ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2562 (วิชาสามัญเก่า)

$$\begin{gathered} \mathrm{จากโจทย์} 4^{\left|3x-1\right|}-2^4 = 6\left(2^{\left|3x-1\right|}\right) \\ 2^{2\left|3x-1\right|}-16 = 6\left(2^{\left|3x-1\right|}\right) \\ {\left(2^{\left|3x-1\right|}\right)}^2-6\left(2^{\left|3x-1\right|}\right)-16 = 0 \end{gathered}$$

$$\begin{gathered} - \mathrm{กำหนดให้} 2^{\left|3x-1\right|} = a \\ \mathrm{จะได้} a^2-6a-16 = 0 \\ \left(a-8\right)\left(a+2\right) = 0 \\ \mathrm{จะได้} a-8 = 0 \mathrm{หรือ} a+2 = 0 \\ a = 8 a = -2 \end{gathered}$$

$$\begin{gathered} - \mathrm{แทน} a = 2^{\left|3x-1\right|} \\ \mathrm{จะได้} 2^{\left|3x-1\right|} = 8 \mathrm{หรือ} 2^{\left|3x-1\right|} = -2 \\ 2^{\left|3x-1\right|} = 2^3 \mathrm{เป็นไปไม่ได้} \\ \left|3x-1\right| = 3 \\ 3x-1 = 3 \mathrm{หรือ} 3x-1 = -3 \\ x = \frac{4}{3} x = \frac{-2}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} \mathrm{ผลบวกของคำตอบ} = \frac{4}{3}+\left(\frac{-2}{3}\right) \\ = \frac{2}{3} \end{gathered}$$