n20311021

其實上年已經整左2份Paper 1 Mock
之前o係資源板都post過
有一份反應都唔錯

今年再整多1份Paper 1 Mock

加埋3份mock集中o係呢個貼

做左CE & DSE都覺得非常淺既話,

不妨試下新口味

Set 1:

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Set 1 Solution: (13/4 amended)

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Set 2:

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Set 2 Solution: (11/4 amended)

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Set 3: (最新Mock卷!) (11/4 amended)

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Set 3 Solution:(11/4 amended)

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覺得題目/ Solution有錯請PM我,
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[ 本帖最後由 n20311021 於 2013-7-4 04:09 PM 編輯 ]

leon592

Very well-set papers!!!
Support

tarya

SUPPORT!!!
IT IS USEFUL!!
THANK YOU!

calvinleeabc

support!!!最後衝刺!

st00141062002

正

yomandy

thanksssss

無從一見

thxxxxxxxxxxxxxxxxxxx

lwt_christine

thzzzz

heartbreakingjk

thank you so muich ;D you're awesome;)

calvinleeabc

set 1 15b 我計到y=3/4x-1/4？

n20311021

你o岩
我個marking 16c^2 - 192c + 49 = 0 係錯既
應該係 16c^2 - 192c - 49 = 0
(4c + 1)(4c - 49) = 0
c = -1/4 or 49/4

faifai219

請問17b既transformation個答案係點解?
(x+2)/3 係咩黎架?

n20311021

Let f(x) = 2x^2 – 8x + 9
Let g(x) = the quadratic equation with roots 3α – 2 and 3β – 2.
g(x) = f((x+2)/3) <--- WHY???

f(x) is transformed to g(x) by 2 steps.
(1) translate 2/3 unit to the left
h(x) = f(x+2/3)
roots of h(x): α – 2/3 and β – 2/3

(2) enlarge along the x-axis by 3 times
g(x)
= h(x/3)
= f(x/3 + 2/3)
= f((x+2)/3)
roots of g(x): 3(α – 2/3) and 3(β – 2/3), i.e. 3α – 2 and 3β – 2.

Think backward.
The two roots give you hints how to transform the graph.

[ 本帖最後由 n20311021 於 2013-4-6 11:48 PM 編輯 ]

thomas3253

我有興趣想知第三份既第16題點做
我無從入手...

n20311021

16(a)
at least 1 study group consists of students from 3 different classes
= 1 study group consists of students from 3 different classes & no restriction on other 4 groups

So 分子 is (5C3)(3C1)^3 (12C3)(9C3)(6C3)(3C3)
分母 is (15C3)(12C3)(9C3)(6C3)(3C3)

16(b)
very simple.
just 1 - answer of (a)
because
all study groups consist of students from at most 2 different classes
= no study group consists of students from 3 different classes