113 高雄聯招
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Re: 113 高雄聯招
第 4 題
1 + 1/x = [x/(x + 1)]^(-1) = [1 - 1/(x + 1)]^(-1) = [1 + 1/-(x + 1)]^(-1)
(1 + 1/x)^(x + 1) = [1 + 1/-(x + 1)]^[-(x + 1)]
-(x + 1) = 2024
x = -2025
第 14 題
2 * [(a_1 + 1)/a_1] * [(a_1 + 2)/(a_1 + 1)] * … * [(a_1 + n)/(a_1 + n - 1)] = n
2 * (1 + n/a_1) = n
n = 2/(1 - 2/a_1)
a_1 = 3 時,n 有最大值 6
所求 = 3 + 4 + … + 8 = 33
第 15 題
z 在射線 y = -(x + 3) (x < -3) 上
|z - 3i| + |z + 6| = z 到 (0,3) 與到 (-6、0) 距離和
最小值 = √(6^2 + 3^2) = 3√5
所求 = √5/15
1 + 1/x = [x/(x + 1)]^(-1) = [1 - 1/(x + 1)]^(-1) = [1 + 1/-(x + 1)]^(-1)
(1 + 1/x)^(x + 1) = [1 + 1/-(x + 1)]^[-(x + 1)]
-(x + 1) = 2024
x = -2025
第 14 題
2 * [(a_1 + 1)/a_1] * [(a_1 + 2)/(a_1 + 1)] * … * [(a_1 + n)/(a_1 + n - 1)] = n
2 * (1 + n/a_1) = n
n = 2/(1 - 2/a_1)
a_1 = 3 時,n 有最大值 6
所求 = 3 + 4 + … + 8 = 33
第 15 題
z 在射線 y = -(x + 3) (x < -3) 上
|z - 3i| + |z + 6| = z 到 (0,3) 與到 (-6、0) 距離和
最小值 = √(6^2 + 3^2) = 3√5
所求 = √5/15
Re: 113 高雄聯招
第 6 題
x + my = 0,過定點 A(0,0)
mx - y - m + 3 = 0,過定點 B(1,3)
易知兩直線垂直
PA^2 + PB^2 = AB^2 = 10
√(PA^2 * PB^2) ≦ (PA^2 + PB^2)/2
PA * PB ≦ 5
第 16 題
令 EF = x,CE = xcosθ
∠DEF = ∠DBE = 60 度
∠BDE = ∠FEC = θ
由正弦定理
BE/sinθ = DE/sin60度
BE = (2/√3)xsinθ
xcosθ + (2/√3)xsinθ = 1
x = 1/[cosθ + (2/√3)sinθ] = √3/(2sinθ + √3cosθ] = √3/[√7sin(θ + α)]
當 sin(θ + α) = π/2 時,x 有最小值
此時 sinθ = cosα = 2/√7
x + my = 0,過定點 A(0,0)
mx - y - m + 3 = 0,過定點 B(1,3)
易知兩直線垂直
PA^2 + PB^2 = AB^2 = 10
√(PA^2 * PB^2) ≦ (PA^2 + PB^2)/2
PA * PB ≦ 5
第 16 題
令 EF = x,CE = xcosθ
∠DEF = ∠DBE = 60 度
∠BDE = ∠FEC = θ
由正弦定理
BE/sinθ = DE/sin60度
BE = (2/√3)xsinθ
xcosθ + (2/√3)xsinθ = 1
x = 1/[cosθ + (2/√3)sinθ] = √3/(2sinθ + √3cosθ] = √3/[√7sin(θ + α)]
當 sin(θ + α) = π/2 時,x 有最小值
此時 sinθ = cosα = 2/√7