ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \underline{\mathrm{พิจารณา} \mathrm{A}} \\ \mathrm{จากโจทย์}\hspace{0.17em} {\log }_{\sqrt{3}}\left(\mathrm{x}-1\right)-{\log }_{\sqrt[3]{3}}\left(\mathrm{x}-1\right)=1 \\ {\log }_{3^{\frac{1}{2}}}\left(\mathrm{x}-1\right)-{\log }_{3^{\frac{1}{3}}}\left(\mathrm{x}-1\right)=1 \\ 2{\log }_3\left(\mathrm{x}-1\right)-3{\log }_3\left(\mathrm{x}-1\right)=1 \end{gathered}$$

$$\begin{gathered} {\log }_3{(\mathrm{x}-1)}^2-{\log }_3{(\mathrm{x}-1)}^3=1 \\ {\log }_3\left[\frac{{\left(\mathrm{x}-1\right)}^2}{{\left(\mathrm{x}-1\right)}^3}\right]=1 \\ {\log }_3\left[\frac{1}{\left(\mathrm{x}-1\right)}\right]=1 \end{gathered}$$
$$\begin{gathered} \frac{1}{\mathrm{x}-1}=3 \\ 1=3\left(\mathrm{x}-1\right) \\ \mathrm{x}=\frac{4}{3} \\ \mathrm{แสดงว่า} \mathrm{A}=\left\{\frac{4}{3}\right\} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณา} \mathrm{B}} \\ \mathrm{จากโจทย์}\hspace{0.17em} \sqrt{\mathrm{x}+1}+\sqrt{\mathrm{x}-1}=2 \\ \sqrt{\mathrm{x}+1}=2-\sqrt{\mathrm{x}-1} \\ {\left(\sqrt{\mathrm{x}+1}\right)}^2={\left(2-\sqrt{\mathrm{x}-1}\right)}^2 \end{gathered}$$
$$\begin{gathered} \cancel{ \mathrm{x}}+1=4-4\sqrt{\mathrm{x}-1}+\left(\cancel{\mathrm{x}}-1\right) \\ 4\sqrt{\mathrm{x}-1}=2 \\ \sqrt{\mathrm{x}-1}=\frac{1}{2} \\ \mathrm{x}-1=\frac{1}{4} \\ \mathrm{x}=\frac{5}{4} \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \mathrm{B}=\left\{\frac{5}{4}\right\} \\ \mathrm{จะได้} \mathrm{A}\cup \mathrm{B}=\left\{\frac{4}{3},\frac{5}{4}\right\} \\ \mathrm{ดังนั้น} \mathrm{สามเท่าของผลคูณของสมาชิกในเซต} \mathrm{A}\cup \mathrm{B} \\ =\cancel{3}\left(\frac{\bcancel{4}}{\cancel{3}}\right)\left(\frac{5}{\bcancel{4}}\right) \\ =5 \end{gathered}$$