已知函数f（x）=lnx,g(x)={（1/2)x^2}+a的图像(a为常数),直线L与函数f(x) g(x)图像相切,

切点F(1, 0)
f'(x) = 1/x, f'(1) = 1/x = 1
g'(x) = x = 1
g(1) = 1/2 + a = f(1) = 0
a = -1/2
g(x) = (x² - 1)/2
h(x) = f(1 + x²) - (x² - 1)/2 = ln(1+ x²) - (x² - 1)/2
h'(x) = 2x/(1 + x²) - x
= x(1 - x²)/(1 + x²) = 0
x = -1, x = 0, x = 1
x < -1: h'(x) > 0
-1 < x < 0: h'(x) < 0
0 < x < 1: h'(x) > 0
x > 1: h'(x) < 0
h(1) = h(-1) = ln2为最大值
h(0) = 1/2为极小值
(i) k > ln2时,f(1 + x²) - g(x) = k无解
(ii) k = ln2时, f(1 + x²) - g(x) = k有二解(x = ±1)
(iii) 1/2 < k < ln2时, f(1 + x²) - g(x) = k有4解
(iv) k = 1/2时, f(1 + x²) - g(x) = k有3解
(v) k < 1/2时, 时, f(1 + x²) - g(x) = k有2解

∵f'(x)=1/x，g'(x)=x
∴f'(1)=1
则直线的k=1
设直线为y=x+b且与f(x)的切点为(1,c)
则1+b=c
ln1=c=0
b=-1
∴直线方程为y=x-1
与g(x)联立
即x-1=(1/2)x^2+a
x^2-2x+2a+2=0
∵△=0
则4-8a-8=0
∴a...

f'(x)=1/x
f'(1)=1
L的斜率为1,且过点(1,0)
L的方程为:y=x-1
g'(x)=x
令g'(t)=1,得t=1
切点为(1,0)
a=-1/2
g(x)=x^2/2-1/2
f(1+x^2)-g(x)=ln(1+x^2)-x^2/2+1/2
令1+x^2=t, 上式=lnt-t/2+1