國立羅東高級工業職業學校 112 學年度專任教師甄試
一、填充題（請在答案紙上作答，每一題 5 分，共 100 分）：
解答：$$令\cases{a=\sqrt[3]{\sqrt 2+1} \\b= \sqrt[3]{\sqrt 2-1}} \Rightarrow \cases{a^3+b^3=2\sqrt 2\\ ab=1}\\ x(x^2-3)=x^3-3x =(a+b)^3-3(a+b) =(a^3+b^3+3ab(a+b))-3(a+b) \\=2\sqrt 2+3(a+b)-3(a+b) =\bbox[red, 2pt]{2\sqrt 2}$$
解答：$$\left({\vec a\cdot \vec b \over |\vec b|^2} \right)\vec b =\left({\vec c\cdot \vec b \over |\vec b|^2} \right)\vec b \Rightarrow \vec a\cdot \vec b=\vec c\cdot \vec b \Rightarrow 2+3k=4+5k \Rightarrow k= \bbox[red,2pt]{-1}$$
解答：$$\sin A:\sin B:\sin C=3:5:7 \Rightarrow a:b:c= 3:5:7 \Rightarrow \cases{a=3k\\ b=5k \\ c=7k} \\ \Rightarrow \cases{\cos A=(b^2+c^2-a^2)/2bc= 13/14\\ \cos B=(a^2+c^2-b^2)/2ac= 11/4\\ \cos C= (a^2+b^2-c^2)/2ab= -1/2} \Rightarrow \cos A:\cos B:\cos C=13:11:\bbox[red, 2pt]{-7}$$
解答：$$1+(1+r)+(1+r+r^2)+ \cdots+ (1+r+\cdots+r^{n-1}) =\sum_{k=1}^n{1-r^k\over 1-r} \\={1\over 1-r}\sum_{k=1}^n(1-r^k) ={1\over 1-r} \left(n-\sum_{k=1}^n r^k\right)=\begin{cases}{1\over 1-r}(n-{r-r^{n+1}\over 1-r}) & r \ne 1\\1+2+\cdots + n= {n(n+1) \over 2} & r=1\end{cases} \\ =\bbox[red, 2pt]{\begin{cases}{n\over 1-r}-{r(1-r^{n}) \over (1-r)^2} & r \ne 1\\ {n(n+1) \over 2} & r=1\end{cases} }$$
解答：$$1-{R\over R+W}-{R\over R+Y}+{R\over R+W+Y} =1-{2\over 2+3}-{2\over 2+5}+{2\over 2+3+5} =\bbox[red,2pt]{18\over 35}\\ \href{https://math.ntnu.edu.tw/~horng/letter/hpm17010.pdf}{公式來源}$$
解答：$$\lim_{n\to \infty}\left( \sqrt{1+2+\cdots +(n+1)} -\sqrt{1+2+\cdots +n}\right) \\=\lim_{n\to \infty}\left( {(\sqrt{1+2+\cdots +(n+1)} -\sqrt{1+2+\cdots +n})(\sqrt{1+2+\cdots +(n+1)} +\sqrt{1+2+\cdots +n}) \over (\sqrt{1+2+\cdots +(n+1)} +\sqrt{1+2+\cdots +n})}\right) \\ =\lim_{n\to \infty}\left( {n+1 \over (\sqrt{(n+1)(n+2)/2} +\sqrt{n(n+1)/2})}\right) ={1\over {1\over \sqrt 2} +{1\over \sqrt 2}} =\bbox[red,2pt]{\sqrt 2\over 2}$$
解答：$$\sum_{k=1}^\infty {k\over k^4+k^2+1} =\sum_{k=1}^\infty {k\over (k^2+k+1)(k^2-k+1)} = {1\over 2} \sum_{k=1}^\infty \left( {1\over k^2-k+1} -{1\over k^2+k+1 }\right) \\ ={1\over 2}\left( 1-{1\over 3}+{1\over 3}-{1\over 7} +{1\over 7}-{1\over 13}+\cdots\right) =\bbox[red, 2pt]{1\over 2}$$
