ข้อสอบ PAT 1 - ตุลาคม 2555

$$\begin{gathered} \underline{\mathrm{เซต} \mathrm{A}} \\ \mathrm{จากโจทย์} \\ \sqrt{3\mathrm{x}+2+2\sqrt{3\mathrm{x}+1}}+\sqrt{3\mathrm{x}+10+6\sqrt{3\mathrm{x}+1}}=14\to \boxed{1} \\ \mathrm{กำหนดให้} \mathrm{a}=\sqrt{3\mathrm{x}+1}\to {\mathrm{a}}^2=3\mathrm{x}+1 \\ {\mathrm{a}}^2-1=3\mathrm{x} \\ -\mathrm{แทน} \mathrm{a} , {\mathrm{a}}^2-1 \mathrm{ใน} \boxed{1} \\ \mathrm{จะได้} \sqrt{\left({\mathrm{a}}^2-1\right)+2+2\mathrm{a}}+\sqrt{\left({\mathrm{a}}^2-1\right)+10+6\mathrm{a}}=14 \\ \sqrt{{\mathrm{a}}^2+2\mathrm{a}+1}+\sqrt{{\mathrm{a}}^2+6\mathrm{a}+9}=14 \\ \sqrt{{\left(\mathrm{a}+1\right)}^2}+\sqrt{{\left(\mathrm{a}+3\right)}^2}=14 \\ \mathrm{โดย} \sqrt{{\mathrm{x}}^2}=\left|\mathrm{x}\right| ; \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} \left|\mathrm{a}+1\right|+\left|\mathrm{a}+3\right|=14 \end{gathered}$$

$$\begin{gathered} \mathbf{โดย} \mathrm{a}=\sqrt{3\mathrm{x}+1} \mathrm{แสดงว่า} \mathrm{a}\ge 0 \\ \mathrm{จะได้} \left(\mathrm{a}+1\right)+\left(\mathrm{a}+3\right)=14 \\ 2\mathrm{a}+4=14 \\ \mathrm{a}=5 \\ \mathrm{โดย} a=\sqrt{3\mathrm{x}+1} ; \\ \mathrm{จะได้} \sqrt{3\mathrm{x}+1}=5 \\ 3\mathrm{x}+1=25 \\ \mathrm{x}=8 \end{gathered}$$

$$\begin{gathered} \mathbf{ตรวจคำตอบ} \mathrm{แทน} \mathrm{x}=8 \mathrm{ใน} \boxed{1} \\ \mathrm{จะได้} \sqrt{3\left(8\right)+2+2\left(5\right)}+\sqrt{3\left(8\right)+10+6\left(5\right)}=14 \\ \sqrt{26+10}+\sqrt{34+30}=14 \\ 6+8=14 \\ 14=14 \\ \mathrm{ดังนั้น} \mathrm{A}=\left\{8\right\} \\ \underline{\mathrm{เซต} \mathrm{B}} \\ \mathrm{จากโจทย์} 2{\mathrm{x}}^2-6\mathrm{x}+11+2\sqrt{{\mathrm{x}}^2-3\mathrm{x}+5}=25\to \boxed{2} \end{gathered}$$

$$\begin{gathered} \mathrm{กำหนดให้} \mathrm{b}=\sqrt{{\mathrm{x}}^2-3\mathrm{x}+5}\to {\mathrm{b}}^2={\mathrm{x}}^2-3\mathrm{x}+5 \\ 2{\mathrm{b}}^2=2{\mathrm{x}}^2-6\mathrm{x}+10 \\ 2{\mathrm{b}}^2-10=2{\mathrm{x}}^2-6\mathrm{x} \\ -\mathrm{แทน} \mathrm{b} , 2{\mathrm{b}}^2-10 \mathrm{ใน} \boxed{2} \\ \mathrm{จะได้} \left(2{\mathrm{b}}^2-10\right)+11+2\mathrm{b}=25 \\ 2{\mathrm{b}}^2+2\mathrm{b}-24=0 \\ {\mathrm{b}}^2+\mathrm{b}-12=0 \\ \left(\mathrm{b}+4\right)\left(\mathrm{b}-3\right)=0 \\ \mathrm{b}=-4,3 \end{gathered}$$
$\mathrm{b}=-4 \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ใช้ไม่ได้}\hspace{0.17em}\hspace{0.17em}\mathrm{เพราะ}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\sqrt{}\ge 0$
$$\begin{gathered} \mathrm{และ} \mathrm{b}=3 \\ \sqrt{{\mathrm{x}}^2-3\mathrm{x}+5}= 3 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} {\mathrm{x}}^2-3\mathrm{x}+5 = 9 \\ {\mathrm{x}}^2-3\mathrm{x}-4 = 0 \\ (\mathrm{x}-4)(\mathrm{x}+1) = 0 \\ \mathrm{x}\hspace{0.17em}=4,-1 \end{gathered}$$ 

$$\begin{gathered} \mathbf{ตรวจคำตอบ} \\ \mathrm{แทน} \mathrm{x}=-1 \mathrm{ใน} \boxed{2} \\ \mathrm{จะได้} 2{\left(-1\right)}^2-6\left(-1\right)+11+2\sqrt{{\left(-1\right)}^2-3\left(-1\right)+5}=25 \\ 2+6+11+2\sqrt{1+3+5}=25 \\ 19+2\left(3\right)=25 \\ 25=25 \end{gathered}$$

$$\begin{gathered} \mathrm{แทน} \mathrm{x}=4 \mathrm{ใน} \boxed{2} \\ \mathrm{จะได้} 2{\left(4\right)}^2-6\left(4\right)+11+2\sqrt{4^2-3\left(4\right)+5}=25 \\ \hspace{0.17em}\hspace{0.17em} 32-24+11+2\sqrt{16-12+5}=25 \\ \hspace{0.17em}\hspace{0.17em} 19+2\left(3\right)=25 \\ \hspace{0.17em}\hspace{0.17em} 25=25 \\ \mathrm{ดังนั้น} \mathrm{B}=\left\{-1,4\right\} \\ \mathrm{หาค่า} \mathrm{ผลบวกของสมาชิก} \mathrm{A}\cup \mathrm{B}=8+\left(-1\right)+4 \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}=11 \end{gathered}$$