ข้อสอบ PAT 1 - กุมภาพันธ์ 2562

$\mathrm{นิยาม} \left|x\right| = \left\{\begin{array}{ll}x& \mathrm{เมื่อ} x\ge 0\\ -x& \mathrm{เมื่อ} x<0\end{array}\right.$

$$\begin{gathered} \mathrm{จากโจทย์} f : y+x = \left|x\right| \\ y = \left|x\right|-x \\ y = \left\{\begin{array}{ll}x-x& ; x\ge 0\\ -x-x& ; x<0\end{array}\right. \\ y = \left\{\begin{array}{ll}0& ; x\ge 0\\ -2x& ; x<0\end{array}\right. \\ \mathrm{ดังนั้น} f\left(x\right) = \left\{\begin{array}{ll}0& ; x\ge 0 \to \boxed{1}\\ -2x& ; x<0 \to \boxed{2}\end{array}\right. \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} g : y-x = \left|x\right| \\ y = \left|x\right|+x \\ y = \left\{\begin{array}{ll}x+x& ; x\ge 0\\ -x+x& ; x<0\end{array}\right.\begin{array}{}\end{array} \\ y = \left\{\begin{array}{ll}2x& ; x\ge 0\\ 0& ; x<0\end{array}\right. \\ \mathrm{ดังนั้น} g\left(x\right) = \left\{\begin{array}{ll}2x& ; x\ge 0 \to \boxed{3}\\ 0& ; x<0 \to \boxed{4}\end{array}\right.\begin{array}{}\end{array} \end{gathered}$$

$$\begin{gathered} \mathrm{พิจารณาข้อความ} \\ ก) \mathrm{หา} f\circ g \\ f\circ g\left(x\right) = f\left(g\left(x\right)\right) \\ = \left\{\begin{array}{ll}f\left(2x\right)& ; x\ge 0 \to 2x\ge 0 \mathrm{เลือก} \boxed{1}\\ f\left(0\right)& ; x<0 \mathrm{เลือก} \boxed{1}\end{array}\right. \\ = \left\{\begin{array}{l}0\\ 0\end{array}\right. \\ = 0 \mathrm{ทุกกรณี} \\ \mathrm{จากโจทย์} g\circ \left(f\circ g\right) = \left(f\circ g\right)\circ g \end{gathered}$$

$$\begin{gathered} -\mathrm{ฝั่งซ้าย} \left(g\circ \left(f\circ g\right)\right)\left(x\right) = g\left(\left(f\circ g\right)\left(x\right)\right) \\ = g\left(0\right) ; \mathrm{เลือก} \boxed{3} \\ = 2\left(0\right) \\ = 0 \\ -\mathrm{ฝั่งขวา} \left(\left(f\circ g\right)\circ g\right)\left(x\right) = \left(f\circ g\right)\left(g\left(x\right)\right) \\ \mathrm{โดย} \left(f\circ g\right)\left(\boxed{}\right) = 0 \mathrm{เสมอ} \\ = 0 \\ \mathrm{ดังนั้น} \mathrm{ข้อ} ก) \mathrm{ถูกต้อง} \end{gathered}$$

$$\begin{gathered} ข) \mathrm{หา} g\circ f \\ \left(g\circ f\right)\left(x\right) = g\left(f\left(x\right)\right) \\ = \left\{\begin{array}{ll}g\left(0\right)& ; x\ge 0 \mathrm{เลือก} \boxed{3}\\ g\left(-2x\right)& ; x<0 \to -2x>0 \mathrm{เลือก} \boxed{3}\end{array}\right. \\ = \left\{\begin{array}{l}2\left(0\right)\\ 2\left(-2x\right)\end{array}\right. \\ = \left\{\begin{array}{ll}0& ; x\ge 0\\ -4x& ; x<0\end{array}\right. \end{gathered}$$

$$\begin{gathered} -\mathrm{ฝั่งซ้าย} \left(\left(g\circ f\right)-f\right)\left(x\right) = \left(g\circ f\right)\left(x\right)-f\left(x\right) \\ = \left\{\begin{array}{ll}0-0& ; x\ge 0\\ -4x-\left(-2x\right)& ; x<0\end{array}\right. \\ = \left\{\begin{array}{ll}0& ; x\ge 0\\ -2x& ; x<0\end{array}\right. \end{gathered}$$

$$\begin{gathered} -\mathrm{ฝั่งขวา} (\left(f\circ g\right)+f)\left(x\right) = (f\circ g)\left(x\right)+f\left(x\right) \\ = 0+f\left(x\right) \\ = f\left(x\right) = \{\begin{array}{ll}0& ; x\ge 0\\ -2x& ; x<0\end{array} \\ \mathrm{ดังนั้น} \mathrm{ข้อ} ข) \mathrm{ถูกต้อง} \end{gathered}$$

$$\begin{gathered} ค) \mathrm{ฝั่งซ้าย} \left(f\circ \left(f\circ g\right)\right)\left(x\right) = f\left(\left(f\circ g\right)\left(x\right)\right) \\ = f\left(0\right) ; \mathrm{เลือก} \boxed{1} \\ \mathrm{ฝั่งขวา} \left(f·g\right)\left(x\right) = f\left(x\right)·g\left(x\right) \\ = \left\{\begin{array}{ll}0·2x& ; x\ge 0\\ -2x·0& ; x<0\end{array}\right. \\ = \left\{\begin{array}{ll}0& ; x\ge 0\\ 0& ; x<0\end{array}\right. \\ = 0 \mathrm{ทุกกรณี} \\ \mathrm{ดังนั้น} \mathrm{ข้อ} ค) \mathrm{ถูกต้อง} \end{gathered}$$