ข้อสอบ PAT 1 - มีนาคม 2560

$$\begin{gathered} \mathrm{จากโจทย์} m = \frac{1}{4}\left(1+\sin \theta \right)cot\theta \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\mathrm{และ} n = \frac{1}{4}\left(1-\sin \theta \right)cot\theta \\ \mathrm{จะได้} m = \frac{1}{4}\left(1+\sin \theta \right)cot\theta = \frac{cot\theta }{4}+\frac{\sin \theta cot\theta }{4} \\ n = \frac{1}{4}\left(1-\sin \theta \right)cot\theta = \frac{cot\theta }{4}-\frac{\sin \theta cot\theta }{4} \end{gathered}$$

$$\begin{gathered} \mathbf{หาค่า}\mathbf{ }\mathit{m}\mathbf{+}\mathit{n} \\ m+n = \left(\frac{cot\theta }{4}+\cancel{\frac{\sin \theta cot\theta }{4}}\right)+\left(\frac{cot\theta }{4}-\cancel{\frac{\sin \theta cot\theta }{4}}\right) \\ \hspace{0.17em}\hspace{0.17em}= \frac{cot\theta }{2} \end{gathered}$$

$$\begin{gathered} \mathbf{หาค่า}\mathbf{ }\mathit{m}\mathbf{-}\mathit{n} \\ m-n = \left(\cancel{\frac{cot\theta }{4}}+\frac{\sin \theta cot\theta }{4}\right)-\left(\cancel{\frac{cot\theta }{4}}-\frac{\sin \theta cot\theta }{4}\right) \\ = \frac{2\sin \theta cot\theta }{4} \\ = \frac{\sin \theta cot\theta }{2} \\ = \frac{\cancel{\sin \theta }}{2}\left(\frac{\cos \theta }{\cancel{\sin \theta }}\right) \\ = \frac{\cos \theta }{2} \end{gathered}$$

$$\begin{gathered} \mathbf{หา}\mathbf{ }\mathit{m}\mathit{n} \\ \mathrm{จะได้} mn = \left[\frac{1}{4}\left(1+\sin \theta \right)cot\theta \right]\left[\frac{1}{4}\left(1-\sin \theta \right)cot\theta \right] \\ = \frac{1}{16}\left(1+\sin \theta \right)\left(1-\sin \theta \right)co{t}^2\theta \\ = \frac{1}{16}\left(1-{\sin }^2\theta \right)co{t}^2\theta \\ \mathbf{ }\mathbf{โดย} 1-{\sin }^2\theta = {\cos }^2\theta \mathrm{และ} cot\theta =\frac{\cos \theta }{\sin \theta } \\ \mathrm{จะได้} mn = \frac{1}{16}{\cos }^2\theta \left(\frac{{\cos }^2\theta }{{\sin }^2\theta }\right) \\ = \frac{1}{16}\frac{{\cos }^4\theta }{{\sin }^2\theta } \end{gathered}$$

$$\begin{gathered} \mathrm{พิจารณาข้อความ} \\ ก) \mathrm{จากโจทย์} {\left(m^2-n^2\right)}^2 = mn \\ \mathrm{หาค่า} {\left(m^2-n^2\right)}^2 = {\left[\left(m-n\right)\left(m+n\right)\right]}^2 \\ = {\left[\left(\frac{\cos \theta }{2}\right)\left(\frac{cot\theta }{2}\right)\right]}^2 \\ = {\left(\frac{\cos \theta }{2}\right)}^2{\left(\frac{cot\theta }{2}\right)}^2 \\ = \frac{1}{16}{\cos }^2\theta co{t}^2\theta \\ \mathrm{โดย} cot\theta = \frac{\cos \theta }{\sin \theta } \\ = \frac{1}{16}\frac{{\cos }^4\theta }{{\sin }^2\theta } \\ = mn \mathrm{ข้อ} (ก) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \left(ข\right) \mathrm{จากโจทย์} \sin \theta = \frac{m-n}{m+n} \\ \mathrm{หาค่า} \frac{m-n}{m+n} = \frac{\left({\displaystyle \frac{\cos \theta }{\cancel{2}}}\right)}{\left({\displaystyle \frac{cot\theta }{\cancel{2}}}\right)} \\ = \frac{\cos \theta }{cot\theta } ; \mathrm{โดย} cot\theta = \frac{\cos \theta }{\sin \theta } \\ = \frac{\cancel{\cos \theta }}{\left({\displaystyle \frac{\cancel{\cos \theta }}{\sin \theta }}\right)} \\ \hspace{0.17em} = \sin \theta \mathrm{ข้อ} \left(ข\right) \mathrm{ถูก} \end{gathered}$$

$$\begin{gathered} \left(ค\right) \mathrm{จากโจทย์} m^2+n^2 = \frac{1}{8}co{t}^2\theta {\cos }^2\theta \\ \mathrm{หาค่า} m^2+n^2 = \left(m^2-2mn+n^2\right)+2mn \\ = {\left(m-n\right)}^2+2mn \\ = {\left(\frac{\cos \theta }{2}\right)}^2+2\left(\frac{1}{16}\frac{{\cos }^4\theta }{{\sin }^2\theta }\right) \\ = \frac{{\cos }^2\theta }{4}+\frac{1}{8}\frac{{\cos }^4\theta }{{\sin }^2\theta } \\ = \frac{{\cos }^2\theta }{4}+\frac{1}{8}co{t}^2\theta {\cos }^2\theta \\ \ne \frac{1}{8}co{t}^2\theta {\cos }^2\theta \mathrm{ข้อ} \left(ค\right) \mathrm{ผิด} \end{gathered}$$