ข้อสอบ PAT 1 - มีนาคม 2555

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3+\mathrm{ax}+\mathrm{b}\to \boxed{1} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=3{\mathrm{x}}^2+\mathrm{a}\to \boxed{2} \\ \mathrm{f}\text{'}\left(\mathrm{a}\right)=3{\mathrm{a}}^2+\mathrm{a} \\ \mathrm{f}\text{'}\left(\mathrm{b}\right)=3{\mathrm{b}}^2+\mathrm{a} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}{\mathrm{L}}_1 \mathrm{และ} {\mathrm{L}}_2 \mathrm{เป็นเส้นสัมผัสเส้นโค้ง} \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{ที่} \mathrm{x}=\mathrm{a} \mathrm{และ} \mathrm{x}=\mathrm{b} \\ \mathrm{จะได้} {\mathrm{m}}_{{\mathrm{L}}_1}={\mathrm{m}}_{{\mathrm{L}}_2} \\ 3{\mathrm{a}}^2\cancel{+\mathrm{a}}=3{\mathrm{b}}^2\cancel{+\mathrm{a}} \\ \cancel{ 3}{\mathrm{a}}^2=\cancel{3}{\mathrm{b}}^2 \\ {\mathrm{a}}^2={\mathrm{b}}^2\to \boxed{3} \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \underset{\mathrm{h}\to 0}{\lim }\frac{9\mathrm{h}}{𝑓\left(1+\mathrm{h}\right)-𝑓\left(1\right)}=1 \\ \underset{\mathrm{h}\to 0}{\lim }\frac{\mathrm{h}}{𝑓\left(1+\mathrm{h}\right)-𝑓\left(1\right)}=\frac{1}{9} \\ \underset{\mathrm{h}\to 0}{\lim }\frac{𝑓\left(1+\mathrm{h}\right)-𝑓\left(1\right)}{\mathrm{h}}=9 \\ \mathrm{f}\text{'}\left(1\right)=9 \end{gathered}$$

$$\begin{gathered} \mathrm{จาก} \boxed{2} \mathrm{จะได้} \mathrm{f}\text{'}\left(1\right)=3{\left(1\right)}^2+\mathrm{a} \\ 9=3+\mathrm{a} \\ \mathrm{a}=6 \\ \mathrm{จาก} \boxed{3} \mathrm{จะได้} \mathrm{b}=-6,6 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{a} \mathrm{และ} \mathrm{b} \mathrm{เป็นจำนวนจริงที่แตกต่างกัน} \\ \mathrm{แสดงว่า} \mathrm{a}=6 , \mathrm{b}=-6 \\ \mathrm{แทน} \mathrm{a} , \mathrm{b} \mathrm{ใน} \boxed{1} \\ \mathrm{จะได้} \mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^3+6\mathrm{x}-6 \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้น} {\int }_0^2𝑓\left(\mathrm{x}\right)d\mathrm{x}={\int }_0^2{\mathrm{x}}^3+6\mathrm{x}-6d\mathrm{x} \\ ={\overline{)\frac{{\mathrm{x}}^4}{4}+3{\mathrm{x}}^2-6\mathrm{x}}}_0^2 \\ =\frac{2^4}{4}+3{\left(2\right)}^2-6\left(2\right) \\ =4 \end{gathered}$$