96學年度指定科目考試試題
數學甲
第壹部分：選擇題一、單選題
解：
$$z=\cos{2\pi \over 7} +i\sin {2\pi \over 7} \Rightarrow |1-z|= \left| 1-\cos{2\pi \over 7} -i\sin {2\pi \over 7}\right| = \sqrt{(1-\cos{2\pi \over 7})^2 + (\sin {2\pi \over 7})^2} \\ =\sqrt{1-2\cos{2\pi \over 7}+\cos^2{2\pi \over 7} + \sin^2 {2\pi \over 7} } = \sqrt{2-2\cos{2\pi \over 7}} = \sqrt{2-2(2\cos^2{\pi \over 7}-1)} \\ = \sqrt{4-4\cos^2{\pi \over 7}} = \sqrt{4(1-\cos^2{\pi \over 7})} =\sqrt{4\sin^2{\pi \over 7}} =2\sin{\pi \over 7}，故選\bbox[red, 2pt]{(1)}$$
解：
解：$$(0,0) \xrightarrow{10次移動} (5+5\log 2, 5+{15\over 2}\log 2) \Rightarrow \cases{5+5\log 2= \log a_1+\log a_3+ \cdots +\log a_9\\ 5+{15\over 2}\log 2= \log a_2+\log a_4+ \cdots +\log a_{10}} \\ \Rightarrow \cases{5+5\log 2= \log (a_1a_3\cdots a_9) =\log (a_1\cdot a_1r^2\cdots a_1r^8) = \log (a_1^5r^{20}) =5\log a_1 +20\log r\\ 5+{15\over 2}\log 2= \log (a_2a_4\cdots a_{10}) =\log (a_1r\cdot a_1r^3 \cdots a_1r^9) = \log (a_1^5r^{25}) = 5\log a_1 +25\log r} \\ \Rightarrow \cases{5+5\log 2= 5\log a_1 +20\log r\cdots (1)\\5+{15\over 2}\log 2= 5\log a_1 +25\log r \cdots(2)} \stackrel{(2)-(1)}{\Longrightarrow} {5\over 2} \log 2=5\log r \Rightarrow r^5=2^{5/2} \Rightarrow r=\sqrt 2 \\ 代回(1) \Rightarrow 5+5\log 2=5\log a_1+10\log 2 \Rightarrow \log a_1=1-\log 2 \Rightarrow a_1=5\\ \Rightarrow (a_1,r) =(5,\sqrt 2)，故選\bbox[red,2pt]{(5)}$$
二、多選題
解：$$(1)\bigcirc: 兩次考試成績適合散佈圖 \\ (2)\times: 相關係數0.016非常接近0，也就是兩者幾乎不相關，不適合用直線來表達兩科成績的關連性 \\(3)\bigcirc: 相關係數 {\sigma(x+5,y+5) \over \sigma(x+5)\sigma(y+5)}= {\sigma(x,y) \over \sigma(x)\sigma(y)}= 0.016\\ (4)\bigcirc: 相關係數 {\sigma(100x,100y) \over \sigma(100x)\sigma(100y)} = {100^2\sigma(x,y) \over 100\sigma(x) 100\sigma(y)} = {\sigma(x,y) \over \sigma(x)\sigma(y)}= 0.016 \\(5) \bigcirc: 相關係數 \cfrac{\sigma({x-\bar x \over s_x},{y-\bar y \over s_y})}{ \sigma({x-\bar x \over s_x})\sigma({y-\bar y \over s_y})} = \cfrac{\sigma({x \over s_x},{y \over s_y})}{ \sigma({x \over s_x})\sigma({y \over s_y})} = \cfrac{{1\over s_xs_y}\sigma(x,y)}{{1\over s_xs_y} \sigma(x)\sigma(y)}= {\sigma(x,y) \over \sigma(x)\sigma(y)}= 0.016 \\，故選\bbox[red,2pt]{(1,3,4,5)}$$
