Question

函数f(x) =log (2)[(ax-1)/(x^2-x+2)+2)]在x属于[1,3]上恒有意义则实数a的取值范围.

Answer 1

x∈[1,3],

(ax-1)/(x²-x+2)+2>0

x²-x+2在R上恒大于0；

∴ax-1+2(x²-x+2)>0;

a>-2x+2-3/x;

2x+3/x≥2√6;

当且仅当2x=3/x时有最小值2√6；即x =√6/2∈[1,3];符合

∴-2x-3/x≤-2√6;

∴a>-2√6+2