ข้อสอบ PAT 1 - กุมภาพันธ์ 2563

$$\begin{gathered} \mathrm{จาก} f\left(x\right)-g\left(x\right) = x^2-2x \dots \overline{)1} \\ \mathrm{แทน} x \mathrm{ด้วย} -x \\ f\left(-x\right)-g\left(-x\right) = {\left(-x\right)}^2-2\left(-x\right) \\ \mathrm{จาก} \left(ก\right) \mathrm{และ} \left(ข\right) \\ -f\left(x\right)-g\left(x\right) = x^2+2x \dots \overline{)2} \\ \overline{)1}+\overline{)2} : -2·g\left(x\right) = 2x^2 \\ g\left(x\right) = -x^2 \mathrm{แทนใน} \overline{)1} \end{gathered}$$

$$\begin{gathered} f\left(x\right)-\left(-x^2\right) = x^2-2x \\ f\left(x\right) = -2x \\ \mathrm{จาก} f\left(10+a\right)-f\left(10-a\right) = g\left(10\right) \\ -2\left(10+a\right)-\left[-2\left(10-a\right)\right] = -{10}^2 \\ -20-2a+20-2a = -100 \\ 100 = 4a \\ a = 25 \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} f\left(g\left(a\right)\right) = f\left(g\left(25\right)\right) \\ = f\left(-{25}^2\right) \\ = f\left(-625\right) \\ = -2\left(-625\right) \\ = 1250 \\ \mathrm{ตอบข้อ} 1. \end{gathered}$$