Question

已知a=(sin2x,cos2x),b(1,根号3)且f(X)=a*b,求f(X)的周期,最大值,单调增区间

Answer 1

f(x)=sin2x+根号3cos2x

=2(1/2*sin2x+2分之根号3cos2x)

=2sin(2x+π/3)

那么最小正周期T=2牌/2=π

最大值f(x)max=2

令2kπ-π/2≤2x+π/3≤2kπ+π/2

则：kπ-5π/12≤x≤kπ+π/12

即单调递增区间为[kπ-5π/12,kπ+π/12]