國立臺南第一高級中學 111 學年度第 2 次教師甄選
第一部分︰填充題
解答：$$\cos 4 \theta = 2\cos^2(2\theta)-1 ={1\over 3} \Rightarrow \cos^2(2\theta)={2\over 3} \Rightarrow \cos(2\theta) ={\sqrt 6\over 3} \Rightarrow 2\cos^2 \theta -1= {\sqrt 6\over 3} \\ \Rightarrow \cos^2\theta = {\sqrt 6+3\over 6} \Rightarrow \sin^2\theta = 1-{\sqrt 6+3\over 6}={3-\sqrt 6\over 6} \Rightarrow \sin^6 \theta +\cos^6\theta \\= \left({3-\sqrt 6\over 6} \right)^3 +\left({3+\sqrt 6\over 6} \right)^3 = \bbox[red,2pt]{3\over 4}$$
解答：$$沒中獎或中獎1次的機率為({2\over 3})^5 +C^5_1{1\over 3}\cdot ({2\over 3})^4= {32\over 243} +{80\over 243} ={112\over 243} \\ \Rightarrow 正確判定=中獎超過1次=1-{112\over 243} = \bbox[red, 2pt]{131\over 243}$$
解答：$$\begin{array} {} \min \{B\} & B & \#\{B\} & A & \#\{A\} & 數量\\\hline 6 & \{6\} & 1 & \{1-5\}的非空子集 & 2^5-1=31 & 31\times 1=31\\ 5 & \{5\},\{5,6\}& 2& \{1-4\}的非空子集 & 2^4-1=15 & 15\times 2=30\\ 4 & \{4\}\cup(\{5,6\}的子集 & 2^2=4 & \{1-3\}的非空子集& 2^3-1=7 & 7\times 4=28\\ 3 & \{3\} \cup (\{4,5,6\}的子集 & 2^3=8 &\{1-2\}的非空子集 & 2^2-1=3 & 3\times 8=24\\ 2 & \{2\} \cup (\{3-6\}的子集 & 2^4=16 & \{1\} & 1 & 1\times 16=16\\\hline\end{array} \\ \Rightarrow 共有31+30+28+24+16= \bbox[red, 2pt]{129}種情形$$
解答：
$$假設\cases{O為原點\\ \overline{OB}=a}，則\cases{B(a,0)\\ P(a,1)}；作\overline{MP}\parallel \overline{OB}及\overline{AQ} \bot\overline{MP}，其中\cases{M \in \overline{OA} \\ Q \in \overline{OB}\\ N=\overline{AQ}\cap \overline{MP}}，見上圖；\\因此\cases{\overline{AN}= 2\sin 30^\circ = 1\\ \overline{NP} =2\cos 30^\circ =\sqrt 3} \Rightarrow \overline{AQ}= 2 \Rightarrow \overline{OQ}= 2/\tan 60^\circ = 2/\sqrt 3\\ \Rightarrow a=\overline{OB}= 2/\sqrt 3+\sqrt 3={5\over 3}\sqrt 3 \Rightarrow \cases{A({2\sqrt 3\over 3},2) \\ B({5\over 3}\sqrt 3,0) \\ P({5\over 3}\sqrt 3,1)} \Rightarrow ({5\over 3}\sqrt 3,1) =\alpha({2\over 3}\sqrt 2,2) +\beta ({5\over 3}\sqrt 3,0)\\ \Rightarrow \cases{2\alpha +5\beta =5\\ 2\alpha =1 } \Rightarrow \cases{\alpha=1/2 \\ \beta=4/5} \Rightarrow (\alpha,\beta )=\bbox[red, 2pt]{(1/2,4/5)}$$
