10. Given that $a, b, c$ are real numbers, when and only when $x \leq 0$ or $x>1$, we have $\sqrt{2 x^{2}+a x+b}>x-c$. Find the range of values for $c$.

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10. Solution: From the problem, we know that the solution to the inequality $\sqrt{2 x^{2}+a x+b}>x-c$ is $x \leq 0$ or $x>1$. Solving the inequality should be divided into two cases:
$$
\left\{\begin{array} { c } 
{ 2 x ^ { 2 } + a x + b \geq 0 } \\
{ 2 x ^ { 2 } + a x + b > ( x - c ) ^ { 2 } }
\end{array} \text { or } \left\{\begin{array}{c}
2 x^{2}+a x+b \geq 0 \\
x-c(x-c)^{2} \text{ has the solution set } B, \text{ and the solution set of the inequality } x-cx-c \text{ is } (A \cap B) \cup(A \cap C)=A \cap(B \cup C). \text{ Since the solution to the inequality } \sqrt{2 x^{2}+a x+b}>x-c \text{ is } x \leq 0 \text{ or } x>1, \text{ it follows that } x=0 \text{ must be the smaller root of the equation } 2 x^{2}+a x+b=0, \text{ i.e., } b=0 \text{ and } -\frac{a}{2}>0.
Similarly, } x=1 \text{ must be the larger root of the equation } 2 x^{2}+a x+b=(x-c)^{2}, \text{ i.e., } a+2 c-c^{2}=-1. \text{ It should also satisfy } c>0, \text{ so } c \in(0,1).
$$