111年臺南女中教甄-數學詳解

國立臺南女子高級中學 111 學年度第一次教師甄選

一、 填充題(每題 4 分，共 60 分)

解答： $$z^{11}+z-1=0 \Rightarrow z^{11} =1-z \Rightarrow |z^{11}|=1 =|1-z| \Rightarrow |z-1|=1\\ 令\cases{O為原點\\ A(1)\\ B(z)}，則\overline{OA} =\overline{AB} =\overline{OB}=1 \Rightarrow \triangle OAB為正\triangle \Rightarrow B=({1\over 2},\pm {1\over 2}\sqrt 3) \Rightarrow z=\bbox[red, 2pt]{{1\over 2}\pm {\sqrt 3\over 2} i}$$

解答： $$假設2021 = m\times q + r，其中\cases{m=除數\\ q=商數\\ r=餘數}，m,q,r\in \mathbb{N}且\,m \gt 1，\Rightarrow 2022= m\times q+r+1；\\ 因此，若\cases{r+1 \lt m \Rightarrow \left[ 2022\over m\right]-\left[ 2021\over m\right] =q-q=0\\[1ex] r+1=m \Rightarrow 2022 = m(q+1) \Rightarrow \cases{\left[ 2022\over m\right]-\left[ 2021\over m\right] =(q+1)-q=1\\ m是2022的因數}}\\ 而2022的因數=1,2,3,6,337,667,1011,2022，扣除1及2022共有6個因數；\\因此 a_{2022}-a_{2021} =\sum_{m=1}^{2022}  \left[ 2022\over m\right] -\sum_{m=1}^{2021}  \left[ 2021\over m\right]\\ =\left[ 2022\over 1\right]-\left[ 2021\over 1\right]+\sum_{m=2}^{2021}  \left(\left[ 2022\over m\right]-\left[ 2021\over m\right] \right) +\left[ 2022\over 2022\right]  =2022-2021+6+1 =\bbox[red, 2pt]{8}$$

解答： $$\cases{L_1: {x-2\over a}= {y-5\over b} ={z-7\over c}\\L_2:{x\over 2}= {y\over 3}={z\over 4}\\ L_3:{x-2\over 4} ={y-3\over 3} ={z-1\over 2}}，因此若\cases{  P\in L_2\\ Q\in L_3 }，則\cases{   P(2s, 3s,4s), s\in \mathbb{R}\\ Q(4t+2, 3t+3, 2t+1), t\in \mathbb{R}} \\ 且\cases{P \in L_1 \\ Q \in L_1} \Rightarrow \cases{{2s-2\over a}= {3s-5\over b} ={4s-7\over c} \\ {4t\over a}= {3t-2\over b} ={2t-6\over c}} \Rightarrow {2s-2\over 4t} ={3s-5\over 3t-2} ={4s-7\over 2t-6} \Rightarrow \cases{s=3\\ t=-2 }\\ \Rightarrow \cases{P(6,9,12)\\ Q(-6,-3,-3)} \Rightarrow \overline{PQ} =\sqrt{12^2+12^2 +15^2} =\bbox[red, 2pt]{3\sqrt{57}}$$

解答： $$1,2,3,4,5,6只能出現在第一張牌、或第二張牌、或第三張牌、或都不在三張牌上；\\因此6個元素都有4種可能，共有4^6 =\bbox[red, 2pt]{4096}$$

解答： $$f(n)={n\over n+1} \Rightarrow (n+1)f(n)-n=0，因此取g(x)=(x+1)f(x)-x，則0-10皆為g(x)=0的根 \\ \Rightarrow g(x) =(x+1)f(x)-x =ax(x-1)\cdots (x-10) \Rightarrow g(-1)= 1= -a\cdot 11! \Rightarrow a=-{1\over 11!}\\ \Rightarrow g(11)= 12f(11)-11 = a\cdot 11! =-1 \Rightarrow f(11)={10\over 12} =\bbox[red, 2pt]{5\over 6}$$

解答： $$令a(z_1-z_2)=b(z_1-z_3) \Rightarrow (a-b)z_1-az_2+bz_3= 0 = z_1-(4+4i)z_2 +(3+4i)z_3=0\\ \Rightarrow \cases{a=4+4i\\ b=3+4i} \Rightarrow z_1-z_3={a\over b}(z_1-z_2) = {4+4i\over 3+4i} (z_1-z_2) \Rightarrow |z_1-z_3| = \left|\cfrac{4+4i}{3+4i}\right||z_1-z_2|\\ = \left| {28-4i\over 25}\right| \times 5 ={20\sqrt 2\over 25} \times 5= \bbox[red, 2pt]{4\sqrt 2}$$

解答： $$聯立方程式有無限多組解 \Rightarrow \begin{vmatrix} \sqrt 2 a-2 & -3 & -3\\ 3 & 1 & -1\\ 13 & 7 & -\sqrt 2a\end{vmatrix}=0，其中a= \sin \theta +\cos \theta\\ \Rightarrow a^2= (\sin \theta+\cos \theta)^2 ={1\over 2} \Rightarrow 1+2\sin\theta \cos \theta ={1\over 2} \Rightarrow \sin 2\theta =-{1\over 2}  \Rightarrow 2\theta =-30^\circ \\ \Rightarrow \theta = \bbox[red, 2pt]{-15^\circ}$$

