Question

设函数f(x)=根号3sinxcosx+cos平方x+a，求f(x)的最小正周期及单调递减区间

Answer 1

f（x）=√3/2sin2x+1/2cos2x+1/2+a=sin（2x+π/6）+1/2+a
得到T=2π/2=π
令2kπ+π/2<=2x+π/6<=2kπ+3π/2
得到kπ+π/6<=x<=kπ+2π/3(k是整数)
所以函数的单调递减区间是[kπ+π/6,kπ+2π/3] (k是整数)

Answer 2

f(x)=√3sinxcosx+cos^2x+a=√3/2sin2x+(1+cos2x)/2+a=sin2xcosπ/6+cos2xsinπ/6+1/2+a=sin（2x+π/6)+1/2+a则f(x)的最小正周期为π。导数f'=2cos(2x+π/6)2kπ+π/2<2x+π/6<2kπ+3π/2时，f'>0时，为增区间2kπ-π/2<2x+π/6<2kπ+π/2时，f<0时，减区间。

Answer 3

f(x)=√3sinxcosx+cos^2x+a=√3/2sin2x+(1+cos2x)/2+a=sin2xcosπ/6+cos2xsinπ/6+1/2+a=sin（2x+π/6)+1/2+a则f(x)的最小正周期为π。导数f'=2cos(2x+π/6)2kπ+π/2<2x+π/6<2kπ+3π/2时，f'>0时，为增区间2kπ-π/2<2x+π/6<2kπ+π/2时，f<0时，减区间。

Answer 4

f(x)=√3sinxcosx+cos^2x+a=√3/2sin2x+(1+cos2x)/2+a=sin2xcosπ/6+cos2xsinπ/6+1/2+a=sin（2x+π/6)+1/2+a则f(x)的最小正周期为π。导数f'=2cos(2x+π/6)2kπ+π/2<2x+π/6<2kπ+3π/2时，f'>0时，为增区间2kπ-π/2<2x+π/6<2kπ+π/2时，f<0时，减区间。