向量m=(acosx,cosx),n=(2cosx,bsinx),f(x)=m·n,且f(0)=2,f(π/3)=1/2+√3/2

题目

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

答案

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