解答： $$\overrightarrow{AP_k}\cdot \overrightarrow{AP_{k+1}} =(\overrightarrow{AB} +\overrightarrow{BP_k}) \cdot (\overrightarrow{AB} +\overrightarrow{BP_{k+1}}) =(\overrightarrow{AB} +k\overrightarrow{BP_1}) \cdot (\overrightarrow{AB} + (k+1)\overrightarrow{BP_{1}}) \\ =\overline{AB}^2+ (2k+1)\overrightarrow{AB} \cdot \overrightarrow{BP_1} +k(k+1) \overline{BP_1}^2 =1+ {2k+1\over n}\cos 120^\circ +{k(k+1)\over n^2} \\ =1- {2k+1\over 2n}  +{k(k+1)\over n^2}  \\ \Rightarrow S_n = \overrightarrow{AB} \cdot \overrightarrow{AP_{ 1}} +\sum_{k=1}^{n-1} \left( \overrightarrow{AP_k}\cdot \overrightarrow{AP_{k+1}} \right)  =\overrightarrow{AB} \cdot (\overrightarrow{AB}+ \overrightarrow{BP_{ 1}}) +\sum_{k=1}^{n-1} \left( 1- {2k+1\over 2n}  +{k(k+1)\over n^2} \right) \\ =1-{1\over 2n} +(n-1)-{n(n-1)+(n-1)\over 2n} +{(n-1)n(2n-1)/6+ n(n-1)/2 \over n^2} =\bbox[red, 2pt]{5n^2-2\over 6n}$$

解答： $$任排:3個謙、3個卑及1個再的排列數={7!\over 3!3! } =140\\ 先排3個謙及1個再，其排列數=4，再將3個卑插入；只有最左邊與再的右邊可以插入卑，\\\qquad 有4種排法，共4\times 4=16;\\ 因此機率為{16\over 140} =\bbox[red, 2pt]{4\over 35}$$

解答： $$\overrightarrow{OP} =(\sin \alpha-\cos\beta,\sin \alpha+2\cos \beta,2\sin \alpha+\cos \beta) =\sin \alpha(1,1,2)+ \cos \beta(-1,2,1)\\ 兩向量\cases{(1,1,2)\\ (-1,2, 1)}張出來的平行四邊形面襎= |(1,1,2)\times (-1,2,1)| = |(-3,-3,3)| =3\sqrt 3\\ 又\cases{0\le \alpha \le \pi/6\\ 0\le \beta \le \pi/3} \Rightarrow \cases{0\le \sin \alpha \le 1/2\\ 1/2\le \cos \beta \le 1} \Rightarrow 欲求面積={1\over 2}\cdot (1-{1\over 2})\cdot 3\sqrt 3 = \bbox[red, 2pt]{{3\over 4}\sqrt 3}$$

解答： $$令k=\log_3 2，則\cases{a= s+\log_3 2= s+k\\ b=s+\log_9 2= s+{1\over 2}k\\ c=s+\log_{27} 2 = s+{1\over 3}k}；\\a,b,c成等比\Rightarrow b^2=ac \Rightarrow (s+{1\over 2}k)^2 = (s+k)(s+{1\over 3}k) \Rightarrow ks+{1\over 4}k^2 = {4\over 3}ks +{1\over 3}k^2 \\\Rightarrow {1\over 3}ks = -{1\over 12}k^2 \Rightarrow s=-{1\over 4}k \Rightarrow 公比={b\over a} = \cfrac{-{1\over 4}k+ {1\over 2}k}{-{1\over 4}k+k} ={1/4\over 3/4} =\bbox[red, 2pt]{1\over 3}$$

解答： $$Q\in L:{x+1\over 1} ={y-k\over -2} ={z+4\over 3} \Rightarrow Q(t-1,-2t+k,3t-4),t \in \mathbb{R}\\ \Rightarrow f(t)=\overline{PQ}^2 = t^2+(2t-k)^2+(3t-1)^2 = 14t^2-(6+4k)t+k^2+1\\ \Rightarrow f(t)的最小值=f({2k+3\over 14}) = {5\over 7}((k-{3\over 5})^2 +{7\over 50}) = {5\over 7}\times {7\over 50} ={1\over 10} (當k={3\over 5})\\ \Rightarrow \overline{PQ}的最小值= \sqrt{1\over 10} =\bbox[red, 2pt]{\sqrt{10}\over 10}$$

解答： $$\cases{(1,0)\in A\cap C\\ (0,1) \in B\cap C} \Rightarrow 剩下一點P \in A\cap B\cap C；\\ 若P\in A\cap B \Rightarrow P({1\over a+1},{1\over a+1})，又P \in C \Rightarrow ({1\over a+1})^2 +({1\over a+1})^2=1 \Rightarrow (a+1)^2 =2\\ \Rightarrow  a= \bbox[red, 2pt]{-1\pm \sqrt 2}$$

