Question

已知向量a=（√3 sinx,m+cosx）,向量b=(cosx,-m+cosx),且f(x)=向量a*向量b

1、求函数f(x)的解析式
2、当x属于[-π/6,π/3]时，f(x)的最小值是-4，求此时f(x)的最大者，并求出相应的x的值

Answer 1

1.
f(x)=√3sinxcosx+(cosx)^2-m^2

=(√3/2)sin2x+(1/2)cos2x+1/2-m^2

=sin2xcosπ/6+cos2xsinπ/6+1/2-m^2

=sin(2x+π/6)+1/2-m^2

2.
-π/6<=x<=π/3，则-π/6<=2x+π/6<=5π/6。

当2x+π/6=-π/6，即x=-π/6时，f(x)取得最小值f(-π/6)=-1/2+1/2-m^2=-1、m^2=4

f(x)=sin(2x+π/6)-7/2

当2x+π/6=π/2，即x=π/6时，f(x)取得最大值f(π/6)=1/2-7/2=-3

.

Answer 2

1）.f(x)=√3/2(sin2x)+cosx²-m²=0.5(sin(2x+30°)）-m²
2).2x+30°∈ [-π/6,5π/6],∴f(x)∈ [-0.25-m²,1-m²]m²=3.75，f(x)max=-2.75,此时x=π/6

Answer 3