115年大理高中教甄-數學詳解

臺北市立大理高中 115 學年度數學科正式教師甄選

一、簡答題 (每題 8 分)

解答:$$|z_1-3i|=|z_1-(2-i)| \Rightarrow \overline{AP}=\overline{AQ}, 其中\cases{A(z_1=x+iy)=(x,y) \\P(3i=0+3i) =(0,3)\\ Q(2-i) =(2,-1)} \\\Rightarrow A在\overline{PQ}的中垂線:x-2y+1=0 上 \\ 又\cases{z_1= re^{i \theta}\\z_2=(-1+\sqrt 3i)z_1=2e^{i120^\circ}\cdot re^{i\theta}=2re^{i(\theta+120^\circ)}  \\z_3= 3i\cdot z_1 =3e^{i90^\circ} \cdot re^{i\theta} =3re^{i(\theta+90^\circ)}} \Rightarrow \triangle ABC面積= {1\over 2} |Im(\bar z_1z_2+ \bar z_2z_3+ \bar z_3z_1)| \\ ={1\over 2} \left| Im \left( re^{-i\theta}\cdot 2re^{i(\theta+120^\circ) } +2re^{-i(\theta+120^\circ)}\cdot 3re^{i(\theta+90^\circ)} +3re^{-i(\theta+90^\circ)} \cdot re^{i\theta} \right)\right|  \\ ={1\over 2} \left| Im \left( 2r^2e^{i(120^\circ) } + 6r^2e^{-i30^\circ}  +3r^2e^{-i 90^\circ}  \right)\right|  ={1\over 2} \left|  \sqrt 3r^2-6r^2 \right| ={6-\sqrt 3\over 2}r^2 \\ 面積要最小\Rightarrow r要最小\Rightarrow r=原點(0,0)至中垂線距離={1\over \sqrt 5} \Rightarrow 最小面積={6-\sqrt 3\over 2}\cdot {1\over 5}\\ =\bbox[red, 2pt]{6-\sqrt 3\over 10}$$

解答:$$9^x+6^x=4^x \Rightarrow  \left( {9\over 4} \right)^x+\left( {6\over 4} \right)^x=1 \Rightarrow \left( {3\over 2} \right)^{2x}+ \left({3\over 2} \right)^{x}-1=0 \\ \Rightarrow \left({3\over 2} \right)^{x} ={-1+\sqrt 5\over 2} \Rightarrow x= \bbox[red, 2pt]{\log_{3/2} \left( {-1+\sqrt 5\over 2} \right)}$$

解答:

$$D,E,F為切點\Rightarrow \cases{ \overline{CE} =\overline{CD} =a \\ \overline{BD}=\overline{BF}=b} \Rightarrow \cases{\overline{BC}= a+b=4 \\ \overline{AE} =\overline{AF}\Rightarrow 5+a=6+b } \Rightarrow \cases{a=5/2\\ b=3/2} \\\Rightarrow a:b =5:3  \Rightarrow \overrightarrow{AD}= {5\over 3+5} \overrightarrow{AB} +{3\over 3+5} \overrightarrow{AC} \Rightarrow (\alpha,\beta) = \bbox[red, 2pt] { \left( {5\over 8},{3\over 8} \right)}$$

解答:$$可能的三角形:\cases{正三角形:(1,1,1),(2,2,2), (3,3,3), (4,4,4)共4種\\ 等腰三角形:(2,2,1),(2,2,3), (3,3,1),(3,3,2), (3,3,4), (4,4,1), (4,4,2),(4,4,3) \\ \qquad 共8種,每種排列數{3!\over 2!}=3 \Rightarrow 共8\times 3=24種\\ 其他:(2,3,4),排列數3!=6} \\ \Rightarrow 機率={4+24+6\over 4^6}={34\over 64} =\bbox[red,2pt] {17\over 32}$$

解答:$$ \left( {1\over 3} \right)^{\sqrt 3\cos \theta} \ge \left( {1\over 3} \right)^{1+\sin \theta} \Rightarrow \sqrt 3\cos \theta \le 1+\sin \theta \Rightarrow \sqrt 3\cos \theta-\sin \theta =2 \left( {\sqrt 3\over 2}\cos \theta-{1\over 2}\sin \theta \right) \le 1 \\ \Rightarrow \cos(\theta+{\pi\over 6}) \le {1\over 2} \Rightarrow {\pi\over 3} \theta+{\pi\over 6} \le {5\pi\over 3} \Rightarrow \bbox[red, 2pt]{{\pi\over 6}\le \theta \le {3\pi\over 2}}$$

