Question

高考数学：已知函数f(x)=ax2-(2a+1)x+lnx,a∈R

Answer 1

f(x)=ax^2-(2a+1)x+lnx,
(1)a=1,f(x)=x^2-3x+lnx,x>0,
f'(x)=2x-3+1/x=(2x^2-3x+1)/x=2(x-1)(x-1/2)/x,

1/2<x<1时f'(x)<0,f(x)是减函数；
0<x<1/2或x>1时f'(x)>0,f(x)是增函数。
f(x)的极大值=f(1/2)=-5/4-ln2,f(x)的极小值=f(1)=-2.
(2)f(x)=2ax^2-2(a+1)x恰有两个不等的实根，
<==>ax^2-x-lnx=0恰有两个不等的正根，
<==>a=1/x+lnx/x^2恰有两个原像，①
a'=-1/x^2+1/x^3-2lnx/x^3=(-x+1-2lnx)/x^3,
0<x<1时a'>0;x>1时a'<0,
∴a的最大值=a(1)=1,
x→0+时a=（x+lnx)/x^2→-∞；x→+∞时a→0：
∴①<==>0<a<1,为所求.
(3)g(x)=e^x-x-1,
g'(x)=e^x-1=0,x=0,
易知g(x)|min=g(0)=0,
∴f(x1)<=g(x2)恒成立,
<==>f(x)<=0,
<==>a(x^2-2x)<x-lnx,②
x=2时②成立；
x>2时②变为a<(x-lnx)/(x^2-2x),记为F(x),
F'(x)=[(1-1/x)(x^2-2x)-(x-lnx)(2x-2)]/(x^2-2x)^2
=[x^2-2x-x+2-2x^2+2x+(2x-2)lnx]/(x^2-2x)^2
=[-x^2-x+2+(2x-2)lnx]/(x^2-2x)^2
=(x-1)(-x-2+2lnx)/(x^2-2x)^2,
设G(x)=-x-2+2lnx,则G'(x)=-1+2/x=(2-x)/x,
0<x<2时G'(x)>0,G(x)是增函数，x>2时G(x)是减函数，
G(x)<=G(2)=-4+2ln2<0,
∴0<x<1时F'(x)>0,F(x)是增函数；x>1时F(x)是减函数，
x→+∞时F(x)→0，
∴a<=0;
0<x<2时②变为a>(x-lnx)/(x^2-2x),
a>F(1)=-1,
综上，-1<a<=0,为所求.

Answer 2