Question

设函数f(x)=√x^2+1-ax期中a>0使函数f(x)在区间【0 +∞】上单调求a的取值范围

Answer 1

证明:首先设x1>x2≥0,则
F(x1)-F(x2)=√(x1^2+1)-ax1-√(x^2+1)+ax2
=(x1^2-x2^2)/[√(x1^2+1)+√(x2^2+1)]-a(x1-x2)
=(x1-x2)[(x1+x2)/(√(x1^2+1)+√(x2^2+1))-a]<0
又因为x1>x2≥0,即x1-x2>0,
所以(x1+x2)/(√(x1^2+1)+√(x2^2+1))-a<0
x1+x2<a(√(x1^2+1)+√(x2^2+1))
a>(x1+x2)/(√(x1^2+1)+√(x2^2+1))
因为(√(x1^2+1)+√(x2^2+1))>x1+x2>0
所以0<(x1+x2)/(√(x1^2+1)+√(x2^2+1)<1
即当a≥1时,a>(x1+x2)/[√(x1^2+1)+√(x2^2+1)]
此时f（x）在[0,＋∞)上是减函数