桃園市立武陵高級中等學校114學年度第一學期第1次正式教師甄選
一、填充題 (每題 7 分；共 70 分)
解答:$$1\cdot 2+ 2\cdot 3+\cdots+ n(n+1) =\sum_{k=1}^n k(k+1) =\sum_{k=1}^n \left( k^2+k \right) ={n(n+1)(2n+1)\over 6}+{{n(n+1)\over 2}} \\={n(n+1)(n+ 2) \over 3} \Rightarrow {1\over \sum_{k=1}^n k(k+1)} ={3\over n(n+1)(n+2)} ={3\over 2} \left( {1\over n(n+1)} -{1\over (n+1)(n+2)} \right) \\ \Rightarrow 原式={3\over 2} \left( ({1\over 1\cdot 2}-{1\over 2\cdot 3})+({1\over 2\cdot 3}-{1\over 3\cdot 4} ) +\cdots \right) ={3\over 2}\cdot {1\over 2}= \bbox[red, 2pt]{3\over 4}$$
解答:$$\cases{A(4,4,6) \in L_1\\ P\in L_1\cap L_2 \\ Q\in L_1\cap L_3 } \Rightarrow \cases{P(2s,3s,4s)\\ Q(t,-2t-3,2t+4)} \Rightarrow \cases{\overrightarrow{AP}=(2s-4,3s-4,4s-6) \\ \overrightarrow{AQ} =(t-4,-2t-7,2t-2)} \\\Rightarrow A,P,Q在一直線上 \Rightarrow \overrightarrow{AP} \parallel \overrightarrow{AQ} \Rightarrow {2s-4\over t-4} ={3s-4\over -2t-7}={4s-6\over 2t-2} \\\Rightarrow \cases{s=1\\ t=-2} \Rightarrow \cases{P(2,3,4) \\Q(-2,1,0)} \Rightarrow \overrightarrow{PQ} =\bbox[red, 2pt]{(-4,-2,-4)}$$
解答:
$$f(x)=|2\cos 3x+1| \Rightarrow 週期為{2\pi\over 3}, 又f(x)=1 \Rightarrow \cases{2\cos 3x=0\\ 2\cos 3x=-2} \Rightarrow \cases{x=\pi/6, \pi/2\\ x=\pi/3}\\ \Rightarrow 在一個週期內有3個解,共有{6\pi\over 2\pi/3}=9個週期, 因此n=3\times 9=\bbox[red, 2pt]{27}個解$$
解答:
$$假設拋物線\Gamma: y^2=4cx \Rightarrow \cases{焦點F(c,0) \\準線L: x=-c} \Rightarrow \cases{d(A,L)= \overline{AF}=1\\ d(B,L)=\overline{BF}=2} \Rightarrow \cases{A(1-c,2\sqrt{c(1-c)}) \\ B(2-c, -2\sqrt{c(2-c)})} \\ \Rightarrow \cases{\overrightarrow{FA} =(1-2c, 2\sqrt{c(1-c)}) \\ \overrightarrow{FB}= (2-2c, -2\sqrt{c(2-c)})} \Rightarrow \overrightarrow{FA} \cdot \overrightarrow{FB}= \overline{FA}\cdot \overline{FB} \cos 120^\circ \\ \Rightarrow (1-2c)(2-2c)-4\sqrt{c^2(1-c)(2-c)} =1\cdot 2\cdot (-{1\over 2})=-1 \Rightarrow 4c^2-6c+3=4\sqrt{c^2(1-c)(2-c)} \\ \Rightarrow (4c^2-6c+3)^2=16c^2(1-c)(2-c) \Rightarrow 28c^2-36c+9=0 \Rightarrow c={9\pm 3\sqrt 2\over 14}\\ 1-2c \gt 0 \Rightarrow c\ne {9+3\sqrt 2\over 14} \Rightarrow c={9- 3\sqrt 2\over 14} \Rightarrow 正焦弦長=4c= \bbox[red, 2pt]{18-6\sqrt 2\over 7}\\註:交點應該是焦點$$
解答:
$$稜長a=6\sqrt 6 \Rightarrow 正四面體面積 =4\times {\sqrt 3\over 4}a^2=216\sqrt 3\\正四面體兩面夾角為\theta \Rightarrow \cos \theta={1\over 3}\Rightarrow \cases{\angle PAQ=\theta\\ \overline{OP}= \overline{OQ}=1} \Rightarrow d=\overline{AP} =\sqrt 2\\ \Rightarrow 正三角形三邊都向內縮d \Rightarrow 收縮後正三角形邊長為a'=a-2\sqrt 3d=4\sqrt 6\\ \Rightarrow 小球可接觸的面積=4\times {\sqrt 3\over 4}a'^2=96\sqrt 3 \Rightarrow 不會被接觸的面積=216\sqrt 3-96\sqrt 3= \bbox[red, 2pt]{120\sqrt 3}$$
