Question

两道高中数学题。求解！帮帮忙！

Answer 1

(1)f(x)=alnx/(x+1)+b/x,
∴f(1)=b,f'(x)=a[(x+1)/x-lnx]/(x+1)^2-b/x^2,
f'(1)=a/2-b,
∴曲线y=f(x)在(1,b)的切线方程是y-b=(a/2-b)(x-1),
即y=(a/2-b)x+2b-a/2,与y=(-1/2)x+3/2重合，
比较得a/2-b=-1/2,
2b-a/2=3/2,
解得a=b=1.
(2)f(x)=lnx/(x+1)+1/x,
f(x)-lnx/(x-1)=1/x-2lnx/(x^2-1)=(x^2-1-2xlnx)/(x^3-x),
设g(x)=x^2-1-2xlnx,x>0,x≠1，则
g'(x)=2x-2lnx-2,
g''(x)=2-2/x=2(x-1)/x,
0<x<1时g''(x)<0,g'(x)是减函数；x>1时g''(x)>0,g'(x)是增函数：
∴g'(x)>g'(1)=0,
∴g(x)是增函数，g(1)=0,
∴x>1时g(x)>0,而x^3-x>0;0<x<1时g(x)<0,而x^3-x<0:
两者都有(x^2-1-2xlnx)/(x^3-x)>0,
∴f(x)>lnx/(x-1).

设甲、乙击中目标的事件分别为A、B,依题意P(A)=P(B)=0.6，
用A'表示非A,余者类推，
(1)P(AB)=P(A)*P(B)=0.6*0.6=0.36.
(2)P(AB'+A'B)=0.6*0.4+0.4*0.6=0.48.
(3)P(A+B)=1-P(A'B')=1-0.4*0.4=0.84.

Answer 2

Answer 3

五、
(1) 0.6×0.6=0.36
(2) 0.6×0.4+0.4×0.6=0.48
(3) 1-0.4×0.4=0.84
六、
(1) f(x)=alnx/(x+1)+b/x
f'(x)=[a(x+1)/x-alnx]/(x+1)^2-b/x^2
f'(1)=[a(1+1)/1-aln1]/(1+1)^2-b/1^2
=a/2-b
∵f(x)在(1,f(1))的切线方程为：x+2y-3=0

∴f'(1)=-1/2
a/2-b=-1/2
b=(a+1)/2
且1+2f(1)-3=0
f(1)=1
aln1/(1+1)+b/1=1

b=1
(a+1)/2=1
a=1
(2) f(x)=lnx/(x+1)+1/x
f(x)-lnx/(x-1)
=lnx/(x+1)+1/x-lnx/(x-1)
=lnx[1/(x+1)-1/(x-1)]+1/x
=2/(x^2-1)lnx+1/x
当0<x<1时

x^2-1<0
2/(x^2-1)<0
lnx<0
2/(x^2-1)lnx>0
2/(x^2-1)lnx+1/x>0
即：f(x)>lnx/(x-1)
当x>1时
x^2-1>0
2/(x^2-1)>0
lnx>0
2/(x^2-1)lnx>0
2/(x^2-1)lnx+1/x>0
即：f(x)>lnx/(x-1)
综上，f(x)>lnx/(x-1)