问题
解答题
定义在区间(-1,1)上的函数f(x)满足: ①对任意x,y∈(-1,1),都有f(x)+f(y)=f(
②当x∈(-1,0)时,f(x)>0.求证: (1)f(0)=0; (2)f(x)在(-1,1)上是减函数; (3)f(
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答案
(1)令x=y=0,
有2f(0)=f(0),
∴f(0)=0;
(2)令-1<x1<x2<1,则
f(x1)-f(x2)=f(x1)+f(-x2)=f(
),x1-x2 1-x1•x2
∵-1<x1<x2<1,
∴x1-x2<0,1-x1•x2>0,
∴-1<
<0,x1-x2 1-x1•x2
∴f(
)>0,x1-x2 1-x1•x2
即f(x1)>f(x2),
∴f(x)在(-1,1)上是减函数;
(3)令y=-x,则f(x)+f(-x)=f(0)=0,
∴f(-x)=-f(x),
∴f(x)为奇函数;
∴-f(
)=f(-1 5
)=f(1 5
) =f(3)+f(-2)=f(3)-f(2),①3-2 1+3×(-2)
-f(
)=f(-1 11
)= =f(1 11
) =f(4)+f(-3)=f(4)-f(3),②4-3 1+4×(-3)
…
-f(
)=f(-1 n2+3n+1
)=f(n+2)+f[-(n+1)]=f(n+2)-f(n+1) ③(n+2)-(n-1) 1+(n+2)•[-(n+1))]
将上式①②…③n个式子累加有
-[f(
)+f(1 5
)+f(1 11
)+…+f(1 19
)]1 n2+3n+1
=f(-
)+f(-1 5
)+f(-1 11
)+…+f(-1 19
)1 n2+3n+1
=f(n+2)-f(2)=f(
),n 1-2(n+2)
又f(x)在(-1,1)上是减函数;
∴f(
)=f(-n 1-2(n+2)
)<f(-n 2n+3)
) =-f(n 2n
)<f(-1 2
) =-f(n 2n
),1 2
∴f(
)+f(1 5
)+f(1 11
)+…+f(1 19
)>f(1 n2+3n+1
)1 2