111年屏東高中教甄-數學詳解

國立屏東高級中學111學年度正式教師甄選

一、填充題(共10題，每題6分，共60分)

解答： $$\cases{f(1012)= f(235+111\times 7)\le f(235)+ 111\times 7=f(235)+777\\ f(1012)= f(235+ 259\times 3) \ge f(235)+259\times 3= f(235)+777}\\ \Rightarrow f(235)+777 \le f(1012)\le f(235)+777 \Rightarrow f(1012)=f(235)+777 = 220+777 =\bbox[red, 2pt]{997}$$

解答： $$a,b為x^2+ 3x+9=0的二根\Rightarrow \cases{a=(-3+3\sqrt 3i)/2\\ b= (-3-3\sqrt 3i)/2} \Rightarrow {a\over b}={-3+3\sqrt 3i\over -3-3\sqrt 3i} = {-1-\sqrt 3i\over 2}=e^{4\pi i/3} \\因此{a\over b} +({a\over b})^2 +\cdots +({a\over b})^{2022} =\cfrac{{a\over b}(1-({a\over b})^{2022})}{1-{a\over b}} =\cfrac{e^{ 4\pi i/3}(1- e^{i2696\pi })}{1- e^{ 4\pi i/3}} = \bbox[red, 2pt]{0}$$

解答：

解答： $$橢圓\Gamma:9x^2+(y-a)^2=9 \Rightarrow x^2+{(y-a)^2\over 9}=1 \Rightarrow P\in \Gamma \Rightarrow P(\cos \theta,3\sin \theta+ a) 代入y=2x^2\\ \Rightarrow 3\sin\theta +a = 2\cos^2\theta =2-2\sin^2\theta \Rightarrow 2\sin^2\theta +3\sin \theta +a-2=0\\ \Rightarrow \sin \theta =\cfrac{-3\pm \sqrt{25-8a}}{4} \Rightarrow  -1 \le \cfrac{-3\pm \sqrt{25-8a}}{4}\le 1 \Rightarrow -1\le \pm \sqrt{25-8a}\le 7\\ \Rightarrow 0\le 25-8a\le 49 \Rightarrow \bbox[red, 2pt]{-3\le a\le {25\over 8}}$$

解答：

$$取\cases{A(0,3)\\ B(-4,3)\\ C(-2\sqrt{10},0)\\ D(0,0)} \Rightarrow \cases{\overrightarrow{AB}=(-4, 0)\\ \overrightarrow{DC} =(-2\sqrt{10},0) \\ \overrightarrow{AD}=(0,-3)\\ \overrightarrow{BC}=(4-2\sqrt{10}-3)} \Rightarrow \cases{\overrightarrow{AB}+ \overrightarrow{DC} =(-4-2\sqrt{10},0)\\ \overrightarrow{AD} +\overrightarrow{BC}=(4-2\sqrt{10},-6)} \\ \Rightarrow (\overrightarrow{AB}+ \overrightarrow{DC})\cdot (\overrightarrow{AD} +\overrightarrow{BC}) =(-2\sqrt{10})^2-4^2=40-16=\bbox[red, 2pt]{24}$$

解答： $$X=11:\cases{第1-10次擲筊:恰有4次聖杯\\ 第11次擲筊出聖杯} \Rightarrow P(X=11)=C^{10}_4\cdot ({1\over 2})^{11} \\X=12:\cases{第1-11次擲筊:恰有4次聖杯\\ 第12次擲筊出聖杯} \Rightarrow P(X=12)=C^{11}_4\cdot ({1\over 2})^{12}\\ 欲求之機率\sum_{n=11}^\infty P(X=n) =1-\sum_{n=5}^{10} P(X=n) =1- {126\over 1024} -{70\over 512}-{35\over 256}-{15\over 128}-{5\over 64}-{1\over 32}\\ =1-{319\over 512}=\bbox[red,2pt]{193\over 512}$$

