Question

求答案。发图就行

Answer 1

(1)|a|=√sin²x + (√3cosx)²
=√sin²x + 3cos²x
=√sin²x + cos²x + 2cos²x
=√1 + 2cos²x
∵-1≤cosx≤1
∴最小值是√3，最小值是1

追答

(2)f(x)=sinx•cosx + √3cosx•(-cosx)
=(1/2)sin2x - √3cos²x
=(1/2)sin2x - √3•[(1+cos2x)/2]
=(1/2)sin2x - (√3/2)cos2x - √3/2
=sin(2x - π/3) - √3/2
∴最小周期是T=2π/2=π
对称轴2x - π/3=kπ + π/2
2x=kπ + 5π/6
∴x=(1/2)kπ + 5π/12，k∈Z