已知向量a=(cosx,-1/2),b=(√3sin,cos2x),x∈R设函数f(x)=向量a·向量b (1)求f(x)在[0,π/2]上的最大值和最小值.

题目

已知向量a=(cosx,-1/2),b=(√3sin,cos2x),x∈R设函数f(x)=向量a·向量b (1)求f(x)在[0,π/2]上的最大值和最小值.

答案

f(x)=向量a.向量b.
f(x)=√3sinccosx-(1/2)cos2x.
=(√3/2)sin2x-(1/2)cos2x.
=sin(2x-π/6).
∵c∈[0,π/2],∴(2x-π/6)∈[-π/6,5π/6].
∵f(x)在x∈[0,π/3]为增函数,在x∈[π/3,π/2]为减函数.∴f(x)在x=π/3处取得最大值1.
f(0)=-sinπ/6=-1/2, f(π/2)=sin(π-π/6=sinπ/6=1/2.
f(0)＜f(π/2).
∴ f(x)max=1
f(x)min=-1/2.