答:
可导必定连续:x=0,f(0)=e^(a*0)-1=1-1=0
f(0)=b+a+2=0………………(1)
求导:f'(x)=bcosx=ae^(ax)
x=0,f'(0)=b=a…………(2)
由(1)和(2)解得:a=b=-1