Question

已知集合M是满足下列性质的所有函数f(x)组成的集合，对于函数f(x),定义域内任意两个不同自变量x1,x2, ...

已知集合M是满足下列性质的所有函数f(x)组成的集合，对于函数f(x),定义域内任意两个不同自变量x1,x2, 均有|f(x1)减f(x2)|小于等于|x1减x2|成立 (1)判断函数f(x)=3x+1是否属于集合M?说明理由 (2)若g(x)=a(x+1/x)在(1,正无穷)上属于M,求实数a的取值范围 急

Answer 1

f(x)=3x+1 取 x1=0，x2=1
|f(x1)-f(x2)|=3 |0-1|=1
3>1
f(x)=3x+1 不属于集合M
取x1>x2>1
|g(x1)-g(x2)|=a[(x1-x2)+(x2-x1)/x1x2]<=x1-x2
(x1-x2)[a-1-(1/x1x2)]<=0
a<=1+(1/x1x2)
x1>x2>1, 0<1/x1x2<1
a<=1

Answer 2

(1)|f(x1)-f(x2)|=3|x1-x2|==>|f(x1)-f(x2)|/ |x1-x2|>1
∴f(x)不属于M
(2设x1,x2在(1,+∞)内
|a(x1-x2+(x2-x1)/(x1x2))|/|x1-x2|=|a(1-1/(x1x2))|
∴-x1x2/(x1x2-1)<=a<= x1x2/(x1x2-1)
x1x2→∞，-1<=a<=1

Answer 3

x1x2→∞，-1<=a<=1