ข้อสอบ PAT 1 - เมษายน 2557

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{c} -\mathrm{x}+\mathrm{a} , \mathrm{x}\le -2\to \boxed{1}\\ -\frac{2}{5}\mathrm{x}+\mathrm{b} ,-2<\mathrm{x}<3\to \boxed{2}\\ {\mathrm{x}}^2-6\mathrm{x}+11 , \mathrm{x}>3\to \boxed{3}\end{array}\right. \\ \hspace{0.17em}\hspace{0.17em} \mathrm{ฟังก์ชัน} \mathrm{f} \mathrm{มีความต่อเนื่องที่} \mathrm{x}=-2 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \mathrm{f}\left(-2\right)=\underset{\mathrm{x}\to -2^{-}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to -2^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right) \\ \mathrm{f}\left(-2\right)=\underset{\mathrm{x}\to -2^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right) \end{gathered}$$​

$$\begin{gathered} \mathrm{เงื่อนไข} \boxed{1}=\mathrm{เงื่อนไข} \boxed{2} \\ -\mathrm{x}+\mathrm{a}=\underset{\mathrm{x}\to -2^{+}}{\lim }(-\frac{2}{5}\mathrm{x}+\mathrm{b}) \\ -\left(-2\right)+\mathrm{a}=-\frac{2}{5}\left(-2\right)+\mathrm{b} \\ 2+\mathrm{a}=\frac{4}{5}+\mathrm{b} \\ \mathrm{a}=\mathrm{b}-\frac{6}{5}\to \boxed{1} \\ \mathrm{จากโจทย์}\hspace{0.17em} \underset{\mathrm{x}\to 3}{\lim }\mathrm{f}\left(\mathrm{x}\right) \mathrm{หาค่าได้} \end{gathered}$$
$$\begin{gathered} \mathrm{จะได้} \underset{\mathrm{x}\to 3^{-1}}{\lim }\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 3^{+}}{\lim }\mathrm{f}\left(\mathrm{x}\right) \\ \mathrm{เงื่อนไข} \boxed{2}=\mathrm{เงื่อนไข} \boxed{3} \\ \underset{\mathrm{x}\to 3^{-}}{\lim }(-\frac{2}{5}\mathrm{x}+\mathrm{b})=\underset{\mathrm{x}\to 3^{+}}{\lim }\left({\mathrm{x}}^2-6\mathrm{x}+11\right) \\ -\frac{2}{5}\left(3\right)+\mathrm{b}=3^2-6\left(3\right)+11 \\ -\frac{6}{5}+\mathrm{b}=2 \\ \mathrm{b}=\frac{16}{5} \mathrm{แทนใน} \boxed{1} \end{gathered}$$
$$\begin{gathered} \mathrm{จะได้} \mathrm{a}=\frac{16}{5}-\frac{6}{5}=2 \\ \mathrm{ดังนั้น} \left|\mathrm{a}+5\mathrm{b}\right|=\left|2+5\left(\frac{6}{5}\right)\right| \\ =18 \end{gathered}$$