已知函数f(x)=(ax+b)除
解:
(1)显然有f(0)=0, 即b=0
又f(1/2)=2/5, 即f(1/2)=2/5, ∴a=1
故f(x)=x/(x²+1)
(2)任取x1,x2∈(-1,1)且x1<x2, x2-x1>0
则
`f(x1)-f(x2)
=x1/(x1²+1)-x2/(x2²+1)
=[x1(x2²+1)-x2(x1²+1)]/[(x1²+1)(x2²+1)]
=[x1x2(x2-x1)+(x1-x2)]/[(x1²+1)(x2²+1)]
=[(x2-x1)(x1x2+1)]/[(x1²+1)(x...全部
解:
(1)显然有f(0)=0, 即b=0
又f(1/2)=2/5, 即f(1/2)=2/5, ∴a=1
故f(x)=x/(x²+1)
(2)任取x1,x2∈(-1,1)且x1<x2, x2-x1>0
则
`f(x1)-f(x2)
=x1/(x1²+1)-x2/(x2²+1)
=[x1(x2²+1)-x2(x1²+1)]/[(x1²+1)(x2²+1)]
=[x1x2(x2-x1)+(x1-x2)]/[(x1²+1)(x2²+1)]
=[(x2-x1)(x1x2+1)]/[(x1²+1)(x2²+1)]
∵x1x2∈(-1,1), x1x2+1>0, x2-x1>0, (x1²+1)(x2²+1)>0
故f(x1)-f(x2)>0
即f(x)单调递增。
(3)原不等式等价于
f(t-1)+f(t)<0
即f(t-1)<-f(t)
由奇函数性质f(-x)=-f(x), 得
f(t-1)<f(-t)
由于f(x)单调递增, 所以只需解以下不等式组
{-1<t-1<1 ==> 0<t<2
{-1<-t<1 ==> -1<t<1
{t-1<-t ==> t<1/2
取交集, 得t∈(0,1/2)。
。收起