f(x)=x^a[cos(1/x)] (x不等于0) =0 (x=0),其导数在x=0处连续,求a的取值范围

f‘(x)=x^a[cos(1/x)] =ax^(a-1)[cos(1/x)] +x^(a-2)[sin(1/x)]=x^(a-2)[axcos(1/x)+[sin(1/x)]
f'(0)=[x^a[cos(1/x)]-0]/x=limx^(a-1)[cos(1/x)]
要使f'(0)存在,a-1>0,这时f'(0)=0
要使导数在x=0处连续,limf‘(x)=0,a-2>0
故a>2