已知f(x)={x−a,x<1lnx,x≥1f(x)=\begin{cases}x - a, & x < 1 \\\ln x, & x \geq 1\end{cases}f(x)={x−a,lnx,x<1x≥1,若函数f(x)f(x)f(x)在x=1x = 1x=1处连续,求aaa的值.
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