三、[] =

       =.

四、[] ,得

      

       =

       所以,极限存在且为.

       得到f (0) += 0+= 1. 如果f (0)f (x)的极值,则= 0

       f (0) = 0,此时= 1,所以,当f (0) = 0时,f (0)f (x)的极小值.

五、[] x = y = 0,得f (0) = f (0) + a f (0),即a f (0) = 0.

       (1) a = 0f (x + y) = f (x)=

              =,得f (x) = C,又,得C = 1,即f (x) = .

       (2) a ¹ 0,则f (0) = 0f (x + y) = f (x) + af (y)

              此时

=

= f (x) + a= f (x) + a,即= f (x) + a,又,得a = 1

所以,= f (x) +,解得f (x) = x.

因此,当a = 0时,f (x) = ;当a = 1时,f (x) = x.

六、[] F(x) =,即证明F(1) > 0.

       .

       由于> 0,即f (x)单调增加,所以,当x > 0时,f (x) > f (0) = 0.

       g(x) =,由于,因此,

       = 2 f (x) - 2 f (x)= 2 f (x)[1 -] > 0 (x > 0),即g(x)单调增加,

       即当x > 0时,g (x) > g (0) = 0.

       所以,当x > 0时,> 0,即F(x)单调增加,因此,F(1) > F(0) = 0

       即原不等式成立.

七、[] 设所求曲线方程为y = y(x),有y(0) = 0.

       =,曲线在P点处的切线为Y - y = (X - x),令X = 0

       得切线跟y轴的交点为(0 , y -x),所以=.

       ,得3+ 2 = 2(x + 1)

       两边求导并整理得2(x + 1)= 1 +,令P =,得

       两边积分得= C(1 + x),即1 += C(1 + x),代入,得= x

       = y =+ C,代入y(0) = 0,得y =.

八、[] 方程lnx - a = 0等价,设f (x) =lnx - a

       ,得x =,所以(0 ,)(, +¥)f (x)的单调区间.

由于=,所以由零点定理及单调性,有

(1) > 0- a < 0,即0 < a <时,方程f (x) = 0在区间(0 ,)(, +¥)

各有一个根.

(2) > 0- a ³ 0,即a £ 0时,方程f (x) = 0在区间(0 ,)内有一个根.

(3) < 0- a £ 0,即a >时,方程f (x) = 0无根.

(4) = 0时,即a =时,方程f (x) = 0有一个根x =.

所以,a的取值范围为a =a £ 0.