ข้อสอบ PAT 1 - มีนาคม 2559

$$\begin{gathered} \mathrm{จากโจทย์} f\left(x\right) = \left\{\begin{array}{ll}x+b-4,& x\mathit{\le }\mathit{ }a \to \boxed{1}\\ x^2+bx+a,& a<x\le b \to \boxed{2}\\ 2bx-a,& x>b \to \boxed{3}\end{array}\right. \\ f \mathrm{เป็นฟังชันก์ต่อเนื่องบนเซตของจำนวนจริง} \end{gathered}$$

$$\begin{gathered} \mathbf{-}\mathbf{โดย}\mathbf{ }\mathit{f}\mathbf{ }\mathbf{ต่อเนื่องที่}\mathbf{ }\mathit{x}\mathbf{=}\mathit{b} \\ \mathrm{จะได้} f\left(b\right) = \underset{x\to b}{\lim }f\left(x\right) \\ \mathbf{โดย} \underset{x\to b}{\lim }f\left(x\right)=\underset{x\to b^{+}}{\lim }f\left(x\right)=\underset{x\to b^{-}}{\lim }f\left(x\right) \end{gathered}$$

$$\begin{gathered} \hspace{0.17em}\hspace{0.17em}\mathrm{ให้} f\left(b\right) = \underset{x\to b^{+}}{lim}f\left(x\right) ; \mathrm{เมื่อ} x\to b^{+} \mathrm{เลือก} f\left(x\right) \mathrm{กรณี} \boxed{3} \\ \hspace{0.17em}{b}^2+b\left(b\right)+a = 2b\left(b\right)-a \\ 2b^2+a = 2b^2-a \\ 2a = 0 \\ a = 0 \end{gathered}$$

$$\begin{gathered} \mathbf{-}\mathbf{โดย}\mathbf{ }\mathit{f}\mathbf{ }\mathbf{ต่อเนื่องที่}\mathbf{ }\mathit{x}\mathbf{=}\mathit{a} \\ \mathrm{จะได้} f\left(a\right)=\underset{x\to a}{\lim }f\left(x\right) \\ \mathbf{โดย} \underset{x\to a}{\lim }f\left(x\right)=\underset{x\to a^{+}}{\lim }f\left(x\right)=\underset{x\to a^{-}}{\lim }f\left(x\right) \end{gathered}$$

$$\begin{gathered} \mathrm{ให้} f\left(a\right)=\underset{x\to a^{-}}{\lim }f\left(x\right) \hspace{0.17em}\hspace{0.17em}; \mathrm{เมื่อ} x\to a^{-} \mathrm{เลือก} \mathrm{f}\left(\mathrm{x}\right) \mathrm{กรณี} \boxed{1} \\ a^2+b\left(a\right)+a=a+b-4 \hspace{0.17em}\hspace{0.17em}\hspace{0.17em} ; \mathrm{แทน} a=0 \\ \mathrm{จะได้} 0+0+0=0+b-4 \\ \hspace{0.17em}\hspace{0.17em}b=4 \\ \mathrm{แทน} a=0 , b=4 \mathrm{ใน} f\left(x\right) \end{gathered}$$

$$\begin{gathered} \mathrm{จะได้} \mathbf{ }\mathit{f}\left(\mathbf{x}\right)\mathbf{=}\left\{\begin{array}{c}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{<}\mathbf{0}\mathbf{\to }\boxed{\mathbf{1}}\\ {\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{\le }\mathbf{4}\mathbf{\to }\boxed{\mathbf{2}}\\ \mathbf{ }\mathbf{ }\mathbf{8}\mathbf{x}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{>}\mathbf{4}\mathbf{\to }\boxed{\mathbf{3}}\end{array}\right. \\ \mathrm{และ} \mathit{f}\mathbf{\text{'}}\left(\mathbf{x}\right)\mathbf{=}\left\{\begin{array}{c}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{<}\mathbf{0}\\ \mathbf{2}\mathbf{x}\mathbf{+}\mathbf{4}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{\le }\mathbf{4}\\ \mathbf{ }\mathbf{ }\mathbf{8}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{>}\mathbf{4}\end{array}\right. \end{gathered}$$

$$\begin{gathered} \mathbf{พิจารณาข้อ}\mathbf{ }\mathbf{\left(}\mathit{ก}\mathbf{\right)} \\ \mathrm{จากโจทย์} \left(f\circ f\right)\left(a-b\right) = a-b \\ \mathrm{แทน} a=0 , b=4 \\ \mathrm{จะได้} \left(f\circ f\right)\left(0-4\right) = 0-4 \\ f\left(f\left(-4\right)\right) = -4 \\ f\left(-4\right) = -4 \\ -4 = -4 \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(ก\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \mathbf{พิจารณาข้อ}\mathbf{ }\mathbf{\left(}\mathit{ข}\mathbf{\right)} \\ \mathrm{จากโจทย์} f\left(a+b\right) = f\left(a\right)+f\left(b\right) \\ \mathrm{แทน} a=0 , b=4 \\ \mathrm{จะได้} f\left(a+b\right) = f\left(a\right)+f\left(b\right) \\ f\left(0+4\right) = f\left(0\right)+f\left(4\right) \\ f\left(4\right) = f\left(0\right)+f\left(4\right) \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{4}^2+4\left(4\right) = 0+\left[4^2+4\left(4\right)\right] \\ 32 = 32 \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(ข\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \mathbf{พิจารณาข้อ}\mathbf{ }\mathbf{\left(}\mathit{ค}\mathbf{\right)} \\ \mathrm{จากโจทย์} f\text{'}\left(f\left(2\right)\right) = f\left(f\text{'}\left(2\right)\right) \\ \mathrm{จะได้} f\text{'}\left[2^2+4\left(2\right)\right] = f\left[2\left(2\right)+4\right] \\ f\text{'}\left(12\right) = f\left(8\right) \\ 8 = 8\left(8\right) \\ 8 = 64 \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(ค\right) \mathrm{ผิด} \\ \mathrm{ดังนั้น} \mathrm{ตอบ} 1) \mathrm{ข้อ} (\mathrm{ก}) \mathrm{และ} \mathrm{ข้อ} (\mathrm{ข}) \mathrm{ถูก} \mathrm{แต่} \mathrm{ข้อ} (\mathrm{ค}) \mathrm{ผิด} \end{gathered}$$