106學年度指定科目考試數學(甲)非選擇題詳解

1. (1)  $A=\displaystyle{\frac{1}{5}}\left[ \begin{matrix} 4 & -3 \\ -3 & 4 \\ \end{matrix} \right]=\left[ \begin{matrix} \displaystyle{\frac{4}{5}} & \displaystyle{\frac{-3}{5}} \\ \displaystyle{\frac{-3}{5}} & \displaystyle{\frac{4}{5}} \\ \end{matrix} \right]$為一旋轉矩陣，$cos\ \theta =\displaystyle{\frac{4}{5}}$、$sin\ \theta =\displaystyle{\frac{3}{5}}$ $\Rightarrow$ $\angle {{P}_{1}}Q{{P}_{3}}=2\angle {{P}_{1}}O{{P}_{2}}$，$sin\ \angle \ {{P}_{1}}O{{P}_{2}}=sin\ 2\theta =\displaystyle{\frac{24}{25}}$
   
   (2)  由(1)小題可作圖：
   
   故$\vartriangle {{P}_{1}}{{P}_{2}}{{P}_{3}}$ $=2\vartriangle O{{P}_{1}}{{P}_{2}}-\vartriangle O{{P}_{1}}{{P}_{3}}$ $=2\cdot \displaystyle{\frac{1}{2}}{{a}^{2}}sin\ \theta -\displaystyle{\frac{1}{2}}{{a}^{2}}\cdot sin\ 2\theta$ $=\displaystyle{\frac{3}{5}}{{a}^{2}}-\displaystyle{\frac{12}{25}}{{a}^{2}}=\displaystyle{\frac{3}{25}}{{a}^{2}}$
   
   (3)  因$\vartriangle {{P}_{1}}{{P}_{2}}{{P}_{3}}$只距離$\overline{O{{P}_{1}}}=a$有關 $\Rightarrow$ ${{a}^{2}}$ $={{x}^{2}}+{{y}^{2}}$ $={{x}^{2}}+{{\left(\displaystyle{\frac{1}{10}}{{x}^{2}}-10 \right)}^{2}}$ $=\displaystyle{\frac{1}{100}}{{x}^{4}}-{{x}^{2}}+100$ $=\displaystyle{\frac{1}{100}}\left( {{x}^{4}}-100x+2500 \right)+75\ge 75$ $\Rightarrow$ $\vartriangle {{P}_{1}}{{P}_{2}}{{P}_{3}}$面積$=\displaystyle{\frac{3}{25}}{{a}^{2}}\ge \displaystyle{\frac{3}{25}}\cdot 75=9$
   
   
2. (1)  令八邊形與$z$軸交於$Q$，$\overset{\rightharpoonup}{O{{P}_{0}}}\cdot \overset{\rightharpoonup}{O{{P}_{4}}}$ $=\left( \overset{\rightharpoonup}{OQ}+\overset{\rightharpoonup}{Q{{P}_{0}}} \right)\cdot \left( \overset{\rightharpoonup}{OQ}+\overset{\rightharpoonup}{Q{{P}_{4}}} \right)$ $={{\left| \overset{\rightharpoonup}{OQ} \right|}^{2}}+\overset{\rightharpoonup}{OQ}\cdot \overset{\rightharpoonup}{Q{{P}_{0}}}+\overset{\rightharpoonup}{OQ}\cdot \overset{\rightharpoonup}{O{{P}_{4}}}+\overset{\rightharpoonup}{Q{{P}_{0}}}\cdot \overset{\rightharpoonup}{Q{{P}_{4}}}$ $={{h}^{2}}+\left( -1 \right)\cdot \left( 1-{{h}^{2}} \right)$ $=2{{h}^{2}}-1$
   
   (2)  ${{P}_{0}}{{P}_{1}}{{P}_{2}}{{P}_{3}}{{P}_{4}}{{P}_{5}}{{P}_{6}}{{P}_{7}}$面積$=8\vartriangle {{P}_{0}}Q{{P}_{1}}$面積$=8\cdot \displaystyle{\frac{1}{2}}\sqrt{1-{{h}^{2}}}\sqrt{1-{{h}^{2}}}\cdot sin45{}^\circ$ $\Rightarrow$ $V\left( h \right)=2\sqrt{2}\left( 1-{{h}^{2}} \right)\cdot h\cdot \displaystyle{\frac{1}{3}}=\displaystyle{\frac{2\sqrt{2}}{3}}\left( h-{{h}^{3}} \right)$
   
   (3)  $\overset{\rightharpoonup}{O{{P}_{0}}}\cdot \overset{\rightharpoonup}{O{{P}_{4}}}$夾角$\le 90{}^\circ$ $\Rightarrow$ $\overset{\rightharpoonup}{O{{P}_{0}}}\cdot \overset{\rightharpoonup}{O{{P}_{4}}}\ge 0$ $\Rightarrow$ $h\ge \displaystyle{\frac{\sqrt{2}}{2}}$，又$0\le h\le 1$ $\Rightarrow$ $\displaystyle{\frac{\sqrt{2}}{2}}\le h\le 1$
   ${V}'\left( h \right)= \displaystyle{\frac{2\sqrt{2}}{3}}\left( 1-3{{h}^{2}} \right)$ $\Rightarrow$ ${V}''\left( h \right)=-4\sqrt{2}h$ $\Rightarrow$ $V$在$h=\displaystyle{\frac{1}{\sqrt{3}}}$有極大值，但$\displaystyle{\frac{1}{\sqrt{3}}}<\displaystyle{\frac{\sqrt{2}}{2}}$，故不合。
   考慮$\displaystyle{\frac{\sqrt{2}}{2}}\le h\le 1$，分別代入計算得$V\left( 1 \right)=0$、$V\left(\displaystyle{\frac{\sqrt{2}}{2}} \right)= \displaystyle{\frac{1}{3}}$，故體積最大值為$\displaystyle{\frac{1}{3}}$