112年台中一中教甄-數學詳解

臺中市立臺中第一高級中等學校 112 學年度第 1 次教師甄選

一、填充題 I：（每格 6 分）

解答：$$(x-1)+(y-1)+(z-1)=22-3 \Rightarrow x+y+z=19非負整數解個數=H^3_{19}\\ 又任一袋個數不能\ge 10,也就是需扣除x+y+z=9的情形，因此H^3_{19}-C^3_1H^3_{9} \\ =C^{21}_{19}-3\cdot C^{11}_9=210-3\cdot 55 =\bbox[red, 2pt]{45}$$

解答：$$令f(x)=x^3-6x^2+13x \Rightarrow \cases{f(\alpha)=6\\ f(\beta)=14}\\ 又f'(x)=3x^2-12x+13 \Rightarrow f''(x)=6x-12=0 \Rightarrow x=2\\ \Rightarrow 圖形y=f(x)的對稱中心為(2,f(2)=10)\\ \Rightarrow \cases{P=(\alpha,6)的對稱點為P'(4-\alpha, 14) \\ Q=(\beta,14)的對稱點為Q'(4-\beta,6)} \Rightarrow \cases{P=Q'\\ Q=P'} \Rightarrow \cases{\alpha =4-\beta\\ \beta=4-\alpha} \Rightarrow \alpha+\beta= \bbox[red, 2pt]4$$

解答：$$\alpha^2+ 5\beta^2+4\gamma^2-2\alpha\beta-8\beta\gamma=0 \Rightarrow (\alpha^2-2\alpha\beta +\beta^2)+ (4\beta^2-8\beta\gamma+4\gamma^2)=0 \\ \Rightarrow (\alpha-\beta)^2 +4(\gamma-\beta)^2=0 \Rightarrow \cases{\angle ABC=90^\circ \\\overline{AB}^2=4\overline{BC}^2} \Rightarrow \overline{BC}=2 \\ \Rightarrow 直角\triangle ABC面積＝{1\over 2}\cdot 2\cdot 4 =\bbox[red,2pt] 4$$

解答：$$假設11根x_i依序為a-5d,a-4d,\dots,a,a+d,\dots,a+5d \Rightarrow \sum_{i=1}^{11} x_i=11a=-6 \Rightarrow a=-{6\over 11}\\ 又(x_1+x_2+\cdots +x_{11})^2= (x_1^2+x_2^2+\cdots +x_{11}^2) +2(x_1x_2+ \cdots +x_{10}x_{11}) \\ \Rightarrow (-6)^2= (11a^2+110d^2)+ 2\cdot 5 \Rightarrow d^2={25\over 121} \Rightarrow d=\bbox[red,2pt]{\pm{5\over 11}}\\ 其中x_1^2+x_2^2+\cdots +x_{11}^2= (x_1^2+x_{11}^2) +(x_2^2+ x_{10}^2)+\cdots +(x_5^2+x_7^2)+(x_6^2) \\ =((a-5d)^2+(a+5d)^2) +((a-4d)^2+(a+4d)^2)+\cdots +((a+d)^2+(a-d)^2)+a^2 \\ =(2a^2+50d^2)+(2a^2+32d^2)+\cdots +(2a^2+2d^2)+a^2\\ =11a^2+110d^2$$

解答：$$f(n)最大 \Rightarrow \cases{f(n)\gt f(n+1)\\ f(n)\gt f(n-1)} \Rightarrow \cases{({4\over 5})^n(n^2+4n)\gt ({4\over 5})^{n+1}(n^2+6n+5) \\ ({4\over 5})^n(n^2+4n)\gt ({4\over 5})^{n-1}(n^2+2n-3)} \\ \Rightarrow \cases{n^2-4n-20\gt 0\\ n^2-6n-15\lt 0} \Rightarrow 2+2\sqrt 6 \lt n\lt 3+2\sqrt 6 \Rightarrow 6.89\lt n\lt 7.89 \\ \Rightarrow n=\bbox[red, 2pt]7$$

解答：$$擬求以(a^2-3),(b^2-3),(c^2-3),(d^2-3)為四根的多項式\\ 取y=x^2-3 \Rightarrow x=\pm \sqrt{y+3}代回原式 \Rightarrow (y+3)^2\pm \sqrt{y+3}=-1 \\ \Rightarrow \left((y+3)^2+1 \right)^2 =(\pm \sqrt{y+3})^2 \Rightarrow (y+3)^4+ 2(y+3)^2+1=y+3\\ \Rightarrow 四根之積只需考慮常數項:3^4+2\cdot 3^2+1-3=97 \\ \Rightarrow (a^2-3)(b^2-3)(c^2-3)(d^2-3)= \bbox[red,2pt]{97}$$