解答:
$$y=mx代入y=3x-x^2 \Rightarrow 3x-x^2=mx \Rightarrow x^2+(m-3)x=0 \Rightarrow x=0, 3-m \\ \Rightarrow \Gamma_1與 \Gamma_2的交點\cases{O(0,0)\\ A(3-m,k)} \Rightarrow \int_0^{3-m} (3x-x^2-mx)\,dx ={1\over 2} \int_0^3 (3x-x^2)\,dx \\ \Rightarrow \left. \left[ {3\over 2}x^2-{1\over 3}x^3-{m\over 2}x^2 \right] \right|_0^{3-m} ={1\over 2} \left. \left[ {3\over 2}x^2-{1\over 3}x^3 \right] \right|_0^3 \Rightarrow {1\over 6}(3-m)^3={9\over 4} \\ \Rightarrow m= 3-{3\over \sqrt[3]2} = \bbox[red, 2pt]{3-{3\over 2}\sqrt[3]4}$$
解答:$$\cases{1^2=1 \Rightarrow \{1\} \\ 2^2=4 \Rightarrow \{4\}, \{1,4\} \\6^2=36 \Rightarrow \{2,3,6\},\{1,2,3,6\}\\ 12^2=144 \Rightarrow \{2,3,4,6\},\{1,2,3,4,6\}} \Rightarrow 共有7種組合\\ 組合數=子集合數=2^7-1(空集合)=128 \Rightarrow 機率= \bbox[red, 2pt]{7\over 127}$$
解答:$$\cases{a^3+3a^2+3a=7\\ b^3+ 3b^2+3b=-9} \Rightarrow \cases{(a+1)^3=8\\ (b+1)^3=-8} \Rightarrow \cases{a+1=2\\ b+1=-2} \Rightarrow a+1+b+1=0 \Rightarrow a+b= \bbox[red, 2pt]{-2}$$
解答:$$\cases{E_1法向量\vec u=(2,-2,1) \\ E_2法向量\vec v=(4,-1,-1)} \Rightarrow L方向向量\vec n=\vec u\times \vec v =(3,6,6) \\ 任取Q(0,-1,-1)\in E_1\cap E_2 \Rightarrow L: {x\over 1}={y+1\over 2}={z+1\over 2} \\假設A在L的投影點為R \Rightarrow R(t,2t-1,2t-1) \Rightarrow \overrightarrow {AR} \bot\vec n \Rightarrow \overrightarrow {AR} \cdot\vec n =0 \Rightarrow t=2 \Rightarrow R(2,3,3) \\ \Rightarrow A對L的對稱點A'=2R-A=(6,5,-1) \Rightarrow 直線L'=\overleftrightarrow{A'B}:{x+1\over -7}={y\over -5}={z-3\over 4} \\ \Rightarrow P=L\cap L' =\bbox[red, 2pt]{({4\over 3},{5\over 3},{5\over 3})} \\註:橢圓光學性質:從焦點發出的光，經過切點後反射會經過另一個焦點$$
解答:$$假設\int_1^3 |f(t)|=k \Rightarrow f(x)=-x+k \Rightarrow \int_1^3 |f(x)|\,dx =\int_1^3 |-x+k|\,dx \Rightarrow \int_1^3 |-x+k|\,dx=k \\\Rightarrow \begin{cases} k\ge 3 \Rightarrow k=\int_1^3(-x+k)\,dx =-4+2k \Rightarrow k=4\\ 1\le k\le 3 \Rightarrow k=\int_1^k(-x+k)\,dx +\int_k^3 (x-k)\,dx \Rightarrow k^2-5k+5=0 \Rightarrow k=(5-\sqrt 5)/2\\ k\le 1 \Rightarrow k=\int_1^3(x-k)dx =4-2k \Rightarrow k=4/3 不符k\le 1\end{cases} \\ \Rightarrow \cases{k=4\Rightarrow f(x)=-x+4 \Rightarrow f(2)=2\\ k=(5-\sqrt 5)/2 \Rightarrow f(x)=-x+(5-\sqrt 5)/2 \Rightarrow f(2)=(1-\sqrt 5)/2} \\ \Rightarrow f(2)=\bbox[red, 2pt]{2 或 {1-\sqrt 5\over 2}}$$
二、計算證明題： (每題 10 分；共 30 分)