解答： $$L_1通過A(0,3) \Rightarrow L_1: y=mx+3 \Rightarrow \cases{B=L_1\cap (L_2:5x=12y) \Rightarrow B=({36\over 5-12m}, {15\over 5-12m})\\ C=L_1\cap (y=0) \Rightarrow C=(-3/m,0)}\\ \triangle OBC周長= \overline{OB}+ \overline{OC}+ \overline{BC} = {39\over 5-12m}+ {-3\over m}-{15\over m(5-12m)}\sqrt{ m^2+ 1}=30\\ \Rightarrow 24m^2-5m-1=\sqrt{m^2+1} \Rightarrow 16m^4-56m^3+36m^2+10m=0\\ \Rightarrow 2m(12m-5)(24m^2 -1)=0 \Rightarrow m=-\sqrt 6/12 (m\lt 0 \Rightarrow 其它不合) \Rightarrow C(6\sqrt 6,0)\\ \Rightarrow \triangle OAC面積 ={1\over 2}\cdot \overline{OA}\cdot \overline{OC} ={1\over 2}\cdot 3\cdot 6\sqrt 6 =\bbox[red,2pt]{9\sqrt 6}$$

解答： $$\left[\cfrac{23x-7}{4} \right]=\cfrac{13x+5}{3} \in \mathbb{Z} \Rightarrow 3\mid 13x+5 \Rightarrow 3\mid 13x+2 \Rightarrow x={3k+1\over 13},k\in \mathbb{Z}\\ \Rightarrow \left[\cfrac{23\cdot {3k+1\over 13}-7}{4} \right]=\cfrac{13 \cdot {3k+1\over 13}+5}{3} \Rightarrow \left[\cfrac{69k-68}{52} \right]=k+2 \Rightarrow k+2\le \cfrac{69k-68}{52}\lt k+3 \\ \Rightarrow 52k+172 \lt 69k \le 52k+ 224 \Rightarrow 172\lt 17k\le 224 \Rightarrow k= 11,12,13\\ \Rightarrow x=\bbox[red,2pt]{{34\over 13}, {37\over 13},{40\over 13}}$$

解答： $$\sum_{0\le i,j\le 111} C^{111}_i\cdot C^{111}_j = {1\over 2}\left( (C^{111}_0+ C^{111}_1+ \cdots +C^{111}_{111})^2 -((C^{111}_1)^2 +(C^{111}_2)^2 + \cdots +(C^{111}_{111})^2 )\right)\\ ={1\over 2}\left( (2^{111})^2-C^{222}_{111}  \right) =2^{221}-{1\over 2} C^{222}_{111}\\ 由於C^{222}_{111} ={222\cdot 221\cdots 112\over 111\cdot 110\cdots 1}，其中\cases{\lfloor {222\over 29} \rfloor =7 \\\lfloor {111\over 29} \rfloor =3} \Rightarrow \cases{分母有3個29的倍數\\ 分子有7-3=4個29的倍數}\\ \Rightarrow C^{222}_{111}是29的倍數，因此只要考慮2^{221}除以29的餘數，\\2^{17}=131072 \equiv 21 \mod 29 \Rightarrow 2^{221} \equiv 21^{13} = (21^3)^4\cdot 21  \equiv 10^4\cdot 21 = 210000 \equiv \bbox[red, 2pt]{11} \mod 29$$

解答： $$代公式:\cfrac{1-({2\over 3})^4}{1-({2\over 3})^7}=\bbox[red, 2pt]{1755\over 2059}\\公式說明:假設\cases{艾莉絲(簡稱A)贏1元的機率p=3/5 \\巴柏(簡稱B)贏1元的機率q=2/5=1-p}，及A、B皆在數線上，其中\cases{A(a):A在位置a\\ B(b):B在位置b};\\ 若硬幣出現正面，A向右移一步；反之，A向左移一步;初始位置\cases{A(4)\\ B(3)}，若\cases{A(4+3)=A(7)代表A贏了\\A(0)代表A輸了}；\\依題意，一旦發生輸贏，遊戲結束，求A(7)的機率;\\先考慮A(4)\to A(0)的機率:P(A_4\to A_0)=(2/3)^4 (不管有沒有經過A(7));\\ 假設P(A_4\to A_7)=Q,則P(A_4 \to A_0)=  Q\cdot P(A_7\to A_0) +(1-Q) \\\Rightarrow (2/3)^4= Q\cdot (2/3)^7+1-Q \Rightarrow Q=\cfrac{1-({2\over 3})^4}{1-({2\over 3})^7}$$