解答：$$\overline{P_kP_{k+1}} ={1\over k} \Rightarrow \overline{P_1P_{k+1}} =1+{1\over 2}+{1\over 3}+\cdots +{1\over k}\\ \Rightarrow \cases{\overline{P_1P_2}=1\\ \overline{P_1P_3}=1+{1\over 2} \\ \overline{P_1P_4}=1+ {1\over 2}+{1\over 3}\\ \cdots\\ \overline{P_1P_{2023}}= 1+{1\over 2}+ {1\over 3}+\cdots {1\over 2022}}\\  \Rightarrow \sum_{k=2}^{2023} \overline{P_1P_k}=2022\cdot 1+2021\cdot {1\over 2}+ 2020\cdot {1\over 3}+\cdots +1\cdot {1\over 2022}\\ \Rightarrow \sum_{k=3}^{2023} \overline{P_2P_k}= 2021\cdot {1\over 2}+ 2020\cdot {1\over 3}+\cdots +1\cdot {1\over 2022} \\\Rightarrow \sum_{m=1}^{2022}\sum_{k=m+1}^{2023} \overline{P_mP_k} =2022\cdot 1+4042\cdot {1\over 2}+6060\cdot {1\over 3}+ \cdots + 2022\cdot {1\over 2022}\\ =2022+2021+2020+\cdots +1= {1\over 2}\cdot 2023\cdot 2022= \bbox[red,2pt]{2045253}$$

解答：

$$假設\cases{A在平面BCDE的垂足A'\\ \overline{CD}中點F}，並令\cases{\overline{AF}=a\\ \angle AFA'=\theta} \Rightarrow \cases{\overline{AA'}=a\sin \theta\\ \overline{ED}= 2a\cos \theta} \\ \Rightarrow \cases{\triangle ACD面積＝{1\over 2}\cdot 2a\cos \theta \cdot a \\ 正方形BCDE面積＝4a^2\cos^2 \theta} \Rightarrow \cases{四角錐表面積＝ 4a^2\cos^2 \theta +4a^2\cos \theta =96\\ 四角錐體積={1\over 3}\cdot 4a^3\cos^2\theta \sin\theta}\\ 取\cases{f(a,\theta)={4\over 3} a^3 \cos^2 \theta\sin \theta \\ g(a,\theta)=a^2\cos^2 \theta +a^2\cos \theta-24},利用\text{Lagrange }算子求f的極值\\ \cases{f_a = \lambda g_a\\ f_\theta =\lambda g_\theta\\ g=0} \Rightarrow \cases{4a^2\cos^2 \theta \sin \theta = \lambda(2a\cos^2 \theta +2a\cos \theta) \\ {4\over 3}a^3( \cos \theta-3\sin^2\theta \cos \theta) =\lambda(2a^2\cos \theta(-\sin \theta)-a^2\sin \theta)}\\ 兩式相除最後可得 3\cos^2\theta +2\cos -1 = 0 \Rightarrow (3\cos \theta-1)(\cos \theta+1)=0 \\ \Rightarrow \cos \theta ={1\over 3}(-1不合，\theta \ne \pi) \Rightarrow g(a,\theta)=a^2\cdot {1\over 9}+a^2\cdot {1\over 3}=24 \Rightarrow a=3\sqrt 6\\ \Rightarrow f(a,\theta)={4\over 3}\cdot 162\sqrt 6\cdot {1\over 9}\cdot {2\sqrt 2\over 3} =\bbox[red, 2pt]{32\sqrt 3}$$

解答：$$\cases{\left[{2\over 3}\right] \to 0 \\\left[{2^2\over 3}\right] \to 1 \\\left[{2^3\over 3}\right] \to 2 \\\left[{2^4\over 3}\right] \to 5 \\\left[{2^5\over 3}\right] \to 0 \\\left[{2^6\over 3}\right] \to 1 \\\left[{2^7\over 3}\right] \to 2 \\\left[{2^8\over 3}\right] \to 5 } \Rightarrow \cases{循環數為4 \\ 每一循環合計1+2+5=8}\\ \Rightarrow 2023=4\times 505+3 \Rightarrow 505\times 8+1+2=4043的個位數字是\bbox[red,2pt]3$$

