ข้อสอบ PAT 1 - ตุลาคม 2555

$$\begin{gathered} \underline{\mathrm{พิจารณา} \mathrm{A}} \\ \mathrm{จากโจทย์}\hspace{0.17em} \hspace{0.17em} \left|2\mathrm{x}+7\right|\le 9 \\ \mathrm{จะได้} -9\le 2\mathrm{x}+7\le 9 \\ -16\le 2\mathrm{x}\le 2 \\ -8\le \mathrm{x}\le 1 \\ \mathrm{แสดงว่า} \mathrm{A}=\left\{-8,-7,-6,...,0,1\right\} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณา} \mathrm{B}} \\ \mathrm{จากโจทย์}\hspace{0.17em} \left|{\mathrm{x}}^2-\mathrm{x}-1\right|>0 \\ \mathrm{ยกกำลัง} 2 \mathrm{ทั้ง} 2 \mathrm{ข้าง} ; \\ \mathrm{จะได้} {\left({\mathrm{x}}^2-\mathrm{x}-1\right)}^2>1^2 \\ {\left({\mathrm{x}}^2-\mathrm{x}-1\right)}^2-1^2>0 \\ \left[\left({\mathrm{x}}^2-\mathrm{x}-1\right)-1\right]\left[\left({\mathrm{x}}^2-\mathrm{x}-1\right)+1\right]>0 \\ \left({\mathrm{x}}^2-\mathrm{x}-2\right)\left({\mathrm{x}}^2-\mathrm{x}\right)>0 \\ \left(\mathrm{x}-2\right)\left(\mathrm{x}+1\right)\mathrm{x}\left(\mathrm{x}-1\right)>0 \end{gathered}$$


$$\begin{gathered} \mathrm{แสดงว่า} \mathrm{B}=\left\{...,-4,-3,-2,3,4,5\right\} \\ \mathrm{ดังนั้น} \\ \mathrm{A}\cap \mathrm{B}=\left\{-8,-7,-6,-5,-4,-3,-2\right\} \mathrm{ข้อ} \mathrm{ก}. \mathrm{ถูก} \\ \mathrm{A}-\mathrm{B}=\left\{-1,0,1\right\} \mathrm{ข้อ} \mathrm{ข}. \mathrm{ถูก} \end{gathered}$$