解答：$$abcd=100的倍數\Rightarrow (a,b,c,d)=(5,5,c,d),cd=4的倍數\\ \begin{array}{}a& b& c& d& 排列數\\\hline 5 & 5 & 1 & 4 & 4!/2 =12\\ 5 & 5 & 2 & 2 & 6\\ 5 & 5 & 2 & 4 & 12\\ 5 & 5 & 2 & 6 & 12\\ 5 & 5 & 3 & 4 & 12\\ 5 & 5 & 4 & 4 & 6\\ 5 & 5 & 4 & 5 & 4!/3!=4\\ 5 & 5 & 4 & 6 & 12\\ 5 & 5 & 6 & 6 & 6\\\hline \end{array} \Rightarrow 共有12\times 5+6\times 3+4 =82 \Rightarrow 機率={82 \over 6^4} =\bbox[red, 2pt]{41\over 648}$$
解答：
$$\angle BAC=120^\circ \Rightarrow \angle BOC=120^\circ \Rightarrow \cos \angle BOC =-{1\over 2}= {1^2+1^2-\overline{BC}^2 \over 2\cdot 1\cdot 1} \Rightarrow \overline{BC}=\sqrt 3\\ 令\cases{\overline{AB}=b\\ \overline{AC}= c}，則\cos \angle BAC=-{1\over 2} ={b^2+c^2-3\over 2bc} \Rightarrow b^2+bc+c^2-3=0\\ \text{利用 lagrange 算子求極值: }令\cases{f(b,c)=2b+3c\\ g(b,c)= b^2+bc+c^2-3}，\cases{f=\lambda g\\ g=0} \Rightarrow \cases{f_b=\lambda g_b\\ f_c= \lambda g_c\\ g=0}\\ \Rightarrow \cases{2 =\lambda (2b+c)\cdots (1) \\ 3=\lambda(b+2c) \cdots(2)\\ b^2+bc+c^2-3=0\cdots(3)}\\ {(1)\over {(2)}} \Rightarrow {2\over 3}={2b+c\over b+2c} \Rightarrow c=4b 代入(3) \Rightarrow b={1\over \sqrt 7} \Rightarrow c={4\over \sqrt 7} \Rightarrow 2b+3c={14\over \sqrt 7} =\bbox[red, 2pt]{2\sqrt 7}$$
解答：$$x^3+ dx^2+e x+1=0的三根為\cases{\alpha=(a+ b\sqrt 2i)/3\\ \beta= (a-b\sqrt 2i)/3 \\ c} \Rightarrow \alpha\beta c=-1 \Rightarrow c\cdot {a^2+2b^2\over 9} =-1\\ \Rightarrow c(a^2+2b^2)=-9 \Rightarrow \begin{array} {} a^2+2b^2 & c & a & b& \\\hline 9 & -1 & \pm 3 & 0\\ & & \pm 1 & \pm 2\\\hdashline 3 & -3 & \pm 1& \pm 1\\\hdashline 1 & -9 & \pm 1& 0\\\hline\end{array} \\ f(x)=x^3 +dx^2 +ex +1 = (x-\alpha)(x-\beta)(x-c) \Rightarrow d+e = f(1)-2 = (1-\alpha)(1-\beta)(1-c) -2\\ = (1-{2\over 3}a+{1\over 9}(a^2+2b^2))(1-c)-2=8-2 = \bbox[red, 2pt]{6}為最大值，此時a=-1,b=\pm 1,c=-3;$$
解答：$$令\cases{f(x,y)=(x^3+1)(y^3+1)\\ g(x,y)= x+y-1}，因此\cases{f= \lambda g\\ g=0} \Rightarrow \cases{f_x=\lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{3x^2(y^3+1)= \lambda \cdots(1)\\ 3y^2(x^3+1)= \lambda \cdots(2)\\ x+y=1 \cdots(3)} \\由(1)及(2) \Rightarrow 3x^2(y^3+1) = 3y^2(x^3+1) \Rightarrow x^2y^2(x-y)-(x^2-y^2)=0 \Rightarrow (x-y)(x^2y^2-1)=0\\ \Rightarrow \cases{x=y \cdots(4)\\ xy=1 \cdots(5)\\ xy=-1 \cdots(6)} \Rightarrow \cases{(4)\cap (3) \Rightarrow x=y=1/2 \\ (5)\cap (3) =\varnothing\\ (6)\cap (3) \Rightarrow (x,y)=({1\pm \sqrt 5\over 2},{1\mp \sqrt 5\over 2})} \Rightarrow \cases{f(1/2,1/2) = 81/64\\ f({1\pm \sqrt 5\over 2},{1\mp \sqrt 5\over 2}) = 4(最大值)}\\ \Rightarrow x^2+y^2 =({1+ \sqrt 5\over 2})^2 +({1- \sqrt 5\over 2})^2 =\bbox[red, 2pt]{3}$$
