Question

f(x)=2sin^2（π/4-x)-根号3cos2x

求f(x)最小正周期及减区间若f(x)<m+2,在x∈[0,π/6]恒成立，求m的范围

Answer 1

pi为π
2sin^2(pi/4-x)=1-cos(pi/2-2x)=1-sin(2x)
f(x)=1-2(1/2sin(2x)+根号3/2cos(2x))=1-2(sin(pi/6)sin(2x)+cos(pi/6)cos(2x))=1-2cos(2x-pi/6)
周期: 2pi/2=pi
cos(2x)增区间为[pi/2+kpi,pi+kpi], 故cos[2(x-pi/12)]的增区间为[pi/2+kpi-pi/12,pi+kpi-pi/12]
所以f(x)减区间为 [5pi/12+kpi，11/12pi+kpi], k=0,1,2...
f(x)在[0,π/6]为增函数，故f(x)<=f(pi/6)=1-2cos(pi/6)=1-根号3<m+2
所以 m>-1-根号3