F4 maths

1.
a)
A' = (-2, 2(1)-4) = (-2, -2)
B' = (4, 2(1)-(-3)) = (4, 5)

b) (i)
Equation of AB :
(y - 4)/(x + 2) = (4 + 3)/(-2 - 4)
(y - 4)/(x + 2) = -7/6
6y - 24 = -7x - 14
7x + 6y - 10 = 0

b) (ii)
Equation of A'B' :
(y - 5)/(x - 4) = (5 + 2)/(4 + 2)
(y - 5)/(x - 4) = 7/6
7x - 28 = 6y - 30
7x - 6y + 2 = 0

c)
AB : 7x + 6y - 10 = 0 ...... [1]
A'B' : 7x - 6y + 2 = 0 ...... [2]

[1] + [2] :
14x - 8 = 0
x = 4/7

[1] - [2] :
12y - 12 = 0
y = 1

Hence, the point of intersection of AB and A'B' = (4/7, 1)


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2.
a)
When f(x) is divided by x - 1, the quotient is 6x² + ax - 19 and the remainderis f(1) = -4
f(x) ÷ (x - 1) = 6x² + ax - 19 ...... remainder = -4
Hence, f(x) = (x - 1)(6x² + ax - 19) - 4

Since x - 3 is a factor of f(x), then f(3) = 0
[(3) - 1] [6(3)² + a(3) - 19)] - 4 = 0
2(54 + 3a - 19) - 4 = 0
3a + 33 = 0
a = -11

f(x) = (x - 1)(6x² - 11x - 19) - 4
f(x) = 6x³ - 11x² - 19x - 6x² + 11x + 19 - 4
f(x) = 6x³ - 17x² -8x + 15

b)
f(x - 1) = 0
6(x - 1)³ - 17(x - 1)² - 8(x - 1) + 15 = 0
6x³ - 18x² + 18x - 6 - 17x² + 34x - 17 - 8x + 8 + 15 = 0
6x³ - 35x² + 44x = 0
x(6x² - 35x + 44) = 0
x(6x - 11)(x - 4) = 0
x = 0 orx = 11/6 orx = 4


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3.
a)
y = -3(x - p)² - 2p + 1
y = (1 - 2p) - 3(x - p)²
Since -3(x - p)² ≤ 0 for all values of x,
Hence, the maximum of y :
1 - 2p = -9
2p - 1 = 9
2p = 10
p = 5

b)
y = -3(x - 5)² - 2(5) + 1
When x = 5, the maximum value of y = -9
Hence, the coordinates of the vertex = (5,-9)


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4.
a)
Put x = 0 into the equation of L1 :
5(0) - 2y - 8 = 0
y = -4
Hence, the y-intercept of L1 = -4

Put x = 0 into the equation of L2 :
2(0) + ky + 12 = 0
y = -12/k
Hence, the y-intercept of L2 = -12/k

L1 and L2 have the same y-intercept :
-12/k = -4
4k = 12
k = 3

b)
Slope of L2 = -2/3
Hence, slope of L3 = -2/3

L3 passes through the point (-2, 0) and with slope of -2/3.
Equation of L3 :
y - 0 = (-2/3)(x + 2)
3y = -2x - 4
2x + 3y + 4 = 0

我寫了大慨的步驟，並且在某一些位寫下要注意的地方，希望你會看得明白。不明白可以再問我。(第一題(C)有兩個做法，看你喜歡哪一個)

第一個做法是要兩條線的x值或y值都一樣才做得下去 (例如：L1的7x，L2都要是7x才能用這個方法)


圖片參考：http://imgcld.yimg.com/8/n/HA00132844/o/20121226120817.jpg