使函数y=cos(wx+π/2)在[0,π/4]为增函数,则w的取值范围

显然w不等于0
原式y=-sin(wx)在[0,pi/4]上为增函数
所以sin(wx)在[0,pi/4]上是减函数
sin的减区间是[2kπ-π/2,2kπ+π/2]
若w>0
则0<=x<=π/4
0<=wx<=wπ/4
要符合[2kπ-π/2,2kπ+π/2]
则此时k=0的一个区间
所以wπ/4<=π/2
w<=2
所以0若w<0
则0<=x<=π/4
wπ/4<=wx<=0
要符合[2kπ-π/2,2kπ+π/2]
则也是k=0的一个区间
所以-π/2<=wπ/4
w>=-2
所以-2<=w<=0
所以-2<=w<0,0