题目

已知向量a=（5根3cosX,cosX),向量b=(sinX,2cosX),其中X属于（π/6,π/2）,设函数f(x)=a.b+|b|平方+3/2
（1)求函数f(x)的值域；（2）若f(X)=8,求函数f(X-π/12）的值

提问时间：2020-10-31

答案

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