104年全國教甄聯招-數學詳解

教育部受託辦理104學年度公立高級中等學校教師甄選

一、單選題

解答：$$a_n= \sqrt{1\times 2}+\sqrt{2\times 3}+\cdots + \sqrt{n(n+1)} =\sum_{k=1}^n\sqrt{k(k+1)} \\ \Rightarrow \sum_{k=1}^n\sqrt{k^2} \lt a_n \lt \sum_{k=1}^n\sqrt{(k+1)^2}\\ \Rightarrow 1+2+\cdots +n \lt a_n \lt 2+3+\cdots +(n+1) \Rightarrow {n(n+1)\over 2} \lt a_n \lt {n(n+3)\over 2} \\ \Rightarrow \lim_{n\to \infty } {n(n+1)\over 2n^2}\lt \lim_{n\to \infty }{a_n\over n^2} \lt \lim_{n\to \infty } {n(n+3)\over 2n^2}\\ \Rightarrow {1\over 2}\lt \lim_{n\to \infty }{a_n\over n^2} \lt  {1\over 2} \Rightarrow \lim_{n\to \infty }{a_n\over n^2} ={1\over 2}，故選\bbox[red,2pt]{(B)}$$

解答：$$888888=13\times 68376 \Rightarrow 2015 \equiv 5 \mod 6 \Rightarrow 2015個8除以13的餘數=5個8除以13的餘數\\88888=6837\times 13+7 ，故選\bbox[red,2pt]{(B)}$$

解答：

$$\overline{AD}為\angle A的角平分線 \Rightarrow {\overline{AB}\over \overline{AC}}= {6\over 4}={\overline{BD}\over \overline{DC}} \Rightarrow  \cases{\overline{BD}=6\sqrt 7/5\\ \overline{CD}=4\sqrt 7/5}\\ 令\overline{AH}=a \Rightarrow \overline{BH}=6-a \Rightarrow \overline{CH}^2=4^2-a^2= (2\sqrt 7)^2-(6-a)^2 \Rightarrow a=2 \\\Rightarrow \cases{\overline{AH}=2\\ \overline{BH}=6-2=4\\\overline{CH}= \sqrt{4^2-2^2} =2\sqrt 3} ;又{\overline{DE}\over \overline{CH}} ={\overline{BD}\over \overline{BC}} ={\overline{BE}\over \overline{BH}}\Rightarrow \cases{\overline{DE}={6\over 5}\sqrt 3 \\\overline{BE}=12/5\\ \overline{EH}=8/5}\\因此可令\cases{A(0,0)\\ B(6,0)\\ C(2,2\sqrt 3)\\ H(2,0)} \Rightarrow \cases{\overleftrightarrow {AD}:y={1\over \sqrt 3}x \\ \overleftrightarrow{CH}:x=2} \Rightarrow P(2,{2\over \sqrt 3}) \Rightarrow \cases{\overrightarrow{BP}=(-4,{2\over \sqrt 3}) \\ \overrightarrow{BA}=(-6,0)\\ \overrightarrow{BC}=(-4,2\sqrt 3)}\\ \Rightarrow \overrightarrow{BP}=\alpha \overrightarrow{BA}+\beta \overrightarrow{BC} \Rightarrow (-4,{2\over \sqrt 3})= (-6\alpha-4\beta,2\sqrt 3\beta) \Rightarrow \cases{\alpha= 4/9\\\beta=1/3} \\ \Rightarrow \alpha+\beta={4+3\over 9} ={7\over 9}，故選\bbox[red,2pt]{(A)}$$

解答：$$f(x)=0的三根為\alpha,\beta,\gamma \Rightarrow f(x)=x^3-3x^2+bx+c =(x-\alpha)(x-\beta)(x-\gamma) 且\alpha+\beta+\gamma=3\\ \Rightarrow f(-1)=-1=(-1-\alpha)(-1-\beta)(-1-\gamma) \Rightarrow (1+\alpha)(1+\beta) (1+\gamma)=-1\\ \Rightarrow \begin{vmatrix} 1+\alpha & 1 & 1\\ 1& 1+\beta & 1\\ 1& 1& 1+\gamma\end{vmatrix} =(1+\alpha)(1+\beta) (1+\gamma)+2-(1+\alpha)-(1+\beta)-(1+\gamma) \\ =-1+2-3-(\alpha+\beta+\gamma)=-2-3=-5，故選\bbox[red,2pt]{(B)}$$

