113年新竹女中教甄-數學詳解

 國立新竹女子高級中學113 學年度第1學期第1次教師甄選

一、填充題：每題5 分，共50 分

解答: $$\cases{|w|=3\\ |z|=4} \Rightarrow \cases{w=3e^{i\theta_1} \\ z=4e^{i\theta_2}} \Rightarrow {w\over z} ={3\over 4} e^{i(\theta_1-\theta_2)} ={3\over 4}\cos(\theta_1-\theta_2)+ {3\over 4}i\sin(\theta_1-\theta_2) \\ \Rightarrow {w-z\over z} ={w\over z}-1={3\over 4}\cos(\theta_1-\theta_2)-1+ {3\over 4}i\sin(\theta_1-\theta_2) \\ \Rightarrow \tan \theta={{3\over 4}\sin (\theta_1-\theta_2) \over {3\over 4} \cos (\theta_1-\theta_2)-1} ={3\sin \alpha\over 3\cos \alpha-4} =k \Rightarrow 4k=3k\cos\alpha-3\sin \alpha \\ \Rightarrow 4k\le \sqrt{9k^2+9} \Rightarrow 16k^2\le 9k^2+9 \Rightarrow \tan^2\theta=k^2\le \bbox[red, 2pt]{9\over 7}$$

解答: $$f(x)=2x+\sqrt{9x^2+16} \Rightarrow f'(x)=2+{9x\over  \sqrt{9x^2+16}} =0 \Rightarrow 81x^2=36x^2+64\\ \Rightarrow x^2={64\over 45} \Rightarrow x=\pm {8\over 3\sqrt 5} =\pm{8\sqrt 5\over 15} \Rightarrow f(-{8\sqrt 5\over 15})=-{16\sqrt 5\over 15}+\sqrt{{64\over 5} +16} ={4\sqrt 5\over 3} \\ \Rightarrow (\alpha,\beta)= \bbox[red, 2pt]{\left(-{8\sqrt 5\over 15}, {4\sqrt 5\over 3}\right)}$$

解答:

$$假設\cases{A(0,0)\\ B(4\sqrt 3, 0) \\ C(2\sqrt 3\cos 60^\circ, 2\sqrt 3 \sin 60^\circ)=(\sqrt 3,3)}，又\overline{AD}是\angle A的角平分線 \Rightarrow {\overline{CD}\over \overline{DB}}={\overline{AC}\over \overline{AB}} \\ \Rightarrow D={1\over 3}(2C+B)=(2\sqrt 3,2) \Rightarrow L=\overleftrightarrow{AD}: \sqrt 3 y=x \Rightarrow P(\sqrt 3t,t)\\ \Rightarrow\cases{ \overrightarrow{PA} = (-\sqrt 3t,-t)\\ \overrightarrow{PB}= (4\sqrt 3-\sqrt 3 t,-t)\\ \overrightarrow{PC}= (\sqrt 3-\sqrt 3 t,3-t)} \Rightarrow (\overrightarrow{PB}+2\overrightarrow{PC} )\cdot \overrightarrow{PA} =(6\sqrt 3-3\sqrt 3 t,6-3t) \cdot (-\sqrt 3t,-t) \\=12t^2-24t =12(t-1)^2-12 \Rightarrow 最小值=\bbox[red, 2pt]{-12}$$

