Question

f(x)=(根号3sinx+cosx)(根号3cosx-sinx)在区间[0,π/2]上的值域是？

Answer 1

• 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]

Answer 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]