Question

设函数f(x)=InX-1/2ax^2-bx

令F（X）=f(x)+1/2ax^2+bx+a/x(0<x<=3)其图象上任意一点P(X0,f(x0))处的切线k<=1/2恒成立，求a的取侄范围
当a=0,b=-1方程2mf(x)=x^2有唯一实数解，求正数m的值

Answer 1

F'(x)=1/x-a/x²<=1/2
a/x²-1/x+1/2>=0
x²-2x+2a>=0
(x-1)²>=1-2a
(x-1)²>=0
所以1-2a=<0,a>=1/2

2mf(x)=x²有唯一实数解
x²-2mlnx-2mx=0有唯一实数解
设g(x)=x²-2mlnx-2mx
g'(x)=(2x²-2mx-2m)/x
g'(x)=0,x²-mx-m=0
m>0,x>0
x1=[m-√(m²+4m)]/2<0舍去
x2=[m+√(m²+4m)]/2
当0<x<x2,g'(x)<0,g(x)在此区间单调减
当x>x2,g'(x)>0,g(x)在此区间单调增
当x=x2,g'(x2)=0,g(x)取最小值g(x2)
则
g(x2)=0,g'(x2)=0
(x2)²-2mlnx2-2mx2=0,(x2)²-mx2-m=0
2mlnx2+mx2+m=0因m>0,2lnx2+x2-1>0
设h(x)=2lnx+x-1
x>0时,h(x)是增函数,h(x)至多有一解
h(1)=0,
2lnx2+x2-1=0,x2=1
[m+√(m²+4m)]/2=1
m=1/2

Answer 2

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