解答: $$假設正方形邊長=4且A為原點,則\cases{A(0,0,0)\\B(3,0,0) \\C(3,3,0)\\ D(0,3,0) \\E(3,2,0)\\ F(2,3,0)} \Rightarrow G(t,t,a) \\ 又\cases{\overline{AG}=3\\ \overline{GE}=2} \Rightarrow\cases{t^2+t^2+a^2=9\\ (t-3)^2+ (t-2)^2+a^2=4} \Rightarrow \cases{t=9/ 5\\ a=3\sqrt 7/  5} \Rightarrow G({9\over   5}, {9\over   5}, {3\sqrt 7\over   5}) \\ \Rightarrow \cases{ \overrightarrow{AF} =(2,3,0) \\ \overrightarrow{AG} =({9\over   5}, {9\over   5}, {3\sqrt 7\over   5})} \Rightarrow \vec n= \overrightarrow{AF} \times \overrightarrow{AG} \parallel(3\sqrt 7,-2\sqrt 7, -3)  \\\Rightarrow \cases{側面AEG:3\sqrt 7x-2\sqrt 7y-3z=0 \\底面AECF: z=0} \Rightarrow \cos \theta={(3\sqrt 7,-2\sqrt 7, -3)  \cdot (0,0,1) \over |(3\sqrt 7,-2\sqrt 7, -3)  |} =-{3\over 10} \\ \Rightarrow \sin \theta= \bbox[red, 2pt]{-{\sqrt{91}\over 10}}$$

解答:

$$丟五次骰子回到原點，也就是走5段，轉4次60度。假設路徑為A\to B\to C\to D\to E\to A，\\其中在B,C,D,E順時針轉60度，路徑如上圖。\\因此延長\cases{\overline{AB}與\overleftrightarrow{CD}交於P \\\overline{AE}與\overleftrightarrow{CD}交於Q}可以得到一個正\triangle APQ,其中\triangle PBC與\triangle ADE也是正\triangle\\假設\cases{ \overline{PB}=a \\ \overline{QD}=b\\ \overline{CD}=x},5段路徑長分別為x+b,a,x,b,x+a,每段路徑長度均需介於1與6之間\\符合條件的樣本: \quad \begin{array}{r} x& a& b& 數量\\\hline 1& 1-5& 1-5 & 25\\ 2& 1-4& 1-4& 16\\ 3& 1-3& 1-3& 9\\ 4& 1-2& 1-2& 4\\ 5& 1& 1& 1\\\hline\end{array} \Rightarrow 合計:55 \Rightarrow 機率=\bbox[red, 2pt]{55\over 6^5}$$

解答: $$\left[\log_2 {1\over x} \right] =\begin{cases}1, & 2\le {1\over x}\lt 4 \Rightarrow {1\over 4}\lt x\le {1\over 2}\\ 3, & 8\le {1\over x}\lt 16 \Rightarrow {1\over 16}\lt x\le {1\over 8}\\ 5, &32\le {1\over x}\lt 64 \Rightarrow {1\over 64}\lt x\le {1\over 32} \\ \dots & \dots\end{cases} \\ \left[ \log_3{1 \over y} \right] =\begin{cases}1,&3\le {1\over y}\lt 9 \Rightarrow {1\over 9}\lt y\le {1\over 3}\\ 3, &27 \le {1\over y}\lt 81 \Rightarrow {1\over 81}\lt y\le {1\over 27}\\ 5,&243 \le {1\over y}\lt 729 \Rightarrow {1\over 729}\lt x\le {1\over 243}\\ \cdots & \cdots \end{cases} \\ \Rightarrow 欲求之面積=\left(({1\over 2}-{1\over 4})+({1\over 8}-{1\over 16}+ ({1\over 32}-{1\over 64})+\cdots \right) \\\qquad \times \left( ({1\over 3}-{1\over 9}) +({1\over 27}-{1\over 81})+ ({1\over 243} -{1\over 729})+ \cdots\right) \\=({1\over 4}+{1\over 16} +{1\over 64}+\cdots )({2\over 9}+{ 2\over 81} +{ 2\over 729} +\cdots) ={1\over 3}\cdot {1\over 4}= \bbox[red, 2pt]{1\over 12}$$