(1)\times: 若\alpha為Q(x)=0的實根，即Q(\alpha)=0 \Rightarrow P(\alpha)=Q(\alpha)(\alpha-3)+2 =0 \Rightarrow \alpha不是P(x)=0的根\\ \qquad\Rightarrow P(x)=0與Q(x)=0沒有共同實根 \\ (2)\times: 若Q(x)=x^4 \Rightarrow P(x)=x^4(x-3)+2 \Rightarrow x=0,3皆是P(x)=2的實根 \\ \qquad \Rightarrow x=3不是唯一的實根 \\(3)\bigcirc: Q(x)為正係數 \Rightarrow Q(4)>0 \Rightarrow P(4) = Q(4)(4-3)+2 = Q(4)+2>0 \Rightarrow P(4)\ne 0 \\\qquad \Rightarrow P(x)不能被(x-4)整除\\ (4)\bigcirc: Q(x)為正係數 \Rightarrow 對所有的x\ge 3 \Rightarrow P(x)=Q(x)(x-3)+2 >0 \\\qquad \Rightarrow P(x)=0沒有大於等於3的實根; 由於P(x)為5次式，一定有實根，所以有小於3的實根\\(5) \times: 若Q(x)= (x+3)^4+2 \Rightarrow P(x)=((x+3)^4+2)(x-3)+2 = (x+3)^4(x-3)+2x-4\\ \qquad \Rightarrow P(x)除以(x+3)(x-3)的餘式為2x-4，並非2 \\故選\bbox[red,2pt]{(3,4)}$$
$$ \cases{{x-3\over 2} ={y-5 \over 3} \cdots(1)\\ {y-5\over 3}=z-4\cdots(2) \\ {x\over a}= z-2 \cdots(3) \\ {y+1 \over 3}=z-2 \cdots(4)},由(2) \Rightarrow {y-5\over 3}+2 =(z-4)+2 \Rightarrow {y+1 \over 3}=z-2 \Rightarrow (2)\equiv (4)\\ 由(1)及(2) \Rightarrow (x,y,z)=(2t+3,3t+5,t+4),t\in R\\ (1)\times: 將(2t+3,3t+5,t+4)代入(3) \Rightarrow {2t+3 \over a}=t+2 \Rightarrow a= {2t+3 \over t+2} = 2-{1\over t+2} \ne 2,\forall t\in R \\(2) \bigcirc: 將(2t+3,3t+5,t+4)代入(3) \Rightarrow t= {2a-3 \over 2-a} 有唯一解\Rightarrow (x,y,z)有唯一解\\ (3)\bigcirc: \cases{{x-3\over 2} ={y-5 \over 3} \\ {x\over a}= z-2 } \Rightarrow (x,y,z)= (2t+3,3t+5,{2t+3 \over a}+2),t\in R \Rightarrow 有無限多組解 \\(4) \times: \cases{{x\over a} =z-2 \\ {y+1\over 3} ={z-2}} \Rightarrow (x,y,z) =(at,3t-1,t+2),t\in R \Rightarrow 有無限多組解 \\(5)\bigcirc: \cases{{x-3\over 2} ={y-5 \over 3} \\ {y-5 \over 3} =z-4} \Rightarrow (x,y,z)= (2t+3,3t+5,t+4),t\in R \Rightarrow 有無限多組解\\故選\bbox[red,2pt]{(2,3,5)}$$
解：$$\cases{A=\begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix} \Rightarrow A^2= \begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix} =I \\ B=\begin{bmatrix}1/2 & -\sqrt 3/2 \\ \sqrt 3/2 & 1/2 \end{bmatrix} =\begin{bmatrix}\cos \pi/3 & -\sin \pi/3 \\ \sin \pi/3 & \cos \pi/3 \end{bmatrix} \Rightarrow \text{逆時針旋轉}60^\circ \Rightarrow B^6=I} \\ (1)\times: \cases{AB= \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix}1/2 & -\sqrt 3/2 \\ \sqrt 3/2 & 1/2 \end{bmatrix} =\begin{bmatrix}1/2 & -\sqrt 3/2 \\ -\sqrt 3/2 & -1/2 \end{bmatrix} \\ BA=\begin{bmatrix}1/2 & -\sqrt 3/2 \\ \sqrt 3/2 & 1/2 \end{bmatrix} \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix} =\begin{bmatrix}1/2 & \sqrt 3/2 \\ \sqrt 3/2 & -1/2 \end{bmatrix}} \Rightarrow AB\ne BA \\(2) \bigcirc: \cases{A^2B= IB =B \\ BA^2=BI =B} \Rightarrow A^2B=BA^2 \\(3) \times: \cases{A^{11}B^3 = (A^2)^5A再旋轉180^\circ =A\begin{bmatrix}\cos \pi & -\sin \pi \\ \sin \pi & \cos \pi \end{bmatrix} =\begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix}-1 & 0 \\ 0 & -1 \end{bmatrix} =\begin{bmatrix}-1 & 0 \\ 0 & 1 \end{bmatrix} \\ B^6A^5 = I (A^2)^2A= IA = A=\begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix}} \\ \qquad \Rightarrow A^{11}B^3 \ne B^6A^5 \\ (4)\bigcirc: \cases{AB^{12}=A(B^6)^2=AI=A \\ A^7= (A^2)^3A =IA =A} \Rightarrow AB^{12}=A^7 \\(5)\bigcirc: (ABA)^{15} =(ABA)(ABA)\cdots (ABA)=ABA^2BA^2B\cdots A^2BA =ABIBIB\cdots IBA =AB^{15}A\\故選\bbox[red, 2pt]{(2,4,5)}$$
解：$$(1) \times: \cases{\lim_{x\to \infty}y= \infty \\ \lim_{x\to -\infty}y= -\infty} \Rightarrow 圖形沒有最高點，也沒有最低點\\ (2)\times: y'=0 \Rightarrow 3x^2+2=0 \Rightarrow 無實數解，即無水平切線\\ (3) \bigcirc: y'=3x^2+2 > 0 \Rightarrow 圖形為嚴格遞增 \Rightarrow 圖形與任一水平線只有一個交點 \\ (4)\bigcirc: (a,b)在圖形上 \Rightarrow b=a^3+2a+3 \Rightarrow -b= -a^3-2a-3 \Rightarrow -b+6=(-a)^3+2(-a)+3 \\ \qquad \Rightarrow (-a,-b+6)在圖形上\\ (5) \bigcirc: 面積= \int_{x=0}^{x=1} x^3+2x +3 \;dx = \left. \left[ {1\over 4}x^4 +x^2 +3x \right] \right|_0^1 = {1\over 4}+1+3 > 4\\，故選\bbox[red,2pt]{(3,4,5)}$$
三、選填題
解：
$$\cfrac{來自第5個工廠的不良品}{所有的不良品} = \cfrac{5 \over 50}{\sum_{k=1}^6 {k\over 50}} = \bbox[red, 2pt]{\cfrac{5}{21}}$$
解：
$$令\cases{A(-2,7) \\ B(-1,6) \\ 圓心O} \Rightarrow 直線\overline{OB}的方向向量為(4,3) \Rightarrow 直線\overline{OB}方程式: {x+1 \over 4} ={y-6 \over 3};\\ O在直線\overline{OB}上 \Rightarrow 令O(4t-1,3t+6) \Rightarrow \overline{OA} =\overline{OB} \Rightarrow \sqrt{(4t+1)^2 +(3t-1)^2} = \sqrt{(4t)^2 +(3t)^2} \\ \Rightarrow 2t+2=0 \Rightarrow t=-1 \Rightarrow O(-5,3) \Rightarrow 圓半徑r= \overline{OB}= \sqrt{(-4)^2+(-3)^2} =5\\ \Rightarrow 圓方程式: (x+5)^2+(y-3)^2 = 5^2 \Rightarrow x^2+y^2 +10x-6y+9=0 \Rightarrow \bbox[red, 2pt]{\cases{a=10 \\b=-6 \\c=9}}$$
解：
