ข้อสอบ PAT 1 - กุมภาพันธ์ 2563

$$\begin{gathered} \mathrm{พิจารณาความต่อเนื่องที่} x=0 \mathrm{จะได้} \\ \underset{x\to 0^{-}}{\lim }f\left(x\right) = f\left(0\right) = \underset{x\to 0^{+}}{\lim }f\left(x\right) \\ \underset{x\to 0^{-}}{\lim }\frac{x}{x-x^2} = \frac{a{\left(0\right)}^2+\left(b-a\right)\left(0\right)-b}{0-1} \\ \underset{x\to 0^{-}}{\lim }\frac{x}{x\left(1-x\right)} = \frac{-b}{-1} \\ \frac{1}{1-0} = b \\ b = 1 \end{gathered}$$

$$\begin{gathered} \mathrm{พิจารณาความต่อเนื่องที่} x=1 \mathrm{จะได้} \\ \underset{x\to 1^{-}}{\lim }f\left(x\right) = f\left(1\right) = \underset{x\to 1^{+}}{\lim }f\left(x\right) \\ \underset{x\to 1^{-}}{\lim } \frac{a{x}^2+\left(1-a\right)x-1}{x-1} = {\left(1+1\right)}^2 \\ \underset{x\to 1^{-}}{\lim }\frac{a{x}^2-ax+x-1}{x-1} = 4 \end{gathered}$$

$$\begin{gathered} \underset{x\to 1^{-}}{\lim }\frac{ax\left(x-1\right)+\left(x-1\right)}{x-1} = 4 \\ \underset{x\to 1^{-}}{\lim }\frac{\left(ax+1\right)\cancel{\left(x-1\right)}}{\cancel{\left(x-1\right)}} = 4 \\ a\left(1\right)+1 = 4 \\ a = 3 \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} f\left(a+b\right) = f\left(3+1\right) = f\left(4\right) \\ = {\left(4+1\right)}^2 = 25 \to \mathrm{ตอบ} 1 \end{gathered}$$