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

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