ข้อสอบ PAT 1 - มีนาคม 2556

$$\begin{gathered} \mathrm{จากโจทย์} \\ {5}^{\left(1+\sqrt{{\mathrm{x}}^2-4\mathrm{x}-1}\right)}+5^{\left(\frac{5+4\mathrm{x}-{\mathrm{x}}^2}{2+\sqrt{{\mathrm{x}}^2-4\mathrm{x}-1}}\right)}=126\to \boxed{1} \\ \mathrm{กำหนดให้} \mathrm{A}={\mathrm{x}}^2-4\mathrm{x}-1 \\ \mathrm{พิจารณา} 5+4\mathrm{x}-{\mathrm{x}}^2=-\left[{\mathrm{x}}^2-4\mathrm{x}-5\right] \end{gathered}$$

$$\begin{gathered} =-[{\mathrm{x}}^2-4\mathrm{x}-1-4] \\ =-[\mathrm{A}-4] \\ =4-\mathrm{A} \\ \mathrm{จาก} \boxed{1} \mathrm{จะได้} 5^{(1+\sqrt{\mathrm{A}})}+5^{\left(\frac{4-\mathrm{A}}{2+\sqrt{\mathrm{A}}}\right)}=126 \end{gathered}$$

$$\begin{gathered} 5·5^{\sqrt{\mathrm{A}}}+5^{\frac{\cancel{\left(2+\sqrt{\mathrm{A}}\right)}\left(2-\sqrt{\mathrm{A}}\right)}{\left(\cancel{2+\sqrt{\mathrm{A}}}\right)}}=126 \\ 5·5^{\sqrt{\mathrm{A}}}+5^{2-\sqrt{\mathrm{A}}}=126 \\ 5·5^{\sqrt{\mathrm{A}}}+\frac{5^2}{5^{\sqrt{\mathrm{A}}}}=126 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{นำ} 5^{\sqrt{\mathrm{A}}} \mathrm{คูณตลอด} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} 5·5^{2\sqrt{\mathrm{A}}}+25=126·5^{\sqrt{\mathrm{A}}}\to \boxed{2} \\ \mathrm{กำหนดให้} \mathrm{B}=5^{\sqrt{\mathrm{A}}} \\ \mathrm{จาก} \boxed{2} \mathrm{จะได้} 5{\mathrm{B}}^2+25=126\mathrm{B} \end{gathered}$$

$$\begin{gathered} 5{\mathrm{B}}^2-126\mathrm{B}+25=0 \\ \left(5\mathrm{B}-1\right)\left(\mathrm{B}-25\right)=0 \\ \mathrm{B}=\frac{1}{5} , 25 \\ \underline{\mathrm{กรณี} 1} \mathrm{ถ้า} \mathrm{B}=\frac{1}{5} \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} 5^{\sqrt{\mathrm{A}}}=\frac{1}{5} \\ 5^{\sqrt{\mathrm{A}}}=5^{-1} \\ \sqrt{\mathrm{A}}=-1 \mathrm{ซึ่งเป็นไปไม่ได้} \\ \mathrm{A}=\varnothing \\ \mathrm{x}=\varnothing \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{กรณี} 2} \mathrm{ถ้า} \mathrm{B}=25 \\ \mathrm{จะได้} 5^{\sqrt{\mathrm{A}}}=25 \\ 5^{\sqrt{\mathrm{A}}}=5^2 \\ \sqrt{\mathrm{A}}=2 \\ \mathrm{A}=4 \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} {\mathrm{x}}^2-4\mathrm{x}-1=\mathrm{A} \\ {\mathrm{x}}^2-4\mathrm{x}-5=0 \\ (\mathrm{x}-5)(\mathrm{x}+1)=0 \\ \mathrm{x}=5 , -1 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{A} \mathrm{แทนเซตคำตอบของสมการ} \\ \mathrm{จะได้} \mathrm{A}=\left\{5,-1\right\} \\ \mathrm{ดังนั้น} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ผลบวกของสมาชิกในเซต} \mathrm{A} =5+\left(-1\right)=4 \end{gathered}$$