Question

已知函数f(x)=根号3sin^2x+sinxcosx,x属于[派/2,派],求零点和最大最小值

Answer 1

sqr(3)是根号3的意思

f(x)=[(sqr(3)/2)*(1-cos2x)]+(1/2)*sin2x=(1/2)*sin2x-(sqr(3)/2)*cos2x+sqr(3)/2

=sin(2x-π/3)+sqr(3)/2 2π/3≤2x-π/3≤5π/3 所以函数的零点是

2x-π/3=π,x=2π/3,

当2x-π/3=2π/3,即x=π/2时,取最大值 0

当2x-π/3=3π/2,即x=11π/6时,取最小值 -1+sqr(3)/2