Question

已知向量a=（cosx,cosx）,b=（更号下3cosx,cosx）且b=0,函数f（x）=2a*b-1.（1）求函数

Answer 1

f(x) = 2a*b - 1 = 2*[√3(cosx)^2+(cosx)^2] - 1 = 2(√3+1)(cosx)^2 = (√3+1)(1+cos2x)

所以f(x)最小正周期T=pi

在(-pi/2+kpi,kpi)单增,在(kpi,kpi+pi/2)单减

(2)cosx=0

tanx=∞

cos2x/[f(x)+1]=-1/1=-1