解答：$$令a_n=甲乙丙三人在n天排打掃，符合同一人不可連兩天打掃的方法數\\ \Rightarrow \cases{a_1=3\\ a_2=3\times 2=6\\ a_n=2a_{n-1}} \Rightarrow a_n=3\cdot 2^{n-1} \Rightarrow a_6=3\cdot 2^5=96\\ 丙沒排到打掃:甲乙甲乙甲乙或乙甲乙甲乙甲，2種排法；\\同理,甲沒排到打掃、乙沒排到打掃都是各有2種排法；\\因此至少每人排到一天的機率為1-{6\over 96} ={90\over 96}=0.9375，故選\bbox[red,2pt]{(C)}$$

解答：$$a_{n+1}= {1\over 3}a_n+2 \Rightarrow a_n={1\over 3}a_{n-1}+2 ={1\over 3}\left( {1\over 3}a_{n-2}+2 \right)+2 = ({1\over 3})^2a_{n-2}+ 2(1+{1\over 3}) \\ =({1\over 3})^3a_{n-3}+ 2(1+{1\over 3} +{1\over 3^2}) =\cdots =({1\over 3})^{n-1}a_{1}+ 2(1+{1\over 3} +{1\over 3^2}+\cdots +{1\over 3^{n-2}}) \\ =({1\over 3})^{n-1}+2(1+{1\over 3} +{1\over 3^2}+\cdots +{1\over 3^{n-2}})\\ \Rightarrow \lim_{n\to \infty}a_n = 0+2\cdot {3\over 2} =3，故選\bbox[red,2pt]{(D)}$$

解答：

$$從S_1出發走3步到S_2:\\S_1\to S_2\to S_2\to S_2 \Rightarrow 機率:1\cdot {1\over 2}\cdot {1\over 2}={1\over 4}\\ S_1\to S_2\to S_3\to S_2  \Rightarrow 機率:1\cdot {1\over 3}\cdot {2\over 3}={2\over 9}\\S_1\to S_2\to S_1\to S_2  \Rightarrow 機率:1\cdot {1\over 6}\cdot 1={1\over 6}\\ 因此機率為{1\over 4}+{2\over 9}+{1\over 6} ={23\over 36}，故選\bbox[red,2pt]{(C)}$$

解答：

$$令\cases{\overline{AB}=a \\ \overline{AD}=b\\ \overline{AC}=c} \Rightarrow \cases{ {a\over \sin C}=2R_3=3k \cdots(1)\\{b\over \sin B}=2R_1=k \cdots(2)\\ {b\over \sin C}=2R_2=2k \cdots(3)\\ {c\over \sin B}=2R_1=3k \cdots(4)} \Rightarrow \cases{\cases{(1)\\ (3)} \Rightarrow \cases{a=3k\sin C\\ b=2k\sin C} \Rightarrow a:b=3:2\\ \cases{(2)\\ (4)} \Rightarrow \cases{b=k\sin B\\ c=3k\sin B} \Rightarrow b:c=1:3} \\ \Rightarrow a:b:c= 3:2:6 ，故選\bbox[red,2pt]{(D)}$$

二、複選題

解答：$$(A)\times: \cases{\Gamma_1的貫軸=2a=\Gamma_2的貫軸 \\\Gamma_1的共軛軸=2b=\Gamma_2的共軛軸}，\Gamma_1的貫軸\ne\Gamma_2的共軛軸 \\(B)\bigcirc: \Gamma_2旋轉90^\circ就是\Gamma_1\\ (C)\bigcirc:都是(0,\pm c)\\ (D)\times: \cases{\Gamma_2的漸近線:y=\pm {a\over b}x \\\Gamma_3的漸近線:y=\pm{b\over a}x}，兩者不同\\故選\bbox[red,2pt]{(BC)}$$

解答：

$$令A'為A(1,1,1)以E:x-2y+z=3為對稱平面的對稱點，\\則直線L:\overleftrightarrow{AA'}的方向向量平行E的法向量\vec n=(1,-2,1) \Rightarrow L:{x-1\over 1} ={y-1\over -2} ={z-1\over 1} \\ \Rightarrow A'(t+1,-2t+1,t+1) \Rightarrow \overline{AA'}的中點A''({1\over 2}t+1,-t+1,{1\over 2}t+1)在E上\\ \Rightarrow {1\over 2}t+1+2t-2+{1\over 2}t+1=3 \Rightarrow t=1 \Rightarrow A'(2,-1,2) \\ \Rightarrow 反射光直線即為\overleftrightarrow{A'B}:\cases{{x-2\over 3}={y+1\over 4}\\z=2} \Rightarrow \cases{{x-2\over 3}-{1\over 3}={y+1\over 4}-{1\over 3}\\z=2} \Rightarrow \cases{{x-3\over 3} ={y-1/3\over 4} \\z=2} \\ \Rightarrow \cases{{x-3\over 4} ={y-1/3\over 16/3} \\z=2} \Rightarrow \cases{a=-1/3\\ b=16/3\\ c=2} \Rightarrow a+b+c=7，，故選\bbox[red,2pt]{(CD)}$$

