[DSE 2021] 大家準備好瘋狂做數學未呢？ [5]

Sigh... 2020-10-27 01:06:45 係, 之後h會keep decreasing, 但用呢個樣就即使h=0都唔會undefined

追捕林鄭月娥老母 2020-10-27 07:12:20 此回覆已被刪除

叫我2047 2020-10-27 08:22:10 可以咁做，因為S^2係continuous 所以你揾左佢等於零果點，即係果點之外的一段(或一塊)必然全大於零或全小於零，咁你take 左一個point証左就OK啦。(純concept，唔駛寫)

好喇，問題係h set 左必然係decreasing，因此你個h domain 由 1.2 r 到 0 先至成立，最好加埋呢句。(HKEAA個MS都有寫)

三九四零五二 2020-10-29 11:47:05 此回覆已被刪除

Sigh... 2020-11-4 22:55:40

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會考都無C 2020-11-4 22:58:16

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尚恩曼德斯 2020-11-4 23:01:30

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Sigh... 2020-11-4 23:09:06

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尚恩曼德斯 2020-11-4 23:30:51

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高遠搖搖板 2020-11-4 23:31:02

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Sigh... 2020-11-4 23:51:14 即係一定要拆咗fractions先至sub u=x-1?
如果我未拆就sub會出:
∫(u^(2)-1)/u^(2)du = ∫1-(1/u^(2)) = u + 1/u + C = (x-1) + 1/(x-1) + C
但正確答案係 x + 1/(x-1) + C

Sigh... 2020-11-4 23:52:31

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即係一定要拆咗fractions先至sub u=x-1?
如果我未拆就sub會出:
∫(u^(2)-1)/u^(2)du = ∫1-(1/u^(2))du = u + 1/u + C = (x-1) + 1/(x-1) + C
但正確答案係 x + 1/(x-1) + C

Sigh... 2020-11-4 23:57:39

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著拖鞋紅衫短袖 2020-11-5 00:01:58 其實你計到 (x+1)+1/(x+1)+C 咪啱囉

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你想似個答案咪寫 x+1/(x+1)+C’

Sigh... 2020-11-5 00:08:30 唔係呀巴打

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Correct Ans: x + [1/(x-1)]+ C
My Ans: (x-1) + [1/(x-1)] + C

著拖鞋紅衫短袖 2020-11-5 00:08:44 1/(x-1)* sorry

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總之你其實計啱，啲constant可以加加減減

Sigh... 2020-11-5 00:11:31 哦!!! 即係將個-1擺落個C度, 出C’ = C-1 咁樣？

著拖鞋紅衫短袖 2020-11-5 00:11:38 Sorry, 我冇追post睇清楚
但你真係計啱呢

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(x-1) + [1/(x-1)] + C
=x + [1/(x-1)]+ （C-1）
=x + [1/(x-1)]+ C‘ （for another arbitrary constant C’=C-1)

著拖鞋紅衫短袖 2020-11-5 00:12:05 Yes

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你明白了

Sigh... 2020-11-5 00:12:58 明白晒, 唔該晒巴打, 我自己concept唔清唔好意思

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著拖鞋紅衫短袖 2020-11-5 00:14:29

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Sigh... 2020-11-5 00:16:46

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著拖鞋紅衫短袖 2020-11-5 00:21:57 學校教緊3D trigo

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我覺得難在佢變化好大

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有咩方法可以master 3D trigo? 同依家啲section b 題目係咪都係3D trigo?

張五常(逃犯) 2020-11-5 00:30:13 做pp 就會發現萬變不離其宗
主要分三類
1. 三面體/四面體 (極少information都要搵到volume, 要記佢哋特別嘅properties)
2. 摺紙擺喺枱度 (extend啲線搵required angle between 2 planes)
3. 太陽&影子 (溝埋bearing出, DSE暫時未出過)

會考都無C 2020-11-5 01:13:33 一樣啫，個減1會被個constant C食咗㗎嘛