ข้อสอบ PAT 1 - เมษายน 2557

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\to \boxed{\mathrm{A}} \\ \mathbf{โดย} \mathrm{f}\left(1\right)=0 \\ \mathrm{จะได้} \mathrm{a}{\left(1\right)}^2+\mathrm{b}\left(1\right)+\mathrm{c}=0 \\ \mathrm{a}+\mathrm{b}+\mathrm{c}=0\to \boxed{1} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{f} \mathrm{มีค่าสูงสุดที่} \mathrm{x}=\frac{1}{3} \\ \mathrm{จะได้} \mathrm{f}\text{'}\left(\frac{1}{3}\right)=0 \\ \mathbf{โดย} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=2\mathrm{ax}+\mathrm{b} \end{gathered}$$​

$$\begin{gathered} \mathrm{จาก} \mathrm{f}\text{'}\left(\frac{1}{3}\right)=0 \\ 2\mathrm{a}\left(\frac{1}{3}\right)+\mathrm{b}=0 \\ \frac{2}{3}\mathrm{a}+\mathrm{b}=0\to \boxed{2} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{F}\left(\mathrm{\alpha },\mathrm{\beta }\right) ={\int }_{\mathrm{\alpha }}^{\mathrm{\beta }}\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x} \\ \mathbf{โดยที่}\mathbf{ }\mathrm{F}\left(0,\mathrm{t}\right)=\mathrm{F}\left(1,\mathrm{t}\right)+1 ; \\ \mathrm{จะได้} {\int }_0^{\mathrm{t}}\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x}={\int }_1^{\mathrm{t}}\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x}+1 \\ {\int }_0^{\mathrm{t}}\left({\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\right)d\mathrm{x}={\int }_0^{\mathrm{t}}\left({\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\right)d\mathrm{x}+1 \\ {\overline{)\left(\frac{\mathrm{a}}{3}{\mathrm{x}}^3+\frac{\mathrm{b}}{2}{\mathrm{x}}^2+\mathrm{cx}\right)}}_0^{\mathrm{t}}={\overline{)\left(\frac{\mathrm{a}}{3}{\mathrm{x}}^3+\frac{\mathrm{b}}{2}{\mathrm{x}}^2+\mathrm{cx}\right)}}_0^{\mathrm{t}}+1 \end{gathered}$$
$$\begin{gathered} \left(\frac{\mathrm{a}}{3}{\mathrm{t}}^3+\frac{\mathrm{b}}{2}{\mathrm{t}}^2+\mathrm{ct}\right)=\left(\frac{\mathrm{a}}{3}{\mathrm{t}}^3+\frac{\mathrm{b}}{2}{\mathrm{t}}^2+\mathrm{ct}\right)-\left(\frac{\mathrm{a}}{3}+\frac{\mathrm{b}}{2}+\mathrm{c}\right)+1 \\ \frac{\mathrm{a}}{3}+\frac{\mathrm{b}}{2}+\mathrm{c}=1\to \boxed{3} \\ \mathrm{จาก} \boxed{1} , \boxed{2} ,\boxed{3} \mathrm{แก้} 3 \mathrm{สมการ} 3 \mathrm{ตัวแปร} \\ \mathrm{จะได้} \mathrm{a}=-3 , \mathrm{b}=2 , \mathrm{c}=1 \\ \mathrm{แทน} \mathrm{a} , \mathrm{b} , \mathrm{c} \mathrm{ใน} \boxed{\mathrm{A}} \\ \mathrm{จะได้} \mathrm{f}\left(\mathrm{x}\right)=-3{\mathrm{x}}^2+2\mathrm{x}+1 \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณา} \left(\mathrm{ก}\right)} \\ \mathrm{จากโจทย์}\hspace{0.17em} \mathrm{F}\left(1,2\right)=\mathrm{F}\left(2,3\right)+10 \\ \mathrm{จะได้} {\int }_1^2\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x}={\int }_2^3\mathrm{f}\left(\mathrm{x}\right)d\mathrm{x}+10 \end{gathered}$$​
$$\begin{gathered} {\int }_1^2\left(-3{\mathrm{x}}^2+2\mathrm{x}+1\right)d\mathrm{x}={\int }_2^3\left(-3{\mathrm{x}}^2+2\mathrm{x}+1\right)d\mathrm{x}+10 \\ {\overline{) (-3\frac{{\mathrm{x}}^3}{3}+2\frac{{\mathrm{x}}^2}{2}+\mathrm{x})}}_1^2={\overline{)(-3\frac{{\mathrm{x}}^3}{3}+2\frac{{\mathrm{x}}^2}{2}+\mathrm{x})}}_2^3+10 \\ {\overline{) (-{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x})}}_1^2={\overline{)(-{\mathrm{x}}^3+{\mathrm{x}}^2+\mathrm{x})}}_2^3+10 \end{gathered}$$
$$\begin{gathered} \left[-{\left(2\right)}^3+2^2+2\right]-\left[-1+1+1\right]=\left[-{\left(3\right)}^3+3^2+3\right]-\left[-{\left(2\right)}^3+2^2+2\right]+10 \\ \left(-8+4+2\right)-1=\left(-27+9+3\right)-\left(-8+4+2\right)+10 \\ -3=-15+2+10 \\ -3=-3 \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ก}\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \underline{\mathrm{พิจารณา} \left(\mathrm{ข}\right)} \\ \mathrm{จากโจทย์} \mathrm{อนุพันธ์ของ} \frac{\mathrm{f}\left(\mathrm{x}\right)}{{\mathrm{x}}^2} \mathrm{เท่ากับ} \left(\frac{-3{\mathrm{x}}^2-2\mathrm{x}-2}{{\mathrm{x}}^3}\right) \\ \mathbf{หาค่า} \frac{\mathrm{f}\left(\mathrm{x}\right)}{{\mathrm{x}}^2} \\ \mathrm{จะได้} \frac{\mathrm{f}\left(\mathrm{x}\right)}{{\mathrm{x}}^2}=\frac{-3{\mathrm{x}}^2-2\mathrm{x}-2}{{\mathrm{x}}^3} \\ =-3+\frac{2}{\mathrm{x}}+\frac{1}{{\mathrm{x}}^2} \\ =-3+2{\mathrm{x}}^{-1}+{\mathrm{x}}^{-2} \end{gathered}$$

$$\begin{gathered} \mathrm{แสดงว่า} \mathrm{อนุพันธ์ของ} \frac{\mathrm{f}\left(\mathrm{x}\right)}{{\mathrm{x}}^2}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{-3{\mathrm{x}}^2-2\mathrm{x}-2}{{\mathrm{x}}^3}\right) \\ =-2{\mathrm{x}}^{-2}-2{\mathrm{x}}^{-3} \\ =\frac{-2}{{\mathrm{x}}^2}-\frac{2}{{\mathrm{x}}^3} \\ =\frac{-2\mathrm{x}-2}{{\mathrm{x}}^3}\ne \frac{-3{\mathrm{x}}^2-2\mathrm{x}-2}{{\mathrm{x}}^3} \\ \mathrm{ดังนั้น} \mathrm{ข้อ} \left(\mathrm{ข}\right) \mathrm{ผิด} \end{gathered}$$