解答：$$(A)\bigcirc: \sigma_X的計算與順序無關，因此\sigma_X不變 \\(B)\times:有兩個(x_i,y_i)改變，因此相關係數隨之變動\\(C)\bigcirc: r={Cov(x,y)\over \sigma_X\sigma_Y} \Rightarrow X,Y互換，r值不變\\(D)\bigcirc: cov(x,y)=cov(x+5,y)且\sigma(X)=\sigma(X+5)，因此斜率不變\\故選\bbox[red,2pt]{(ACD)}$$

解答：$$(A)\bigcirc: (a,b,c)=(1,2,3-6),(1,3,4-6),(1,4,5-6),(1,5,6);(2,3,4-6),(2,4,5-6),(2,5,6);\\\qquad (3,4,5-6),(3,5,6);(4,5,6);共有(4+3+2+1) +(3+2+1)+(2+1)+1=20種\\ \qquad \Rightarrow 機率為20/6^3=5/54\\(B)\bigcirc: 算法同(A)，共有\sum_{n=1}^6 \sum_{k=1}^n k=21+15+10+6+3+1=56種，機率為56/216=7/27\\ (C)\bigcirc: (a,b,c)=(6,4,1),(6,3,2),(5,5,1),(5,4,2),(5,3,3),(4,4,3)及其排列，\\\qquad 共有6+6+3+6+3+3=27種，機率為27/216=1/8\\ (D)\bigcirc: (a,b,c)=(a,a,c),(a,c,c)，共有36+36=72種，再扣除重複的(a,a,a)有6種，共66種\\\qquad 機率為66/216=11/36\\ 故選\bbox[red,2pt]{(ABCD)}$$

第二部分：綜合題

一、填充題

解答：$$a+\log_2 3,a+\log_4 3,a+\log_8 3成等比\Rightarrow (a+\log_4 3)^2 = (a+\log_2 3)(a+\log_8 3)\\ \Rightarrow (a+{1\over 2}\log_2 3)^2 = (a+\log_2 3)(a+{1\over 3}\log_2 3) \\\Rightarrow a^2+a\log_2 3+{1\over 4}(\log_2 3)^2=a^2+ {4\over 3}a\log_2 3 +{1\over 3}(\log_2 3)^2\\ \Rightarrow {1\over 12}(\log_2 3)^2+ {1\over 3}a\log_2 3=0 \Rightarrow {1\over 12}\log_2 3(\log_2 3+4a)=0  \Rightarrow a=-{1\over 4}\log_2 3 \\ \Rightarrow 公比={a+\log_4 3 \over a+\log_2 3} = {-{1\over 4}\log_2 3+{1\over 2}\log_2 3 \over -{1\over 4}\log_2 3+\log_2 3} ={{1\over 4}\log_2 3\over {3\over 4}\log_2 3} = \bbox[red, 2pt]{1\over 3}$$

解答：$$\lim_{n\to \infty}{1\over n^2}\sum_{k=1}^n \sqrt{n^2-k^2} =\lim_{n\to \infty}{1\over n}\sum_{k=1}^n \sqrt{1-({k\over n})^2} = \int_0^1\sqrt{1-x^2}\;dx\\ =\int_0^{\pi/2} \sqrt{1-\sin^2\theta} \cos\theta \;d\theta =\int_0^{\pi/2} \cos^2\theta \;d\theta ={1\over 2}\int_0^{\pi/2} \cos 2\theta+1\;d\theta ={1\over 2}\left.\left[ {1\over 2}\sin 2\theta+\theta \right]\right|_0^{\pi/2} \\ ={1\over 2}\cdot {\pi\over 2} =\bbox[red,2pt]{\pi \over 4}$$

解答：$$第1種填法:a[x][y]=x+(y-1)\times 29,1\le x\le 29,1\le y\le 17\\第2種填法:b[x][y]=y+(x-1)\times 17,1\le x\le 29,1\le y\le 17\\a[x][y]=b[x][y] \Rightarrow x+29(y-1)=y+17(x-1) \Rightarrow 28y=16x+12 \Rightarrow 14y=8x+6\\ \Rightarrow (x,y)=(1,1), (8,5),(15,9),(22 ,13),(29,17)\\ \Rightarrow a[1][1]+a[8][5]+a[15][9]+a[22][13]+a[29][17]=1+124 +247 +370+ 493= \bbox[red,2pt]{1235}$$註:題目倒數第二行的\(\color{blue}{38}\)應該是\(\color{blue}{34}\)；

