Question

设函数fx=根号3sinx·cosx+cos²x+2m-1（x∈R）（2）若x∈[-π/6,π/3]fx的最小值为5,求常数m的值

设函数fx=根号3sinx·cosx+cos²x+2m-1（x∈R）（
（1）求fx的最小正周期及fx的单调区间
2）若x∈[-π/6,π/3]fx的最小值为5,求常数m的值

Answer 1

(1)f(x)=√3sinx·cosx+cos²x+2m-1
=1/2*(√3 *2*sinx·cosx+2cos²x)+2m-1
=1/2*(√3 *sin2x+cos2x+1)+2m-1
=1/2*(√3 *sin2x+cos2x)+2m-1/2
=√3/2 *sin2x+1/2*cos2x+2m-1/2
故f(x)=sin(2x+π/6)+2m-1/2
最小正周期：2π/2=π
单调增区间：-π/2+2kπ<2x+π/6≤π/2+2kπ
-2π/3+2kπ<2x≤π/3+2kπ
即 -π/3+kπ<x≤π/6+kπ

单调减区间：π/2+2kπ<2x+π/6≤π+2kπ
π/3+2kπ<2x≤5π/6+2kπ
即 π/6+kπ<x≤5π/12+kπ
(2) 当x∈[-π/6,π/3，f(x)最小值为5
根据(1)可知
当x=-π/6时， f(x)有最小值
即f(-π/6)=sin(-π/3+π/6)+2m-1/2=5
即 -1/2+2m-1/2=5
m=3