已知函数f(x)的定义域为[0,1]且同时满足:①对任意x∈[0,1]总有f(x)≥2;②f(1)=3;③若x1≥0,x2≥0且x1+x2≤1,则有f(x1+x2)=f(x1)+f(x2)-2. (I)求f(0)的值; (II)求f(x)的

问题描述:

已知函数f(x)的定义域为[0,1]且同时满足:①对任意x∈[0,1]总有f(x)≥2;②f(1)=3;③若x1≥0,x2≥0且x1+x2≤1,则有f(x1+x2)=f(x1)+f(x2)-2.
(I)求f(0)的值;
(II)求f(x)的最大值;
(III)设数列{an}的前n项和为Sn,且Sn=−

1
2
(an−3)(n∈N*),求f(a1)+f(a2)+…+f(an).

(Ⅰ)令x1=x2=0,
由③知f(0)=2f(0)-2⇒f(0)=2;
(Ⅱ)任取x1x2∈[0,1],且x1<x2
则0<x2-x1≤1,∴f(x2-x1)≥2
∴f(x2)-f(x1)=f[(x2-x1)+x1]-f(x1
=f(x2-x1)+f(x1)-2-f(x1)=f(x2-x1)-2≥0
∴f(x2)≥f(x1),则f(x)≤f(1)=3.
∴f(x)的最大值为3;
(Ⅲ)由Sn=−

1
2
(an−3)知,
n=1时,a1=1;当n≥2时,an=−
1
2
an+
1
2
an−1

an
1
3
an−1(n≥2),又a1=1,∴an
1
3n−1

f(an)=f(
1
3n−1
)=f(
1
3n
+
1
3n
+
1
3n
)=f(
2
3n
)+f(
1
3n
)−2

=3f(
1
3n
)−4=3f(an+1)−4

f(an+1)=
1
3
f(an)+
4
3

f(an+1)−2=
1
3
(f(an)−2)

又f(a1)-2=1∴f(an)−2=(
1
3
)n−1,∴f(an)=(
1
3
)n−1+2

f(a1)+f(a2)++f(an)=2n+
3
2
1
3n−1
.