Question

已知向量a=(5√3cosx,cosx),b=(sinx,2cosx),设函数f(x)=a*b+|

已知向量a=(5√3cosx,cosx),b=(sinx,2cosx),设函数f(x)=a*b+|b|^2+3/2
若f（θ/2）=9 求f（θ＋ π/3）的值
求使f（x）≥5成立的x的取值集合

Answer 1

f(x)=5√3sinxcosx+2cos^2x+sin^2x+4cos^2x+3/2
=5√3/2sin2x+6cos^2+sin^2x+3/2
=5√3/2sin2x+6cos^2x-3+3+sin^2x-1/2+1/2+3/2
=5√3/2sin2x+3cos2x-1/2cos2x+5
=5√3/2sin2x+5/2cos2x+5
=5sin(2x+π/6)+5

(1)
2x+π/6=π/2+2kπ（k∈Z）
x=π/6+kπ
2x+π/6=3π/2+2kπ（k∈Z）
x=2π/3+kπ
f（x）在[π/6+kπ，2π/3+kπ]单调递减
f（x）的值域为[5/2,10]

(2)
x∈[π/6，π/2]
2x+π/6∈[π/2,7π/6]
5sin(2x+π/6)+5=8
sin(2x+π/6)=3/5
cos(2x+π/6)=-4/5

sin[2(x-π/12)+π/6]
=sin[2x-π/6+π/6]
=sin(2x+π/6)cosπ/6-cos(2x+π/6)sinπ/6
=3/5*√3/2-(-4/5)*1/2
=3√3/10+4/10
=(3√3+4)/10

Answer 2

答：

向量a=(5√3cosx,cosx),b=(sinx,2cosx),设函数f(x)=a*b+|b|^2+3/2

f(x)=5√3cosxsinx+cosx*2cosx+(sinx)^2+(2cosx)^2+3/2
=(5√3/2)sin2x+5(cosx)^2+1+3/2
=(5√3/2)sin2x+(5/2)(cos2x+1)+5/2
=(5√3/2)sin2x+(5/2)cos2x+5
=5sin(2x+π/6)+5
因为：
f(θ/2)=5sin(θ+π/6)+5=9
所以：sin(θ+π/6)=4/5
所以：
f(θ+π/3)
=5sin(2θ+2π/3+π/6)+5
=5sin(2θ+5π/6)+5
=5sin[2(θ+π/6)+π/2]+5
=5cos[2(θ+π/6)]+5
=5-10*[sin(θ+π/6)]^2+5
=10-10*(4/5)^2
=18/5
所以：f(θ+π/3)=18/5

f(x)=5sin(2x+π/6)+5>=5
sin(2x+π/6)>=0
所以：
2kπ<=2x+π/6<=2kπ+π
kπ-π/12<=x<=kπ+5π/12，k属于Z