111年武陵高中教甄-數學詳解

桃園市立武陵高級中學111學年度第一學期第1次正式教師甄選

一、 填充題 A(每題 6%共 36%)

解答：

$$由上圖可知圓半徑r=a-1 \Rightarrow x^2+(y-a)^2 = r^2 = (a-1)^2 \Rightarrow x^2 =(a-1)^2-(y-a)^2 \cdots(1)\\ 又y=\left| {x^2 \over 2}-1\right| \Rightarrow y ={x^2 \over 2}-1, x\ge \sqrt 2 \Rightarrow x^2=2y+2 代入(1) \Rightarrow 2y+2=(a-1)^2-(y-a)^2 \\ \Rightarrow y^2+(2-2a)y+2a+1 =0 恰有一根\Rightarrow (2-2a)^2-4(2a+1) =0 \Rightarrow 4a(a-4)=0\\ \Rightarrow a= \bbox[red,2pt]{4} (\because a\gt 1, a=0不合)$$

解答：$$\left( \begin{vmatrix} a & b\\ x_1 & y_1\end{vmatrix},  \begin{vmatrix} b & c\\ y_1 & z_1\end{vmatrix},  \begin{vmatrix} c & a\\ z_1 & x_1\end{vmatrix} \right) =(c,a,b)\times (z_1,x_1,y_1 ) = (1,2,3) \Rightarrow (a,b,c)\bot (2,3,1) \\ \Rightarrow (a,b,c) \cdot (2,3,1)= 2a+3b+c=0\cdots(1);\; 同理，(a,b,c)\cdot (5,6,4)= 5a+6b +4c =0\cdots(2)\\ 由(1)及(2)可得\cases{a=-2b\\ b=c} \Rightarrow a:b:c = 2:-1:-1 \Rightarrow ax+ by+cz =0 \Rightarrow 2x-y-z=0\\ 柯西不等式((x-1)^2 +(y+2)^2 +(z-3)^2) (2^2+(-1)^2 +(-1)^2) \ge (2(x-1)-(y+2)-(z-3))^2\\   \Rightarrow ((x-1)^2 +(y+2)^2 +(z-3)^2) \cdot 6 \ge (2x-y-z-1)^2 =(0-1)^2 =1\\ \Rightarrow (x-1)^2 +(y+2)^2 +(z-3)^2 \ge {1\over 6} \\因此x^2+ y^2 +z^2 -2x+4y-6z = (x-1)^2+(y+2)^2+(z-3)^2 -14 \ge {1\over 6}-14 =\bbox[red,2pt]{-{83\over 6}}$$

解答：$$假設\cases{a_n:含偶數個9且長度為n的字串個數\\ b_n:含奇數個9且長度為n的字串個數} \Rightarrow a_n+b_n =10^{n}; \\初始值\cases{a_1=9，也就是字串為0,1,\dots,8 ，都是含0個九\\ b_1=1，字串為9，含一個九};\\考慮兩種情形:\cases{長度為n-1且含偶數個九的字串，在其後加上0,或1,...,或8 \\長度為n-1且含奇數個九的字串，在其後加上一個九 }\\ 也就是a_n= 9a_{n-1}+ b_{n-1} = 9a_{n-1}+(10^{n-1}-a_{n-1}) \Rightarrow a_n= 8a_{n-1}+ 10^{n-1}\\ = 8^2a_{n-2}+ 8\times 10^{n-2}+ 10^{n-1} =\cdots = 8^{n-1}a_1 +8^{n-2}\times 10 +8^{n-3}\times 10^2+\cdots +10^{n-1}\\ =9\cdot 8^{n-1}+ 10^{n-1} \sum_{k=0}^{n-2} ({8\over 10})^k  =9\cdot 8^{n-1}+ 5\cdot 10^{n-1}\left(1- ({8\over 10})^{n-1}\right) \\=9\cdot 8^{n-1}+ {1\over 2}  10^{n}\left(1- ({8\over 10})^{n-1}\right) =\bbox[red, 2pt]{{1\over 2}(10^n+8^n)}$$

