Question

已知向量a=（cosx，sinx），向量b=（根号3倍的cosx，cosx)若f（x）=向量a向量b+根号3求函数f（x）的最小正

求f（x）的最小正周期和图象的对称轴方程；求f（x，在区间〔-5兀/12，兀/12）上的值域

Answer 1

f（x）=向量a向量b+根号3
=√3cos²x+sinxcosx+√3
=(√3/2)(cos2x+1)+(1/2)sin2x+√3
=(1/2)sin2x+(√3/2)cos2x+(3/2)√3
=sin(2x+π/3)+(3/2)√3
所以最小正周期T=2π/2=π
对称轴方程2x+π/3=2kπ+π/2
x=kπ+π/12 k∈∈Z
x∈[-5π/12, π/12]
2x+π/3∈[-7π/6, -π/6]
f(x)最大=f(-5π/6)=sin(-7π/6)+(3/2)√3
=sin(5π/6)+(3/2)√3
=sin(π/6)+(3/2)√3
=1/2+(3/2)√3
f(x)最小=f(-π/6)=sin(-π/2)+(3/2)√3
=-1+(3/2)√3
所以值域f(x)∈[-1+(3/2)√3, 1/2+(3/2)√3]

Answer 2

是不是叫求最小值？f(x)=根号3cos"x+sinxcosx+根号3;=根号3/2 cos2x+1/2 sin2x+3根号3/2 ;=sin(2x+60')+3根号3/2;所以最小值为3根号3/2 -1