解答:$${x^2 \over A^2}-{y^2\over B^2}=1 \equiv {x^2\over 12}-{y^2\over 15}=1 \Rightarrow \cases{A=2\sqrt 3\\ B=\sqrt{15}} \Rightarrow C=\sqrt{27}=3\sqrt 3 \Rightarrow \cases{F_1(-3\sqrt 3,0) \\F_2(3\sqrt 3,9) \\ 離心率e=C/A=3/2} \\ \Rightarrow \cases{\overline{PF_1}= ex+A=ea+A\\ \overline{PF_2}= ex-A =ea-A} \Rightarrow \cases{\triangle PF_1F_2面積={1\over 2} \overline{F_1F_2}\cdot b=Cb \\ \triangle PF_1F_2周長=2ea +2C} \\\Rightarrow 內切圓半徑r={面積\over 半周長} ={Cb\over ea+C} ={C\cdot B\sqrt{a^2/A^2-1}\over ea+C} ={C\cdot B \sqrt{{1\over A^2}-{1\over a^2}} \over e+C/a} \\ \Rightarrow \lim_{a\to \infty}r(a)= \lim_{a\to \infty} {C\cdot B \sqrt{{1\over A^2}-{1\over a^2}} \over e+C/a} ={C\cdot B\cdot (1/A) \over e} ={CB\over Ae}= {CB\over C} =B=\bbox[red, 2pt]{\sqrt{15}}$$
解答:
$$假設\cases{球心A(0,0,0)\\ 球半徑R=2\sqrt 3\\ 稜長s=3}, 分別討論六個面與球相交的情形\\ \textbf{Type I }與A相鄰的三個面, 分別是x=0,y=0, z=0. 三個面情形相同, 以z=0為例\\\qquad z=0代入球:x^2+y^2+z^2=12 \Rightarrow C_1:x^2+y^2=12 \Rightarrow \cases{C_1與x=3的交點P_1(3,\sqrt 3)\\ C_1與y=3的交點P_2(\sqrt 3, 3)} \\ \Rightarrow \cases{P_1與z=0的\theta_1:\tan \theta_1=y/x=\sqrt 3/3 \Rightarrow \theta_1=\pi/6\\ P_2與z=0的\theta_2: \tan\theta_2=3/\sqrt 3 =\sqrt 3\Rightarrow \theta_2=\pi/3} \Rightarrow 夾角={\pi\over 3}-{\pi\over 6}={\pi\over 6} \\ \Rightarrow 弧長=2\sqrt 3\times {\pi\over 6}={\sqrt 3\pi\over 3} \Rightarrow 三個面共有3\times{\sqrt 3\pi\over 3}=\sqrt 3\pi \\\textbf{Type II }與A不相鄰的三個面, 分別是x=3,y=3, z=3. 三個面情形相同, 以x=3為例 \\ \Rightarrow x=3代入球\Rightarrow y^2+z^2=3,也就是以B(3,0,0)為圓心, 半徑\sqrt 3的圓\\ 0\le y,z\le 3 \Rightarrow 截出的曲線剛好是四分之一圓\Rightarrow 弧長=\sqrt 3\times {\pi\over 2} \\ \Rightarrow 三個面共有3\times \sqrt 3\times {\pi\over 2} ={3\sqrt 3\pi\over 2}\\ 因此總弧長=\sqrt 3\pi +{3\sqrt 3\pi\over 2} = \bbox[red, 2pt]{5\sqrt 3\pi\over 2}$$
解答:$$假設2n位數的數字依序為d_1,d_2,\dots,d_{2n}, 並令\cases{A= \{d_1,d_2,\dots, d_n\}\\ B= \{d_{n+1}, d_{n+2},\dots, d_{2n}\}} \\ 及\cases{S_A= 集合A的數字和\\ S_B= 集合B的數字和} \Rightarrow \triangle=S_A-S_B \\ 若從A中取出a, 從B中取出b進一交換,新的集合加總\cases{S_A'=S_A-a+b\\ S_B'=S_B-b+a}\\ \Rightarrow 新的差值\triangle '=S_A'-S_B'= (S_A-S_B)-2(a-b) =\triangle -2(a-b) \\ \Rightarrow 一次交換的差值減少2(a-b) \Rightarrow 只要a\gt b, 每次交換後\triangle 會減少\\ 證明:若\triangle \gt 9\Rightarrow S_A\gt S_B, 若\max(A)\le \min(B) \Rightarrow S_A\le S_B, 與 \triangle \gt 9矛盾\\ \qquad 一定存在a\in A, b\in B使得a\gt b, 因此只要\triangle \gt 9, 一次交換就可達成且\triangle 嚴格遞減\\ \qquad 又a-b\le 9 \Rightarrow \triangle -\triangle '\le 2(a-b)=18, 當\triangle \gt 9才需要交換, 即\triangle \gt 9 \Rightarrow \triangle' \gt 9-18=-9 \\\qquad \Rightarrow |\triangle '|\lt 9, 不會減過頭.\\ 經由以上交換步驟(數字有限, 步驟有限), \triangle 逐步遞減, 最後滿足 |\triangle |\lt 9, \bbox[red, 2pt]{故得證}$$
解題僅供參考，其他教甄試題及詳解
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