ข้อสอบ PAT 1 - มีนาคม 2564

$$\begin{gathered} \mathrm{จากโจทย์} f\left(x\right) = \left\{\begin{array}{ll}x+5& \mathrm{เมื่อ} x>a \to \overline{)1}\\ x+1& \mathrm{เมื่อ} x\le a \to \overline{)2}\end{array} \mathrm{และ} a>0\right. \\ \underset{x\to a^{-}}{\lim }\left(g\circ f\right)\left(x\right)-\underset{x\to a^{+}}{\lim }\left(f\circ g\right)\left(\sqrt{x}\right) = 2 \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{หาค่า}} \underset{x\to a^{-}}{\lim }\left(g\circ f\right)\left(x\right) \\ \mathrm{โดย} f\left(a^{-}\right) \mathrm{จะได้สมการ} \overline{)2} ; x\to a^{-} \mathrm{คือ} x\le a \\ \mathrm{จาก} \overline{)2} f\left(x\right) = x+1 \\ \mathrm{จะได้} \underset{x\to a^{-}}{\lim }\left(g\circ f\left(x\right)\right) = \underset{x\to a^{-}}{\lim }g\left(f\left(x\right)\right) \mathrm{โดย} f\left(x\right) = x+1 \\ = \underset{x\to a^{-}}{\lim }g\left(x+1\right) \\ = g\left(a+1\right) \to \overline{)3} \\ \mathrm{โดย} g\left(x\right)=x^2 \mathrm{จะได้} g\left(a+1\right) = {\left(a+1\right)}^2 ; \mathrm{แทนใน} \overline{)3} \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \overline{)3} \underset{x\to a^{-}}{\lim }\left(g\circ f\right)\left(x\right) = {\left(a+1\right)}^2 \\ \underline{\mathrm{หาค่า}} \underset{x\to a^{+}}{\lim }\left(f\circ g\right)\left(\sqrt{x}\right) \\ \mathrm{โดย} g\left(x\right)=x^2 \mathrm{จะได้} g\left(\sqrt{x}\right)={\left(\sqrt{x}\right)}^2 = x \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \underset{x\to a^{+}}{\lim }\left(f\circ g\right)\left(x\right) = \underset{x\to a^{+}}{\lim }f\left(g\left(\sqrt{x}\right)\right) \\ = \underset{x\to a^{+}}{\lim }f\left(x\right) \to \overline{)4} \\ \mathrm{โดย} f\left(a^{+}\right) \mathrm{จะได้สมการ} \overline{)1} ; x\to a^{+} \mathrm{คือ} x>a \\ \mathrm{จาก} \overline{)1} f\left(x\right) = x+5 \\ \mathrm{จาก} \overline{)4} \underset{x\to a^{+}}{\lim }\left(f\circ g\right)\left(\sqrt{x}\right) = \underset{x\to a^{+}}{\lim }\left(x+5\right) \\ = a+5 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \underset{x\to a^{-}}{\lim }\left(g\circ f\right)\left(x\right)-\underset{x\to a^{+}}{\lim }\left(f\circ g\right)\left(\sqrt{x}\right) = 2 \\ \mathrm{จะได้} {\left(a+1\right)}^2-\left(a+5\right) = 2 \\ a^2+2a+1-\left(a+5\right) = 2 \\ a^2+a-6 = 0 \\ \left(a+3\right)\left(a-2\right) = 0 \\ a = -3, 2 \\ \mathrm{จากโจทย์} a>0 \\ \mathrm{ดังน้ัน} a=2 \end{gathered}$$