ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{f}\left(\mathrm{x}\right)=\left\{\begin{array}{ll}\frac{1}{\mathrm{x}}& , \left|\mathrm{x}\right|<\frac{1}{2}\to \boxed{1}\\ \frac{1}{2}+\frac{1}{\mathrm{x}}& , \left|\mathrm{x}\right|\ge \frac{1}{2}\to \boxed{2}\end{array}\right. \\ \mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(-\frac{1}{3}\right)\right)\right)=? \\ \underline{\mathrm{หาค่า}} \mathrm{f}\left(-\frac{1}{3}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{โดย} \left|-\frac{1}{3}\right|=\frac{1}{3}<\frac{1}{2} \mathrm{เลือกสมการ} \boxed{1} \\ \mathrm{จะได้} \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}} \\ \mathrm{แทน} \mathrm{x}=-\frac{1}{3}; \mathrm{f}\left(-\frac{1}{3}\right)=\frac{1}{\left(-{\displaystyle \frac{1}{3}}\right)}=-3 \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{หาค่า}} \mathrm{f}\left(\mathrm{f}\left(-\frac{1}{3}\right)\right)=\mathrm{f}\left(-3\right) \\ \mathrm{โดย} \left|-3\right|=3\ge \frac{1}{2} \mathrm{เลือกสมการ} \boxed{2} \\ \mathrm{จะได้} \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{2}+\frac{1}{\mathrm{x}} \\ \mathrm{แทน} \mathrm{x}=-3 \mathrm{f}\left(-3\right)=\frac{1}{2}+\frac{1}{\left(-{\displaystyle 3}\right)}=\frac{1}{2}-\frac{1}{3}=\frac{1}{6} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{หาค่า}} \mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(-\frac{1}{3}\right)\right)\right)=\mathrm{f}\left(\frac{1}{6}\right) \\ \mathrm{โดย} \left|\frac{1}{6}\right|=\frac{1}{6}<\frac{1}{2} \mathrm{เลือกสมการ} \boxed{1} \\ \mathrm{จะได้} \mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{x}} \\ \mathrm{แทน} \mathrm{x}=\frac{1}{6} \mathrm{f}\left(\frac{1}{6}\right)=\frac{1}{{\displaystyle \frac{1}{6}}}=6 \\ \mathrm{ดังนั้น} \mathrm{f}\left(\mathrm{f}\left(\mathrm{f}\left(-\frac{1}{3}\right)\right)\right)=6 \end{gathered}$$