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:$$888888=13\times 68376 \Rightarrow 2015 \equiv 5 \mod 6 \Rightarrow 2015個8除以13的餘數=5個8除以13的餘數\\88888=6837\times 13+7 ,故選\bbox[red,2pt]{(B)}$$
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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:$$a_{n+1}= {1\over 3}a_n+2 \Rightarrow a_n={1\over 3}a_{n-1}+2 ={1\over 3}\left( {1\over 3}a_{n-2}+2 \right)+2 = ({1\over 3})^2a_{n-2}+ 2(1+{1\over 3}) \\ =({1\over 3})^3a_{n-3}+ 2(1+{1\over 3} +{1\over 3^2}) =\cdots =({1\over 3})^{n-1}a_{1}+ 2(1+{1\over 3} +{1\over 3^2}+\cdots +{1\over 3^{n-2}}) \\ =({1\over 3})^{n-1}+2(1+{1\over 3} +{1\over 3^2}+\cdots +{1\over 3^{n-2}})\\ \Rightarrow \lim_{n\to \infty}a_n = 0+2\cdot {3\over 2} =3,故選\bbox[red,2pt]{(D)}$$
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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:
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:$$令\cases{a=x_1-1\\ b=x_2-x_1\\ c=x_3-x_2 \\d=20-x_3} \Rightarrow \cases{0\le a\\ 4\le b \\ 5\le c \\ 0\le d\\a+b+c+d =19},再令\cases{e=b-4\\ f=c-5}\\,則本題相當於求a+e+f+d=19-4-5=10的非負整數解,共有H^4_{10}\\ \Rightarrow 機率為{H^4_{10}\over C^{20}_3} ={C^{13}_{10}\over C^{20}_3} ={286\over 1140}= \bbox[red,2pt]{143\over 570}$$
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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。
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您好:請問第4題的第2行(a+1)(b+1)(c+1)是不是應該為1?謝謝
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