國立臺南第二高級中學 113 學年度第一次教師甄選
一、填充題（每題 5 分，共 60 分）
解答：$$G是重心\Rightarrow \overrightarrow{AG}={1\over 3} \overrightarrow{AB} + {1\over 3}\overrightarrow{AC} ,又\cases{\overline{BC}=42\\ \overline{GI}=2\\ \overline{GI} \parallel \overline{BC}} \Rightarrow \overrightarrow{GI}= {1\over 21} \overrightarrow{BC}\\ 因此 \overrightarrow{AI} = \overrightarrow{AG} +\overrightarrow{GI} ={1\over 3} \overrightarrow{AB} + {1\over 3}\overrightarrow{AC} + {1\over 21} \overrightarrow{BC} ={1\over 3} \overrightarrow{AB} + {1\over 3}\overrightarrow{AC} + {1\over 21} \left( -\overrightarrow{AB} + \overrightarrow{AC} \right) \\\qquad ={6\over 21}\overrightarrow{AC} +{8\over 21} \overrightarrow{AC} \Rightarrow \cases{x=6/21=2/7\\ y=8/21} \Rightarrow (x,y)= \bbox[red,2pt]{\left({2\over 7},{8\over 21} \right)}$$
$$假設\cases{\alpha =\log_2 x \ge 0 \\ \beta=\log_2 y \ge 0} \Rightarrow \alpha^2+ \beta^2 =\log_2(4x^4)+ \log_2(8x^8) =2+4\log_2 x+3+8\log_2 x\\ \qquad =5+ 12\alpha \Rightarrow (\alpha-6)^2+\beta^2= 41 為一橢圓，但僅限第一象限(如圖)\\\Rightarrow \cases{\alpha =\sqrt{41} \cos \theta+6\\ \beta=\sqrt{41} \sin \theta} \Rightarrow \log_2(xy)= \log_2 x+\log_2 y=\alpha+\beta =\sqrt{41}(\cos \theta+\sin \theta)+6 \\=\sqrt{82} \sin(\theta+{\pi/4})+6 \Rightarrow \theta={\pi\over 2}時有極大值6+\sqrt{82}\\ 當\alpha=0時，\beta=\sqrt 5 \Rightarrow \alpha+\beta=\sqrt 5為極小值 \Rightarrow M+m=\bbox[red, 2pt]{6+\sqrt{82}+\sqrt 5}$$
解答：$$|z|=1 \Rightarrow z=\cos \theta+i\sin \theta =e^{i\theta} \Rightarrow z^3=e^{3 i\theta} \Rightarrow \overline{z^3} =\cos 3\theta-i\sin3\theta \\ \Rightarrow \left| -1+\sqrt 3i-z\right| =\left| {(-1+\sqrt 3i)^2 \over 2} -\overline{z^3} \right|\\ \Rightarrow |(-1-\cos \theta)+(\sqrt 3-\sin\theta)i| = \left| (-1-\cos 3\theta) +(\sin 3\theta -\sqrt 3)i \right| \\ \Rightarrow (-1-\cos \theta)^2+(\sqrt 3-\sin \theta)^2=(-1-\cos 3\theta)^2+(\sin 3\theta-\sqrt 3)^2 \\ \Rightarrow 2\cos \theta-2\sqrt 3\sin \theta+ 5= 2\cos 3\theta-2\sqrt 3\sin 3\theta +5 \Rightarrow \cos 3\theta-\cos \theta=\sqrt 3(\sin 3\theta-\sin \theta) \\ \Rightarrow -2\sin 2\theta \sin \theta= 2\sqrt 3\cos 2\theta \sin \theta \Rightarrow \sin \theta(\sqrt 3\cos 2\theta+ \sin 2\theta)=0 \\\Rightarrow \cases{\sin \theta=0 \Rightarrow \theta =0 \Rightarrow z=1\\ \tan 2\theta =-\sqrt 3 \Rightarrow \theta=\pi/3 \Rightarrow z={1\over 2}+{\sqrt 3\over 2}i} \Rightarrow z=\bbox[red, 2pt]{1,{1\over 2}+{\sqrt 3\over 2}i}$$
