數學3題~複數、極值

1
令f(x)=x^3-3x+1=(x-α)(x-β)(x-γ)
PA*PB*PC= | i-α | * | i-β | * | i-γ |
= | f(i) |
= |-4i+1 |
=17^(1/2)
2
令A=2^x,B=2^y
x+y=1 => AB=2
A^2+B^3
=(A^2/3)+(A^2/3)+(A^2/3)+(B^3/2)+(B^3/2)
>= 5*[(A^2/3)^3(B^3/2)^2]^(1/5)
=5*[(AB)^6/108]^(1/5)
=5*[16/27]^(1/5)

3
可以將(sinθ)^2=1-(cosθ)^2代入再令x=cosθ得到
[(sinθ)^2][(cosθ)^3]=x^3-x^5=f(x)
微分f'(x)=3x^2-5x^4
令f'(x)=0,x=0,(3/5)^(1/2),-(3/5)^(1/2)
最大值(6/125)*(15)^(1/2)
最小值-(6/125)*(15)^(1/2)
或者是用第二小題的技巧

1 α+β+γ=0,αβγ=-1
(線段PA)x(線段PB)x(線段PC)
=|(i-α)(i-β)(i-γ)|
=|(i^2-(α+β)i+αβ)(i-γ)|
=|(-1-γi-1/γ)(i-γ)|
=|1-(1+γ^2+1/γ)i|
=|1-(1/γ)(γ+γ^3+1)i|
=|-1-(1/γ)(γ+3γ)i|
=|-1-4i|
=√17
2 Since xy>0 , x+y=1, so x,y>0. Then the minimum of 4^x + 8^y should be 4. To find the local minimum
Let f=4^x + 8^y,g=x+y-1
Then L=4^x + 8^y+λ(x+y-1)
∂L/∂x=(4^x)(ln4)+λ
∂L/∂y=(8^y)(ln8)+λ
∂L/∂λ=x+y-1
Set them equal to 0
Then (4^x)(ln4)=(8^y)(ln8)
=>(4^x)/(8^y)=3/2
=>2^(2x-3y)=3/2
=>2^(5x-3)=3/2
(5x-3)ln2=ln(3/2)
x=[ln3/ln2+2]/5
The local minimum 4^x + 8^y is 4.5032
3 Let f=[(sinθ)^2][(cosθ)^3]
df/dθ
=(sin^2θ)(3cos^2θ)(-sinθ)+(cos^3θ)(2sinθ)(cosθ)
=-3sin^3θcos^2θ+2cos^4θsinθ
Let df/dθ=0
=>2cos^4θsinθ-3sin^3θcos^2θ=0
=>sinθcos^2θ(2cos^2θ-3sin^2θ)=0
=>sinθ=0 or tan^2θ=2/3
So the minimum of [(sinθ)^2][(cosθ)^3] is -1 and the maximum value of [(sinθ)^2][(cosθ)^3] is (6/25)(√3/5)