ข้อสอบคณิต O-NET - 2559

$$\begin{gathered} \mathrm{จากโจทย์} A \mathrm{จะได้} \\ \left|x+1\right| \le 2 \\ -2 \le x+1 \le 2 \\ -3 \le x \le 1 \\ \mathrm{ดังนั้น} A = \left[-3, 1\right] \end{gathered}$$

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

$$\begin{gathered} \mathrm{จากแต่ละตัวเลือก} \\ 1. \left[-3, 1\right] \cap \left\{0, 1\right\} = \left\{0\right\} \\ \left\{0, 1\right\} = \left\{0\right\} \to \mathrm{ผิด} \\ 2. \left[-3, 1\right] \cup \left\{0, 1\right\} = \left\{0, 1\right\} \\ \left[-3, 1\right] = \left\{0, 1\right\} \to \mathrm{ผิด} \\ 3. \left\{0, 1\right\} - \left[-3, 1\right] = \varnothing \\ \varnothing = \varnothing \to \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} 4. \left[-3, 1\right] - \left\{0, 1\right\} = \left[-3, 1\right] \\ [-3, 0) \cup (0, 1) = \left[-3, 1\right] \to \mathrm{ผิด} \\ 5. A\text{'} \cup B\text{'} = \left(1, \infty \right) \\ \left(A\cap B\right)\text{'} = \left(1, \infty \right) \\ \left\{0, 1\right\}\text{'} = \left(1, \infty \right) \to \mathrm{ผิด} \end{gathered}$$