 1 − 5 1+ 5 9 The inequality holds for all < x < . The initial conditions were − < x or − 2 < x . The 2 2 2 √ √ 1 − 5 1+ 5 intersection of these three conditions occurs when < x < . 2 2 9 Case (2) is ≤ x ≤ − 2. For all x satisfying these conditions, x + 2 < 0 and 2 x + 9 > 0. Then the 2 following operations will change the direction of the inequality: 2 x + 1 3 x + 4

x + 2 2 x + 9 2 2 ⇒ 2 x + 11 x + 9 < 3 x + 10 x + 8 2 ⇒ 0 < x − x − 1 √ √ 1 − 5 1+ 5 − 9 The inequality holds for all x < and < x . The initial condition was ≤ x ≤ − 2. Hence 2 2 2 − 9 the intersection of these conditions yields all x such that ≤ x ≤ − 2. Then all possible cases of x are 2 √ √ 1 − 5 1+ 5 − 9 ≤ x ≤ − 2 ∪ < x < . 2 2 2 2 f ( x ) 1 2