已知向量a=(cos(2x-π/3),sin(2x-π/3)),b=(cos(2x-π/3),3/2).(1)若a⊥b且
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已知向量a=(cos(2x-π/3),sin(2x-π/3)),b=(cos(2x-π/3),3/2).(1)若a⊥b且x∈(0,π)
求x的值.(2)c=(0,1),f(x)=a•(b+λc)(λ∈R,若存在x属于(7π/24,π/2)使得f(x)=0,求λ的取值范围
求x的值.(2)c=(0,1),f(x)=a•(b+λc)(λ∈R,若存在x属于(7π/24,π/2)使得f(x)=0,求λ的取值范围
(1)
ab=cos(2x-π/3)^2+3sin(2x-π/3)/2=1-sin(2x-π/3)^2+3sin(2x-π/3)/2=0
解得sin(2x-π/3)=-1/2或2(舍去)
又x∈(0,π)所以2x-π/3∈(-π/3,5π/3)
所以2x-π/3=-π/6或7π/6,所以x=π/12或3π/4
(2)
b+λc=(cos(2x-π/3),3/2+λ)
f(x)=a•(b+λc)=cos(2x-π/3)^2+(3/2+λ)sin(2x-π/3)
=-sin(2x-π/3)^2+(3/2+λ)sin(2x-π/3)+1
若存在x属于(7π/24,π/2)使得f(x)=0,则
f(7π/24)*f(π/2)
ab=cos(2x-π/3)^2+3sin(2x-π/3)/2=1-sin(2x-π/3)^2+3sin(2x-π/3)/2=0
解得sin(2x-π/3)=-1/2或2(舍去)
又x∈(0,π)所以2x-π/3∈(-π/3,5π/3)
所以2x-π/3=-π/6或7π/6,所以x=π/12或3π/4
(2)
b+λc=(cos(2x-π/3),3/2+λ)
f(x)=a•(b+λc)=cos(2x-π/3)^2+(3/2+λ)sin(2x-π/3)
=-sin(2x-π/3)^2+(3/2+λ)sin(2x-π/3)+1
若存在x属于(7π/24,π/2)使得f(x)=0,则
f(7π/24)*f(π/2)
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