解得{■(a=1"," @b="-" 1)┤或{■(a=9/4 "," @b="-" 8/3 "." )┤

　　∵a,b∈Z,∴f(x)=x+1/(x"-" 1).

(2)证明在曲线y=f(x)上任取一点(x_0 "," x_0+1/(x_0 "-" 1)).

　　由f'(x0)=1-1/("(" x_0 "-" 1")" ^2 ),知过此点的切线方程为

　　y-(x_0^2 "-" x_0+1)/(x_0 "-" 1)=[1"-" 1/("(" x_0 "-" 1")" ^2 )](x-x0).

　　令x=1,得y=(x_0+1)/(x_0 "-" 1),即切线与直线x=1的交点为(1"," (x_0+1)/(x_0 "-" 1));

　　令y=x,得y=2x0-1,即切线与直线y=x的交点为(2x0-1,2x0-1);

　　直线x=1与直线y=x的交点为(1,1).

　　∴三条直线围成的三角形面积为

　　1/2 |(x_0+1)/(x_0 "-" 1) "-" 1||2x0-1-1|

　　=1/2 |2/(x_0 "-" 1)||2x0-2|=2.

　　故所围成的三角形面积为定值2.

B组

1.直线y=kx+1与曲线y=x3+ax+b相切于点A(1,3),则2a+b的值为(　　)

A.2 B.-1 C.1 D.-2

解析:由条件可知,点A(1,3)在直线y=kx+1上,则k=2.

　　∵点A在曲线y=x3+ax+b上,

　　∴a+b+1=3,即a+b=2.

　　由y=x3+ax+b,得y'=3x2+a,

　　∴3+a=k=2.

　　∴a=-1,b=3.∴2a+b=1.

答案:C

2.导学号88184025函数f(x)=(x-a)(x-b)(x-c)(a,b,c是两两互不相等的常数),则a/(f"'(" a")" )+b/(f"'(" b")" )+c/(f"'(" c")" )=　　　　　.

解析:∵f(x)=x3-(a+b+c)x2+(ab+bc+ca)x-abc,∴f'(x)=3x2-2(a+b+c)x+ab+bc+ca.

　　∴f'(a)=(a-b)(a-c),同理f'(b)=(b-a)(b-c),f'(c)=(c-a)(c-b).

　　代入原式,得a/(f"'(" a")" )+b/(f"'(" b")" )+c/(f"'(" c")" )=0.

答案:0