1.解： f(x)=根号3sin2x+cos2x+1+m =2sin(2x+π/6)+1+m 因为0= 所以0=<2x+π/6<=π+π/6 当2x+π/6=π/2时有最大值3+m=6 m=3所以f(x)=2sin(2x+π/6)+42.因为x∈R当 2x+π/6=-π/2+2kπ (k∈Z)时有最小值 2 x=-π/3+kπ (k∈Z)