解答: $$若最小元素是a,則有C^{2024-a}_{999}挑法,其中a=1,2,\dots 1025(=2024-999)\\ 因此平均值為{\sum_{a=1}^{1025} a\cdot C^{2024-a}_{999} \over \sum_{a=1}^{1025}   C^{2024-a}_{999}} ={C^{2025}_{1001} \over C^{2024}_{1000}} = \bbox[red, 2pt]{2025\over 1001}\\ 分母:C^{2024}_{1000} = {C^{2023}_{1000}} +C^{2023}_{999} =  C^{2022}_{1000}+ C^{2022}_{999}+C^{2023}_{999} = C^{2021}_{1000}+ C^{2021}_{999}+ C^{2022}_{999}+C^{2023}_{999} = \cdots \\=C^{999}_{999} +C^{1000}_{999} +\cdots + C^{2022}_{999}+C^{2023}_{999} \Rightarrow \sum_{a=1}^{1025} C^{2024-a}_{999} =C^{2024}_{1000}\\分子:\sum_{a=1}^{1025} a\cdot C^{2024-a}_{999} =\sum_{a=1}^{1025} C^{2024-a}_{999} +\sum_{a=1}^{1024} C^{2023-a}_{999}  +\sum_{a=1}^{1023} C^{2022-a}_{999} +\cdots  +\sum_{a=1}^{2} C^{1001-a}_{999} +\sum_{a=1}^{1} C^{1000-a}_{999} \\=C^{2024}_{1000}+ C^{2023}_{1000} +C^{2022}_{1000} +\cdots+C^{1001}_{1000} +C^{1000}_{1000} =C^{2005}_{1001}$$

解答: $$\lim_{n\to \infty} \sum_{k=n}^{3n-1} {n^2\over a_k} =\lim_{n\to \infty} \sum_{k=n}^{3n-1} {n^2\over k^3} =\lim_{n\to \infty} \sum_{k=n}^{3n-1} {1/n\over (k/n)^3} =\int_1^3{1\over x^3}\,dx =\left. \left[ -{1\over 2x^2} \right] \right|_1^3 = \bbox[red, 2pt]{4\over 9}$$

解答:

$$\cases{A(1,\sqrt 3)\\ B(1,-\sqrt 3) \\ P\in \overline{AB}} \Rightarrow P(1,t),-\sqrt 3\le t\le \sqrt 3 \Rightarrow \overline{OP} =\sqrt{1+t^2} \Rightarrow 1\le \overline{OP} \le 2\\ Q(x,y)\in \overrightarrow{OP} \Rightarrow y=tx \Rightarrow   \cases{\overline{OP}=\sqrt{1+y^2/x^2}\\ \overline{OQ} =\sqrt{x^2+y^2} } \Rightarrow (1+{y^2\over x^2})(x^2+y^2) =4^2 \\ \Rightarrow (x^2+y^2)^2-(4x)^2=0 \Rightarrow (x^2+4x+y^2)(x^2-4x+y^2  ) =((x+2)^2+y^2-4) ((x-2)^2+y^2-4)=0 \\ 由於Q(x,y)\in \overrightarrow{OP} \Rightarrow x\ge 0\Rightarrow  (x+2)^2+y^2=4不合,因此(x-2)^2+y^2=4 \\ 又1\le \overline{OP} \le 2 \Rightarrow 2\le \overline{OQ} \le 4 \Rightarrow x\ge 1 \Rightarrow Q點軌跡為一圓,但x\ge 1，即{2\over 3}圓\\ \Rightarrow 軌跡長={2\over 3}2\pi\cdot 2= \bbox[red, 2pt]{{8\over 3}\pi}$$

解答: $$E(X)=\int_0^\infty \left( 1-\prod_{i=1}^{N} (1-e^{-p_ix})\right)\,dx = \int_0^\infty \left( 1- (1-e^{-x/2}) (1-e^{-x/3}) (1-e^{-x/6})\right)\,dx \\=\int_0^\infty \left(e^{-x/6}+e^{-x/3}- e^{-2x/3} -e^{-5x/6} +e^{-x} \right)\,dx \\=0-(-6-3+{3\over 2}+{6\over 5}-1) = \bbox[red, 2pt]{73\over 10}, \href{https://mat.uab.cat/matmat_antiga/PDFv2014/v2014n02.pdf}{公式來源}$$

========================== END ===============================