设向量a=(cos(x/2),sin(x/2)),向量b=(sin(3x/2),cos(3x/2)),x∈[0,π/2].(1)求a·b及|a+b|;(2)若函数f(x)=a·b+(√2)|a+b|,求f(x)的最小值,最大值.

设向量a=(cos(x/2),sin(x/2)),向量b=(sin(3x/2),cos(3x/2)),x∈[0,π/2].(1)求a·b及|a+b|;(2)若函数f(x)=a·b+(√2)|a+b|,求f(x)的最小值,最大值.
设向量a=(cos(x/2),sin(x/2)),向量b=(sin(3x/2),cos(3x/2)),x∈[0,π/2].
(1)求a·b及|a+b|;
(2)若函数f(x)=a·b+(√2)|a+b|,求f(x)的最小值,最大值.

设向量a=(cos(x/2),sin(x/2)),向量b=(sin(3x/2),cos(3x/2)),x∈[0,π/2].(1)求a·b及|a+b|;(2)若函数f(x)=a·b+(√2)|a+b|,求f(x)的最小值,最大值.
(1)a.b=cos(x/2)sin(3x/2)+cos(3x/2)*sin(x/2)=sin(3x/2+x/2)=sin2x
|a+b|=根号(a^2+b^2+2ab)=根号(2+2sin2x)=根号(2+2sin2x)
(2)f(x)=sin2x+2根号(1+sin2x)
根号(1+sin2x)=t t属于[1,根号2]
f(x)=2t+t^2-1=t^2+2t-1
对称轴为-1 所以[1,根号2]单调递增
f(x)最大值为 f(根号2)=1+2根号2
f(x)最小值为 f(1)=1+2-1=2
望采纳

(1)a*b=sin(3x/2+x/2)=sin2x,|a+b|=根号(2+2sin2x) (2)f(x)=sin2x+2根号(1+sin2x),设t=根号(1+sin2x),因为x属于[0,派/2],所以t的范围是[0,根号2],sin2x=t^2-1,f(x)=t^2-1+2t=(t+1)^2-2,范围是[-1,1+2根号2]

(1)a·b=(cos(x/2),sin(x/2))。(sin(3x/2),cos(3x/2))
=cos(x/2)sin(3x/2)+sin(x/2))cos(3x/2))
=sin(3x/2+x/2)=sin2x
a+b=(cos(x/2)+sin(3x/2)，sin(x/2))+cos(3x/2))
|a+b|^2=(cos(x/2)+sin(3x/2))^2...

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(1)a·b=(cos(x/2),sin(x/2))。(sin(3x/2),cos(3x/2))
=cos(x/2)sin(3x/2)+sin(x/2))cos(3x/2))
=sin(3x/2+x/2)=sin2x
a+b=(cos(x/2)+sin(3x/2)，sin(x/2))+cos(3x/2))
|a+b|^2=(cos(x/2)+sin(3x/2))^2+(sin(x/2))+cos(3x/2))^2
=2sin2x+2
|a+b|=√2*(cosx+sinx)
(2)f(x)=a·b+(√2)|a+b|
=sin2x+2(cosx+sinx)
=(cosx+sinx)^2+2(cosx+sinx)-1
=(cosx+sinx+1)^2-2
又cosx+sinx=√2sin(x+π/4），x∈[0,π/2].
所以√2/2≤cosx+sinx≤1
因此f(x)的最小值是-1/2+√2
f(x)的最大值是2

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