解答：$$\cases{f(a,b)=\sqrt{4a+{1\over 3}}+\sqrt{6b+{1\over 5}} \\ g(a,b)=a+b-2} \Rightarrow \cases{f_a=\lambda g_a\\ f_b= \lambda g_b\\ g=0} \Rightarrow \cases{{2\over \sqrt{4a+{1\over 3}}} =\lambda \\ {3\over \sqrt{6b+{1\over 5}}} =\lambda} \\ \Rightarrow {2\over \sqrt{4a+{1\over 3}}} ={3\over \sqrt{6b+{1\over 5}}} \Rightarrow {4\over 4a+{1\over 3}} ={9\over 6b+{1\over 5}} \Rightarrow 36a-24b+{11\over 5}=0 \\ \Rightarrow 36a-24(2-a)+{11\over 5}=0 \Rightarrow a={229\over 300} \Rightarrow b={371\over 300} \\ \Rightarrow f({229\over 300},{371\over 300}) =\sqrt{{229\over 75}+{1\over 3}} + \sqrt{{371\over 50}+{1\over 5}} ={1\over 5} \left( {1\over 3}+{1\over 2}\right) \sqrt{762} = \bbox[red, 2pt]{\sqrt{762}\over 6}$$
解答：$$f(x)= \int_0^x g(t)\,dt +1 \Rightarrow f(0)=1\\ g(x)=12x^2-6x+ \int_0^1[f(t)+g'(t)]\,dt \Rightarrow g(0)=\int_0^1[f(t)+g'(t)]\,dt \;為一常數\\ \Rightarrow f(x)=\int_0^x [12t^2-6t+ g(0)]\,dt +1=4x^3-3x^2+xg(0)+1 \\ \Rightarrow g(x)=12x^2-6x+ \int_0^1 f(t) \,dt +\int_0^1 g'(t)\,dt \\\qquad = 12x^2-6x+ \int_0^1 (4t^3-3t^2+ tg(0)+1)\,dt +g(1)-g(0)\\ \qquad = 12x^2-6x+\left. \left[ t^4-t^3 +{1\over 2}g(0)t^2 +t\right] \right|_0^1+ g(1)-g(0) \\\qquad =12x^2-6x+1 +g(1)-{1\over 2}g(0) \\\Rightarrow g(x) =12x^2-6x+1 +g(1)-{1\over 2}g(0) \Rightarrow g(1)=7+g(1)-{1\over 2}g(0) \\ \Rightarrow 7-{1\over 2}g(0)=0 \Rightarrow g(0)= \bbox[red, 2pt]{14}$$
解答：$$\lim_{n\to \infty}{(\sqrt 1+\sqrt 8+ \sqrt{27}+ \cdots +\sqrt{n^3})^2 \over n^5} =\lim_{n\to \infty} {1\over n^5}{\left(\sum_{k=1}^n k^{3/2}\right)^2} \\ = \left( \lim_{n\to \infty} {1\over n}\sum_{k=1}^n \left({k\over n} \right)^{3/2}\right)^2 =\left( \int_0^1 x^{3/2}\,dx \right)^2 =\left({2\over 5} \right)^2 =\bbox[red, 2pt]{4\over 25}$$
解答：$$丟一次骰子點數為偶數的機率=2/6=1/3,奇數機率為2/3\\ 假設a_n為擲骰子n次點數和為偶數的機率,則a_n=前n-1次點數和為偶數且第n次也為偶數或\\前n-1次點數和為奇數且第n次也為奇數.因此a_n={1\over 3}a_{n-1} +{2\over 3}(1-a_{n-1}),n\ge 2\\ \Rightarrow a_n={2\over 3}-{1\over 3}a_{n-1} \Rightarrow a_n-{1\over 2}=-{1\over 3}(a_{n-1}-{1\over 2})\\ 取b_n=a_n-{1\over 2} \Rightarrow b_n=-{1\over 3}b_{n-1},b_1=a_1-{1\over 2}={1\over 3}-{1\over 2}=-{1\over 6} \\ \Rightarrow b_n=b_1(-{1\over 3})^{n-1} =-{1\over 6}(-{1\over 3})^{n-1} \Rightarrow a_n={1\over 2}-{1\over 6}(-{1\over 3})^{n-1} ={1\over 2}(1+(-{1\over 3})\cdot (-{1\over 3})^{n-1} \\ \Rightarrow a_n=\bbox[red, 2pt]{{1\over 2}\left(1+(-{1\over 3})^n \right)}$$
