美夢成真教甄討論區

請參考附件

計算第 3 題
PC/sin(30∘+ θ) = 3/sin60∘
PC = 3sinθ + √3cosθ

RC/sin(120∘- θ) = 4/sin60∘
RC = (4/3)√3sinθ + 4cosθ

PR = [3 + (4/3)√3]sinθ + (4 + √3)cosθ
疊合後，可得 PR 最大值 = √[(100 + 48√3)/3]

△PQR 面積最大值 = 12 + (25/3)√3

計算第 2 題
(1) (x^2 + x + 1)f(x) = 2x^302 + x^100 + x^61 + a
x = ω = (-1 + √3i)/2 代入，可得
0 = 2ω^2 + ω + ω + a
a = 2

(2) f(x) = (x^2 - x + 1)q(x) + (ax + b)
2x^302 + x^100 + x^61 + 2 = (x^4 + x^2 + 1)q(x) + (ax + b)(x^2 + x + 1)
x = -ω 代入，可得
2ω^2 + 2 = 2aω^2 - 2bw
ω^2 + 1 = aω^2 - bw
-ω = -aω - a - bω
a = 0，b = 1
所求之餘式為 1

(3) f(x) = (x^2 + x + 1)k(x) + (cx + d)
2x^302 + x^100 + x^61 + 2 = [(x^2 + x + 1)^2]k(x) + (cx + d)(x^2 + x + 1)
微分，可得
604x^301 + 100x^99 + 61x^60 = 2(x^2 + x + 1)(2x + 1)k(x) + [(x^2 + x + 1)^2]k'(x) + c(x^2 + x + 1) + (cx + d)(2x + 1)
x = ω 代入，可得
604ω + 100 + 61 = (cω + d)(2ω + 1) = 2cω^2 + (c + 2d)ω + d
604ω + 161 = (-c + 2d)ω + (-2c + d)
c = 94，d = 349
所求之餘式為 94x + 349