Question

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Answer 1

f(x)=coswx(√3sinwx-coswx)+k
=√3sinwxcoswx-(coswx)^2+k
=(√3/2)sin2wx-(1/2)cos2wx-1/2+k
=sin2wxcosπ/6-cos2wxsinπ/6-1/2+k
=sin(2wx-π/6)-1/2+k
(1)最小正周期为T=2π/2w=π/2，则w=2。
f(x)=sin(4x-π/6)-1/2+k。
0<=x<=π/4，则-π/6<=4x-π/6<=5π/6。
当4x-π/6=-π/6，即x=0时，f(x)取得最小值为f(0)=sin(-π/6)-1/2+k=-1+k=-2，则k=-1。
f(x)=sin(4x-π/6)-3/2。
(2)2kπ-π/2<=4x-π/6<=2kπ+π/2，则kπ/2-π/12<=x<=kπ/2+π/6。
所以，f(x)的单调递增区间是：[kπ/2-π/12,kπ/2+π/6]，k为正整数。