107學年度指定科目考試數學(甲)非選擇題詳解

1. (1)  令立方體邊長為$a$ $\Rightarrow $四面體$ABDE$體積$=\displaystyle{\frac{1}{6}}{{a}^{3}}$，以$BDE$為底、$A$到$\vartriangle BDE$距離為高，則$\vartriangle BDE$面積$\times $高$\times \displaystyle{\frac{1}{3}}$ $=\displaystyle{\frac{1}{6}}{{a}^{3}}$，又$\vartriangle BDE$邊長$=\sqrt{2}a$ $\Rightarrow $ $\displaystyle{\frac{\sqrt{3}}{4}}{{\left( \sqrt{2}a \right)}^{2}}\times $高$\times \displaystyle{\frac{1}{3}}$ $=\displaystyle{\frac{1}{6}}{{a}^{3}}$ $\Rightarrow $高$=\displaystyle{\frac{1}{\sqrt{3}}}a$，又$\overline{AG}=\sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}=\sqrt{3}a=3\cdot \displaystyle{\frac{1}{\sqrt{3}}}a$，故$A$點到平面$BDE$的距離是對角線$\overline{AG}$長度的三分之一。
   
   (2)  $\overset{\rightharpoonup}{AG}\cdot \overset{\rightharpoonup}{BD}$ $=\left( \overset{\rightharpoonup}{AB}+\overset{\rightharpoonup}{BC}+\overset{\rightharpoonup}{CG} \right)\cdot \overset{\rightharpoonup}{BD}$ $=\overset{\rightharpoonup}{AB}\cdot \overset{\rightharpoonup}{BD}+\overset{\rightharpoonup}{BC}\cdot \overset{\rightharpoonup}{BD}+\overset{\rightharpoonup}{CG}\cdot \overset{\rightharpoonup}{BD}$ $=\sqrt{2}a\cdot a\cdot \left( -\displaystyle{\frac{\sqrt{2}}{2}} \right)+\sqrt{2}a\cdot a\cdot \displaystyle{\frac{\sqrt{2}}{2}}+0$ $=0$
   $\overset{\rightharpoonup}{AG}\cdot \overset{\rightharpoonup}{BE}$ $=\left( \overset{\rightharpoonup}{AB}+\overset{\rightharpoonup}{BC}+\overset{\rightharpoonup}{CG} \right)\cdot \overset{\rightharpoonup}{BE}$ $=\overset{\rightharpoonup}{AB}\cdot \overset{\rightharpoonup}{BE}+\overset{\rightharpoonup}{BC}\cdot \overset{\rightharpoonup}{BE}+\overset{\rightharpoonup}{CG}\cdot \overset{\rightharpoonup}{BE}$ $=\sqrt{2}a\cdot a\cdot \left( -\displaystyle{\frac{\sqrt{2}}{2}} \right)+0+\sqrt{2}a\cdot a\cdot \displaystyle{\frac{\sqrt{2}}{2}}$ $=0$
   因$\overset{\rightharpoonup}{AG}\bot \overset{\rightharpoonup}{BE}$、$\overset{\rightharpoonup}{AG}\bot \overset{\rightharpoonup}{BD}$ $\Rightarrow $ $\overset{\rightharpoonup}{AG}\bot $平面$BDE$
   
   (3)  根據距離公式$=\displaystyle{\frac{\left| \ 4\ +\ 4\ -6+7\ \right|}{\sqrt{{{2}^{2}}+{{2}^{2}}+{{\left( -1 \right)}^{2}}}}}$ $=3$
   
   (4)  因$\left| \overset{\rightharpoonup}{AG} \right|$為$A$到$BDE$距離的$3$倍且$\overset{\rightharpoonup}{AG}\bot $平面$BDE$ $\Rightarrow $ $\left| \overset{\rightharpoonup}{AG} \right|$ $=3\times 3$ $=9$ $\Rightarrow $ $\overset{\rightharpoonup}{AG}$ $=\pm 3\left( 2，2，-1 \right)$ $=\left( 6，6，-3 \right)$或$\left( -6，-6，3 \right)$得$G$ $=A+\overset{\rightharpoonup}{AG}$ $=\left( 8，8，3 \right)$或$\left( -4，-4，9 \right)$。又$A$、$G$在$2x+2y-z=-7$異側，代入檢查得$G=\left( -4，-4，9 \right)$
   
   
2. (1)  ${f}'(x)=-3{{x}^{2}}-6x$，解${f}'(x)=0$得$x=-2$、$x=0$
   ${f}''(x)=-6x-6$，解${f}''(x)=0$得$x=-1$
   故$f(x)$在$\left( 0，f\left( 0 \right) \right)$與$\left( -2，f\left( -2 \right) \right)$有極值，其中$f(0)=3$、$f(-2)=-1$，作圖如下：
   (2)  利用勘根定理
   
   得$-3<{{a}_{1}}<-2<{{a}_{2}}<-1<0<{{a}_{3}}<1$
   
   (3)  由圖可知$f(x)={{a}_{3}}$有3個實根、$f(x)={{a}_{2}}$與$f(x)={{a}_{1}}$都只有一實根
   
   (4)  $f\left( f\left( x \right) \right)=0$即求$f\left( x \right)={{a}_{1}}$、$f\left( x \right)={{a}_{2}}$、$f\left( x \right)={{a}_{3}}$的實根個數(因為$f\left( {{a}_{1}} \right)=f\left( {{a}_{2}} \right)=f\left( {{a}_{3}} \right)=0$)，由(3)小題可知：$f\left( f\left( x \right) \right)=0$有$5$個實根。