解答：$$旋轉矩陣A=\begin{bmatrix}\cos 60^\circ & -\sin 60^\circ \\ \sin 60^\circ & \cos 60^\circ \end{bmatrix} =\begin{bmatrix}1/2 & -\sqrt 3/2 \\ \sqrt 3/2 & 1/2 \end{bmatrix} \Rightarrow A\begin{bmatrix} x\\ y\end{bmatrix} =\begin{bmatrix} x'\\ y'\end{bmatrix} \Rightarrow \begin{bmatrix} x\\ y\end{bmatrix} =A^{-1}\begin{bmatrix} x'\\ y'\end{bmatrix} \\ \Rightarrow \begin{bmatrix} x\\ y\end{bmatrix} =\begin{bmatrix}1/2 & \sqrt 3/2 \\ -\sqrt 3/2 & 1/2 \end{bmatrix} \begin{bmatrix} x'\\ y'\end{bmatrix}=\begin{bmatrix} (x'+ \sqrt 3y')/2\\(-\sqrt 3x'+ y')/2\end{bmatrix} 代回原橢圓方程式 \\ \Rightarrow {(x'+\sqrt 3y')^2\over 12} +{ (-\sqrt 3x'+y')^2\over 16}=1 \Rightarrow \bbox[red, 2pt]{13x'^2+ 2\sqrt 3x'y' +15y'^2= 48}$$
解答：$${x^2\over 9}+{y^2\over 4}=1 \Rightarrow {2x\over 9}+{yy'\over 2}=0 \Rightarrow y'=-{4x\over 9y} \\ 假設切點為P(3\cos \theta,2\sin \theta) \Rightarrow 切線斜率y'(3\cos \theta,2\sin \theta) =-{12\cos \theta \over 18\sin \theta} =-{2\cos \theta \over 3\sin \theta} \\ \Rightarrow 切線L:y=-{2\over 3}\cot \theta(x-3\cos \theta)+ 2\sin \theta \Rightarrow L與坐標軸的交點\cases{A(0,{2\over \sin\theta})\\ B({3\over \cos \theta},0)} \\ 柯西不等式：\left( ({2\over \sin \theta})^2 +({3\over \cos \theta})^2\right) (\sin^2 \theta+ \cos^2 \theta) \ge (2+3)^2 \\\Rightarrow \overline{AB}^2={{4\over \sin^2 \theta} +{9\over \cos^2 \theta}} \ge 5^2 \Rightarrow \overline{AB}的最小值= \bbox[red, 2pt ]5$$
解答：$$此題相當於求兩曲線\Gamma_1,\Gamma_2與直線x+y=k(斜率=-1)切點的坐標和\\令\cases{\Gamma_1: y=(1-x)^2 \Rightarrow y'=-2(1-x)=-1 \Rightarrow x=1/2 \Rightarrow y(1/2)=1/4 \Rightarrow x+y=3/4\\ \Gamma_2: y=1-x^2 \Rightarrow y'=-2x =-1 \Rightarrow x=1/2 \Rightarrow y(1/2)=3/4 \Rightarrow x+y=5/4} \\ \Rightarrow (M,m)= \bbox[red,2pt]{({5\over 4},{3\over 4})}$$
解答：$$稜長=a \Rightarrow \cases{A(0,0,0)\\ B(a/\sqrt 2,0, a/\sqrt 2)\\ C(0,a/\sqrt 2, a/\sqrt 2)\\ D(a/\sqrt 2,a/\sqrt 2,0)} \Rightarrow 平面E=\triangle BCD: x+y+z=\sqrt 2a \\ \Rightarrow 正四面體的高h=d(A,E)={\sqrt 2a\over \sqrt 3} =\bbox[red,2pt]{\sqrt 6 a\over 3}$$
解答：$$兩歪斜線L_1,L_2的距離d=1 \Rightarrow 稜長=\sqrt 2 \Rightarrow 體積={\sqrt 2\over 12} \times (\sqrt 2)^3 =\bbox[red ,2pt]{1\over 3}$$
解答：
