Question

已知函数f(x)=mx+n/1+x²是定义在[-1/2,1/2]上的奇函数,且f(-1/4)=8/17(1)确定函数解析式。

（2）用定义证明函数f(x)在[-1/2,1/2]上是减函数。（3）若实数t满足f(3t)+f（t+1)＜0.求t的取值范围

Answer 1

已知函数f(x)=(mx+n)/(1+x²)是定义在[-1/2,1/2]上的奇函数,且f(-1/4)=8/17.
(1)确定函数解析式｡
f(-1/2)=(-m/2+n)/(1+1/4)
f(1/2)=(m/2+n)/(1+1/4)
f(-1/2)=-f(1/2)
(-m/2+n)/(1+1/4)=-(m/2+n)/(1+1/4)
-m/2+n=-m/2-n
n=0
f(x)=mx/(1+x²)
f(-1/4)=(-m/4)/(1+1/16)=-4m/17
f(-1/4)=8/17
-4m/17=8
m=-2
f(x)=-2x/(1+x²)

(2)用定义证明函数f(x)在[-1/2,1/2]上是减函数｡
1/2>x1>x2>-1/2
f(x1)-f(x2)=-2x1/(1+x1²)+2x2/(1+x2²)
=2(x2/(1+x2²)-x1/(1+x1²)
=2[x2(1+x1²)-x1(1+x2²)]/[(1+x2²)(1+x1²)]
(1+x2²)(1+x1²)>0
x2(1+x1²)-x1(1+x2²)
=x2+x1²x2-x1-x2²x1
=(x2-x1)-x1x2(x2-x1)
=(x2-x1)(1-x1x2)
x2-x1<0
x1x2<1/4 1-x1x2>0
f(x1)-f(x2)<0
f(x)在 [-1/2，1/2] 减

(3)若实数t满足f(3t)+f(t+1)<0.求t的取值范围
-1/2≤3t≤1/2 -1/2/≤t+1≤1/2
-1/6≤t≤1/6 -3/2≤t≤-1/2
不成立
题目有误