113新竹市四校聯招
想請問2,5,7,8,19
113新竹四校聯招
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113新竹四校聯招
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Re: 113新竹四校聯招
第 2 題
x = sinA,y = cosA
x^2 + y^2 = 1
令 t = x + y = (a + b)/c > 1
xy = [(x + y)^2 - (x^2 + y^2)]/2 = (t^2 - 1)/2
13xy = 15(x + y) - 15
13 * (t^2 - 1)/2 = 15t - 15
解 t
第 5 題
0 < n!/n^n ≦ 1/n
夾擠......
第 7 題
lim{∫[t/(1 + t^3)]dx / x^2} (t 從 0 積到 x) (x → 0)
利用羅必達及微積分基本定理
= lim{[x/(1 + x^3)] / (2x)} (x → 0)
= lim[1/(2 + 2x^3)] (x → 0)
= 1/2
第 8 題
1 + 1/n^2 + 1/(n + 1)^2
= [n^2(n + 1)^2 + (n + 1)^2 + n^2]/[n^2(n + 1)^2]
= [n^4 + 2n^3 + 3n^2 + 2n + 1]/[n^2(n + 1)^2]
= (n^2 + n + 1)^2/[n^2(n + 1)^2]
= [(n^2 + n + 1)/(n^2 + n)]^2
= [1 + 1/n - 1/(n + 1)]^2
所求 = 99 + 1 - 1/100 = 9999/100
題目出錯了
第 19 題
令 ∠BAP = θ,∠CAP = 60∘- θ
AP/sin75∘= BP/sinθ
AP = BPsin75∘/sinθ
AP/sin45∘= PC/sin(60∘- θ)
AP = PCsin45∘/sin(60∘- θ)
BPsin75∘/sinθ = PCsin45∘/sin(60∘- θ)
2sin75c/sinθ = sin45∘/sin(60∘- θ)
2sin(60∘- θ)sin75∘ = sinθsin45∘
2sin(60∘- θ)cos15∘ = sinθsin45∘
θ = 45∘
∠APB = 60∘
x = sinA,y = cosA
x^2 + y^2 = 1
令 t = x + y = (a + b)/c > 1
xy = [(x + y)^2 - (x^2 + y^2)]/2 = (t^2 - 1)/2
13xy = 15(x + y) - 15
13 * (t^2 - 1)/2 = 15t - 15
解 t
第 5 題
0 < n!/n^n ≦ 1/n
夾擠......
第 7 題
lim{∫[t/(1 + t^3)]dx / x^2} (t 從 0 積到 x) (x → 0)
利用羅必達及微積分基本定理
= lim{[x/(1 + x^3)] / (2x)} (x → 0)
= lim[1/(2 + 2x^3)] (x → 0)
= 1/2
第 8 題
1 + 1/n^2 + 1/(n + 1)^2
= [n^2(n + 1)^2 + (n + 1)^2 + n^2]/[n^2(n + 1)^2]
= [n^4 + 2n^3 + 3n^2 + 2n + 1]/[n^2(n + 1)^2]
= (n^2 + n + 1)^2/[n^2(n + 1)^2]
= [(n^2 + n + 1)/(n^2 + n)]^2
= [1 + 1/n - 1/(n + 1)]^2
所求 = 99 + 1 - 1/100 = 9999/100
題目出錯了
第 19 題
令 ∠BAP = θ,∠CAP = 60∘- θ
AP/sin75∘= BP/sinθ
AP = BPsin75∘/sinθ
AP/sin45∘= PC/sin(60∘- θ)
AP = PCsin45∘/sin(60∘- θ)
BPsin75∘/sinθ = PCsin45∘/sin(60∘- θ)
2sin75c/sinθ = sin45∘/sin(60∘- θ)
2sin(60∘- θ)sin75∘ = sinθsin45∘
2sin(60∘- θ)cos15∘ = sinθsin45∘
θ = 45∘
∠APB = 60∘
Re: 113新竹四校聯招
第 3 題
兩邊微分
f(x^2) * 2x = x/√(1 + x^2)
x 用 √2 代入,可得 f(2)
第 11 題
a_0 = 1
a_1 = 7a_0 + 3 = 7 + 3
a_2 = 7a_1 + 3 = 7^2 + 7 * 3 + 3
a_2 = 7a_2 + 3 = 7^3 + 7^2 * 3 + 7 * 3 + 3
:
:
a_n = 7^n + 7^(n - 1) * 3 + 7^(n - 2) * 3 + .... + 7 * 3 + 3
= 7^n + 3 * [7^(n - 1) + 7^(n - 2) + .... + 7 + 1]
= 7^n + 3 * (7^n - 1)/(7 - 1)
= (3/2)7^n - 1/2
......
第 18 題
a^2 + b^2 = c^2
a(1 - x^2) - 2√2bx + c(1 + x^2) = 0
(c - a)x^2 - 2√2bx + (c + a) = 0
α + β = 2√2b/(c - a)
αβ = (c + a)/(c - a)
12 = α^2 + β^2 = (α + β)^2 - 2αβ = [2√2b/(c - a)]^2 - 2(c + a)/(c - a)
= [8b^2 - 2(c^2 - a^2)]/(c - a)^2
= 6b^2/(c - a)^2
2(c - a)^2 = b^2
2c^2 - 4ac + 2a^2 = c^2 - a^2
c = 3a or c = a (不合)
b = 2√2a
第 22 題
R = a√(bc)/(b + c) = 2RsinA√(bc)/(b + c)
sinA = (b + c)/[2√(bc)] ≦ 1
(√b - √c)^2 ≦ 0
b = c
sinA = 1
∠A = 90 度
答案給錯,應是 (A)
兩邊微分
f(x^2) * 2x = x/√(1 + x^2)
x 用 √2 代入,可得 f(2)
第 11 題
a_0 = 1
a_1 = 7a_0 + 3 = 7 + 3
a_2 = 7a_1 + 3 = 7^2 + 7 * 3 + 3
a_2 = 7a_2 + 3 = 7^3 + 7^2 * 3 + 7 * 3 + 3
:
:
a_n = 7^n + 7^(n - 1) * 3 + 7^(n - 2) * 3 + .... + 7 * 3 + 3
= 7^n + 3 * [7^(n - 1) + 7^(n - 2) + .... + 7 + 1]
= 7^n + 3 * (7^n - 1)/(7 - 1)
= (3/2)7^n - 1/2
......
第 18 題
a^2 + b^2 = c^2
a(1 - x^2) - 2√2bx + c(1 + x^2) = 0
(c - a)x^2 - 2√2bx + (c + a) = 0
α + β = 2√2b/(c - a)
αβ = (c + a)/(c - a)
12 = α^2 + β^2 = (α + β)^2 - 2αβ = [2√2b/(c - a)]^2 - 2(c + a)/(c - a)
= [8b^2 - 2(c^2 - a^2)]/(c - a)^2
= 6b^2/(c - a)^2
2(c - a)^2 = b^2
2c^2 - 4ac + 2a^2 = c^2 - a^2
c = 3a or c = a (不合)
b = 2√2a
第 22 題
R = a√(bc)/(b + c) = 2RsinA√(bc)/(b + c)
sinA = (b + c)/[2√(bc)] ≦ 1
(√b - √c)^2 ≦ 0
b = c
sinA = 1
∠A = 90 度
答案給錯,應是 (A)