题目

已知函数f(x)=2cos²x+2根号3sinxcosx.求 （1）函数f（x）的周期.（2）函数f（x）的单调递减区间.
（3）函数f（x）在区间[0,π/2]上的单调区间及最值.

提问时间：2020-07-16

答案

f(x)=2cos²x+2根号3sinxcosx
= (cos2x+1) + √3sin2x
= √3sin2x + cos2x + 1
= 2(sin2xcosπ/6+cos2xsinπ/6) +1
= 2sin(2x+π/6) + 1
最小正周期 = 2π/2 = π
2x+π/6∈(2kπ+π/2,2kπ+3π/2),其中k∈Z时单调减
所以单调减区间：x∈（kπ+π/6,kπ+2π/3),其中k∈Z
x∈[0,π/2]时：
2x+π/6∈【π/3,4π/3】
其中,2x+π/6∈【π/3,π/2）时单调增,2x+π/6∈【π/2,4π/3）时单调减
∴单调增区间：x∈∈【π/12,π/6）,单调减区间x∈【π/6,7π/12）
2x+π/6=π/2时,最大值f(x)max = 2+1 = 3
2x+π/6=4π/3时,最小值f(x)min = 2*(-√3/2)+1 = 1-√3