$$假設\cases{\overleftrightarrow{AB}為x軸 \\ \overline{AB}中點O為原點} \Rightarrow \cases{O(0,0)\\ A(-1/2,0)\\ B(1/2,0)\\ C({1\over \sqrt 2}-{1\over 2},{1\over \sqrt 2}) =({\sqrt 2-1\over 2},{\sqrt 2\over 2})} \\ 又\angle OAP=\theta \Rightarrow \angle AOP=\alpha= 180^\circ-2\theta \Rightarrow P(-{1\over 2}\cos \alpha,-{1\over 2} \sin \alpha)\\ 因此直線L=\overleftrightarrow{AC}: x-y+{1\over 2}=0 \Rightarrow d(P,L)={|{\sqrt 2\over 2} \sin(\alpha-45^\circ)+{1\over 2}| \over \sqrt 2}\\ d(P,L)最大值={\sqrt 2+1\over 2\sqrt 2} \Rightarrow \triangle APC面積最大={\sqrt 2+1\over 4\sqrt 2} ={2+\sqrt 2\over 8}，\\此時\alpha=135^\circ \Rightarrow \theta= (180^\circ-135^\circ)\div 2= 22.5^\circ ={\pi \over 8} \Rightarrow (\theta_0,M) = \bbox[red, 2pt]{({\pi \over 8},{2+\sqrt 2\over 8})}$$
解答：$$A是三位數 \Rightarrow A=10^{2+\log r},1\le r\lt 10，又B是四位數且\log B的尾數是\log A的3倍 \Rightarrow B=10^{3+3\log r}\\ 10000\gt B\gt 8000 \Rightarrow 4\gt \log B\gt 3+3\log 2 \Rightarrow 4\gt 3+3\log r\gt 3+ 3\log 2\\ \Rightarrow 1\gt 3\log r\gt 3\log 2 \Rightarrow 2\lt r\lt \sqrt[3]{10} \approx2.15 \Rightarrow r=2.1(B必須是整數) \\ \Rightarrow A=100\times 2.1=\bbox[red, 2pt]{210}$$
解答：$$本題\bbox[cyan,2pt]{送分}$$
解答：$$\sin^2 9^\circ +\sin^2 36^\circ+ 2\sin 9^\circ \sin 36^\circ\sin 45^\circ ={1-\cos 18^\circ\over 2}+ {1-\cos 72^\circ\over 2} +\sin 36^\circ(2\sin 9^\circ\sin 45^\circ) \\ =1-{1\over 2}(\cos 18^\circ+ \cos 72^\circ)+ \sin 36^\circ(-\cos 54^\circ+\cos 36^\circ)\\=1-{1\over 2}(\cos 18^\circ+ \cos 72^\circ)- \sin 36^\circ \cos 54^\circ+ {1\over 2}\sin 72^\circ\\ =1-{1\over 2}(\cos 18^\circ+ \cos 72^\circ)- \sin 36^\circ \sin 36^\circ+ {1\over 2}\cos 18^\circ=1-{1\over 2}\cos 72^\circ-\sin^2 36^\circ \\=1-{1\over 2}\cos 72^\circ-{ 1-\cos 72^\circ\over 2} =\bbox[red, 2pt]{1\over 2}$$
解答：$$假設六位數中000000-999999，各位數字分別是a_1,a_2,\dots,a_6 \Rightarrow a_1+a_2+ \dots+ a_6 \le 10 \\ \Rightarrow 共有H^7_{10}=8008個(含0，但不含1000000，互相抵消)，需扣除a_i=10的情形\\，因此共有8008-6= \bbox[red, 2pt]{8002}個$$
解答：$$假設x=100 \Rightarrow \cases{101=x+1\\ 102=x+2\\ 7242409= 7x^3+ 24x^2+24x+9}\\ 因此我們可以取f(x)=(7x^3+ 24x^2+24x+9)^{10} =(x+1)(x+2)p(x)+ax+b \\ \Rightarrow \cases{f(-1)=2^{10}=1024=-a+b\\ f(-2)= 1 =-2a+b} \Rightarrow \cases{a=1023\\ b=2047} \\\Rightarrow 7242409^{10}= (101\times 102)p(100)+1023\times 100+2047 \equiv 104347 \mod 101\cdot 102 \\ 而104347 = 101\cdot 102 \cdot 10+1327 \Rightarrow 餘數為\bbox[red, 2pt]{1327}$$
