Question

高考数学复习：已知函数f(x)=x-1/x+1+2alnx(a∈R)

Answer 1

f(x)=x-1/x+1+2alnx,x>0,a∈R,
f(1)=1,
f'(x)=1+1/x^2+2a/x,
(I)f(x)在点(1,1)处的切线是y=b,
∴f'(1)=2+2a=0,a=-1.b=1,
∴a+b=0.
(II)f(x)有两个极值点，
<==>f'(x)有两个互异的零点，
<==>x^2+2ax+1=0有不等的正根:0<x1<x2，①
△/4=a^2-1>0,a<0,
∴a<-1.
由①，x1x2=1,x2>1,
设g(x)=2xlnx+1-x^2(x>1),则
g'(x)=2lnx+2-2x,
g''(x)=2/x-2<0,g'(x)是减函数，
g'(x)<g(1)=0,g(x)是减函数，
g(x)<g(1)=0,
∴2xlnx<x^2-1,
(lnx2-lnx1)/(x2-x1)=2lnx2/(x2-1/x2)=2x2lnx2/(x2^2-1)<1
AB的斜率k=[f(x2)-f(x1)]/(x2-x1)=1+1/(x1x2)+2a(lnx2-lnx1)/(x2-x1)>2+2a,
∴k/2-1>a.