ข้อสอบ PAT 1 - เมษายน 2557

$$\begin{gathered} \underline{\mathrm{พิจารณา} \mathrm{A}} \\ \mathrm{จากโจทย์}\hspace{0.17em} {\log }_6\left(3·4^{\mathrm{x}}+2·9^{\mathrm{x}}\right)=\mathrm{x}+{\log }_{6}5 \\ \mathrm{โดย} \mathrm{x}={\mathrm{xlog}}_{6}6={\log }_6{6}^{\mathrm{x}} \end{gathered}$$
$$\begin{gathered} \mathrm{จะได้} {\log }_6\left(3·4^{\mathrm{x}}+2·9^{\mathrm{x}}\right)={\log }_6{6}^{\mathrm{x}}+{\log }_{6}5 \\ {\log }_6\left(3·4^{\mathrm{x}}+2·9^{\mathrm{x}}\right)={\log }_6\left[5·6^{\mathrm{x}}\right] \\ 3·4^{\mathrm{x}}+2·9^{\mathrm{x}}=5·6^{\mathrm{x}} \\ 3·{\left(2^2\right)}^{\mathrm{x}}+2·{\left(3^2\right)}^{\mathrm{x}}=5·{\left(2·3\right)}^{\mathrm{x}} \\ 3·{\left(2^{\mathrm{x}}\right)}^2+2·{\left(3^{\mathrm{x}}\right)}^2=5·2^{\mathrm{x}}·3^{\mathrm{x}} \\ \mathrm{กำหนดให้} \mathrm{A}=2^{\mathrm{x}} \mathrm{และ} \mathrm{B}=3^{\mathrm{x}} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} 3{\mathrm{A}}^2+2{\mathrm{B}}^2=5\mathrm{AB} \\ 3{\mathrm{A}}^2-5\mathrm{AB}+2{\mathrm{B}}^2=0 \\ \left(3\mathrm{A}-2\mathrm{B}\right)\left(\mathrm{A}-\mathrm{B}\right)=0 \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} 3\mathrm{A}-2\mathrm{B}=0 \mathrm{หรือ} \mathrm{A}-\mathrm{B}=0 \\ 3\mathrm{A}=2\mathrm{B} \mathrm{A}=\mathrm{B} \\ \frac{\mathrm{A}}{\mathrm{B}}=\frac{2}{3} 2^{\mathrm{x}}=3^{\mathrm{x} } \\ \frac{2^{\mathrm{x}}}{3^{\mathrm{x}}}=\frac{2}{3} \frac{2^{\mathrm{x}}}{3^{\mathrm{x}}}=1 \\ {\left(\frac{2}{3}\right)}^{\mathrm{x}}=\frac{2}{3} {\left(\frac{2}{3}\right)}^{\mathrm{x}}={\left(\frac{2}{3}\right)}^0 \\ \mathrm{x}=1 \mathrm{x}=0 \\ \mathrm{ดังนั้น} \mathrm{A}=\left\{0,1\right\} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณา} \mathrm{B}} \\ \mathrm{จากโจทย์} \mathrm{x}+\sqrt{1-{\mathrm{x}}^2}=1+2\mathrm{x}\sqrt{1-{\mathrm{x}}^2}\to \boxed{1} \\ \mathrm{พิจารณา} \mathrm{x}+\sqrt{1-{\mathrm{x}}^2} \mathrm{ลองยกกำลัง} 2 \mathrm{ทั้งพจน์} \\ \mathrm{จะได้} {\left[\mathrm{x}+\sqrt{1-{\mathrm{x}}^2}\right]}^2={\mathrm{x}}^2+2\mathrm{x}\sqrt{1-{\mathrm{x}}^2}+{\left(\sqrt{1-{\mathrm{x}}^2}\right)}^2 \\ =1+2\mathrm{x}\sqrt{1-{\mathrm{x}}^2} \end{gathered}$$

$$\begin{gathered} \mathrm{กำหนดให้} \mathrm{M}=\mathrm{x}+\sqrt{1-{\mathrm{x}}^2} \\ \mathrm{แทน} \mathrm{M} \mathrm{ใน} \boxed{1} \\ \mathrm{จะได้} \mathrm{M}={\mathrm{M}}^2 \\ {\mathrm{M}}^2-\mathrm{M}=0 \\ \mathrm{M}\left(\mathrm{M}-1\right)=0 \\ \mathrm{M}=0,1 \end{gathered}$$​

$$\begin{gathered} \mathrm{จะได้} \mathrm{M}=0 \mathrm{หรือ} \mathrm{M}=1 \\ \mathrm{x}+\sqrt{1-{\mathrm{x}}^2}=0 \mathrm{x}+\sqrt{1-{\mathrm{x}}^2}=1 \\ \sqrt{1-{\mathrm{x}}^2}=-\mathrm{x} \sqrt{1-{\mathrm{x}}^2}=1-\mathrm{x} \\ {\left[\sqrt{1-{\mathrm{x}}^2}\right]}^2={\left(-\mathrm{x}\right)}^{2 } {\left[\sqrt{1-{\mathrm{x}}^2}\right]}^2={\left(1-\mathrm{x}\right)}^2 \\ 1-{\mathrm{x}}^2={\mathrm{x}}^2 1-{\mathrm{x}}^2=1-2\mathrm{x}+{\mathrm{x}}^2 \\ 2{\mathrm{x}}^2=1 2{\mathrm{x}}^2-2\mathrm{x}=0 \\ {\mathrm{x}}^2=\frac{1}{2} {\mathrm{x}}^2-\mathrm{x}=0 \\ \mathrm{x}=\pm \frac{1}{\sqrt{2}} \mathrm{x}\left(\mathrm{x}-1\right)=0 \\ \mathrm{x}=0,1 \end{gathered}$$
$$\begin{gathered} \mathrm{แทนค่า} \mathrm{x} \mathrm{ทั้ง} 4 \mathrm{ตัว} \mathrm{ใน} \boxed{1} \mathrm{ซึ่ง} \mathrm{x}=\frac{1}{\sqrt{2}} \mathrm{ทำให้สมการไม่เป็นจริง} \\ \mathrm{ดังนั้น} \mathrm{B}=\left\{-\frac{1}{\sqrt{2}},0,1\right\} \\ \mathrm{จะได้} \mathrm{A}\cup \mathrm{B}=\left\{-\frac{1}{\sqrt{2}},0,1\right\} \\ \mathrm{ดังนั้น} \mathrm{จำนวนสมาชิกของเซต} \mathrm{A}\cup \mathrm{B} \mathrm{เท่ากับ} 3 \mathrm{ตัว} \end{gathered}$$