6. (Mathematics of Intermediate Level, Issue 1, 2005 Olympiad Training Problem) If the inequality $x|x-a|+b<0$ (where $b$ is a constant) holds for any value of $x$ in $[0,1]$, find the range of real numbers $a$.

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The inequality $x|x-a|+b<0$ (where $b$ is a constant) must hold for all $x$ in the interval $[0,1]$. Determine the range of real numbers $a$.
6. Obviously $b1+b$.
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In summary, when $-1 \leqslant b<2 \sqrt{2}-3$, the range of $a$ is $(1+b, 2 \sqrt{-b})$; when $b<-1$, the range of $a$ is $(1+b, 1-b)$. Obviously, $b$ cannot be greater than or equal to $2 \sqrt{2}-3$.