数学OG18 391求解

ZNan · 发表于 2018-10-3 05:21:42

If q, s, and t are all different numbers, is q < s < t ?

t − q = |t − s| + |s − q|
t > q

答案是A

解释是：

It is given that t − q = |t − s| + |s − q|, which can be rewritten without absolute values in four mutually exclusive and collectively exhaustive cases by making use of the algebraic definition of absolute value. Recall that |x| = x if x > 0, and |x| = –x if x < 0. Thus, for example, if t − s < 0, then |t − s| = –(t − s).
Case 1: t > s and s > q. In this case, t − s > 0 and s − q > 0, and so t − q = |t − s| + |s − q| is equivalent to t − q = (t − s) + (s − q), which is an identity. Therefore, the case for which t > sand s > q is consistent with the given information and the assumption t − q = |t − s| + |s − q|.
Case 2: t > s and s < q. In this case, t − s > 0 and s − q < 0, and so t − q = |t − s| + |s − q| is equivalent to t − q = (t − s) − (s − q), or s = q, which is not consistent with the assumption that q, s, and t are all different numbers. Therefore, the case for which t > s and s < q is not consistent with the given information and the assumption t − q = |t − s| + |s − q|.
Case 3: t < s and s > q. In this case, t − s < 0 and s − q > 0, and so t − q = |t − s| + |s − q| is equivalent to t − q = –(t − s) + (s − q), or t = s, which is not consistent with the assumption that q, s, and t are all different numbers. Therefore, the case for which t < s and s > q is not consistent with the given information and the assumption t − q = |t − s| + |s − q|.
Case 4: t < s and s < q. In this case, t − s < 0 and s − q < 0, and so t − q = |t − s| + |s − q| is equivalent to t − q = –(t − s) − (s − q), or t = q, which is not consistent with the assumption that q, s, and t are all different numbers. Therefore, the case for which t < s and s < q is not consistent with the given information and the assumption t − q = |t − s| + |s − q|.

The only case that is consistent with the given information and the assumption t − q = |t − s| + |s − q| is Case 1. Therefore, it follows that t > s and s > q, and this implies q < s < t; SUFFICIENT.

DS题不是只能唯一确认是正确答案吗？为什么官方说明却是可以多种情况分析？OG的答案有些时候只能唯一证明，有些时候又可以多情况分析。。。

SKYWORKS · 发表于 2018-10-3 06:08:35

I think you can ignore the explanation from OG, it makes a little more complicated. The only way to make t-q = abs (t-s)+ abs (s-q) is when t-s>0, and s-q>0. then you can get t>s, s>q. hence t>s>q, sufficient.

kluivert · 发表于 2018-10-3 07:48:55

The only way to make t-q = abs (t-s)+ abs (s-q) is when t-s>0, and s-q>0.

please elaborate on this part ....honestly,i got no clue

kluivert · 发表于 2018-10-3 07:53:10

OG的解释一点问题没有，case1 已经成立了

OG只是一个一个的排除CASE 2 ,3 ,4 不会造成 t − q = |t − s| + |s − q|
如果2，3，4中有一个造成t − q = |t − s| + |s − q|

那么 q < s < t 就不成立了

OG的方法挺费时的，希望高手出来解答...下面的高手是说
The only way to make t-q = abs (t-s)+ abs (s-q) is when t-s>0, and s-q>0.但没说为啥..俺数学不好，不会推

郑布拉希莫维奇 · 发表于 2018-10-3 08:07:50

|t − s| = t - s 或 s - t
|s − q| = s - q 或 q - s
想要得到|t − s| + |s − q| = t - q，只有|t − s| = t - s，|s − q| = s - q

kluivert · 发表于 2018-10-3 08:54:39

郑布拉希莫维奇发表于 2018-10-3 08:07
|t − s| = t - s 或 s - t
|s − q| = s - q 或 q - s
想要得到|t − s| + |s − q| = t - q，只有|t − s ...

为什么呢？
我自己做了下，感觉必须要按OG的去推才行
是我太笨了

郑布拉希莫维奇 · 发表于 2018-10-3 09:39:02

kluivert 发表于 2018-10-3 08:54
为什么呢？
我自己做了下，感觉必须要按OG的去推才行
是我太笨了

如果想要|t − s| + |s − q| = t - q，就得把s约掉，不是等于t-s+s-q, 就是等于s-t-s+q, 第一个是t-q, 第二个是q-t.
只能是t-q呗

irissunflower · 发表于 2018-10-3 10:15:33

不用草稿纸的算法：
条件一：首先肯定t>p嘛，当s大于这两个数的时候，s=t，排除；当s小于这两个数，s=p，排除；只有当s在这两个数中间的时候，等式成立。

吳縣大 · 发表于 2018-10-3 10:47:42

kluivert 发表于 2018-10-3 08:54
为什么呢？
我自己做了下，感觉必须要按OG的去推才行
是我太笨了

emotional or self-blame is no help

我做這題的想法是這樣的

首先，絕對值就是距離，所以一定是非負的

所以 t-q = |t-s| + |s-q| 必然是正的，意思就是 t > q ，在數線上 t 在 q 的右邊，t-q 就是 t 到 q 的距離

A地到B地的距離 = A地到C地的距離加上 C地到B地的距離 --> 表示C地正好位在A地到B地連線中的某處

所以 t-q = |t-s| + |s-q| 意思就是說 s 這點在 t 跟 q 的中間，綜合起來就是 q < s < t 了

kluivert · 发表于 2018-10-3 11:23:06

吳縣大发表于 2018-10-3 10:47
emotional or self-blame is no help

我做這題的想法是這樣的

牛一划线结果就出来了厉害谢谢

[ESR] 数学OG18 391求解

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