Question

已知f(x)=ln(x+1),g(x)=1/2ax^2+bx，（1）若b=2,且h(x)=f(x-1)-g(x)存在单调递减区间，求a的取值范围

已知f(x)=ln(x+1),g(x)=1/2ax^2+bx，（1）若b=2,且h(x)=f(x-1)-g(x)存在单调递减区间，求a的取值范围（2）a=0,b=1时,试判断f(x)与g(x)图像的公共点的个数并说明理由

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Answer 1

1.
h(x)=f(x-1)-g(x)=lnx-(1/2)ax²-2x,x>0
h'(x)=1/x-ax-2=-(ax²+2x-1)/x
i)当a=0,h'(x)=-(2x-1)/x
知x∈(1/2,+∞),h'(x)<0,h(x)单调减少
ii)a≠0,只需讨论ax²+2x-1=0的判别式△=4(1+a),[x1+x2=-a/2,x1x2=1/a,x1<x2]
当a<-1,△<0,恒有h'(x)>0,知不存在x使得h'(x)<0,h(x)单调减少
当a=-1,h'(x)=(x-1)²/x≥0,且h'(x)不恒为0,知h(x)在定义域内单调增加,知不存在单调减少区间
当-1<a<0,△>0,令h'(x)=0,x>0.仅有一解x=x2,知x∈(0,x2),有h'(x)<0,h(x)单调减少
当a>0,△>0,令h'(x)=0,x>0.x无解,此时恒有h'(x)<0知x∈(0,+∞),h(x)单调减少
综上要使得h(x)存在单调减区间,a的取值范围为a∈(-1,+∞)
2.
证法1
记f(y)=xlnx+ylny-(x+y)ln[(x+y)/2],0<x<y,把x看做常数对y求导
f'(y)=lny+1-ln[(x+y)/2]-1=lny-ln[(x+y)/2]>lny-ln[(y+y)/2]=0
则f(y)在y>x>0上单调增加,又f(y)可在y=x处连续有
f(y)>f(x)=2xlnx-2xlnx=0,y>x>0
即xlnx+ylny-(x+y)ln[(x+y)/2]>0
亦即xlnx+ylny>(x+y)ln[(x+y)/2],0<x<y
证毕.
证法2
原命题等价于[(xlnx)+(ylny)]/2>[(x+y)/2]ln[(x+y)/2]
即证 [f(x)+f(y)]/2>f[(x+y)/2],其中f(x)=xlnx,x>0
即证f(x)为凹函数
只需证其二阶导数f"(x)>0
∵f'(x)=lnx+1
f"(x)=1/x>0
∴命题成立.

Answer 2

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