$$假設底面正方形邊長為x, 紙盒的高為y,見上圖;\\由題意知: \cases{灰色面積= 4xy+x^2=432 \\ 容量 = x^2y}，由算機不等式\cfrac{x^2+2xy+2xy}{3} \ge \sqrt[3]{x^2\times 2xy \times 2xy} \\\Rightarrow \cfrac{x^2+4xy}{3} \ge \sqrt[3]{4x^4y^2} \Rightarrow \cfrac{432}{3} \ge \sqrt[3]{4x^4y^2} \Rightarrow 144^3 \ge 4x^4y^2 \Rightarrow \sqrt{144^3 \over 4} \ge x^2y \\ \Rightarrow 864 \ge x^2y \Rightarrow 容量最大為864，此時 x^2=2xy，即x=2y \Rightarrow 容量=x^2y= 4y^2\cdot y = 4y^3=864\\ \Rightarrow y=6 \Rightarrow x=2\times 6=\bbox[red, 2pt]{12}$$
$$假設底面正方形邊長為x, 紙盒的高為y,見上圖;\\由題意知: \cases{灰色面積= 4xy+x^2=432 \\ 容量 = x^2y}，由算機不等式\cfrac{x^2+2xy+2xy}{3} \ge \sqrt[3]{x^2\times 2xy \times 2xy} \\\Rightarrow \cfrac{x^2+4xy}{3} \ge \sqrt[3]{4x^4y^2} \Rightarrow \cfrac{432}{3} \ge \sqrt[3]{4x^4y^2} \Rightarrow 144^3 \ge 4x^4y^2 \Rightarrow \sqrt{144^3 \over 4} \ge x^2y \\ \Rightarrow 864 \ge x^2y \Rightarrow 容量最大為864，此時 x^2=2xy，即x=2y \Rightarrow 容量=x^2y= 4y^2\cdot y = 4y^3=864\\ \Rightarrow y=6 \Rightarrow x=2\times 6=\bbox[red, 2pt]{12}$$
解：
$$(1)f(x)=x^3-6x^2-x+30=(x-3)(x-5)(x+2) \Rightarrow f(x)=0的解為 \bbox[red, 2pt]{x=3,5,-2} \\(2) \cos \angle ACB = {a^2+b^2 - \overline{AB}^2 \over 2ab} \Rightarrow -{1\over 2}= {9+25-\overline{AB}^2 \over 30} \Rightarrow \overline{AB} = 7\\ \Rightarrow \overline{DE} = \overline{AE} +\overline{BD}-\overline{AB} =5+3-7=1 \Rightarrow \cfrac{\triangle CDE}{ \triangle ABC} = \cfrac{\overline{DE}}{\overline{AB}} = \cfrac{1}{7} \Rightarrow \triangle CDE={1\over 7}\triangle ABC \\ = {1\over 7}\times {1\over 2}\times \overline{BC}\times \overline{AC}\sin \angle ACB= {1\over 14}\times 15\times {\sqrt 3\over 2} = \bbox[red, 2pt]{{15 \over 28}\sqrt 3}$$
解：$$(1)\cases{A(-2,7,15) \\ B(1,16,3) \\ C(10,7,3)} \Rightarrow \cases{\overrightarrow{AB}= (3,9,-12) \\ \overrightarrow{AC} =(12, 0,-12)} \Rightarrow \vec n= \overrightarrow{AB} \times \overrightarrow{AC} = (-108,-108,-108) \\ \Rightarrow 過A且法向量為\vec n的平面方程式: -108(x+2)-108(y-7)-108(z-15)=0\\ \Rightarrow \bbox[red, 2pt]{x+y+z=20}\\ (2) 外心P(a,b,c) \Rightarrow \cases{\overline{PA} =\overline{PB} =\overline{PC} \\ P在\triangle ABC平面上} \\\Rightarrow \cases{(a+2)^2 +(b-7)^2 +(c-15)^2 =(a-1)^2+(b-16)^2+(c-3)^2 \\ (a-1)^2+(b-16)^2+(c-3)^2 =(a-10)^2+(b-7)^2+(c-3)^2 \\ a+b+c=20 } \\ \Rightarrow \cases{a+3b-4c=-2\\ a-b=-6 \\ a+b+c=20} \Rightarrow (a,b,c)= \bbox[red,2pt]{(3,9,8)}$$
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第五題的詳解應該有錯喔 Q(x)必是四次式
回覆刪除且係數均為正數，(2)Q(x)的常數項不可能為-4
謝謝提醒，已修正替換另一個例子!
刪除算幾不等式 不是柯西不等式
回覆刪除選擇C
刪除已修訂, 謝謝!
刪除