Question

若函数f(x)=(sinx+cosx)2-2cos2x-m在[0,]上有零点,则实数m的取值范围为(　　)

(A)[-1,] (B)[-1,1]

(C)[1,] (D)[-,-1]

 

Answer 1

A

【解析】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-)≤,

故当-1≤m≤时,f(x)在[0,]上有零点.