ข้อสอบ PAT 1 - กุมภาพันธ์ 2562

$$\begin{gathered} \mathrm{จากโจทย์} (f\circ g^{-1})\left(x\right) = 2 \\ \mathrm{จะได้} f\left(g^{-1}\right(x\left)\right) = 2 \to \boxed{1} \\ \mathrm{จากโจทย์} f\left(x\right) = \{\begin{array}{ll}x-2& ; x\le 4\\ 3x-10& ; x>4\end{array} \\ \mathrm{จะได้} f(g^{-1}\left(x\right)) = \{\begin{array}{ll}{g}^{-1}\left(x\right)-2& ; g^{-1}\left(x\right)\le 4\\ 3g^{-1}\left(x\right)-10& ; g^{-1}\left(x\right)>4\end{array} \end{gathered}$$

$$\begin{gathered} -\mathrm{โดย} g^{-1}\left(x\right)\le 4 -\mathrm{โดย} g^{-1}\left(x\right)>4 \\ f\left(g^{-1}\left(x\right)\right) = g^{-1}\left(x\right)-2 f\left(g^{-1}\left(x\right)\right) = 3g^{-1}\left(x\right)-10 \\ -\mathrm{จาก} \boxed{1} \mathrm{จะได้} \\ 2 = g^{-1}\left(x\right)-2 2 = 3g^{-1}\left(x\right)-10 \\ g^{-1}\left(x\right) = 4 g^{-1}\left(x\right) = 4 \\ \mathrm{ขัดแย้งกับเงื่อนไข} \\ \mathrm{แสดงว่า} g^{-1}\left(x\right) = 4 \to \boxed{2} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} g\left(x\right) = \{\begin{array}{ll}x+2& ; x<1\\ \frac{1}{2}\left(x+5\right)& ; x\ge 1\end{array} \\ \mathrm{จะได้} y = \{\begin{array}{ll}x+2& ; x<1\\ \frac{1}{2}\left(x+5\right)& ; x\ge 1\end{array} \end{gathered}$$

$$\begin{gathered} \mathrm{หา} y^{-1} \mathrm{เปลี่ยน} y\to x , x\to y \\ \mathrm{จะได้} x = \{\begin{array}{ll}y+2& ; y<1\\ \frac{1}{2}\left(y+5\right)& ; y\ge 1\end{array} \\ x = \{\begin{array}{ll}{g}^{-1}\left(x\right)+2& ; g^{-1}\left(x\right)<1\\ \frac{1}{2}\left[g^{-1}\left(x\right)+5\right]& ; g^{-1}\left(x\right)\ge 1\end{array} \end{gathered}$$

$$\begin{gathered} - \mathrm{โดย} g^{-1}\left(x\right)<1 - \mathrm{โดย} g^{-1}\left(x\right)\ge 1 \\ x = g^{-1}\left(x\right)+2 x = \frac{1}{2}[g^{-1}\left(x\right)+5] \\ {g}^{-1}\left(x\right) = x-2 g^{-1}\left(x\right) = 2x-5 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{2} g^{-1}\left(x\right) = 4 \\ \mathrm{จะได้} g^{-1}\left(x\right) = 2x-5 \\ 4 = 2x-5 \\ x = 4.5 \end{gathered}$$