國立彰化女子高級中學 114 學年度第一次教師甄試
一、填充題(每題 5 分，答案都須化為最簡分數或有理化，且完全正確才給分，共 75 分)
解答:$${x+y\over x^2-xy+y^2}={3\over 7} \Rightarrow 7x+7y=3x^2-3xy+3y^2 \Rightarrow 3x^2-(3y+7)x+3y^2-7y=0 \\ \Rightarrow 判別式(3y+7)^2-12(3y^2-7y)\ge 0 \Rightarrow 27y^2-126y-49\le 0 \\ \Rightarrow {21- 14\sqrt 3\over 9}\le y\le {21+ 14\sqrt 3\over 9} 且y\in \mathbb N \Rightarrow y=1,2,\dots,5 \\ \Rightarrow \cases{y=1 \Rightarrow 3x^2-10x-4=0 \Rightarrow x\not \in \mathbb N\\ y=2 \Rightarrow 3x^2-13x-2=0 \Rightarrow x\not \in \mathbb N \\ y=3 \Rightarrow 3x^2-16x+6=0 \Rightarrow x\not \in \mathbb N\\ y=4 \Rightarrow 3x^2-19x+ 20=0 \Rightarrow x=5\\ y=5 \Rightarrow 3x^2-22x+40=0 \Rightarrow x=4} \Rightarrow (x,y)= \bbox[red, 2pt]{(4,5), (5,4)}$$
解答:$$x^2-{y^2\over 3}=1 \Rightarrow \cases{a=1\\ b=\sqrt 3} \Rightarrow c=2 \Rightarrow 離心率e={c\over a}=2\\ 假設\cases{A(x_A,y_A) \\ B(x_B,y_B)} \Rightarrow \cases{ \overline{AF}=ex_A-a=2x_A -1\\ \overline{BF}= ex_B-a=2x_B-1} \Rightarrow \overline{AF}+ \overline{BF}=2(x_A+ x_B)-2=16 \\ \Rightarrow x_A+ x_B=9 \cdots(1) \\ L: y=m(x-3) 代入雙曲線求交點 \Rightarrow x^2-{m^2(x-3)^2 \over 3} =1 \Rightarrow (3-m^2)x^2+6m^2 x-9m^2-3=0 \\ \Rightarrow 兩根之和=x_A+ x_B={6m^2\over m^2-3}=9 \Rightarrow m^2=9 \Rightarrow m= \bbox[red, 2pt]{\pm 3}$$
解答:$$取\cases{O(0,0,0) \\A(a,0,0) \\ B(0,b,0) \\C(0,0,c)} \Rightarrow \cases{\overrightarrow{AB} =(-a,b,0) \\ \overrightarrow{AC} =(-a,0,c)} \Rightarrow \triangle ABC面積= {1\over 2} ||(-a,b,0) \times (-a,0,c) || \\={1\over 2}||(bc,ac,ab)|| ={1\over 2}\sqrt{a^2b^2+ b^2c^2 +c^2a^2} =1 \Rightarrow a^2b^2+ b^2c^2 +c^2a^2=4 \\ \Rightarrow 平面E=\triangle ABC: {x\over a}+{y\over b}+{z\over c}=1 \Rightarrow h=d(O,E)={abc \over 2} \\ \Rightarrow 三角錐體積V={1\over 3}\cdot 1\cdot {abc\over 2} ={abc\over 6} \\算幾不等式: {a^2b^2+ b^2c^2 +c^2a^2\over 3}\ge \sqrt[3]{a^4b^4c^4} =(abc)^{4/3} \Rightarrow (abc)^{4/3}\le {4\over 3} \Rightarrow abc \le \left( {4\over 3} \right)^{3/4} \\ \Rightarrow V\le {1\over 6}\left( {4\over 3} \right)^{3/4}={1\over 6}\left( { 2^4\cdot 2^2 \cdot 3\over 3^4} \right)^{1/4} =\bbox[red, 2pt]{\sqrt[4]{12} \over 9}$$