解答：$$\begin{array}{} 總局數 & 勝者排列 & 機率 & 小計\\\hline 2 & 甲甲& ({3\over 4})^2= {9\over 16}\\ & 乙乙& ({1\over 4})^2 ={1\over 16} & {5\over 8}\\\hline 4 & 甲乙甲甲 & ({3\over 4})^3\cdot {1\over 4}= {27\over 256} \\ & 乙甲甲甲 & {27\over 256} \\ & 乙甲乙乙 & ({1\over 4})^3\cdot {3\over 4}= {3\over 256} \\ & 甲乙乙乙& {3\over 256} & {15\over 64}\\\hdashline 6 & 甲乙甲乙XX\\ & 甲乙乙甲XX\\ & 乙甲乙甲XX\\ & 乙甲甲乙XX & & 4\cdot ({3\over 4})^2({1\over 4})^2\\\hline \end{array} \\ \Rightarrow 期望值=2\cdot {5\over 8}+ 4\cdot {15\over 64} +6\cdot 4\cdot ({9\over 256}) =\bbox[red,2pt]{97\over 32}\\ 註:比賽六局不一定要分出勝負
$$

解答：$$\omega^{503}=1 \Rightarrow (1-\omega)(1+\omega+\omega^2+\cdots +\omega^{502})=0 \Rightarrow \sum_{k=0}^{502}\omega^k =0\\ 又\cfrac{1}{\omega^k-1} +\cfrac{1}{\omega^{503-k}-1} =\cfrac{1}{\omega^k-1} +\cfrac{1}{\omega^{-k}-1} =\cfrac{1}{\omega^k-1} +\cfrac{\omega^k}{1-\omega^{k}}=-1\\ 原式=\sum_{k=1}^{502}{\omega^{2k}\over \omega^k-1} =\sum_{k=1}^{502} \left( {\omega^{2k}-1\over \omega^k-1} +{1\over w^k-1} \right) =\sum_{k=1}^{502} \left(  \omega^{k}+1 +{1\over w^k-1} \right)\\ =-1+502+\sum_{k=1}^{502}{1\over w^k-1}\\ =501+\left({1\over \omega-1} +{1\over \omega^{502}-1}\right) +\left({1\over \omega^2-1} +{1\over \omega^{501}-1}\right) +\cdots +\left({1\over \omega^{251}-1} +{1\over \omega^{252}-1}\right) \\ =501+(-1)\times 251= \bbox[red,2pt]{250}$$

解答：$$假設\cases{A(0,0,0)\\ G(3,3,3)} \Rightarrow \cases{\overline{AG}中點M(3/2,3/2,3/2)\\ \overrightarrow{AG} =(3,3,3)} \Rightarrow 通過M且法向量為\overrightarrow{AG}的平面E:x+y+z=9/2\\ 對任意單位立方體頂點(x,y,z)及其對角頂點(x+1,y+1,z+1)，若兩頂點位於平面E的異側，\\需滿足: x+y+z \lt {9\over 2} \lt (x+1)+(y+1)+ (z+1) \Rightarrow x+y+z=2,3,4 \\ \Rightarrow \begin{array}{} x+y+z & (x,y,z) & 排列數\\\hline 2 & (1,1,0) & 3\\ & (2,0,0) & 3\\\hdashline 3 & (1,1,1) & 1\\\hdashline 4 & (3,1,0) & 6\\ & (2,2,0) & 3 \\ & (2,1,1) & 3\\\hline\end{array} \Rightarrow 共有\bbox[red,2pt]{19}個小立方體被E切成兩塊$$

二、 填充題 B(每題 8%共 32 分)

解答：$$取\cases{f(\theta,\alpha)=a\\ g(\theta,\alpha)=b\\ |\vec u|=|\vec v|= |\vec w|=r} \Rightarrow \cases{\vec u\cdot \vec v=r^2 \cos\theta\\ \vec u\cdot \vec w= r^2\cos(\alpha) =\vec u\cdot (a\vec u +b\vec v) =ar^2 +br^2\cos\theta} \\ \Rightarrow r^2\cos\alpha = ar^2+br^2\cos\theta \Rightarrow \cos \alpha = a+b\cos\theta \cdots(1)\\ 又\vec v\cdot \vec w =r^2\cos (\alpha-\theta) = \vec v\cdot (a\vec u+b\vec v) = ar^2\cos\theta +br^2 \Rightarrow \cos (\alpha-\theta) = a\cos \theta +b \cdots(2)\\ 因此(1)+(2) \Rightarrow \cos \alpha +\cos(\alpha-\theta) =a+b+ \cos\theta(a+b) =(a+b)(\cos \theta+1)\\ \Rightarrow f(\theta,\alpha)+g(\theta,\alpha) =a+b= \bbox[red, 2pt]{\cfrac{\cos \alpha +\cos(\alpha-\theta)}{\cos \theta+1}}$$

