题目

已知函数f(x)=根号3sinxcosx-cos^2+a,x属于R求f(x)的最小正周期和单调区间

提问时间：2020-11-02

答案

f(x)=√3sinxcosx-cos²x+a
=(√3/2)(2sinxcosx)-(1/2)(2cos²x -1) +a -1/2
=(√3/2)sin(2x)-(1/2)cos(2x) +a-1/2
=sin(2x-π/6) +a-1/2
最小正周期T=2π/2=π
a-1/2为常数,2kπ-π/2≤2x-π/6≤2kπ+π/2 (k∈Z)时,即kπ-π/6≤x≤kπ+π/3 (k∈Z)时,
sin(2x-π/6)单调递增,sin(2x-π/6) +a-1/2单调递增,f(x)单调递增
kπ+π/3≤x≤kπ+5π/6 (k∈Z)时,f(x)单调递减.
f(x)的单调递增区间为[kπ-π/6,kπ+π/3] (k∈Z),单调递减区间为[kπ+π/3,kπ+5π/6] (k∈Z)