函数f(x)=(ax+b)/（x²-1）是定义在(-1,1)上单调递减的奇函数,求实数a的取值范围

因f(x)是定义在(-1,1)的奇函数,因此f(0)=0,即b/(-1)=0,b = 0,所以f(x)=ax/(x²-1).
设x1>x2,f(x1)-f(x2)=ax1/(x1²-1) - ax2/(x2²-1)=(ax1x2^2-ax1 -ax2x1^2 + ax2)/[(x1²-1)(x2²-1)]
=[ax1x2(x2-x1)+a(x2-x1)]/[(x1²-1)(x2²-1)]=a(x2-x1)(x1x2+1)/[(x1²-1)(x2²-1)] < 0(因f(x)单调递减)
因 x1,x2∈(-1,1),所以x1x2 > -1,x1x2+1> 0,(x1²-1)(x2²-1) > 0,x1 - x2 > 0,
因此,a > 0