114 復興高中
版主: thepiano
Re: 114 復興高中
第 10 題
作 GD 垂直 BC 於 D,GE 垂直 AC 於 E
6 = GD + GE = (1/3)(AC + BC)
AC + BC = 18
又 AC^2 + BC^2 = 15^2
可求出 △ABC = (1/2) * AC * BC = 99/4
作 GD 垂直 BC 於 D,GE 垂直 AC 於 E
6 = GD + GE = (1/3)(AC + BC)
AC + BC = 18
又 AC^2 + BC^2 = 15^2
可求出 △ABC = (1/2) * AC * BC = 99/4
Re: 114 復興高中
第 4 題
y = (x - x^3)/(1 + 2x^2 + x^4)
= x(1 - x^2)/(1 + x^2)^2
= (1/2) * [(2x)/(1 + x^2)] * [(1 - x^2)/(1 + x^2)]
令 x = tanθ,tan2θ = (2x)/(1 - x^2)
y = (1/2)sin2θcos2θ = (1/4)sin4θ
當 sin4θ = 1 時,y 有最大值 1/4
y = (x - x^3)/(1 + 2x^2 + x^4)
= x(1 - x^2)/(1 + x^2)^2
= (1/2) * [(2x)/(1 + x^2)] * [(1 - x^2)/(1 + x^2)]
令 x = tanθ,tan2θ = (2x)/(1 - x^2)
y = (1/2)sin2θcos2θ = (1/4)sin4θ
當 sin4θ = 1 時,y 有最大值 1/4
Re: 114 復興高中
第 4 題
另解
(1) x = 0,1,-1,(x - x^3)/(1 + 2x^2 + x^4) = 0
(2) x ≠ 0,1,-1
(x - x^3)/(1 + 2x^2 + x^4)
= (1/x - x)/(1/x^2 + 2 + x^2)
= (1/x - x)/[(1/x - x)^2 + 4)]
= 1/[(1/x - x) + 4/(1/x - x)]
1/x - x < 0,1/[(1/x - x) + 4/(1/x - x)] < 0
1/x - x > 0,1/[(1/x - x) + 4/(1/x - x)] ≦ 1/(2√4) = 1/4
另解
(1) x = 0,1,-1,(x - x^3)/(1 + 2x^2 + x^4) = 0
(2) x ≠ 0,1,-1
(x - x^3)/(1 + 2x^2 + x^4)
= (1/x - x)/(1/x^2 + 2 + x^2)
= (1/x - x)/[(1/x - x)^2 + 4)]
= 1/[(1/x - x) + 4/(1/x - x)]
1/x - x < 0,1/[(1/x - x) + 4/(1/x - x)] < 0
1/x - x > 0,1/[(1/x - x) + 4/(1/x - x)] ≦ 1/(2√4) = 1/4