解答：$$2x^6-3x^5+4x^4-3x^3+4x^2-3x+2=(x^2+x+1)(x^2-x+1)(2x^2-3x+2)=0\\ \Rightarrow x=\cases{(-1\pm \sqrt 3i)/2\\ (1\pm \sqrt 3i)/2 \\ (3\pm \sqrt 7i)/4} \Rightarrow 第一象限的根：\bbox[red,2pt]{{1+\sqrt 3i\over 2}, {3+\sqrt 7i \over 4}}$$
解答：$$本題\bbox[cyan,2pt]{送分}$$
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解答：$$A是三位數 \Rightarrow A=10^{2+\log r},1\le r\lt 10，又B是四位數且\log B的尾數是\log A的3倍 \Rightarrow B=10^{3+3\log r}\\ 10000\gt B\gt 8000 \Rightarrow 4\gt \log B\gt 3+3\log 2 \Rightarrow 4\gt 3+3\log r\gt 3+ 3\log 2\\ \Rightarrow 1\gt 3\log r\gt 3\log 2 \Rightarrow 2\lt r\lt \sqrt[3]{10} \approx2.15 \Rightarrow r=2.1(B必須是整數) \\ \Rightarrow A=100\times 2.1=\bbox[red, 2pt]{210}$$
解答：$$本題\bbox[cyan,2pt]{送分}$$
解答：$$\sin^2 9^\circ +\sin^2 36^\circ+ 2\sin 9^\circ \sin 36^\circ\sin 45^\circ ={1-\cos 18^\circ\over 2}+ {1-\cos 72^\circ\over 2} +\sin 36^\circ(2\sin 9^\circ\sin 45^\circ) \\ =1-{1\over 2}(\cos 18^\circ+ \cos 72^\circ)+ \sin 36^\circ(-\cos 54^\circ+\cos 36^\circ)\\=1-{1\over 2}(\cos 18^\circ+ \cos 72^\circ)- \sin 36^\circ \cos 54^\circ+ {1\over 2}\sin 72^\circ\\ =1-{1\over 2}(\cos 18^\circ+ \cos 72^\circ)- \sin 36^\circ \sin 36^\circ+ {1\over 2}\cos 18^\circ=1-{1\over 2}\cos 72^\circ-\sin^2 36^\circ \\=1-{1\over 2}\cos 72^\circ-{ 1-\cos 72^\circ\over 2} =\bbox[red, 2pt]{1\over 2}$$
解答：$$假設六位數中000000-999999，各位數字分別是a_1,a_2,\dots,a_6 \Rightarrow a_1+a_2+ \dots+ a_6 \le 10 \\ \Rightarrow 共有H^7_{10}=8008個(含0，但不含1000000，互相抵消)，需扣除a_i=10的情形\\，因此共有8008-6= \bbox[red, 2pt]{8002}個$$
解答：$$假設x=100 \Rightarrow \cases{101=x+1\\ 102=x+2\\ 7242409= 7x^3+ 24x^2+24x+9}\\ 因此我們可以取f(x)=(7x^3+ 24x^2+24x+9)^{10} =(x+1)(x+2)p(x)+ax+b \\ \Rightarrow \cases{f(-1)=2^{10}=1024=-a+b\\ f(-2)= 1 =-2a+b} \Rightarrow \cases{a=1023\\ b=2047} \\\Rightarrow 7242409^{10}= (101\times 102)p(100)+1023\times 100+2047 \equiv 104347 \mod 101\cdot 102 \\ 而104347 = 101\cdot 102 \cdot 10+1327 \Rightarrow 餘數為\bbox[red, 2pt]{1327}$$
解答：$$2x^6-3x^5+4x^4-3x^3+4x^2-3x+2=(x^2+x+1)(x^2-x+1)(2x^2-3x+2)=0\\ \Rightarrow x=\cases{(-1\pm \sqrt 3i)/2\\ (1\pm \sqrt 3i)/2 \\ (3\pm \sqrt 7i)/4} \Rightarrow 第一象限的根：\bbox[red,2pt]{{1+\sqrt 3i\over 2}, {3+\sqrt 7i \over 4}}$$
解答：$$本題\bbox[cyan,2pt]{送分}$$
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