Question

高中数学导数题 ，如何求下题单调区间

Answer 1

1.f'(x)=1/x-2a/(x+1)^2=[x^2+(2-2a)x+1]/[x(x+1)^2],
(1-a)^2-1=a^2-2a<=0,即0<=a<=2时f'(x)>=0,f(x)的定义域(0,+∞)是增区间；
a<0或a>2时设x1=a-1-√(a^2-2a),x2=a-1+√(a^2-2a),
x12.由x>0且x≠1时,lnx/(x-1)>a/(x+1)恒成立,得
a<(x+1)lnx/(x-1),记为h(x),则
h'(x)={(x-1)[lnx+(x+1)/x]-(x+1)lnx}/(x-1)^2
=(x-1/x-2lnx)/(x-1)^2,
设F(x)=x-1/x-2lnx,x>0,则F'(x)=(x-1)^2/x>=0,∴F(x)是增函数,F(1)=0,
0x>1时F(x)>0,h'(x)>0,h(x)是增函数,
x→1时h(x)→lnx+(x+1)/x(罗必达法则)→2,
∴a<=2,为所求.