解答：$$假設\cases{f(x)=x^{2024}-2x^{427}+ 3x^{113}-4x \\ p(x)= x^3+2x^2+2x+1} \\ p(x)= x^3+x^2+ x^2+2x+1 =x^2(x+1)+(x+1)^2=(x+1)(x^2+x+1)\\ 假設\omega為x^2+x+1=0的根，則\omega^3=1 \Rightarrow f(\omega) = (\omega^3)^{674}\cdot \omega^2-2(\omega^3)^{142} \cdot \omega+3(\omega^3)^{37}\cdot \omega^2-4\omega \\ \Rightarrow f(\omega)=4\omega^2-6\omega =4(-\omega-1)-6\omega=-10\omega-4 \\ \Rightarrow f(x)=p(x)Q(x)+ a(x^2+x+1)-10x-4 \\ \Rightarrow f(-1)=1+2-3+4=a(1-1+1)+10-4 \Rightarrow 4=a+6 \Rightarrow a=-2 \\\Rightarrow 餘式： -2(x^2+x+1)-10x-4= \bbox[red, 2pt]{-2x^2-12x-6}$$
解答：
解答：$$假設C為原點,則\cases{A(0,2\sqrt 3,2\sqrt 3)\\ B(0,4\sqrt 3,0)\\ C(0,0,0)\\ D(4,0,0)} \Rightarrow \cases{\overrightarrow{AD}=(4,-2\sqrt 3,-2\sqrt 3)\\ \overrightarrow{BC} =(0,-4\sqrt 3,0)} \\\Rightarrow \cases{\vec n= \overrightarrow{AD} \times \overrightarrow{BD}=(-24,0,-16\sqrt 3) \\ \overrightarrow{CD}=(4,0,0)} \Rightarrow d( \overleftrightarrow{AD},\overleftrightarrow{BC}) = \overrightarrow{CD}在\vec n上的投影\\ ={|\vec n\cdot \overrightarrow{CD}| \over |\vec n|} ={96\over 8\sqrt{21}}= \bbox[red, 2pt] {{4\over 7}\sqrt{21}}$$
解答：$$f(p)=(p+q)^{10}=\sum_{k=0}^{10}p^kq^{10-k}C^{10}_k \Rightarrow f'(p)=10(p+q)^9= \sum_{k=1}^{10}p^{k-1}q^{10-k}kC^{10}_k \\ \Rightarrow g(p)=pf'(p)=10p(p+q)^9 = \sum_{k= 1}^{10}p^{k}q^{10 -k}kC^{10}_k \\\Rightarrow g'(p)=10(10p+q) (p+q)^8 = \sum_{k= 1}^{10}p^{k-1}q^{10 -k}k^2C^{10}_k\\ \Rightarrow pg'(p)= 10p(10p+q)(p+q)^8 =\sum_{k= 1}^{10}p^{k}q^{10 -k}k^2C^{10}_k \\ \sum_{k=1}^{10}3^{k-1}\cdot 2^{9-k}\cdot k^2C^{10}_k = {1\over 6}\sum_{k=1}^{10}3^{k}\cdot 2^{10-k}\cdot k^2C^{10}_k = {1\over 6}\cdot 10\cdot 3\cdot 32\cdot 5^8 \\ = 10\cdot 2^4\cdot 5^8 =10\cdot (2\cdot 5)^4\cdot 5^4=\bbox[red, 2pt]{625 00000}$$
解答：$${a+b\over a}={\sin B\over \sin B-\sin A}={b\over b-a} \Rightarrow b^2-a^2=ab \Rightarrow a^2+ab-b^2=0 \\ \Rightarrow a={-b+b\sqrt5\over 2} =b\left( {\sqrt 5-1\over 2}\right) \cdots(1) \\ \cos(A-B)+\cos C=1-\cos 2C \Rightarrow \cos(A-B)-\cos (A+B)=2\sin^2 C \\ \Rightarrow 2\sin A\sin B=2\sin^2 C\Rightarrow ab=c^2 \Rightarrow c=\sqrt{ab} =\sqrt{b^2\cdot {\sqrt 5-1 \over 2}} =b \sqrt{\sqrt 5-1\over 2}\cdots(2) \\\text{By (1) and (2), we have } {a+c\over b}= \bbox[red, 2pt]{ \left( {\sqrt 5-1\over 2}\right)+\sqrt{\sqrt 5-1\over 2}}$$