解答： $$f(x)\ge x \Rightarrow y=f(x)圖形為凹向上；又兩圖形\cases{y=(1+x^2)/2\\ y=x} 恰交於B(1,1)\\ 因此y=f(x)的頂點為(-1,0)並經過(1,1) \Rightarrow y=f(x)=a(x+1)^2 且f(1)=1\Rightarrow a={1\over 4} \\ \Rightarrow f(x)={1\over 4}(x+1)^2 \Rightarrow f(4)= \bbox[red, 2pt]{25\over 4}$$

解答： $$令u=4-x^2\Rightarrow du =-2xdx \Rightarrow \int_0^2 x\sqrt{4-x^2}\,dx = \int_4^0 -{1\over 2}\sqrt u\,du =\left. \left[ -{1\over 3}u^{3/2} \right]\right|_4^0 = \bbox[red, 2pt]{8\over 3}$$

二、 計算證明題(需詳列計算、證明過程， 否則不予計分。共 40 分)

解答： $$\omega = \cos{2\pi \over 111} +i\sin {2\pi \over 111} \Rightarrow \omega^{111}=1 \Rightarrow \omega^{111}-1=0 \Rightarrow (\omega-1) \cdot \sum_{k=0}^{110} \omega^k =0 \Rightarrow \sum_{k=0}^{110} \omega^k =0\\ 因此\sum_{k=1}^{110}{\omega^{2k} \over \omega^k-1} =\sum_{k=1}^{110} \left(\omega^k+1 +{1 \over \omega^k-1} \right)= -1+110 +\sum_{k=1}^{110}{1 \over \omega^k-1} \cdots(1)\\ 取f(x)= \sum_{k=0}^{110} x^{k} =\prod_{k=1}^{110}(x-\omega^k) \Rightarrow f'(x)=\sum_{k=1}^{110} kx^{k-1} = \sum_{m=1}^{110}\prod_{k=1,k\ne m}^{110}(x-\omega^k) \\ \Rightarrow g(x)= {f'(x)\over f(x)} = \cfrac{\sum_{k=1}^{110} kx^{k-1}}{\sum_{k=0}^{110} x^{k}}= \sum_{k=1}^{110} {1\over x-\omega^k} \Rightarrow g(1)=\cfrac{(111\times 110)\div 2}{111} =55 =\sum_{k=1}^{110} {1\over 1-\omega^k} \\ \Rightarrow \sum_{k=1}^{110} {1\over \omega^k-1} =-55 代入(1) \Rightarrow \sum_{k=1}^{110}{\omega^{2k} \over \omega^k-1} =-1+110-55 = \bbox[red, 2pt]{54}$$

解答： $$假設三次函數f(x)=ax^3 +bx^2 +cx +d =a(x+{b\over 3a})^3 +{ 3ac-b^2\over 3a}(x+{b\over 3a})+ { 27a^2d +2b^3-9abc\over 27a^2}\\ 因此圖形y=f(x)可透過平移的方式變成y=g(x)= ax^3+px，而原圖形上A、B、C三點平移後變為A'、B'、C'\\ 假設\cases{A'(s,g(s))=(s,as^3+ps)\\ C'(t,g(t))= (t,at^3+pt)} \Rightarrow B'=(A'+C')\div 2 = ({s+t\over 2},{as^3+at^3+ps+pt \over 2})\\ B'也在y=g(x)上 \Rightarrow a({s+t\over 2})^3+p({s+t\over 2}) ={as^3+at^3+ps+pt \over 2} \Rightarrow 4a(s^3+t^3)=a(s+t)^3 \\ \Rightarrow 3s^3-3s^2t-3st^2+3t^3=0 \Rightarrow 3(s+t)(s-t)^2=0 \Rightarrow s+t=0(A',C'相異\Rightarrow s\ne t)\\ \Rightarrow B'(0,0) \Rightarrow B'為y=g(x)的反曲點(對稱中心) \Rightarrow B為y=f(x)的反曲點(對稱中心)，\bbox[red, 2pt]{故得證}$$

解答： $$6n+8m要最大，也就是正奇數及正偶數要從最小的值開始累加；\\ \cases{n個最小的正奇數和=1+3+ \cdots +(2n-1) =n^2 \\ m個最小正偶數和= 2+4+\cdots + 2m= m^2+m} \Rightarrow n^2+m^2+m \le 1000\\ \Rightarrow n^2 +(m+{1\over 2})^2  \le 1000{1\over 4}\\ 由柯西不等式 \left( n^2 +(m+{1\over 2})^2\right)(6^2+8^2) \ge (6n+8m+4)^2 \Rightarrow 1000{1\over 4}\cdot 100 \ge (6n+8m+4)^2\\ \Rightarrow 316.XX \ge 6n+8m+4 \Rightarrow 6n+8m的最大值=316-4= \bbox[red, 2pt]{312}$$

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解題僅供參考，其他教甄試題及詳解