解答:$$\alpha,\beta, \gamma 為g(x)=0的三根 \Rightarrow g(x)=(x-\alpha)( x-\beta) (x-\gamma) \Rightarrow \cases{ \alpha+ \beta+\gamma=4\\ \alpha\beta+ \beta\gamma+ \gamma\alpha= -1\\\alpha\beta\gamma =-3}\\又 \cases{f(x)= x^4-2x^3-9x^2+7 \\ g(x)= x^3-4x^2-x+3} \Rightarrow f(x)=g(x)(x+2)-x+1 \Rightarrow \cases{f( \alpha)=1-\alpha\\ f(\beta)=1-\beta\\ f(\gamma) =1-\gamma} \\ \Rightarrow {f(\alpha) \over f(\beta) f(\gamma)} +{f(\beta) \over f(\gamma) f(\alpha)} +{f(\gamma) \over f(\alpha) f(\beta)} ={1-\alpha\over (1-\beta)(1-\gamma)} +{1-\beta\over (1-\gamma)( 1-\alpha)}+ {1-\gamma\over (1-\alpha)(1-\beta)} \\={ (1-\alpha)^2+ (1-\beta)^2+ (1-\gamma)^2\over (1-\alpha) (1-\beta)( 1-\gamma)} ={\alpha^2+ \beta^2+ \gamma^2-2(\alpha+ \beta+ \gamma)+3 \over g(1)} \\={ 4^2-2(-1)-2\cdot 4+3\over -1} = \bbox[red, 2pt]{-13}$$
解答:$$假設\cases{F_1(2,4) \\F_2(7,11)} \Rightarrow 橢圓在 P 點的切線，會是\angle F_1PF_2的外角平分線,而法線則是內角平分線 \\ \cases{\overrightarrow{PF_1}=(-3,-1) \\ \overrightarrow{PF_2} =(2,6)} \Rightarrow \cases{|\overrightarrow{PF_1}|= \sqrt{10} \\ |\overrightarrow{PF_2}| =2\sqrt{10}} \Rightarrow \cases{\vec e_1 =\overrightarrow{PF_1}/|\overrightarrow{PF_1}|= {1\over \sqrt{10}}(-3,-1) \\ \vec e_2 = \overrightarrow{PF_2}/|\overrightarrow{PF_2}| ={1\over \sqrt{10}}(1,3)} \\ \Rightarrow 切線(外角平分線)的方向向量\vec u \parallel (\vec e_1-\vec e_2)\Rightarrow \vec u \parallel {1\over \sqrt{10}}(-4,-4) \Rightarrow \vec u=(1,1) \\ \Rightarrow 切線斜率=1 \Rightarrow 切線方程式: y=(x-5)+5 \Rightarrow \bbox[red, 2pt]{x-y=0}$$
解答:$$\cases{P\in C_1 \Rightarrow P(\cos \alpha, \sin \alpha) \\ Q\in C_2 \Rightarrow Q(2\cos \beta+4, 2\sin \beta)}\;又\alpha=2\beta \Rightarrow P(\cos 2\beta, \sin 2\beta), 且\cases{P(1,0) =(\cos 0^\circ, \sin 0^\circ) \\Q(6,0) =(2\cos 0^\circ+4, 2\sin 0^\circ)} \\ \Rightarrow \overline{PQ} ^2 =(\cos 2\beta -2\cos \beta-4)^2+ (\sin 2\beta-2\sin \beta)^2 \\=21-4(\cos \beta\cos 2\beta +\sin \beta \sin 2\beta) +16\cos \beta-8\cos 2\beta =21 -4\cos(2\beta-\beta)+16\cos \beta-8\cos 2\beta \\ =21+12\cos \beta-8(2\cos^2 \beta-1) =29-16(\cos^2 \beta-{3\over 4}\cos \beta+{9\over 64}) +{9\over 4} \\={125\over 4}-16(\cos \beta-{3\over 8})^2 \ge {125\over 4} \Rightarrow \overline{PQ}最大值=\sqrt{125\over 4} = \bbox[red, 2pt]{5\sqrt 5\over 2}$$