解答：$$\cases{A(0,0,0)\\ B(a,a,0)\\ C(a,0,a)\\ D(0,a,a)} \Rightarrow \cases{\triangle ABC 重心M(2a/3,a/3,a/3)\\ \triangle ACD重心N(a/3,a/3, 2a/3)} \\\Rightarrow \overline{MN} =\sqrt{{a^2\over 9} +0+{a^2\over 9}} ={\sqrt 2\over 3}a=2  \Rightarrow a=3 \sqrt 2 \\ 又\cases{ABCD 重心G(2a/4,2a/4,2a/4)=({3\sqrt 2\over 2}, {3\sqrt 2\over 2}, {3\sqrt 2\over 2})\\ E:平面BCD:x+y+z=2a=6\sqrt 2} \\\Rightarrow 球半徑r=d(G,E)= \left|{9\sqrt 2/2-6\sqrt 2\over \sqrt 3}\right|={\sqrt 6\over 2} \\ \Rightarrow 球體積={4\over 3}\pi r^3= {4\over 3}\pi\cdot {6\sqrt 6\over 8} =\bbox[red,2pt]{\sqrt 6\pi}$$

解答：$$\overrightarrow{OP}=(3\cos \alpha+\sin \beta,2\cos \alpha+4\sin \beta)= \cos\alpha(3,2)+\sin \beta(1,4) =\cos\alpha\cdot  \vec u+\sin \beta \cdot\vec v\\ 由\vec u及\vec v所展開的平行四邊形面積=\begin{Vmatrix} 3& 2\\ 1& 4\end{Vmatrix} =10，其中\cases{\vec u=(3,2)\\ \vec v=(1,4)},而\cases{0\le \cos\alpha \le \sqrt 3/2\\ 0\le \sin \beta \le \sqrt 3/2} \\ 因此P點所形成的面積=10\cdot {\sqrt 3\over 2}\cdot {\sqrt 3\over 2} = \bbox[red,2pt]{15\over 2}$$

解答：$$任抽二球=(x,y)，可得獎金如下表：\\ \begin{array}{cc|r}x & y & 獎金\\\hline 1 & 6 & 7\\ 6 & 2& 8\\ 5 & 9 & 14\\ 9 & 7 & 16\\ 4 & 8 &12 \\ 8 & 3 & 11 \\\hdashline 1 & 5 & 6  \\ 5 & 4 & 9\\ 6 & 9 & 15 \\ 9 & 8 & 17 \\2 & 7 & 9 \\ 7 & 3& 10\\\hline \end{array}\\ \Rightarrow 總獎金=7+8+\cdots +10=134 \Rightarrow 期望值=134\times {1\over C^9_2}=\bbox[red,2pt]{67\over 18} $$

解答：$$令\cases{a=x_1-1\\ b=x_2-x_1\\ c=x_3-x_2 \\d=20-x_3} \Rightarrow \cases{0\le a\\ 4\le b \\ 5\le c \\ 0\le d\\a+b+c+d =19}，再令\cases{e=b-4\\ f=c-5}\\，則本題相當於求a+e+f+d=19-4-5=10的非負整數解，共有H^4_{10}\\ \Rightarrow 機率為{H^4_{10}\over C^{20}_3} ={C^{13}_{10}\over C^{20}_3} ={286\over 1140}= \bbox[red,2pt]{143\over 570}$$

解答：

$$\cos \angle A= {(\sqrt 3)^2 +(3\sqrt 3)^2-(\sqrt{21})^2 \over 2\times \sqrt 3\times 3\sqrt 3} ={9\over 18}={1\over 2} \Rightarrow \angle A=60^\circ\\ \cases{正\triangle ACD:\cases{\overline{AC}=3\sqrt 3 \Rightarrow \overline{AG_3}=3\\\angle G_3AC=60^\circ \div 2=30^\circ} \\ 正\triangle ABE:\cases{\overline{AB}=\sqrt 3  \Rightarrow \overline{AG_1}=1\\ \angle G_1AB =60^\circ \div 2=30^\circ}} \Rightarrow \cos \angle G_1AG_3={\overline{AG_1}^2+ \overline{AG_3}^2-\overline{G_1G_3}^2 \over 2\times \overline{AG_1} \times \overline{AG_3}} \\ \Rightarrow \cos (30^\circ+60^\circ +30^\circ)=\cos 120^\circ =-{1\over 2}={10-\overline{G_1G_3}^2\over 6} \Rightarrow \overline{G_1G_3}= \sqrt{13} \\\Rightarrow 正\triangle G_1G_2G_3面積= \bbox[red,2pt]{{13\over 4}\sqrt 3}$$

