【正确答案】由f(χ)在χ=0处可导,得f(χ)在χ=0处连续.由表达式知,f(χ)在χ=0右连续.于是,f(χ)在χ=0连续

(sinχ+2ae
χ)=2a=f(0)

2a=-2b,即a+b=0.
又f(χ)在χ=0可导

f′
+(0)=f′
-(0).在a+b=0条件下,f(χ)可改写成

于是f′
+(0)=[9arctanχ+2b(χ-1)
3]′|
χ=0=[

+6b(χ-1)
2]|
χ=0=9+6b,
f′
-(0)=(sinχ+2ae
χ)′|
χ=0=1+2a.
因此f(χ)在χ=0可导