解答:$$一局A贏:\cases{A正面1、B正面0 \Rightarrow 機率={1\over 2} \cdot {1\over 2} ={1\over 4} \\A正面2、B正面0 \Rightarrow 機率={1\over 4} \cdot {1\over 2} ={1\over 8} \\A正面2、B正面1 \Rightarrow 機率={1\over 4} \cdot {1\over 2} ={1\over 8}} \Rightarrow 機率合計{1\over 2} \\ 二局A贏:第一局平手\cases{A,B都正面0 \Rightarrow 機率={1\over 4} \cdot {1\over 2}={1\over 8}\\ A,B都正面1 \Rightarrow 機率={1\over 2}\cdot {1\over 2}= {1\over 4}} \Rightarrow 機率合計{3\over 8} \\\qquad 且第二局A贏(機率={1\over 2}) \Rightarrow 機率為{1\over 2} \cdot {3\over 8} ={3\over 16} \\ 三局A贏:\cases{第一局平手、第二局平手、第三局A贏:機率= {3\over 8} \cdot {3\over 8}\cdot {1\over 2}={9\over 128} \\ 第一局B贏、第二局A贏、第三局A贏:機率={1\over 8} \cdot {1\over 8}\cdot {1\over 2} ={1\over 128}} \Rightarrow 機率合計{5\over 64} \\ 因此前三區A贏的機率={1\over 2}+{3\over 8}+{5\over 64} = \bbox[red, 2pt]{49\over 64}$$
解答:$$y=x^2 \Rightarrow y'=2x \Rightarrow 若切點為(x_0,y_0),則切線斜率為2x_0\\ 因此假設兩切點\cases{A(a,a^2) \\ B(b,b^2)} \Rightarrow 兩切線\cases{L_1: y=2a(x-a)+a^2\\ L_2: y=2b(x-b)+b^2} \\\Rightarrow P= L_1\cap L_2=({a+b\over 2}, ab) =(x,y) \Rightarrow (a-b)^2= 4x^2-4y\\ \overline{AB}^2 =(a-b)^2 +(a^2-b^2)^2= (a-b)^2+ (a-b)^2(a+b)^2 =12 \\ \Rightarrow 4x^2-4y+(4x^2-4y)4x^2 =12 \Rightarrow y={4x^4+x^2-3\over 4x^2+1} \\ P在第三或第四象限 \Rightarrow y\lt 0 \Rightarrow {4x^4+x^2-3\over 4x^2+1} \lt 0 \Rightarrow 4x^4+x^2-3\lt 0 \Rightarrow (4x^2-3)(x^2+1) \lt 0\\ \Rightarrow 4x^2-3 \lt 0 \Rightarrow \bbox[red, 2pt]{- {\sqrt 3\over 2} \lt x\lt {\sqrt 3\over 2}}$$
解答:$$\cases{V'(t)=t/2+4\\ V(0)=100} \Rightarrow V(t) = \int \left( {t\over 2}+4 \right)\,dt ={1\over 4}t^2+4t+C \Rightarrow V(0)=C= 100 \\ \Rightarrow V(t)={1\over 4}t^2+4t+100 \Rightarrow 第1次達到120 \text{ cm}^3 \Rightarrow V(t)=100 \Rightarrow t^2+16t-80=0 \\ \Rightarrow (t+20)(t-4) =0 \Rightarrow t=4\\ 當t\ge 4時，V'(t)=t/2+4-13=t/2-9 \Rightarrow V(t) =\int \left( {t\over 2}-9 \right)\,dt={t^2\over 4}-9t+C\\ \Rightarrow V(4)=120=4-36+C \Rightarrow C=152 \Rightarrow V(t) ={t^2\over 4}-9t+152, t\ge 4\\ V(t)=80 \Rightarrow {t^2\over 4}-9t+152=80 \Rightarrow t^2-36t+288=0 \Rightarrow (t-12)(t-24)=0 \Rightarrow t=12\\ 當t\ge 12時，\cases{V'(t)=t/2+4 \\ V(12)=80} \Rightarrow V(t)={t^2\over 4}+4t-4 \\ 第二次V(t)=120 \Rightarrow t^2+16t-496=0 \Rightarrow t=\bbox[red, 2pt]{-8+4\sqrt{35}}$$
