我来试试吧...
1.a=0,f(x)=lnx+1,故f(x)在(0,∞)上↑
2.a≠0,f(x)=(a+1)lnx+ax²+1
f'(x)=(a+1)/x+2ax=[2ax²+(a+1)]/x
f"(x)=2a-(a+1)/x²
1.-1