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已知向量a=(cos3x/2,sin3x/2),b=(cosx/2,-sinx/2),且x∈[0,π/2]
已知向量a=(cos3x/2,sin3x/2),b=(cosx/2,-sinx/2),且x∈[0,π/2]
若f(x)=a*b-2λ|a+b|的最小值为-3/2,求λ的值
人气:269 ℃ 时间:2019-12-01 13:23:40
解答
=cos3x/2cosx/2-som3x/2sinx/2=cos2x
|a+b|^2=(cos3x/2+cosx/2)^2+(sin3x/2-sinx/2)^2=
(cos3x/2 ^2+sin3x/2^2)+ (cosx/2^2+sinx/2^2)+2(cos3x/2cosx/2-sin3x/2sinx/2)
=2+2cos2x=4cos^2x-2+2=4cos^2x
|a+b|=2cosx
f(x)=a*b-2λ|a+b|=2cos^2x-1+4λcosx=2(cosx+λ)^2-2λ^2-1
00f(x)递增
所以f(x)min=2(0+λ)^2-2λ^2-1=-1≠-3/2
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