Question

已知平面向量a=(根号3sinx,cosx),b=(cosx,cosx),x€(0,pai)

Answer 1

f(x)=a*b

=√3sinx*cosx+cos²x

=√3/2*2sinx*cosx+cos²x-1/2+1/2

=√3/2*sin2x+1/2(2cos²x-1)+1/2

=√3/2*sin2x+1/2*cos2x+1/2

=sin(2x+π/6)+1/2

∵ x∈(0,π)

∴ (2x+π/6)∈(π/6,13π/6)

则 当2x+π/6=π/2,即x=π/6时,f(x)取得最大值,有

f(π/6)=sinπ/2+1/2=3/2

因此,f(x)的最大值3/2,此时x值为π/6