解答：$$為了讓數字簡化,將\triangle ABC貼在x-y平面上,因此取\cases{A(0,4\sqrt 3,0)\\ B(-6,-2\sqrt 3,0)\\ C(6,-2\sqrt 3,0)} \Rightarrow \cases{H(0,0,0)\\ O(0,0,6)}\\ \Rightarrow \overrightarrow{OB} \times \overrightarrow{OC} =24(0,-3,\sqrt 3) \Rightarrow \triangle OBC平面:-3y+\sqrt 3z=6\sqrt 3 \Rightarrow K(0,-{\sqrt 3\over 2},{9\over 2})\\ \Rightarrow P={A+5K\over 6} =(0,{\sqrt 3\over 4},{15\over 4}) ,假設E所截的三角形為\triangle DEF\\ \Rightarrow d(E,\triangle ABC)={15\over 4} \Rightarrow {\triangle DEF \over \triangle ABC}={ (\overline{OH}-15/4)^2 \over \overline{OH}^2}={(6-15/4)^2\over 6^2} ={9\over 64}\\ 又\triangle ABC={\sqrt 3\over 4}12^2=36\sqrt 3 \Rightarrow \triangle DEF={9\over 64}\times 36\sqrt 3=\bbox[red,2pt]{{81\over 16}\sqrt 3}$$

二、填充題 II：（每格 8 分）

解答：$${a-1\over (a-1)^4} ={1\over (a-1)^3}+{2\over (a-1)^4},因此我們要求以{1\over a-1},{1\over b-1},{1\over c-1}為三根的多項式\\ 令x_1=x-1 \Rightarrow x=x_1+1代入原式x^3-4x+1=0 \Rightarrow (x_1+1)^3-4(x_1+1)+1=0\\ \Rightarrow x_1^3+3x_1^2-x_1-2=0,再令x_2={1\over x_1} \Rightarrow x_1={1\over x_2}代入:{1\over x_2^3}+ {3\over x_2^2}-{1\over x_2}-2=0 \\ \Rightarrow 2x_2^3+x_2^2-3x_2-1=0 \equiv x^3+{1\over 2}x^2-{3\over 2}x-{1\over 2}=0\\就是以\alpha={1\over a-1},\beta={1\over b-1},\gamma={1\over c-1}為三根的多項式\\ 欲求之{a+1\over (a-1)^4} +{b+1\over (b-1)^4}+ {c+1\over (c-1)^4} =\left({1\over (a-1)^3}+ {1\over (b-1)^3}+ {1\over (c-1)^3}\right)\\\qquad +2\left({1\over (a-1)^4}+ {1\over (b-1)^4}+ {1\over (c-1)^4}\right) =(\alpha^3+\beta^3 +\gamma^3)+2(\alpha^4+ \beta^4+ \gamma^4) \cdots(1) \\  令f(x)=x^3+{1\over 2}x^2-{3\over 2}x-{1\over 2} \Rightarrow f'(x)=3x^2+x-{3\over 2}\\ 再利用長除法計算f'(x)/f(x)={3\over x}-{1\over 2}\cdot {1\over x^2}+{13\over 4}\cdot {1\over x^3}-{7\over 8}\cdot {1\over x^4}+{81 \over 16}x^5+\cdots \\ \Rightarrow \cases{\alpha^3+\beta^3+ \gamma^3 = -7/8\\ \alpha^4+ \beta^4+ \gamma^4 =81/16} \Rightarrow (\alpha^3+\beta^3 +\gamma^3)+2(\alpha^4+ \beta^4+ \gamma^4)=-{7\over 8}+2\cdot {81\over 16}=\bbox[red, 2pt]{37\over 4}$$

解答：$$\left(\sqrt{a+x}+\sqrt{a-x}\right)^2=a^2 \Rightarrow 2a+ 2\sqrt{a^2-x^2}=a^2 \Rightarrow \left( 2\sqrt{a^2-x^2} \right)^2 =(a^2-2a)^2 \\ \Rightarrow 4a^2-4x^2 =a^4-4a^3+4a^2 \Rightarrow x^2={4a^3-a^4\over 4} \Rightarrow (4a^3-a^4)\ge 0 \Rightarrow a^3(a-4)\le 0 \\ \Rightarrow 0\le a\le 4 \Rightarrow \cases{a=1 \Rightarrow \sqrt{1+x}+\sqrt{1-x}=1無實數解\\ a=2,3,4 \Rightarrow x有實數解} \\ \Rightarrow a之總和=2+3+4=\bbox[red, 2pt]9$$

