Question

已知a=(sinx/2,cosx/2),b=(2,1),且a//b

已知向量a=(cos3/2x,sin3/2x),b=(cosx/2,-sinx/2),且x∈[π/2,3π/2]
1.求|a+b|的取值范围2.求函数f（x）=a·b－|a+b|的最小值,并求此时x的值

Answer 1

1.|a|=|b|=1,
ab=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)=cos(3x/2+x/2)=cos2x,
(a+b)^2=a^2+2ab+b^2=2+2cos2x=4(cosx)^2,
∴|a+b|=-2cosx,x∈[π/2,3π/2],
它的取值范围是[0,2].
2.f(x)=cos2x+2cosx=2(cosx)^2+2cosx-1=2(cosx+1/2)^2-3/2,
cosx的值域是[-1,0],f(x)的最小值=-3/2,这时cosx=-1/2,x=2π/3或4π/3.