已知向量a=(cos3x/2,sin3x/2）,b=（cosx/2,—sinx/2）,且x∈[0,π

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(1)因为向量a=(cos3x/2,sin3x/2）,b=（cosx/2,—sinx/2）,所以：
|a|＝|b|＝1
且a*b=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)
=cos(3x/2 +x/2)
=cos2x
则|a+b|²=|a|²+2a*b+|b|²
=2+2cos2x
=2(1+cos2x)
=4cos²x
因为x∈[0,π/2],所以：
|a+b|=2cosx
(2)由（1）可得：
f（x）=a·b-2λ│a+b│
＝cos2x－2λ*2cosx
=2cos²x-4λcosx-1
=2(cosx-λ)²-2λ²-1
因为x∈[0,π/2],所以cosx∈[0,1]
若λ======以下答案可供参考======
供参考答案1:
ab=cos(3x/2)*cos(x/2)-sin(3x/2)*sin(x/2)
=cos[(3x+x)/2]
=cos(2x).
a+b=(cos(3x/2)+cos(x/2),sin(3x/2)-sin(x/2)),
|a+b|=√[(cos(3x/2)+cos(x/2))^2+(sin(3x/2)-sin(x/2))^2]
=√[2(1+cos2x)]
=2*|cosx|,
因为,x∈[-π/3,π/4]。则有,cosx>0, 即, |a+b|=2*|cosx|=2cosx.
2.若f（x）=a*b-|a+b|。则有,
f(x)=cos2x-2cosx,
=2cos^2x-1-2cosx
=2(cosx-1/2)^2-3/2.
而,x∈[-π/3,π/4]。则有,
1)当X=0时,cos0=1,则f(x)=2(1-1/2)^2-3/2=-1.
2)当X=π/4时,cosπ/4=√2/2,则f(x)=2*(√2/2-1/2)^2-3/2=-√2.
则,f(x)最大值=-1,f(x)最小值=-√2.