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

Answer 2

f(x)=[(√3/2)*(1-cos2x)]+(1/2)*sin2x
=(1/2)*sin2x-(√3/2)*cos2x+√3/2
=sin(2x-π/3)+√3/2
∵ 2π/3≤2x-π/3≤5π/3
∴f(x)的零点是
2x-π/3=π,x=2π/3.
当2x-π/3=2π/3,即x=π/2时,f(x)取得最大值 0.
当2x-π/3=3π/2,即x=11π/6时,f(x)取得最小值 √3/2 -1.

Answer 3

f(x) = sqrt(3) (1-cos2x)/2 + 1/2 sin2x = sqrt(3)/2 + sin(2x - pi/3)
x 属于（pi/2,pi), 2x 属于 (pi, 2pi)