【正确答案】正确答案:(1)S
1
(c)=cf(c),S
2
(c)=∫
c
1
f(t)dt=一∫
1
c
f(t)dt,即证明S
1
(c)=S
2
(c),或cf(c)+∫
1
c
f(t)dt=0.令φ(x)=x∫
1
x
f(t)dt,φ(0)=φ(1)=0,根据罗尔定理,存在f∈(0,1),使得φ"(c)=0,即cf(f)+∫
1
c
f(t)dt=0,所以S
1
(c)=S
2
(c),命题得证. (2)令h(x)=xf(x)一∫
x
1
f(t)dt,因为h"(x)=f(x)+xf"(x)>0,所以h(x)在[0,1]上为单调函数,所以(1)中的c是唯一的.
【答案解析】