ข้อสอบ PAT 1 - มีนาคม 2558

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{กำหนดเอกภพสัมพัทธ์คือ} \left\{\mathrm{x}\in \mathrm{R}|1<\mathrm{x}<2\right\} \\ \mathrm{จะได้} \mathrm{U}=\left(1,2\right) \\ \underline{\mathrm{หา} \mathrm{P}\left(\mathrm{x}\right)} \\ \mathrm{จากโจทย์} \mathrm{P}\left(\mathrm{x}\right) \mathrm{แทน} 3{\mathrm{x}}^2-4\mathrm{x}-4<0 \\ \mathrm{จะได้} 3{\mathrm{x}}^2-4\mathrm{x}-4<0 \\ \left(3\mathrm{x}+2\right)\left(\mathrm{x}-2\right)<0 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=(\frac{-2}{3},2) \\ \mathrm{ดังนั้น} \mathrm{เซตคำตอบของ} \mathrm{P}\left(\mathrm{x}\right) \mathrm{คือ} (\frac{-2}{3},2) \end{gathered}$$​

$$\begin{gathered} \underline{\mathrm{หา} \mathrm{Q}\left(\mathrm{x}\right)} \\ \mathrm{จากโจทย์}\hspace{0.17em}\mathrm{Q}\left(\mathrm{x}\right) \mathrm{แทน} {\mathrm{x}}^2>\left|{\mathrm{x}}^2-4\right| \\ \mathrm{จะได้} {\mathrm{x}}^2>\left|{\mathrm{x}}^2-4\right| \\ \left|{\mathrm{x}}^2-4\right|<{\mathrm{x}}^2 \\ -{\mathrm{x}}^2<{\mathrm{x}}^2-4<{\mathrm{x}}^2 \\ \mathrm{จะได้} -{\mathrm{x}}^2<{\mathrm{x}}^2-4 \mathrm{และ} {\mathrm{x}}^2-4<{\mathrm{x}}^2 \\ 4<2{\mathrm{x}}^2 \cap -4<0 \\ {\mathrm{x}}^2-{\left(\sqrt{2}\right)}^2>0 \mathrm{เป็นจริงทุกกรณี} \\ \left(\mathrm{x}-\sqrt{2}\right)\left(\mathrm{x}+\sqrt{2}\right)>0 \mathrm{x}=\mathrm{R} \end{gathered}$$

$$\begin{gathered} \mathrm{x}=\left(-\infty ,-\sqrt{2}\right)\cup \left(\sqrt{2},\infty \right) \\ \mathrm{ดังนั้น} \mathrm{เซตคำตอบของ} \mathrm{Q}\left(\mathrm{x}\right) \mathrm{คือ} \left[\left(-\infty ,-\sqrt{2}\right)\cup \left(\sqrt{2},\infty \right)\cap \left(\mathrm{x}=\mathrm{R}\right)\right] \\ =\left(-\infty ,-\sqrt{2}\right)\cup \left(\sqrt{2},\infty \right) \\ \mathrm{แสดงว่า} \mathrm{U}=\left(1,2\right) \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{P}\left(\mathrm{x}\right)=(\frac{-2}{3},2) \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{Q}\left(\mathrm{x}\right)=\left(-\infty ,-\sqrt{2}\right)\cup \left(\sqrt{2},\infty \right) \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณาข้อ} \left(\mathrm{ก}\right)} \\ \mathrm{จากโจทย์}\hspace{0.17em}\forall \mathrm{x}\left[\mathrm{P}\left(\mathrm{x}\right)\right]\to \exists \mathrm{x}\left[\mathrm{P}\left(\mathrm{x}\right)\wedge \mathrm{Q}\left(\mathrm{x}\right)\right] \mathrm{มีค่าความจริงเป็น}\underline{\mathrm{จริง}} \\ \mathrm{โดย} \forall \mathrm{x}\left[\mathrm{P}\left(\mathrm{x}\right)\right]=\mathrm{ค่า} \mathrm{x} \mathrm{ทุกตัวใน} \mathrm{U} \mathrm{อยู่ใน} \mathrm{P}\left(\mathrm{x}\right) \\ \mathrm{จะได้} \forall \mathrm{x}\left[\mathrm{P}\left(\mathrm{x}\right)\right]\equiv \mathrm{T} \\ \mathrm{โดย} \exists \mathrm{x}\left[\mathrm{P}\left(\mathrm{x}\right)\wedge \mathrm{Q}\left(\mathrm{x}\right)\right]=\mathrm{ค่า} \mathrm{x} \mathrm{บางตัวใน} \mathrm{U} \mathrm{อยู่ใน} \left[\mathrm{P}\left(\mathrm{x}\right)\wedge \mathrm{Q}\left(\mathrm{x}\right)\right] \end{gathered}$$

$$\begin{gathered} \hspace{0.17em}\mathrm{P}\left(\mathrm{x}\right)\wedge \mathrm{Q}\left(\mathrm{x}\right)=\left(\sqrt{2},2\right) \\ \mathrm{เช่น} \mathrm{x}=1.5 \mathrm{ใน} \mathrm{U} \mathrm{อยู่ใน} \left(\sqrt{2},2\right) \\ \mathrm{จะได้} \exists \mathrm{x}\left[\mathrm{P}\left(\mathrm{x}\right)\wedge \mathrm{Q}\left(\mathrm{x}\right)\right]\equiv \mathrm{T} \\ \mathrm{แสดงว่า} \forall \mathrm{x}\left[\mathrm{P}\left(\mathrm{x}\right)\right]\to \exists \mathrm{x}\left[\mathrm{P}\left(\mathrm{x}\right)\wedge \mathrm{Q}\left(\mathrm{x}\right)\right]\equiv \mathrm{T}\to \mathrm{T} \\ \equiv \mathrm{T} \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ก}\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณาข้อ} \left(\mathrm{ข}\right)} \\ \mathrm{จากโจทย์}\hspace{0.17em}\exists \mathrm{x}\left[\mathrm{Q}\left(\mathrm{x}\right)\right]\to \forall \mathrm{x}\left[\mathrm{P}\left(\mathrm{x}\right)\right] \mathrm{มีค่าความจริงเป็น}\underline{\mathrm{เท็จ}} \\ \mathrm{โดย} \exists \mathrm{x}\left[\mathrm{Q}\left(\mathrm{x}\right)\right]=\mathrm{ค่า} \mathrm{x} \mathrm{บางตัว} \mathrm{ใน} \mathrm{U} \mathrm{อยู่ใน} \mathrm{Q}\left(\mathrm{x}\right) \\ \mathrm{เช่น} \mathrm{x}=1.5 \mathrm{ใน} \mathrm{U} \mathrm{อยู่ใน} \left(-\infty ,-\sqrt{2}\right)\cup \left(\sqrt{2},\infty \right) \\ \mathrm{จะได้} \exists \mathrm{x}\left[\mathrm{Q}\left(\mathrm{x}\right)\right]\equiv \mathrm{T} \\ \mathrm{แสดงว่า} \exists \mathrm{x}\left[\mathrm{Q}\left(\mathrm{x}\right)\right]\to \forall \mathrm{x}\left[\mathrm{P}\left(\mathrm{x}\right)\right]\equiv \mathrm{T}\to \mathrm{T} \\ \equiv \mathrm{T} \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ข}\right) \mathrm{ผิด} \end{gathered}$$