二、計算證明題(共5題，每題8分，共40分)

解答： $$a_1=a_2=\cdots =a_{109}=1,a_{110}=111，重複此數列a_1,a_2,\dots,a_{110}, 22 次，\\共110\times 22= 2420項，總和為220\times 22=4840； 剩下80項均為1，則總和為4840+80= \bbox[red,2pt]{4920}$$

解答： $$105=3\times 5\times 7 \Rightarrow 1-105中與105互質的數字有\\105-\left(\lfloor{105\over 3}\rfloor +\lfloor{105\over 5}\rfloor +\lfloor{105\over 7}\rfloor -\lfloor{105\over 3\times 5}\rfloor  -\lfloor{105\over 5\times 7}\rfloor  -\lfloor{105\over 7\times 3}\rfloor + \lfloor{105\over 3\times 5\times 7}\rfloor  \right)\\ =105-57=48個; 將這48個數字依小至大排列為a_i,i=1-48 \Rightarrow a_1=1,a_2=2, a_3=4, a_4=8\\,\dots, a_{48}=104,a_{49}=105+a_1,a_{50}=105 +a_2,\dots,a_{96}=105+a_{48},\dots\\ 而1204= 48\times 25+4 \Rightarrow a_{1204}= 105\times 25+a_4= 2625+8 =\bbox[red, 2pt]{2633}$$

解答： $$取\cases{u =(\pi-x)^m\\ dv =x^n\,dx} \Rightarrow \cases{du =-m(\pi-x)^{m-1} \,dx\\ v={1\over n+1}x^{n+1}}\\ \Rightarrow \int_0^\pi (\pi-x)^m x^n\,dx = \left.\left[{1\over n+1}(\pi-x)^{m}x^{n+1} \right] \right|_0^\pi+{m\over n+1} \int_0^\pi (\pi-x)^{m-1}x^{n+1}\,dx \\= 0+{m\over n+1} \int_0^\pi (\pi-x)^{m-1}x^{n+1}\,dx ={m\over n+1} \int_0^\pi (\pi-x)^{m-1}x^{n+1}\,dx\\ 同理,取\cases{u =(\pi-x)^{m-1}\\ dv =x^{n+1}\,dx} \Rightarrow \cases{du =-(m-1)(\pi-x)^{m-2}\,dx\\ v={1\over n+2}x^{n+2}} \\ \Rightarrow {m\over n+1} \int_0^\pi (\pi-x)^{m-1}x^{n+1}\,dx\\=\left. \left[{m\over (n+1)(n+2)}(\pi-x)^{m-1}x^{n+2}\right] \right|_0^\pi +{m(m-1)\over (n+1)(n+2)}\int_0^\pi (\pi-x)^{m-2}x^{n+2}\,dx \\ ={m(m-1)\over (n+1)(n+2)}\int_0^\pi (\pi-x)^{m-2}x^{n+2}\,dx =\cdots ={m!\over (n+1)(n+2)\cdots (n+m)} \int_0^\pi (\pi-x)^0 x^{m+n}\,dx \\ =\cfrac{m!}{(m+n)!/n!} \left.\left[{1\over m+n+1}x^{m+n+1} \right]\right|_0^\pi =\cfrac{m!n!}{(m+n)!\cdot (m+n+1)} \cdot \pi ^{m+n+1}\\ =\bbox[red, 2pt]{\cfrac{\Gamma(m+1)\Gamma(n+1)}{\Gamma(m+n+2)}\pi^{m+n+1}}, 其中\Gamma(k+1)=k!$$

