已知函数f(x)=alnx+bx^2图象上点P(1,f(1))处的切线方程为2x-y-3=0.

f'(x)=a/x+2bx
f(1)=b
f'(1)=a+2b
切线为：y=(a+2b)(x-1)+b=(a+2b)x-a-b
对比方程系数得：a+2b=2, a+b=3
故解得：a=4, b=-1,
f(x)=4lnx-x^2