（1）f(x)=ax³+bx²+cx(a≠0,x∈R）為奇函數(shù),
∴b=0,f'(x)=3ax^2+c
f(x)在x=1處取得極大值2,
∴f(1)=a+c=2,
f'(1)=3a+c=0,
解得a=-1,c=3,f(x)=-x^3+3x.
(2)g(x)=-x^2+3+(k+1)lnx,x>0.
g'(x)=-2x+(k+1)/x=(k+1-2x^2)/x,
k0,g(x)↑；x>√[(k+1)/2]時(shí)g'(x)0),
h'(x)=-2x-1+3/x=-2(x-1)(x+3/2)/x,
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