已知函数f(x)=((a+1)x-3)/(x-1),不等式f(x)1恒成立,求a的取值范围
问题描述:
已知函数f(x)=((a+1)x-3)/(x-1),不等式f(x)1恒成立,求a的取值范围
答
f(x)=((a+1)x-3)/(x-1),
不等式f(x)
即:
((a+1)x-3)/(x-1) > x-a
因为x>1,两边同乘(x-1)
(a+1)x-3 > (x-1)(x-a)=x2-(a+1)x +a
x2-2(a+1)x x2-2(a+1)x +1 (x-a-1)2 =0
解得:
a
答
((a+1)x-3)/(x-1)<x-a→[((a+1)x-3)-(x-a)(x-1)]/(x-1)<0→(-x^2+2(a+1)x-a-3)/(x-1)<0对任意x>1恒成立∴-x^2+2(a+1)x-a-3<0,x>1恒成立令g(x)=-x^2+2(a+1)x-a-3当对称轴x=a+1>1,即a>0时,则要求g(a+1)=a^2-a-2...