解答：$$\cases{\Gamma_1: y_1=x^2-2x+2 \Rightarrow y_1'=2x-2\\ \Gamma_2: y_2=-x^2+ax+b \Rightarrow y_2'=-2x+a} \\ \Rightarrow \cases{y_1=y_2 \Rightarrow 2x^2-(a+2)x+2-b=0 \Rightarrow x={a+2\pm \sqrt{(a+2)^2-8(2-b)}\over 4}\\ y_1'\times y_2'=-1 \Rightarrow 4x^2-(2a+4)x+2a-1 =0 \Rightarrow x={ 2a+4\pm \sqrt{(2a+4)^2-16(2a-1)}\over 8}}\\ 兩根相等\Rightarrow (a+2)^2-8(2-b)= (a+2)^2-4(2a-1) \Rightarrow 4a+ 8b-12=-4a+8 \\ \Rightarrow a+b={5\over 2} 代入算幾不等式: {a+b\over 2} \ge \sqrt{ab} \Rightarrow ab \le ({5\over 4})^2 ={25\over 16} \Rightarrow ab最大值為\bbox[red,2pt]{25\over 16}$$

解答：

$$y^4-2xy^2+ 2x^2-4=0 \Rightarrow (y^2-x)^2 +x^2-4=0 \Rightarrow y^2-x= \pm \sqrt{4-x^2} \\  \Rightarrow \cases{y_1 = \sqrt{x-\sqrt{4-x^2}} \Rightarrow  \sqrt 2\le x\le 2 \\y_2= \sqrt{x+ \sqrt{4-x^2}} \Rightarrow -\sqrt 2\le x\le 2} \\\Rightarrow 旋轉體積=\pi \left(\int_{-\sqrt 2}^2 y_2^2 \,dx -\int_{\sqrt 2}^2 y_1^2\,dx \right)  = \pi \left(\int_{-\sqrt 2}^2 x+\sqrt{4-x^2} \,dx -\int_{\sqrt 2}^2 x-\sqrt{4-x^2}\,dx \right) \\=\pi \left(\left.\left[ {1\over 2}x^2 +{1\over 2} x\sqrt{4-x^2} +2 \sin^{-1}{x\over 2}\right] \right|_{-\sqrt 2}^2 -\left.\left[ {1\over 2}x^2 -{1\over 2} x\sqrt{4-x^2} -2 \sin^{-1}{x\over 2}\right] \right|_{\sqrt 2}^2\right)\\ =\pi \left(\left(2+{3\over 2}\pi \right) -(2-{\pi\over 2})\right) = \bbox[red,2pt]{2\pi^2}$$

解答：$$5c-3a \le b\le 4c-a \Rightarrow \cases{5{c\over a}\le {b\over a}\le 4{c\over a}-1 \cdots(1) \\ 5c-3a\le 4c-a \Rightarrow c\le 2a\cdots(2)}，將(2)代入(1) \Rightarrow {b\over a} \le 4\cdot 2-1=7\\ 又c\ln b\ge a+c\ln c \Rightarrow  a \le c\ln b-c\ln c =c\ln{b\over c} \Rightarrow {b\over a}\ge {b\over c}\cdot {1\over \ln (b/c)}\\ 現在要求{b/c\over \ln(b/c)}的極值，令f(x)=x/\ln x，則f'(x)=0 \Rightarrow {1\over \ln x}-{1\over (\ln x)^2}=0 \Rightarrow \ln x=1 \Rightarrow x=e\\ \Rightarrow 極值為f(e)=e \Rightarrow {b\over a}\ge e，因此e\le x\le 7 \Rightarrow x\in \bbox[red,2pt]{[e,7]}$$ 

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解題僅供參考，其他教甄試題及詳解