若定义在$R$上的函数$f\left(x\right)$满足$f\left(0\right)=-1$,其导函数满足$f^{`}\left(x\right)>k>1$，则下列结论中一定错误的是

$f\left(\frac{1}{k}\right)<\frac{1}{k}$ $f\left(\frac{1}{k}\right)>\frac{1}{k-1}$ $f\left(\frac{1}{k-1}\right)<\frac{1}{k-1}$ $f\left(\frac{1}{k-1}\right)>\frac{k}{k-1}$