Question

求答案详解

Answer 1

f(x)=4cosxsin(x+π/6)-1
=4cosx(√3/2sinx+1/2cosx)-1
=2√3sinxcosx+2cosxcosx-1
=√3sin2x+cos2x
=2sin(2x+π/6)
则最小正周期为T=2π/2=π
-π/6≤x≤π/4
-π/6≤2x+π/6≤2π/3
所以在区间的最大值为2 ，x=π/6
在区间的最小值为-1，x=-π/6

Answer 2

aB=(F-u2mg)/m=(3-0.1*1*10)/1=2 A恰好能追上B说明追上时AB速度相等：vA-aAt=aBt vAt-0.5aAt*t=0.5aBt*t 0.75 aA=u1mg/m=0.4*1*