解答：$$令P(s,t)在雙曲線上,且位於第一象限 \Rightarrow G=\triangle PF_1F_2的重心=(s/3,t/3) \\ 由於\overleftrightarrow{GI}垂直x軸,I=({s\over 3},r),I是內心 \Rightarrow r=內切圓半徑\\ 又\href{https://web.math.sinica.edu.tw/math_media/d371/37107.pdf}{內切圓圓心必位在過其貫軸頂點且垂直其貫軸的直線上} \Rightarrow {s\over 3}=a=3 \Rightarrow s=9\\ 將P(9,t)代入雙曲線 \Rightarrow {9^2 \over 9}-{t^2\over 16}=1 \Rightarrow t=8\sqrt 2 \Rightarrow \triangle PF_1F_2面積={1\over 2}\cdot 10\cdot 8\sqrt 2=40\sqrt2\\ \triangle PF_1F_2 三邊長\cases{a=\overline{F_1F_2}=10\\ b=\overline{PF_1} =\sqrt{14^2+(8\sqrt 2)^2}=18\\ c=\overline{PF_2}=\sqrt{4^2+(8\sqrt 2)^2}=12} \Rightarrow r={2\triangle 面積\over a+b+c} ={80\sqrt 2\over 40} =\bbox[red,2pt]{2\sqrt 2}$$

解答：$$\cases{A(2,3) \\B(-9,6) \\O(0,0) \\ P在圓上} \Rightarrow \cases{\overline{OA}=\sqrt{13} \\ \overline{OB}=3\sqrt{13} \\ \overline{OP} =2\sqrt{13}\\ \overrightarrow{OA}\cdot \overrightarrow{OB} =0 \Rightarrow \angle BOA=90^\circ} \\ 假設\angle AOP=\theta \Rightarrow \cos \theta ={\overline{OA}^2+ \overline{OP}^2-\overline{AP}^2\over 2\cdot \overline{OA}\cdot \overline{OP}} ={13+52-\overline{AP}^2 \over 2\cdot \sqrt{13}\cdot 2\sqrt{13}} ={65-\overline{AP}^2\over 52} \\ \Rightarrow \overline{AP} =\sqrt{65-52\cos\theta}=\sqrt{13}\sqrt{5-4\cos \theta} \\ 同理,假設\angle BOP=\phi \Rightarrow \cos \phi = {117+52-\overline{BP}^2 \over 2\cdot 3\sqrt{13}\cdot 2\sqrt{13}}  \Rightarrow \overline{BP} =\sqrt{13} \sqrt{13-12\cos\phi}\\ 因此欲求之t=3\overline{AP}-2\overline{BP}= \sqrt{13}\left( \sqrt{45-36\cos\theta} -\sqrt{52-48\cos\phi}\right) \\= \sqrt{13}\left( \sqrt{(3\cos\theta-6)^2+ (3\sin \theta)^2} -\sqrt{(4\cos\phi-6)^2+ (4\sin \phi)^2}\right) \\ =\sqrt{13}(\overline{RT}-\overline{ST}),其中\cases{R(3\cos \theta,3\sin \theta)\\ S(4\cos \phi, 4\sin \phi)\\ T(6,0)}, 即\cases{R在半徑3的圓上\\ S在半徑4的圓上}\\t值要最小，即R，S，T在一直線上，因此最小的t=\sqrt{13}\cdot (-\overline{RS}) =\bbox[red, 2pt]{-5\sqrt{13}} \\ 註：\angle BOA=90^\circ \Rightarrow \phi-\theta=90^\circ \Rightarrow \angle ROS=90^\circ \Rightarrow \overline{RS}=\sqrt{\overline{OR^2}+\overline{OS}^2}=5$$

解答：$${1\over \sqrt[3]{1000}}+\int_{1001}^{3376} {1\over \sqrt[3]{k-1}}\,dk\gt \sum_{k=1000}^{3375}{1\over \sqrt[3]{k}} \gt \int_{1000}^{3375}{1\over \sqrt[3]{k}}\,dk+{1\over \sqrt[3]{3375}} \\ \cases{左式= {1\over 10}+ 187.5=187.6 \\ 右式=187.5+{1\over 15}=187.5666} \Rightarrow \sum_{k=1000}^{3375}{1\over \sqrt[3]{k}}  \approx \bbox[red,2pt]{187.6}$$

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