ข้อสอบคณิต 2 - มีนาคม 2562

$$\begin{gathered} \mathrm{กรณี} a<1 \mathrm{จะได้} \\ f\left(a\right) = \left|a-1\right|+1 \\ 10 = \left|a-1\right|+1 \\ 9 = \left|a-1\right| \\ 9, -9 = a-1 \\ a = \xcancel{10 }, -8 \end{gathered}$$

$$\begin{gathered} \mathrm{กรณี} a\ge 1 \mathrm{จะได้} \\ f\left(a\right) = {\left(a-1\right)}^2+1 \\ 10 = {\left(a-1\right)}^2+1 \\ 9 = {\left(a-1\right)}^2 \\ 3 , -3 = a-1 \\ a = 4 ,\xcancel{ -2} \end{gathered}$$

$$\begin{gathered} \mathrm{รวม} 2 \mathrm{กรณี} \mathrm{จะได้คำตอบ} \mathrm{คือ} -8, 4 \\ \mathrm{ดังนั้น} -8+4 = -4 \to \mathrm{ตอบ} 1 \end{gathered}$$