Question

若函數f(x)=(sinx+cosx)²-2cos²x-m在[0,]上有零點,則實數m的取值范圍為(　　)

A．[-1,]	B．[-1,1]
C．[1,]	D．[-,-1]

Answer 1

A

f(x)=(sinx+cosx)²-2cos²x-m
="1+sin" 2x-2cos²x-m
="1+sin" 2x-1-cos 2x-m
=sin(2x-)-m.
∵0≤x≤,∴0≤2x≤π,∴-≤2x-≤,
∴-1≤sin(2x-)≤,
故當-1≤m≤時,f(x)在[0,]上有零點.