Question

已知函数f(x)=lg[a/(x^2+1)],在定义域内存在x0使得f(x0+1)=f(x0)+f(1),求a的取值范围

Answer 1

lg[a/｛（x0+1）^2+1)｝=lg[a/(x0^2+1)] + lg[a/(1+1)]

=lg｛a/(x0^2+1)] * [a/(1+1)] ｝

a/｛（x0+1）^2+1）｝=a/(x0^2+1)] * [a/(1+1)] 有实根.化简用二次方程判别.