解答：$$A=\begin{bmatrix} -1 & 4 & 2 \\ -1 & 3 & 1 \\-1 & 2 & 2 \end{bmatrix} = \begin{bmatrix}2 & 1 & 2 \\ 1 & 0 & 1 \\0 & 1 & 1 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 2 \end{bmatrix} \begin{bmatrix}1 & -1 & -1 \\ 1 & -2 & 0 \\-1 & 2 & 1 \end{bmatrix} \\ \Rightarrow A^n = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 2^n \end{bmatrix} \begin{bmatrix}1 & -1 & -1 \\ 1 & -2 & 0 \\-1 & 2 & 1 \end{bmatrix} = \bbox[red, 2pt]{ \begin{bmatrix}3- 2^{n+1} & 2^{n+2}-4 & 2^{n+1}-2 \\-2^n+1 & 2^{n+1}-1 & 2^n-1 \\1-2^n & 2^{n+1}-2 & 2^n \end{bmatrix}}$$
解答：$$\cases{(a-b)^2=a^2-2ab +b^2\ge 0\\-1\le \sin x\le 1} \Rightarrow a^2-2ab\sin x+b^2\ge 0, \forall x \\ 取x=\theta+30^\circ \Rightarrow a^2-2ab(\theta+30^\circ)+b^2\ge 0 \Rightarrow a^2-\sqrt 3ab\sin \theta -ab \cos \theta+b^2\ge 0 \\ \Rightarrow a^2-ab\cos \theta+b^2 \ge \sqrt 3ab\sin \theta=2\sqrt 3 \cdot {1\over 2}ab\sin \theta =2\sqrt 3\triangle \\ \Rightarrow 2a^2-2ab\cos \theta+2 b^2\ge 4\sqrt 3\triangle \Rightarrow a^2+b^2+(a^2-2ab\cos \theta+b^2)\ge 4\sqrt 3\triangle \\ \Rightarrow a^2+b^2+c^2 \ge 4\sqrt 3\triangle \Rightarrow {a^2+b^2 +c^2\over \triangle }\ge 4\sqrt 3. \;\bbox[red, 2pt]{QED.}$$
解答：
$$\cases{\sqrt{x^4-4x^2-12x+25} =\sqrt{(x^2-4)^2+(2x-3)^2} =\overline{PA}\\ \sqrt{x^4+2x^2+1} =\sqrt{(x^2-1)^2+(2x-0)^2} =\overline{PB}},其中\cases{P(2x,x^2)\\ A(3,4)\\ B(0,1)} \\ 而P\in \Gamma:y=({x\over 2})^2,\Gamma 為一拋物線,B剛好為焦點,準線L:y=-1 \\ 因此\overline{PB}=d(P,L) \Rightarrow 當\overleftrightarrow{PA}成一垂直線時, \overline{PA} +\overline{PB}=d(A,L)=\bbox[red, 2pt]5為最小值$$
二、計算證明題（每題10 分，共40 分）（需將演算過程寫在答案卷上，並註明題號）
$$P(x,y) \Rightarrow \overrightarrow{OP}=(x,y)=t^2\overrightarrow{OA}+ t\overrightarrow{OB} =(\sqrt 3t^2-t,t^2+\sqrt 3 t) \\ \Rightarrow \begin{bmatrix}x \\y \end{bmatrix}= \begin{bmatrix} \sqrt 3 & -1 \\1 & \sqrt 3 \end{bmatrix} \begin{bmatrix}t^2 \\t \end{bmatrix} = \begin{bmatrix} \sqrt 3/2 & -1/2 \\1/2 & \sqrt 3/2 \end{bmatrix} \begin{bmatrix}t^2/2 \\t/2 \end{bmatrix} =\begin{bmatrix}\cos 30^\circ & -\sin 30^\circ \\\sin 30^\circ & \cos 30^\circ \end{bmatrix} \begin{bmatrix}x' \\y' \end{bmatrix}\\ \qquad 其中\;\Gamma: y'^2=2x' 為一拋物線\\ 因此P點軌跡就是圖形\Gamma 逆時針旋轉30^\circ,而直線L: y=-{1\over \sqrt 3}x逆時針旋轉30^\circ 就是x軸\\ 欲求之圖形面積就上圖紅色區域，而藍色區域就是旋轉前的區域，兩者面積相等\\而藍色區域較易計算，藍色區域為兩圖形\cases{x=-\sqrt 3y\\ x=y^2/2}所夾區域，交點為\cases{O(0,0)\\ A(6,-2\sqrt 3)} \\ \Rightarrow 面積=\int_{-2\sqrt 3}^y \left(-\sqrt 3y-{1\over 2}y^2 \right)\,dy =\bbox[red, 2pt]{2\sqrt 3}$$
解答：$$這是有名的歐拉定理，證明過程可參考維基百科$$
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解題僅供參考，其他歷年試題及詳解
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