113 松山高中
版主: thepiano
Re: 113 松山高中
第 2 題
f(x) * f(1/x + f(x)) = 1
f(1) * f(1 + f(1)) = 1
令 f(1) = k
k * f(1 + k) = 1
f(1 + k) = 1/k
f(1 + k) * f(1/(1 + k) + f(1 + k)) = 1
f(1/(1 + k) + 1/k) = k = f(1)
1/(1 + k) + 1/k = 1
k = (1 + √5)/2 或 (1 - √5)/2
當 k = (1 + √5)/2,f(1 + k) = 2/(1 + √5) < 1 < k = f(1),不合
f(x) * f(1/x + f(x)) = 1
f(1) * f(1 + f(1)) = 1
令 f(1) = k
k * f(1 + k) = 1
f(1 + k) = 1/k
f(1 + k) * f(1/(1 + k) + f(1 + k)) = 1
f(1/(1 + k) + 1/k) = k = f(1)
1/(1 + k) + 1/k = 1
k = (1 + √5)/2 或 (1 - √5)/2
當 k = (1 + √5)/2,f(1 + k) = 2/(1 + √5) < 1 < k = f(1),不合
Re: 113 松山高中
第 7 題
作 AM 垂直平面 BCD 於 M;GN 垂直平面 BCD 於 N
設 AM = h,則 GN = h/3
作 DE 垂直 BC 於 E
設 BC = x,NE = (1/3)ME = (1/3)(1/3)DE = (1/9)DE
DN = (8/9)DE = (8/9)(√3/2)x = (4√3/9)x
GN = √(DG^2 - DN^2) = √[1 - (16/27)x^2]
A-BCD 體積 = (h/3)(√3/4)x^2 = √(3/16)x^4 - (1/9)x^6]
微分可知,當 x^2 = 9/8 時,體積有最大值 9/32
作 AM 垂直平面 BCD 於 M;GN 垂直平面 BCD 於 N
設 AM = h,則 GN = h/3
作 DE 垂直 BC 於 E
設 BC = x,NE = (1/3)ME = (1/3)(1/3)DE = (1/9)DE
DN = (8/9)DE = (8/9)(√3/2)x = (4√3/9)x
GN = √(DG^2 - DN^2) = √[1 - (16/27)x^2]
A-BCD 體積 = (h/3)(√3/4)x^2 = √(3/16)x^4 - (1/9)x^6]
微分可知,當 x^2 = 9/8 時,體積有最大值 9/32