HMMT 二月 1998 · 代数 · 第 8 题

英文解析

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 withoutchanging 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
   ⇒ 0 > x − x − 12
   √ √
   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
   Case (2) is ≤ x ≤ − 2. For all x satisfying these conditions, x + 2 < 0 and 2 x + 9 > 0. Then the 9
   following operations will change the direction of the inequality:2
   x + 1 3 x + 42
   x + 2 2 x + 9>
   2 2
   ⇒ 2 x + 11 x + 9 < 3 x + 10 x + 8
   ⇒ 0 < x − x − 12
   √ √
   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 )
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