1. Use the Remainder Theorem to find the remainder in each of the following
a. (x^(678)+2x-9)/(x+1)
b.(x^(1000)+3x^(999)+x-5)/(1+x)
c.(3x^(100)-27x^(98)+x^(2)-4)/(x-3)
d.(2k^(3)x^(3)+kx+4)/(kx+2), where k≠0
2. Let f(x)=x^(3)+2ax^(2)-5a^(2)x-6a^(3)
a. show that x-2a is a fator of f(x)
b. Hence, factorize x^(3)+2x^(2)-5x-6
3. Andy is going to hold a concert at the Hong Kong Coliseum. Suppose the auditorium can accommodate at most 12000 people. He estimates that if the price for each ticket is $160, then all the tickets will be sold. For every increase of $1 in the ticket price, the number of tickets sold will decrease by 50.
a. Let $p be the price of each ticket and q be the number of tickets sold. Expess q in trems of p.
b. What should the price of each ticket be in order to get the maximum income?
4. If the expression (a+3i)(-2+3i) is a real number, find the value of the real number a.
5. If (2-i)(a+bi)=5+20i, where a and b are real numbers, find the values of a and b.
6. Let k be a constant. If the quadratic equation x^(2)+14x+k=0 has no real roots, find the range of values of k.
極急!!! F.4 Mathematics
2014-01-08 6:19 am
回答 (3)
2014-01-08 7:36 am
✔ 最佳答案
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
參考: 賣女孩的火柴
2014-01-09 2:45 am
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僋匾井儈厄
僋匾井儈厄
2014-01-08 7:44 am
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～
http://hk.knowledge.yahoo.com/question/question?qid=7013122800198
重覆的那一題我引用了你的答案，已經寫明鳴謝你了～
http://hk.knowledge.yahoo.com/question/question?qid=7013122800198
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