100全國高中職

[johncai]

請問一下綜合題第2跟第8題
謝謝

[thepiano]

第 2 題
這是 IMO 1963 的考題
前幾天松山工農才剛考過，多一個負號而已

cos(6π/7) - cos(5π/7) + cos(4π/7)
= - [cos(π/7) + cos(3π/7) + cos(5π/7)]

2 * sin(π/7) * [cos(π/7) + cos(3π/7) + cos(5π/7)]
= sin(2π/7) + sin(4π/7) - sin(2π/7) + sin(6π/7) - sin(4π/7)
= sin(6π/7)
= sin(π/7)

cos(π/7) + cos(3π/7) + cos(5π/7) = 1/2

cos(6π/7) - cos(5π/7) + cos(4π/7) = - 1/2


第 8 題
96 埔里高工考過

微分後可知 x = 7/4 時，有最大值 5/2

也可以用柯西不等式
[(8 - 4x) + (4x - 3)](1/4 + 1) ≧ [√(2 - x) + √(4x - 3)]^2
√(2 - x) + √(4x - 3) ≦ 5/2
等號成立於 (8 - 4x) / (1/4) = (4x - 3) / 1，x = 7/4

[thepiano]

第 9 題
好像也看過 ......
考柯西和算幾不等式等號成立的條件
所求 = 14

這次進複試要很高分吧？

[8y383249]

想請問選擇8和填充5,7題,第7題我只得到|a-b|=(根號3)|a|,=>|a|=2,再來就不會了,麻煩各位高手了

[lingling02]

(1):? 想請教選擇7和8和9(第9是一直用餘弦去作嗎@@式子好複雜丫 )
(2)填充第3
(3)另外想請教填充第一是否有較簡易的算法
我是假設餘數為ax^5+bx^4+cx^3+dx^2+ex+f
代x=i,x=-i,x^2=-2,x^2=2去解abcdef

[tsaisy]

填充第1：試著用x^2=t代人替換，應該會快一點

[thepiano]

選擇
第 7 題
k * k! = (k + 1)! - k!
n = 51! - 1
n ≡ -1 ≡ 49 (mod 50)


第 8 題
f(x) = - f(x + 3/2) = f(x + 3/2 + 3/2) = f(x + 3)
f(3) = f(0) = -2
f(2) = f(-1) = 1

f(x) 之圖形對稱於 (-3/4，0)，注意：其圖形為點對稱圖形
故 f(-3/4 + x) = - f(-3/4 - x)

x 用 7/4 代入 f(-3/4 + x) = - f(-3/4 - x) 得
f(1) = - f(-5/2)

x 用 -5/2 代入 f(x) = - f(x + 3/2)
f(-5/2) = - f(-1) = -1

f(1) = 1

f(1) + f(2) + f(3) = 0

所求 = f(2011) = f(1) = 1


第 9 題
∠DAB = θ，∠CAD = 2θ
tan2θ = 3/4
易知 tanθ = 1/3

令 BD = a
(a + 3) / 4 = tan3θ = (3/4 + 1/3) / (1 - 3/4 * 1/3) = 13/9
a = 25/9

CD / BD = 3 / (25/9) = 27/25



填充
第 1 題
令 y = x^2
原題改為 y^10 + 1 除以 (y + 1)(y^2 - 4) 之餘式

令 y^10 + 1 = Q(y)(y + 1)(y^2 - 4) + ay^2 + by + c
y = -1，2，-2 代入上式

a - b + c = 2
4a + 2b + c = 1025
4a - 2b + c = 1025

a = 341，b = 0，c = -339

所求 = 341y^2 - 339 = 341x^4 - 339


第 3 題
x^6 - 1 = (x - 1)(x^5 + x^4 + x^3 + x^2 + x + 1)
f(x)∣x^6 - 1
x^6 ≡ 1 ( mod f(x) )
f(x^6) ≡ f(1) ≡ 6 ( mod f(x) )


第 5 題
令 ∫g(x)dx (從 0 積到 2) = a
f(x) = x + a + 1
∫f(x)dx (從 0 積到 1) = 1/2 + a + 1 = a + 3/2
g(x) = 2x - 3 + a + 3/2 = 2x + a - 3/2
a = ∫g(x)dx (從 0 積到 2) = 4 + 2(a - 3/2)
a = -1
g(x) = 2x - 5/2
......


第 7 題
易知 |α| = 2
4α^2 - 2αβ + β^2 = 0
(2α/β)^2 - (2α/β) + 1 = 0
2α/β = cos(π/3) ± isin(π/3)

2|α| / | β| = 1
| β| = 4

△OAB = (1/2) * 2 * 4 * sin(π/3)

[thepiano]

填充第 8 題類似題

求 √(x + 27) + √(13 - x) + √x 的最大值和最小值

√(x + 27) + √(13 - x) + √x = √(x + 27) + √{13 + 2√[x(13 - x)]} ≧ √27 + √13 = 3√3 + √13

[√(x + 27) + √(13 - x) + √x]^2 ≦ (1 + 1/3 + 1/2)[x + 27 + 3(13 - x) + 2x] = 121

√(x + 27) + √(13 - x) + √x ≦ 11

[lingling02]

多謝鋼琴老師解惑

選擇
第 8 題
f(x) = - f(x + 3/2) = f(x + 3/2 + 3/2) = f(x + 3)
f(3) = f(0) = -2
f(2) = f(-1) = 1

x 用 -1/2 代入 f(x) = - f(x + 3/2)
f(-1/2) = - f(1)
x 用 -1 代入 f(x) = - f(x + 3/2)
f(-1) = - f(1/2) = 1

f(1) = 1===>就是轉不過來為何....f(1)=1

f(1) + f(2) + f(3) = 0

所求 = f(2011) = f(1) = 1

[Herstein]

填充第7題 為什麼|α|=2?
再請教 選擇第 4. 5題