向量m=(acosx,cosx),n=(2cosx,bsinx),f(x)=m·n,且f(0)=2,f(π/3)=1/2+√3/21）求f(x)的最大值与最小值2）若f(α)=0,α∈(0,2π),求α的值

向量m=(acosx,cosx),n=(2cosx,bsinx),f(x)=m·n,且f(0)=2,f(π/3)=1/2+√3/21）求f(x)的最大值与最小值2）若f(α)=0,α∈(0,2π),求α的值
向量m=(acosx,cosx),n=(2cosx,bsinx),f(x)=m·n,且f(0)=2,f(π/3)=1/2+√3/2
1）求f(x)的最大值与最小值
2）若f(α)=0,α∈(0,2π),求α的值

向量m=(acosx,cosx),n=(2cosx,bsinx),f(x)=m·n,且f(0)=2,f(π/3)=1/2+√3/21）求f(x)的最大值与最小值2）若f(α)=0,α∈(0,2π),求α的值
m.n
=(acosx,cosx).(2cosx,bsinx)
= 2a(cosx)^2+ bsinxcosx = f(x)
f(0) = 2a = 2
=> a = 1
f( π/3) = 2(1/4) + b(√3/4) = 1/2+√3/2
b(√3/4) = √3/2
b = 2
f(x) = 2(cosx)^2 + 2sinxcosx
f'(x) = -4cosxsinx + 2(-(sinx)^2 +(cosx)^2)
= -2sin2x+ 2cos2x = 0
sin2x = cos2x
tan2x = 1
x = π/8 or 5π/8
f''(x) = -4cos2x- 4sin2x
f''(π/8) < 0 ( max)
f''(5π/8) > 0 (min)
f(x) = 2(cosx)^2 + 2sinxcosx
= cos2x+1 + sin2x
maxf(x) =f(π/8)
= √2/2 + 1 +√2/2
= 1+√2
minf(x) = f(5π/8)
= -√2/2 +1 -√2/2
= 1-√2
f(α)=0
=>cos2α+1 + sin2α = 0
(√2/2) (sin2α + cos2α) = - (√2/2)
(2α+π/4) = (7π/4) or 5π/4
α = 3π/4 or π/2

1)
f(x)=m·n=2a(cosx)²+bcosxsinx=acos2x+a+b/2sin2x
f(0)=2a=2 ∴a=1
f(π/3)=-1/2+1+√3b/4=1/2+√3/2 ∴b=2
f(x)=cos2x+1+sin2x=√2sin(2x+π/4)
f(x)的最大值√2最小值-√2
2)f(α)=√2sin(2a+π/4)=0 α∈(0,2π),α=3/8π

1) f(x)=m·n=2a(cosx)²+bcosxsinx=a+acos2x+b/2sin2x
由 f(0)=2a=2 得 a=1
由 f(π/3)=-1/2+1+√3b/4=1/2+√3/2 得 b=2
f(x)=1+cos2x+sin2x=1+√2sin(2x+π/4)
f(x)的最大值为1+√2，最小值为1-√2
2)因为0<α<2π，...

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1) f(x)=m·n=2a(cosx)²+bcosxsinx=a+acos2x+b/2sin2x
由 f(0)=2a=2 得 a=1
由 f(π/3)=-1/2+1+√3b/4=1/2+√3/2 得 b=2
f(x)=1+cos2x+sin2x=1+√2sin(2x+π/4)
f(x)的最大值为1+√2，最小值为1-√2
2)因为0<α<2π，∴π/4<2α+π/4<17π/4
由f(α)=0得 sin(2α+π/4)=-√2/2，
故2α+π/4可能的取值有四个：π+π/4，2π-π/4，3π+π/4，4π-π/4
故α有四π/2，3π/4，3π/2及7π/4

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1）第一步：f(x)=2acos^2+bsinx*cosx
第二步：将f(0)=2& f(π/3)=1/2+√3/2 带入，得a=1 b=2
第三步：化简f(x)=cos2x+sin2x-1=√2sin(2x+π/4)-1
f(x)max=√2-1 f(x)min=-√2-1
2) 若f(α)=0，则√2sin(2x+π/4)=1，所以sin(2x+π/4)=√2/2 故a =π