ข้อสอบ PAT 1 - มีนาคม 2558

$$\begin{gathered} \mathrm{จากโจทย์} \mathrm{f}\left(\mathrm{x}\right)=12\mathrm{x}-9{\mathrm{x}}^2 \\ \mathrm{หาจุดวิกฤตของฟังก์ชัน} \mathrm{จาก} \mathrm{f}\text{'}\left(\mathrm{x}\right)=0 \\ 12-18\mathrm{x}=0 \\ \mathrm{x}=\frac{12}{18}=\frac{2}{3} \\ \mathrm{ตรวจสอบค่าสูงสุด}-\mathrm{ต่ำสุดของฟังก์ชัน} \\ \mathrm{f}\text{'}\left(\mathrm{x}\right)=12-18\mathrm{x} \\ \mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)=-18\to \mathrm{f}\text{'}\text{'}\left(\mathrm{c}\right)=-18 \\ \mathrm{f}\text{'}\text{'}\left(\frac{2}{3}\right)=-18<0 \end{gathered}$$
$$\begin{gathered} \mathrm{แสดงว่า} \mathrm{x}=\frac{2}{3} \mathrm{ทำให้} \mathrm{f}\left(\frac{2}{3}\right) \mathrm{เกิดค่าสูงสุดของฟังก์ชัน} \\ \mathrm{จากโจทย์} \mathrm{เมื่อ} \mathrm{a} \mathrm{เป็นจำนวนจริง} \mathrm{f}\left(\mathrm{a}\right) \mathrm{มีค่ามากที่สุด} \\ \mathrm{จะได้} \mathrm{a}=\frac{2}{3} \\ \mathrm{จากโจทย์} \mathrm{sin\theta }=\mathrm{a} \\ \mathrm{จะได้} \mathrm{sin\theta }=\frac{2}{3} \\ \mathrm{วาดรูปสามเหลี่ยมมุมฉาก} \end{gathered}$$

$$\begin{gathered} \mathrm{จากรูป} \mathrm{sec\theta }=\frac{1}{\mathrm{cos\theta }}=\frac{3}{\sqrt{5}} \\ \mathrm{cot\theta }=\frac{1}{\mathrm{tan\theta }}=\frac{\sqrt{5}}{2} \\ \mathrm{จากโจทย์}\hspace{0.17em} \frac{\left({\cot }^2\mathrm{\theta }\right)\left(\mathrm{sec\theta }-1\right)}{1+\mathrm{sin\theta }}+\frac{\left({\sec }^2\mathrm{\theta }\right)\left(\mathrm{sin\theta }-1\right)}{1+\mathrm{sec\theta }} \mathrm{เท่ากับ}? \\ \mathrm{จะได้} \frac{\left({\cot }^2\mathrm{\theta }\right)\left(\mathrm{sec\theta }-1\right)}{1+\mathrm{sin\theta }}+\frac{\left({\sec }^2\mathrm{\theta }\right)\left(\mathrm{sin\theta }-1\right)}{1+\mathrm{sec\theta }} \end{gathered}$$

$$\begin{gathered} =\frac{\left({\displaystyle \frac{\sqrt{5}}{2}}{)}^2\right({\displaystyle \frac{3}{\sqrt{5}}}-1)}{1+{\displaystyle \frac{2}{3}}}+\frac{\left({\displaystyle \frac{3}{\sqrt{5}}}{)}^2\right({\displaystyle \frac{2}{3}}-1)}{1+{\displaystyle \frac{3}{\sqrt{5}}}} \\ =\frac{\left({\displaystyle \frac{5}{4}}\right)\left({\displaystyle \frac{3-\sqrt{5}}{\sqrt{5}}}\right)}{{\displaystyle \frac{5}{3}}}+\frac{\left({\displaystyle \frac{9}{5}}\right)\left({\displaystyle -\frac{1}{3}}\right)}{{\displaystyle \frac{3+\sqrt{5}}{\sqrt{5}}}} \\ =\frac{5\sqrt{5}\left(3-\sqrt{5}\right)}{20}·\frac{3}{5}-\left(\frac{3}{5}\right)\left(\frac{\sqrt{5}}{3+\sqrt{5}}\right) \end{gathered}$$

$$\begin{gathered} =\frac{3\sqrt{5}\left(3-\sqrt{5}\right)}{20}-\left(\frac{3}{5}\right)\left(\frac{\sqrt{5}}{3+\sqrt{5}}\right)\left(\frac{3-\sqrt{5}}{3-\sqrt{5}}\right) \\ =\frac{3\sqrt{5}\left(3-\sqrt{5}\right)}{20}-\frac{3\sqrt{5}\left(3-\sqrt{5}\right)}{20} \\ =0 \\ \mathrm{ค่าของ} \frac{\left({\cot }^2\mathrm{\theta }\right)\left(\mathrm{sec\theta }-1\right)}{1+\mathrm{sin\theta }}+\frac{\left({\sec }^2\mathrm{\theta }\right)\left(\mathrm{sin\theta }-1\right)}{1+\mathrm{sec\theta }} \mathrm{เท่ากับ} 0 \end{gathered}$$