極急!!! F.4 Mathematics

1.
a.
Let f(x) = x⁶⁷⁸ + 2x -9

The remainder
= f(-1)
= (-1)⁶⁷⁸ + 2(-1) -9
= 1 - 2 - 9
= -10

b.
Let f(x) = x¹⁰⁰⁰ + 3x⁹⁹⁹ + x - 5

The remainder
= f(-1)
= (-1)¹⁰⁰⁰ + 3(-1)⁹⁹⁹ + (-1) - 5
= 1 - 3 - 1 - 5
= -8

c.
Let f(x) = 3x¹⁰⁰ - 27x⁹⁸ + x² - 4

The remainder
= f(-1)
= 3(3)¹⁰⁰ - 27(3)⁹⁸ + 3² - 4
= 3¹⁰¹ - (3)¹⁰¹ + 9 - 4
= 5

d.
Let f(x) = 2k³x³ + kx + 4

The remainder
= f(-2/k)
= 2k³(-2/k)³ + k(-2/k) + 4
= -16 - 2 + 4
= -14


2.
a.
f(x) = x³ + 2ax² - 5a²x - 6a³

f(2a)
= (2a)³ + 2a(2a)² - 5a²(2a) - 6a³
= 8a³ + 8a³ - 10a³ - 6a³
= 0

By Factor Theorem, (x - 2a) is a factor of f(x).

b.
(x - 2a) is a factor of x³ + 2ax² - 5a²x - 6a³

Put a= 1 :
(x - 2) is a factor of x³ + 2x² - 5x - 6

Use long division,
(x³ + 2x² - 5x - 6) ÷ (x - 2) = x² + 4x + 3

Hence, x³ + 2x² - 5x - 6
= (x - 2)(x² + 4x + 3)
= (x - 2)(x + 1)(x + 3)


3.
a.
When p = 160 + n ...... [1]
then q = 12000 - 50n ...... [2]

[1]*50:
50p = 8000 - 50n ...... [3]

[2] + [3]:
q + 50p = 20000
q = 20000 - 50p

b.
Let $I be the total income.

I = pq
I = p(20000 - 50p)
I = -50(p² - 400p)
I = -50(p² - 400p + 200²) + 50*200²
I = -50(p - 200)² + 2000000

For any real value of p, -50(p - 200)² ≤ 0
Hence, I = -50(p - 200)² + 2000000 ≤ 2000000
When p = 200, I gets the maximum value.

To get the maximum income, price of each ticket = $200


4.
(a + 3i)(-2 + 3i)
= a(-2) + a(3i) + (3i)(-2) + (3i)(3i)
= -2a + 3ai - 6i - 9
= -(2a + 9) + (3a - 6)i

For a real number, (3a - 6)i = 0
3a - 6 = 0
a = 2


5.
(2 - i)(a + bi) = 5 + 20i
2a + 2bi - ai - bi² = 5 + 20i
(2a + b) + (-a + 2b) = 5 + 20i

2a + b = 5 ...... [1]
-a + 2b = 20 ...... [2]

[2]*2:
-2a + 4b = 40 ...... [3]

[1] + [3]:
5b = 45
b = 9

Put b = 9 into [1]:
2a + (9) = 5
2a = -4
a = -2


6.
x² + 14x + k = 0 has no realroots.
Then, discriminant Δ < 0
(14)² - 4(1)(k) < 0
196 - 4k < 0
49 - k < 0
k > 49

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fooks，多謝你的解答，上一次 GP 那一題發問者問了兩次，但你只答了一題。

重覆的那一題我引用了你的答案，已經寫明鳴謝你了～
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