解答：

$$三平面E_1、E_2、E_3相交情形如上圖所示，其中E_1與E_2相交於一直線L(上圖紅色直線);\\因此我們可以找一平面E_4，通過L且平行E_3，則E_4為E_1與E_2的線性組合；\\ 即E_4: \alpha(a_1 x+ b_1y+ c_1z-d_1)+ \beta(a_2 x+ b_2y+ c_2z-d_2)=0, \\又E_3\parallel E_4 \Rightarrow {a_3\over \alpha a_1+\beta a_2} ={b_3\over \alpha b_1+\beta b_2} ={c_3\over \alpha c_1+\beta c_2} \ne {d_3\over \alpha d_1+\beta d_2} \\ 假設\cases{{a_3\over \alpha a_1+\beta a_2} ={b_3\over \alpha b_1+\beta b_2} ={c_3\over \alpha c_1+\beta c_2}= t\\ {d_3\over \alpha d_1+\beta d_2}=s}, 其中s\ne t\\ \Rightarrow \triangle_x= \begin{vmatrix} d_1 & b_1 & c_1\\ d_2 & b_2 & c_2\\ d_3 & b_3 & c_3\end{vmatrix} =\begin{vmatrix} d_1 & b_1 & c_1\\ d_2 & b_2 & c_2\\ s(\alpha d_1+ \beta d_2) & t(\alpha b_1+\beta b_2) & t(\alpha c_1+ \beta c_2)\end{vmatrix} \\=\begin{vmatrix} d_1 & b_1 & c_1\\ d_2 & b_2 & c_2\\ (t-s)(\alpha d_1+\beta d_2)  & 0 & 0\end{vmatrix} =(b_1c_2-b_2c_1) (t-s)(\alpha d_1+\beta d_2)\\ 同理可得\cases{\triangle_y= (a_2c_1-a_1c_2)(t-s)(\alpha d_1+\beta d_2) \\ \triangle_z = (a_1b_2-a_2b_1)(t-s)(\alpha d_1+\beta d_2)};\\ 若\triangle_x =\triangle_y =\triangle_z=0 \Rightarrow {a_1\over a_2} ={b_1\over b_2} ={c_1\over c_2} \Rightarrow E_1\parallel E_2，矛盾(E_1與E_2相交於一直線)\\因此\triangle_x,\triangle_y,\triangle_z 不全為0，\bbox[red,2pt]{故得證}$$

解答： $$假設數列\langle a_n\rangle 為{x_0\over x_1},{x_1\over x_2},\dots,{x_{n-1}\over x_n}由大至小排序後的數列，且\Pi_{k=1}^n a_k=1\\ \text{依 Chebyshev's sum inequality}，兩數列\langle a_n^{n-1}\rangle ,\langle a_n \rangle 滿足以下不等式:\\ {1\over n} \sum_{k=1}^n \left(a_k^{n-1} \cdot a_k\right) \ge \left({1\over n}\sum_{k=1}^n a_k^{n-1}\right)\left({1\over n}\sum_{k=1}^n a_k \right) \cdots(1)\\ 再由算幾不等式:{1\over n}\sum_{k=1}^n a_k^{n-1} \ge \left(\Pi_{k=1}^n a_n^{n-1} \right)^{1/n} =1; 因此式(1) \Rightarrow {1\over n} \sum_{k=1}^n a_k^{n} \ge {1\over n}\sum_{k=1}^n a_k\\ 即({x_0\over x_1})^n + ({x_1\over x_2})^n,\dots,({x_{n-1}\over x_n})^n \ge ({x_0\over x_1}) +({x_1\over x_2})+ \cdots +({x_{n-1}\over x_n})$$

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計算證明題僅公告第3題，解題僅供參考，其他教甄試題及詳解