14. Set $A=\left\{(x, y) \mid y=x^{2}+m x+2\right\}, B=\{(x, y) \mid x-y+1=0$ and $0 \leqslant x \leqslant 2\}$. If $A \cap B=\varnothing$, find the range of values for $m$.
14. Solve $A \cap B=\varnothing \Leftrightarrow\left\{\begin{array}{l}y=x^{2}+m x+2 \\ y=x+1(0 \leqslant x \leqslant 2)\end{array}\right.$ has no solution, i.e., $x^{2}+(m-1) x+1=0$ has no solution in $[0,2]$.
Because
or
$$
\begin{array}{c}
\Delta<0 \\
f(0)>0, \quad f(2)>0
\end{array}\right. \\
\left\{\begin{array}{l}
\Delta \geqslant 0 \\
-\frac{m-1}{2}<0 \\
f(0)>0, \quad f(2)>0
\end{array}\right.
\end{array}
$$

From (1) we get $-1-1\}$.