解答：

$$\sqrt{x^4-3x^2-6x+13} -\sqrt{x^4-x^2 +1} =\sqrt{(x^2-2)^2+ (x-3)^2}- \sqrt{(x^2-1)^2+x^2} \\ =\overline{PA}-\overline{PB}，其中\cases{P(x^2,x) \in \Gamma:x=y^2\\ A(2,3)\\ B(1,0)}\\ 要使此距離差最大，可取L:\overleftrightarrow{AB}與\Gamma 的交點，且距B較近，如上圖；\\此時，\overline{PA}-\overline{PB}=\overline{AB} = \sqrt{1+9}=\bbox[red,2pt]{\sqrt{10}}$$

二、計算證明題

解答：

(1)$$(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))\\= (a^3+ab^2+ac^2-(a^2b+abc+ca^2)) \\\qquad +a^2b+b^3+bc^2-(ab^2+b^2c+abc)\\\qquad +(a^2c+ b^2c+c^3-(abc+bc^2+c^2a))\\= a^3+b^3+c^3-3abc\\ \Rightarrow a^3+b^3+c^3-3abc = \bbox[red,2pt]{(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))}$$(2)$$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-(ab+bc+ca)) \\={1\over 2}(a+b+c)((a-b)^2+(b-c)^2 +(c-a)^2) \ge 0 \Rightarrow a^3+b^3+c^3-3abc \ge 0\\ 令\cases{A=a^2\\ B=b^3\\ C=c^3} \Rightarrow A+B+C-3\cdot \sqrt[3]A \cdot \sqrt[3]B \cdot \sqrt[3]C \ge 0\Rightarrow {A+B+C\over 3} \ge \sqrt[3]{ABC}，\bbox[red,2pt]{故得證}$$(3)$${(a+1)+(b+2)+(c+3)\over 3} ={18+6\over 3}=8\ge \sqrt[3]{(a+1)(b+2)(c+3)} \\ \Rightarrow 8^3=512 \ge (a+1)(b+2)(c+3) \Rightarrow (a+1)(b+2)(c+3) \bbox[red,2pt]{最大值為512}\\，此時a+1=b+2=c+3 \Rightarrow \cases{a=c+2\\ b=c+1} 代入a+b+c=18 \Rightarrow \bbox[red,2pt]{\cases{a=7\\ b=6\\c=5}}$$註:第(3)題應該是求\(\color{blue}{最大值}\)，而不是最小值。

解答：

$$\overline{BC}^2-\overline{AB}^2 =\overline{AC} \times \overline{AB} \Rightarrow \overline{BC}^2 = \overline{AB}(\overline{AB} +\overline{AC}) \Rightarrow {\overline{BC}\over \overline{AB}}={\overline{AB}+\overline{AC} \over \overline{BC}}\\ 因此在\overleftrightarrow{BA}上找一點D，使得\overline{AD}=\overline{AC}(見上圖)，則\triangle ABC \sim \triangle CBD \Rightarrow \angle C=\angle D = \alpha\\ 又\overline{AD}=\overline{AC} \Rightarrow \angle ACD=\angle D=\alpha \Rightarrow \angle D+\angle B+\angle BCD =54+3\alpha=180\\ \Rightarrow \alpha= \angle C=\bbox[red,2pt]{ 42^\circ}$$

 解答：$$\cases{x+{1\over y}=4\\ y+{1\over z}=1\\ z+{1\over x}={7\over 3}} \Rightarrow (x+{1\over y})(  y+{1\over z})( z+{1\over x})=4\cdot 1\cdot {7\over 3}={28\over 3} \\\Rightarrow xyz+(x+{1\over y})+(y+{1\over z})+(z+{1\over x}) +{1\over xyz}={28\over 3} \Rightarrow xyz+4+ 1+ {7\over 3}+{1\over xyz}={28\over 3} \\\Rightarrow xyz+{1\over xyz}=2 \Rightarrow (xyz)^2-2(xyz)+1=0 \Rightarrow xyz=\bbox[red,2pt]{1}$$

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註:學校未公告計算證明題答案，解題僅供參考；其他教甄試題及詳解