Question

已知函数f(x)=2sinxcosx+2√3cos^2x

（1）求函数f（x）的最小正周期（2）当x属于【π/4，3π/4】时，求函数f（x）的最大值与最小值

Answer 1

f(x)=2sinxcosx+2√3cos^2x=sin2x+√3cos2x+√3=2sin(2x+π/3)+√3
最小正周期=π
2kπ-π/2<2x+π/3<2kπ+π/2
kπ-5π/12<x<kπ+π/12递增
kπ+π/12<x<kπ+7π/12递减 ， 【π/4，3π/4】=【3π/12，9π/12】
最小值f(7π/12)=2-√3 , 最大值f(π/4)=√3

Answer 2

f(x)=2sinxcosx-2根号3cos^2x+根号3
=sin2x-√3(1+cos2x)+√3
=sin2x-√3cos2x
=2[(1/2)sin2x-(√3/2)cos2x]
=2sin(2x-π/3)
所以递增区间为2x-π/3∈[2kπ-π/2, 2kπ+π/2]
即x∈[kπ-π/12, kπ+5π/12]
递减区间为2x-π/3∈[2kπ+π/2, 2kπ+3π/2]
即x∈[kπ+5π/12, kπ+11π/6]

Answer 3

解
f(x)=2sinxcosx+2√3cos²x-√3
=sin2x+√3(2cos²x-1)
=sin2x+√3cos2x
=2(1/2sin2x+√3/2cos2x)
=2(sin2xcosπ/3+cos2xsinπ/3)
=2sin(2x+π/3)
T=2π/2=π是f(x)的最小正周期
∵-1≤sin(2x+π/3)≤1
∴f(x)的最小值为:2×(-1)=-2
(2)f(A)=2sin(2A+π/3)=1
∴2A+π/3=π/6+kπ
或2A+π/3=5π/6+kπ
∵A∈(0,π/2)
∴A=60
∵AB*AC=√2
∴cbcosA=√2
即cb=2√2
∴s=1/2cbsinA=√2×√3/2=√6/2