解答:$$假設面方形BCDE的中心為原點 \Rightarrow \cases{A(0,0,h) \\B(3,3,0) \\C(-3,3,0) \\D(-3,-3,0) \\E(3,-3,0)} \Rightarrow \overline{AB}^2 =\overline{OA}^2 + \overline{OB}^2 \Rightarrow 5^2=h^2+(3\sqrt 2)^2 \\ \Rightarrow h=\sqrt 7 \Rightarrow A(0,0,\sqrt 7)\\ \textbf{Case I }A在F左側、BCDE 在 F右側: d(A,F) =d(F, 平面BCDE) \Rightarrow F: z={\sqrt 7\over 2}\\ \textbf{Case II }BC在F左側、ADE 在 F右側: \overleftrightarrow{BC} \parallel 平面ADE \Rightarrow F法向量=平面ADE法向量=(0,\sqrt 7,-3) \\ \qquad \Rightarrow F=過原點且法向量為(0,\sqrt 7,-3) \Rightarrow F:\sqrt 7y-3z= 0\\ \textbf{Case III }CD在F左側、ABE 在 F右側 \Rightarrow F: \sqrt 7y+3z=0 \\\textbf{Case IV }DE在F左側、ABC 在 F右側 \Rightarrow F: \sqrt 7x-3z=0 \\\textbf{Case V }BE在F左側、ACD 在 F右側 \Rightarrow F: \sqrt 7x+3z=0 \\五個平面構成一個倒立的四角錐:\cases{頂點O(0,0,0)\\ P(3/2,3/2, \sqrt 7/2) \\ Q(3/2,-3/2, \sqrt 7/2) \\R(-3/2, 3/2, \sqrt 7/2) \\ S(-3/2,-3/2, \sqrt 7/2)} \Rightarrow \cases{底面為正方形PQRS, 面積=3\cdot 3= 9\\ d(O, 平面PQRS) =\sqrt 7/2} \\ \Rightarrow V= {1\over 3} \cdot 9\cdot {\sqrt 7\over 2} = {3\sqrt 7\over 2} \Rightarrow (n,V) = \bbox[red, 2pt]{\left( 5, {3\sqrt 7\over 2}\right)}$$
解答:$$在八、卦、山三字順序維持下，將「巍巍峨峨八卦山」七個字任排，有{7! \over 2!2!3!} =210種\\ 接著將彰、女、中三字插入七個字的八個空隙中，有C^8_3=56種，因此總共有210\cdot 56=11760種\\ 「巍巍峨峨八卦山」七個字中，巍巍相鄰有{6!\over 2!3!} =60，此時有七個空隙(巍巍綁在一起算一個)\\選三個空隙插入彰、女、中，有C^7_3=35，因此有60\times 35=2100;\\ 同理，峨峨相鄰也是有2100\\巍巍相鄰且峨峨相鄰的條件下，「巍巍峨峨八卦山」排列數為{5!\over 3!}=20，\\在六個空隙中插入彰、女、中，有C^6_3=20，因此有20\times 20=400種\\ 最後的排列數為\;11760-2100-2100+400= \bbox[red, 2pt]{7960}$$
解答:$$O為\triangle ABC 外心 \Rightarrow |\overrightarrow{OA}| =|\overrightarrow{OB}| =|\overrightarrow{OA}| =r= 外接圓半徑\\2\overrightarrow{AO} \cdot \overrightarrow{BC}+ 3\overrightarrow{BO}\cdot \overrightarrow{CA}+ 5\overrightarrow{CO} \cdot \overrightarrow{AB} =2\overrightarrow{AO} \cdot (\overrightarrow{BO} +\overrightarrow{OC})+ 3\overrightarrow{BO}\cdot (\overrightarrow{CO} +\overrightarrow{OA})+ 5\overrightarrow{CO} \cdot (\overrightarrow{AO} +\overrightarrow{OB}) \\=2\overrightarrow{OA} \cdot \overrightarrow{OB}-2\overrightarrow{OA}\cdot \overrightarrow{OC} +3\overrightarrow{OB} \cdot \overrightarrow{OC} -3\overrightarrow{OB}\cdot \overrightarrow{OA} +5\overrightarrow{OC} \cdot \overrightarrow{OA}- 5\overrightarrow{OC}\cdot \overrightarrow{OB} \\=-\overrightarrow{OA}\cdot \overrightarrow{OB}+ 3\overrightarrow{OA}\cdot \overrightarrow{OC}-2 \overrightarrow{OB}\cdot \overrightarrow{OC} = \vec 0 \\ \Rightarrow 3\overrightarrow{OA}\cdot \overrightarrow{OC}= \overrightarrow{OA}\cdot \overrightarrow{OB} +2 \overrightarrow{OB}\cdot \overrightarrow{OC} \\\Rightarrow 3|\overrightarrow{OA}|| \overrightarrow{OC}|\cos \angle AOC =|\overrightarrow{OA}|| \overrightarrow{OB}|\cos \angle AOB +2 |\overrightarrow{OB}|| \overrightarrow{OC}| \cos \angle BOC \\ \Rightarrow 3\cos \angle AOC= \cos \angle AOB+2 \cos \angle BOC \Rightarrow 3\cos (2\angle B) =\cos(2\angle C) +2\cos(2\angle A) \\ \Rightarrow 3(1-2\sin^2 \angle B) =1-2\sin^2 \angle C+2(1-2\sin^2 \angle A) \Rightarrow 3\sin^2 \angle B=\sin^2\angle C+ 2\sin^2 \angle A \\ \Rightarrow 3\cdot \left( {b\over 2r} \right)^2 =\left( {c\over 2r} \right)^2+2\cdot \left( {a\over 2r} \right)^2 \Rightarrow 3b^2=c^2+2a^2 \\ \Rightarrow \cos \angle B={a^2+c^2-b^2 \over 2ac} ={a^2+c^2-(c^2+2a^2)/3 \over 2ac} ={{a^2\over 3}+{2c^2\over 3}\over 2ac} \\={a^2+ 2c^2\over 6ac} \ge{ 2\sqrt{2a^2c^2}\over 6ac} ={2\sqrt 2\over 6} = \bbox[red, 2pt]{\sqrt 2\over 3}$$
解答:$$擲出點數為1的機率為p={1\over 6}，不是1的機率為q=1-p={5\over 6}\\ 在前k-1次投擲中，恰出現一次點數1,機率為{k-1 \choose 1}pq^{k-2};第k次投擲出現點數1, 機率為p\\ 因此P(X=k) ={k-1\choose 1}pq^{k-2} \cdot p ={k-1\choose 1} {1\over 6^2}\cdot \left( {5\over 6} \right)^{k-2} =(k-1) {1\over 36}\cdot \left( {5\over 6} \right)^{k-2}\\ 令f(k)= (k-1) {1\over 36}\cdot \left( {5\over 6} \right)^{k-2} \Rightarrow f'(k)= {1\over 36} \left( {5\over 6} \right)^{k-2} \left( 1+(k-1) \ln{5\over 6} \right) =0 \\ \Rightarrow 1+(k-1) \ln{5\over 6}=0 \Rightarrow k=1-{1\over \ln(5/6)} \approx 6.5\\ 由於k\in \mathbb N \Rightarrow f(6)=f(7)={5^5\over 6^6} \Rightarrow k= \bbox[red, 2pt]{6或7}\\註:\log{5\over 6} =\log 5-\log 6=(1-\log 2)-(\log 2+\log 3) =1-2\cdot 0.301-0.4771=-0.0794\\ \Rightarrow \ln {5\over 6} ={\log(5/6) \over \log e} ={-0.0794\over 0.434} \Rightarrow 1-{1\over \ln(5/6)} =1+{0.434\over 0.0794} \approx 6.5$$
