✔ 最佳答案
1.
設 f(x) = 2x³ - x² + kx + 5
f(x) ÷ (x + 2) 的餘數 = -9
f(-2) = -9
2(-2)³ - (-2)² + k(-2) + 5 = -9
-16 - 4 - 2k + 5 = -9
2k = -6
k = -3
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2.
(a)
二分之一的寫法是1/2,四分之一的寫法是 1/4。
等比數列應為: -2, 1, -1/2, 1/4, ......
首項 a = -2
公比 r = -1/2
(b)
數列的無限項和 S∞
= a/[1 - r]
= -2/[1 - (-1/2)]
= -2 ÷ (3/2)
= -2 x (2/3)
= -4/3
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3.
(a)
f(2) = 4
3(2)² + k(2) - 10 = 4
12 + 2k - 10 = 4
2k = 2
k = 1
(b)
3x² + x - 10 = 0
(3x - 5)(x + 2) = 0
x = 5/3 或 x = -2
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4.
(a)
P(x) = (x - a)(x - b)(x + 3)
當 P(x) 除以 (x - 1) 時,所得的餘數為 12。
所以 P(1) = 12
(1 - a)(1 - b)(1 + 3) = 12
(1 - a)(1 - b) = 3
[-(a - 1)] [-(b - 1)] = 3
(a - 1)(b - 1) = 3
(b)
因為 a 及 b 為正整數(a < b),且 (a - 1)(b - 1) ≠ 0
因此 (a - 1) 及 (b - 1) 亦為正整數,且 (a - 1) < (b - 1)
"3" 為質數,因式分解 3 = 1 × 3
比對 3 = (a - 1)(b - 1)
a - 1 = 1 及 b - 1 = 3
a = 2 及 b = 4
(c)
P(x) = 0
(x - 2)(x - 4)(x + 3) = 0
x = 2 或 x = 4 或 x = -3
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5.
(a)
∠A = 180° - (30° + 50°) = 100°
BC² = AB² + AC² - 2 × AB × AC × cos∠A (餘弦定律)
BC² = 60² + 70² - 2 × 60 × 70 × cos100°
BC = 99.8 m
(b)
sin∠A/BC = sin∠C/AB
sin100°/99.8 = sin∠C/60
sin∠C = sin100° × (60/99.8)
sin∠C = 36.3°
50° - 36.3° = 13.7°
由 C 測得 B 的羅盤方位角 = S13.7° W