F.4 MATHS

1)
a)
x-intercept : Sub y = 0 into L; x= -6
y-intercept : Sub x = 0 into L; y = 3

b)
By equation of L:
x - 2y + 6 = 0
y = (1/2) x + 3
So, slope of L = (1/2)

Note that L perpendicular to PQ at Q

So, (Slope of PQ) x (Slope of L) = -1
So, Slope of PQ = -2

Equation of PQ:
(y+3) = (-2) (x+2)

Ans: 2x + y + 7 =0

c) Interception of L and PQ
Sub equation L into 2x + y + 7 = 0
Get answers: Q (-4, 1)

d) SQRT( (-2 +4)^2 + (-3 - 1)^2 )
SQRT (20)
Ans: 2 x SQRT(5)

2014-02-26 00:38:41 補充：
2. 如果 P (2, 1), 甘 PS y = 1/2 x 並唔合理

3. Delta of x^2+ax+12=0

(a)^2 - 4 (1) (12) >= 0
a >= 6.92
So, possible values of a >= 7, for a is integer

4.
Length of AB = 2 x 35 = 70 m
Let length of BC be x
Time taken for Ken to reach C = [35 + (x/2) - 10] seconds
cont...

2014-02-26 00:43:26 補充：
Length of AC = SQRT( AB^2 + BC^2)
[35 + (x/2) - 10] x 2 = SQRT ( 70^2 + x^2)
x = 24
Length of BC is 24 m, AB is 70 m

5. Let one root be Q, such that another root is 2Q
Sum of roots = 3Q = 2k-5 -----------(1)
Product of roots = 2Q^2 = k^2-5k+56/9 -----------(2)
(2) / (1):
Q = (2k-5)/3

2014-02-26 00:51:17 補充：
b) Sub k to find answer

6) 題目不合理

7)
a)
Let f(x) = (x^2008+3)
Sub f(1) => 4
By Reminder Theorem, remainder is 4
b) two days after Saturday, So it's Monday

又傻豬左啦～

20 分 問 七題，有人答你的機會很低。

相反，5 分問兩題，分四次問，大把人爭住答～
你在點數計算上亦無吃虧～

原因好簡單：
20 分問七題，如果人地無咁多時間，或者只係識答其中三題，咁佢答唔答你好呢？

2014-02-22 21:11:09 補充：
我係意見欄 KO 左 第 7 題先啦～

#7

(a) Let f(x) = x²⁰⁰⁸ + 3, the required remainder is f(-1) = 1 + 3 = 4

2014-02-22 21:11:31 補充：
(b) From part (a), you know that when 6²⁰⁰⁸ + 3 is divided by 7, the remainder is 4.

If today is Saturday, then after (6²⁰⁰⁸ + 3 - 4) days, it is still Saturday;

2014-02-22 21:11:50 補充：
but after (6²⁰⁰⁸ + 3) days, that would be 4 days after Saturday, that is Wednesday.

2014-02-22 22:24:14 補充：
你咁好禮貌，我繼續答你吧．．．

3. If the roots of x² + ax + 12 = 0 are integers, what are the possible values of a?

Let the two roots be α and β.

{ α + β = -a
{ αβ = 12

The two roots are integers, we list all the factors of 12:
1, 2, 3, 4, 6, 12, -1, -2, -3, -4, -6, -12

2014-02-22 22:25:38 補充：
The pair of (α, β) can only be the following (or reverse x and y)
(1, 12), (2, 6), (3, 4), (-1, -12), (-2, -6), (-3, -4)

Therefore α + β can only be
13, 8, 7, -13, -8, -7

a can only be
-13, -8, -7, 13, 8, 7

2014-02-22 22:27:36 補充：
5.
If one root of the equation x² - (2k-5)x + k² - 5k + 56/9 = 0, where k is a constant, is twice the other root.

(a) Let the two roots be α and 2α.

α + 2α = 3α = 2k-5
Therefore, one root is α = (2k-5)/3

2014-02-22 22:29:42 補充：
(b) From (a), one root is (2k-5)/3, therefore, if we put x = (2k-5)/3, the equation would be satisfied.　＜－－　重要 concept！！！

[(2k-5)/3]² - (2k-5)[(2k-5)/3] + k² - 5k + 56/9 = 0

(2k-5)²/9 - (2k-5)²/3 + k² - 5k + 56/9 = 0

-2(2k-5)²/9 + k² - 5k + 56/9 = 0

-2(2k-5)² + 9k² - 45k + 56 = 0

2014-02-22 22:30:37 補充：
-2(4k² - 20k + 25) + 9k² - 45k + 56 = 0

-8k² + 40k - 50 + 9k² - 45k + 56 = 0

k² - 5k + 6 = 0

(k - 2)(k - 3) = 0

k = 2 or k = 3