f(x)=(√3sinx+cosx)(√3cosx-sinx)
=2sinxcosx+√3(cos²x-sin²x)
=sin2x+√3cos2x
=2(½sin2x+√3/2cos2x)
=2sin(2x+⅓π)
x∈[0,π/2]
2x+⅓π=½π→x=π/12 f(x)取得最大值2
x=½π时 f(½π)=2sin(π+⅓π)=-2sin(⅓π)=-√3为最小值
f(x)∈[-√3,2]
f(x)=(√3sinx+cosx)(√3cosx-sinx)
=3sinxcosx-√3sin²x+√3cos²x-sinxcosx
=2sinxcosx+√3(cos²x-sin²x)
=sin2x+√3cos2x
=2[½sin2x+(√3/2)cos2x]
=2sin(2x+π/3)
x∈[0，π/2]，则π/3≤2x+π/3≤4π/3，-√3/2≤sin(2x+π/3)≤1
-√3≤2sin(2x+π/3)≤2
f(x)的值域为[-√3，2]
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