解答：$$f(n)=\cfrac{(\sqrt 1+\sqrt 2+\cdots +\sqrt n)^2(1^3+2^3 +\cdots +n^3)}{ (\sqrt[3] 1+\sqrt[3]2 +\cdots +\sqrt[3] n )^3 (1^2 +2^2 +\cdots +n^2)} =\cfrac{\left(\sum_{k=1}^n \sqrt k \right)^2}{\left(\sum_{k=1}^n \sqrt[3] k \right)^3} \cdot \cfrac{\sum_{k=1}^n k^3}{\sum_{k=1}^n k^2}\\ =\cfrac{\left(\sqrt n\sum_{k=1}^n \sqrt {k\over n} \right)^2}{\left(\sqrt[3] n\sum_{k=1}^n \sqrt[3] {k\over n} \right)^3} \cdot \cfrac{\left({n(n+1)\over 2} \right)^2}{n(n+1)(2n+1)\over 6} =\cfrac{\left( \sum_{k=1}^n \sqrt {k\over n} \right)^2}{\left( \sum_{k=1}^n \sqrt[3] {k\over n} \right)^3} \cdot \cfrac{3n(n+1)}{2(2n+1)} \\=\cfrac{n^2\left( \sum_{k=1}^n {1\over n}\sqrt {k\over n} \right)^2}{n^3\left( \sum_{k=1}^n {1\over n} \sqrt[3] {k\over n} \right)^3} \cdot \cfrac{3n(n+1)}{2(2n+1)} =\cfrac{ \left( \sum_{k=1}^n {1\over n}\sqrt {k\over n} \right)^2}{ \left( \sum_{k=1}^n {1\over n} \sqrt[3] {k\over n} \right)^3} \cdot \cfrac{3(n+1)}{2(2n+1)} \\ \Rightarrow \lim_{n\to \infty }f(n)=\cfrac{\left(\int_0^1 \sqrt x\right)^2}{\left(\int_0^1 \sqrt[3] x\right)^3} \cdot \lim_{n\to \infty }\cfrac{3(n+1)}{2(2n+1)} =\cfrac{\left({2\over 3}\right)^2}{\left({3\over 4}\right)^3} \cdot {3\over 4} =\bbox[red,2pt]{64\over 81}$$
解答：
$$f(x)=(x+1)(x-1)(2x+a(a+3)) \Rightarrow f(x)=0的三根為-1,1,及-a(a+3)/2\\ 又 f(x)\ge 0, \forall |x|\le 1，因此y=f(x)的圖形為右上左下，第三根-{a(a+3)\over 2}\ge 1 \\ \Rightarrow a^2+3a+2 \le 0 \Rightarrow \bbox[red, 2pt]{-2\le a\le -1}$$
解答：$$p=({1\over 1}-{1\over 2})+ ({1\over 3}-{1\over 4})+\cdots +({1\over 2021}-{1\over 2022})\\ =({1\over 1}+{1\over 3}+\cdots +{1\over 2021})-({1\over 2}+{1\over 4}+\cdots +{1\over 2022}) \\ =({1\over 1}+{1\over 2}+{1\over 3}+\cdots +{1\over 2021}+ {1\over 2022})-2({1\over 2}+{1\over 4}+\cdots +{1\over 2022}) \\ =({1\over 1}+{1\over 2}+{1\over 3}+\cdots +{1\over 2021}+ {1\over 2022}) -({1\over 1}+{1\over 2}+\cdots +{1\over 1011}) \\ ={1\over 1012}+{1\over 1013}+\cdots +{1\over 2022}\\ =({1\over 1012}+{1\over 2022}) +({1\over 1013} +{1\over 2021}) +\cdots +({1\over 1516}+{1\over 1518})+{1\over 1517}\\ ={3034\over 1012\times 2022} +{3034\over 1013\times 2021} +\cdots +{3034\over 1516\times 1518}+{3034\over 2\times 1517\times 1517}\\ q={1\over 1012\times 2022} +{1\over 1013\times 2021}+\cdots+{1\over 1516\times 1518} +{1\over 1517\times 1517}+{1\over 1518\times 1516}+\cdots {1\over 2022\times 1022}\\ =2\left( {1\over 1012\times 2022} +{1\over 1013\times 2021}+\cdots+{1\over 1516\times 1518}+{1\over 2\times 1517\times 1517}\right)\\ \Rightarrow {p\over q} ={3034\over 2}=\bbox[red, 2pt]{1517}$$
解答：
$$由\cases{A(6,3)\\ B(2,6) } \Rightarrow \cases{C(-1,2)\\ D(3,-1)\\ 邊長=\overline{AB}=5} \Rightarrow 正方形對角線交點P(5/2,5/2) \Rightarrow \overline{AP}=5/\sqrt 2\\ \Rightarrow 圖形|x|+|y| \lt 5\sqrt 2 四個頂點A'B'C'D'(上圖藍色區域)與正方形ABCD有相同面積(上圖棕色)；\\現在我們要旋轉A'B'C'D再平移讓A'B'C'D'與ABCD重疊；\\旋轉的角度就是\overline{AC}與x軸的角度，即 \theta = \tan^{-1}1/7 \\\Rightarrow \cases{\cos \theta =7/5\sqrt 2\\ \sin \theta =1/5\sqrt 2} \Rightarrow \cases{A'(5/\sqrt 2,0) \to A''(7/2,1/2)\\ B'(0,5/\sqrt 2) \to B''(-1/2,7/2}\\ \Rightarrow \overleftrightarrow{A''B''}: 6x+8y=25 \Rightarrow A''B''C''D'' 區域|7x+y|+|x-7y|\lt 25；\\ 中心點(0,0)平移至P(5/2,5/2) \Rightarrow |7(x-5/2)+y-5/2|+|x-5/2-7(y-5/2)|\lt 25 \\ \Rightarrow |7x+6y-20|+|x-7y+15|=25 \Rightarrow |{7\over 25}x +{1\over 25}y-{20\over 25}|+|{1\over 25}x-{7\over 25}y+{15\over 25}|\lt 1 \\ \Rightarrow |c_1|+|c_2|= {20\over 25}+{15\over 25}= \bbox[red, 2pt]{7\over 5}$$
$$\cases{L_1=\overleftrightarrow{AC}:{x-3\over -2} ={y+1\over 2} ={z+7\over 1} \\L_2=\overleftrightarrow{HF}:{x\over 1} ={y \over 4} ={z \over -3} } \Rightarrow \cases{L_1方向向量 \vec u=(-2,2,1)\\ L_2 方向向量\vec v=(1,4,-3)} \Rightarrow \vec n =\vec u\times \vec v=(-10,-5,-10)\\ 又\cases{P\in L_1\\ Q\in L_2} \Rightarrow \cases{P(-2t+3,2t-1,t-7)\\ Q(s,4s,-3s)} \Rightarrow \overrightarrow {PQ}=(s+2t-3,4s-2t+1,-3s-t+7) \parallel \vec n \\ \Rightarrow {s+2t-3\over 2}= {4s-2t+1\over 1}= {-3s-t+7\over 2} \Rightarrow \cases{s=1\\ t=2} \Rightarrow \cases{P(-1,3,-5)\\ Q(1,4,-3)} \Rightarrow \overline{PA}=6 =\overline{PB}=\overline{PC}\\ 假設L_1與L_2的夾角為\theta \Rightarrow \cos\theta = {\vec u\cdot \vec v\over |\vec u||\vec v|} = {1\over \sqrt{26}} \Rightarrow \cases{{1\over 26}=\cfrac{6^2+6^2-\overline{AB}^2}{ 2\cdot 6\cdot 6} \\ -{1\over 26}= \cfrac{6^2+6^2-\overline{AD}^2}{2\cdot 6\cdot 6}} \\\Rightarrow \cases{\overline{AB}^2 = 72(\sqrt{26}-1)/\sqrt{26}\\ \overline{AD}^2 =72(\sqrt{26}+1)/\sqrt{26}} \Rightarrow ABCD面積= \sqrt{{72(\sqrt{26}-1)\over \sqrt{26}}\cdot {72(\sqrt{26}+1)\over \sqrt{26}}} =\sqrt{72^2\cdot 5^2\over 26} \\ ={72\cdot 5\over \sqrt{26}} =\bbox[red, 2pt]{{180\over 13}\sqrt{26}}$$
$$\cases{O(0,0)\\ A(x,y)\\ B(m,n) } \Rightarrow \cases{\overrightarrow{OA} =(x,y)\\ \overrightarrow{OB}=(m,n)} \Rightarrow \overrightarrow{OA} +\overrightarrow{OB} =(x+m,y+n)\\ 因此(m-4)^2 +(n-4)^2=4 \Rightarrow B在圓C上，其中圓C:(x-4)^2 +(y-4)^2 =4\\又 \sqrt{m^2+n^2 }+\sqrt{x^2+y^2} =\sqrt{(m+x)^2 +(n+y)^2} \Rightarrow |\overrightarrow{OB}| +|\overrightarrow{OA}| = |\overrightarrow{OB} + \overrightarrow{OA}| \\ \Rightarrow O、A、B三點在一直線上 \Rightarrow A 在上圖正方形紅線範圍內\\先求過原點O(0,0)的切線: y=ax \Rightarrow (x-4)^2 +(ax-4)^2 =4 \Rightarrow (a^2+1)-(8a+8)x +28=0\\ \Rightarrow 判別式=0 \Rightarrow 64(a+1)^2-112(a^2+1)=0 \Rightarrow a={4\pm \sqrt 7\over 3} \\ \Rightarrow S=\left(y={4+\sqrt 7\over 3}x \right) \cap (y=1) \Rightarrow S=({4-\sqrt 7\over 3},1) \Rightarrow \bbox[red, 2pt]{{4-\sqrt 7\over 3}\le x\le 1}$$
解答：$$由於a_n\in \mathbb{N}，因此逐一試試a_1，只要能找出a_1，其他a_n就能被推算出；\\由 a_i\lt a_j, i\lt j \Rightarrow a_{a_1}= 4,a_1只能是1,2,3,4 \Rightarrow \cases{a_1=1 \Rightarrow a_{a_1} =a_1=1\ne 4，矛盾\\ a_1=2 \Rightarrow a_{a_1} =a_2= 4\\ a_1= 3\Rightarrow a_{a_1}= a_3=4，嚴格遞增數列，a_2無解\\ a_1=4 \Rightarrow a_{a_1} =a_4= 4 \Rightarrow a_1=a_4=4，違反嚴格遞增 } \\ 因此a_1=2 \Rightarrow a_2= 4 \Rightarrow a_{a_2} =2\times 4 \Rightarrow a_4=8 \Rightarrow a_{a_4} =4\times 4 \Rightarrow a_{8}= \bbox[red, 2pt]{16}$$
第二部分︰計算題
解答：$$略:繼續加油$$
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老師你好 關於填充七 你剛好答案對 但是數對不合 因為b不等於0
回覆刪除a=-3,b=0,c=-1 →(a^2+2b^2)=9,(1-c)=2 →(1+2+1)(2)-2=6 沒錯啊!!!
刪除因為題目規定b不等於0
刪除喔! 對耶，已修訂，謝謝!!!
刪除