解答:$$若前一次在甲袋，操作後仍在甲袋的情形:\\操作前\cases{甲:1紅1黑\\ 乙:2黑} \Rightarrow \cases{ 甲抽紅球放乙,再從乙抽紅球放甲:機率=(1/2)(1/3)=1/6 \\甲抽黑球放乙, 再從乙抽黑球放甲:機率=(1/2)(1/1)=1/2} \Rightarrow 合計:{2\over 3} \\ \Rightarrow p_n={2\over 3}p_{n-1} +{1\over 3}(1-p_{n-1}) \Rightarrow \bbox[red, 2pt]{p_n={1\over 3}p_{n-1}+{1\over 3}} \\ \lim_{n\to \infty} p_n= p \Rightarrow p={1\over 3}p+{1\over 3} \Rightarrow p= \bbox[red, 2pt]{1\over 2} $$

解答:$$假設\cases{x_n= (3+\sqrt 7)^n\\ y_n= (3-\sqrt 7)^n} \Rightarrow x_n+y_n = \sum_{k=0}^n {n\choose k} \left( 3^{n-k}(\sqrt 7)^k+3^{n-k}(-\sqrt 7)^k \right) \\= 2\sum_{k=0}^{\lfloor n/2 \rfloor} {n\choose 2k} 3^{n-2k}(\sqrt 7)^{2k} = 2\sum_{k=0}^{\lfloor n/2 \rfloor} {n\choose 2k} 3^{n-2k}7^{k} 為一偶數 \Rightarrow x_n+y_n=2M, M\in \mathbb N \\ 2 \lt \sqrt 7\lt 3 \Rightarrow 0\lt 3-\sqrt 7 \lt 1 \Rightarrow 0\lt (3-\sqrt 7)^n=y_n\lt 1 \Rightarrow x_n=2M-y_n \\\Rightarrow 2M-1\lt x_n\lt 2M \Rightarrow x_n的整數部分=2M-1為一奇數, \bbox[red, 2pt]{故得證}$$

解答:$$\int_2^x f(t)\,dt =2x^3-6x+a-1 \Rightarrow f(x)=6x^2-6 \\ f(x)=0 \Rightarrow 6(x^2-1)=0 \Rightarrow x=\pm 1 \Rightarrow 繞x軸旋轉體積V= \pi \int_{-1}^1 [6x^2-6]^2\,dx = \bbox[red, 2pt]{{192\over 5}\pi}$$

 

解答:$$一、1車2羊:\cases{選中車的機率1/3\\ 選中羊的機率2/3} \Rightarrow \cases{一開始選中車，換門必輸\\ 一開始選羊，換門必贏} \\\qquad \Rightarrow 換門贏的機率是2/3, 不換贏的機率是1/3 \Rightarrow 應該換門\\ 二、1車3羊: \cases{選中車的機率1/4\\ 選中羊的機率3/4} \Rightarrow  不換門，勝率1/4; 換門勝率:{3\over 4}\times {1\over 2}={3\over 8} \Rightarrow 應該換門\\ 三、1車n-1羊:\cases{選中車的機率1/n\\ 選中羊的機率(n-1)/n} \Rightarrow 不換門勝率:{1\over n};換門勝率:{n-1\over n}\times {1\over n-2}\\ \qquad \Rightarrow {n-1\over n(n-2)} \gt {1\over n} \Rightarrow 應該要換門$$

解答:$$\ell_1=3\times 1=3 \Rightarrow A_1= {\sqrt 3\over 4} \ell^2 = {\sqrt 3\over 4} \Rightarrow \ell_n={4\over 3}\ell_{n-1} = 3\cdot \left( {4\over 3} \right)^{n-1} \Rightarrow \bbox[red, 2pt]{\cases{\ell_1=3\\ \ell_n=3\cdot ({4\over 3})^{n-1}, n\ge 2}}\\ 在步驟(n-1)時的總邊數為 N_{n-1}=3\times 4^{n-2} \Rightarrow 步驟n時，新增的小正三角形邊長為 \left( {1\over 3} \right)^{n-1} \\ \Rightarrow 新增的小正三角形面積= {\sqrt 3\over 4} \cdot \left( {1\over 3} \right)^{2(n-1)}= {\sqrt 3\over 4} \cdot \left( {1\over 9} \right)^{n-1} \\ \Rightarrow 步驟n所增加的總面積= 3\times 4^{n-2} \times  {\sqrt 3\over 4} \cdot \left( {1\over 9} \right)^{n-1}  ={\sqrt 3\over 12} \left( {4\over 9} \right)^{n-2} \\ \Rightarrow \bbox[red, 2pt]{ \cases{A_1= \sqrt 3/4\\ A_n= A_{n-1}+{\sqrt 3\over 12} \left( {4\over 9} \right)^{n-2}, n\ge 2 }}$$

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