【正确答案】正确答案:(I)当x≠0时按求导法则得

当x=0时按导数定义得

(II)由于f(x)一f(0)=一x
2
(2+sin

)<0(x≠0),即f(x)<f(0),于是由极值的定义可知x=0是f(x)的极大值点. (III)令x
N
=

(N=±1,±2,±3,…),则sin

=0,cos

=(一1)
n
,于是

(Ⅳ)对

δ>0,当n为

负奇数且|n|充分大时x
n
∈(一δ,0),f'(x
n
)<0

f(x)在(一δ,0)不单调上升;当n为正偶数且n充分大时x
n
∈(0,δ),f'(x
n
)>0