解答:$$假設圖n頂點數字為a_n \Rightarrow a_1=1^2+2^2 \Rightarrow a_2= 1^2+ 2\cdot 2^2+ 2\cdot 3^2+3^2 \\ \Rightarrow a_3=1^2+3\cdot 2^2+ 3\cdot 3^2+4^2 \Rightarrow \cdots \Rightarrow a_n= \sum_{k=0}^n {n\choose k}(k+1)^2 = \sum_{k=0}^n {n\choose k}\left( k(k-1)+3k+1 \right) \\= n(n-1) 2^{n-2}+3n\cdot 2^{n-1}+ 2^n = \bbox[red, 2pt]{(n+1)(n+4)2^{n-2}} \\ 註:f(x) =(1+x)^n = \sum_{k=0}^n {n\choose k}x^k \Rightarrow f'(x)= n(1+x)^{n-1} =\sum_{k=0}^n k{n\choose k}x^{k-1} \\\Rightarrow f''(x) =n(n-1) (1+x)^{n-2}=\sum_{k=0}^n k(k-1){n\choose k}x^{k-2} \Rightarrow \cases{\sum_{k=0}^n {n\choose k} =2^n \\ \sum_{k=0}^n k{n\choose k} =n\cdot 2^{n-1} \\ \sum_{k=0}^n k(k-1){n\choose k} =n(n-1)2^{n-2}}$$
二、計算證明題(共 25 分)
解答:$$\textbf{(1) }\lim_{n\to \infty} {3\over n^2} \left(\sqrt{4n^2+ 9\times 1^2} +\sqrt{4n^2+ 9\times 2^2} + \cdots +\sqrt{4n^2+ 9\times (n-1)^2}\right)\\ \quad =\lim_{n\to \infty} \sum_{k= 1}^{n-1} {3\over n^2}\sqrt{4n^2+9\times k^2} =\lim_{n\to \infty} \sum_{k= 1}^{n-1} {3\over n}\sqrt{4+ (3k/n)^2} =\int_0^3 \sqrt{4+x^2}\,dx, 故選\bbox[red, 2pt]{(C)} \\也可以 = \lim_{n\to \infty} \sum_{k=1}^{n-1} {3\over n} \sqrt{4 + 9 \left(\frac{k}{n}\right)^2} = \int_0^1 3 \sqrt{4 + 9x^2} ,dx =\int_0^3 \sqrt{4+u^2}\,du \;(取u=3x) \\=\int_0^3 \sqrt{4+ x^2}\,dx \; 結果相同 \\\textbf{(2) }取x=2\tan \theta \Rightarrow dx =2\sec^2 \theta\,d\theta \Rightarrow \int_0^3 \sqrt{4+x^2}\,dx = 4\int_0^{\tan^{-1}(3/2)} \sec^3 \theta\,d\theta \\\quad \cases{u=\sec \theta\\ dv=\sec^2 \theta\, d\theta} \Rightarrow \cases{du =\tan \theta\sec \theta \,d\theta\\ v=\tan \theta} \Rightarrow I= \int \sec^3\theta\,d\theta=\sec \theta \tan \theta- \int \tan^2\theta \sec \theta\,d\theta \\=\sec \theta \tan \theta-\int(\sec^2 \theta-1) \sec \theta\,d\theta= \sec \theta\tan \theta-I+ \int \sec \theta \,d\theta \\\Rightarrow 2I=\sec \theta\tan \theta + \int \sec \theta \,d\theta =\sec \theta \tan \theta+ \ln |\tan \theta+\sec \theta|+C \\ \Rightarrow 4 \int_0^{ \tan^{-1} (3/2)} \sec^3 \theta\,d\theta =2 \left. \left[ \sec \theta\tan \theta+ \ln |\tan \theta+\sec \theta|\right] \right|_0^{\tan^{-1} (3/2)} = \bbox[red, 2pt]{{3 \sqrt{13} \over 2}+ 2\ln{3+\sqrt{13}\over 2}}$$
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解題僅供參考，其他教甄試題及詳解
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