Question

已知mx^2-2x+1+m>0,当x∈【-1,1】时,恒成立,求实数m的取值范围。最好有详细的过程。

Answer 1

mx^2-2x+1+m>0
m(x^2+1)>2x-1
m>(2x-1)/(x^2+1)
设f(x)=(2x-1)/(x^2+1),下面证明f(x)在∈【-1,1】是增函数.
设x1,x2∈【-1,1】，且有x1<x2,则
f(x1)-f(x2)
=(2x1-1)/(x1^2+1)-[(2x2-1)/(x2^2+1)]
=[(2x1-1)(x2^2+1)-(2x2-1)(x1^2+1)/[(x1^2+1)(x2^2+1)]
=[2x1*x2^2+2x1-x2^2-1-(2x1^2*x2+2x2-x1^2-1)]/[(x1^2+1)(x2^2+1)]
=[2x1x2(x2-x1)+2(x1-x2)+(x1^2-x2^2)]/[(x1^2+1)(x2^2+1)]
=[2x1x2(x2-x1)-2(x2-x1)-(x1+x2)(x2-x1)]/[(x1^2+1)(x2^2+1)]
={[2x1x2-2-(x1+x2)](x2-x1)}/[(x1^2+1)(x2^2+1)]
由于x1,x2∈【-1,1】，且有x1<x2，则有2x1x2-2-(x1+x2)<0,x2-x1>0,
故f(x1)-f(x2)<0
因此f(x)在∈【-1,1】是增函数.则其最大值为f(1)=1/2
已知mx^2-2x+1+m>0,当x∈【-1,1】时,恒成立,即m>(2x-1)/(x^2+1)，当x∈【-1,1】时,恒成立，
因此只需要m>maxf(x)即可，故m>f(1)=1/2
实数m的取值范围为m>1/2