國立中興大學附屬高級中學115學年度第1次正式教師甄試
第壹部分、填充題(共 10 題， 占 60 分)
解答:$$|z|=1 \Rightarrow z= \cos \theta+ i\sin \theta \Rightarrow z^2+2z-2= \cos 2\theta+i\sin 2\theta+2\cos \theta+i2\sin \theta-2 \\ \Rightarrow | z^2+2z-2|= |\cos2\theta+2\cos \theta-2+i(\sin 2\theta+2\sin \theta)| \\= \sqrt{(\cos 2\theta+2\cos \theta-2)^2+(\sin 2\theta+2\sin \theta)^2} = \sqrt{9-4\cos \theta-4\cos 2\theta} \\=\sqrt{-8\cos^2\theta-4\cos \theta+13} = \sqrt{-8(\cos\theta+ {1\over 4})^2+{27\over 2}} \Rightarrow 最大值為\sqrt{27\over 2}= \bbox[red, 2pt]{3\sqrt 6\over 2}$$
解答:$$f(x)=(x^2+x+1)^5= \sum_{n=0}^{10} a_nx^n \Rightarrow g(x)=xf(x)= x(x^2+x+1)^5= \sum_{n=0}^{10} a_nx^{n+1} \Rightarrow \\ \Rightarrow g'(x)=(x^2+x+1)^5+5x(x^2+x+1)^4(2x+1) =\sum_{n=0}^{10}(n+1)a_n x^n \\ \Rightarrow g'(1) =3^5+ 15\cdot 3^4=11a_{10}+ 10a_9+ \cdots +a_0 \\ \Rightarrow 11a_{10}+ 10a_9+ \cdots+ a_0=3^5(1+5) =2\cdot 3^6= \bbox[red, 2pt]{1458}$$
解答:$$假設\cases{\overline{BC} =a\\ \overline{AC} =b\\ \overline{AB}=c } \Rightarrow \cases{ \overrightarrow{AB} \cdot \overrightarrow{AC} =bc\cos A =bc \cdot \displaystyle {b^2+c^2-a^2 \over 2bc}=54\\\overrightarrow{BA} \cdot \overrightarrow{BC} =ac\cos B= ac\cdot \displaystyle {a^2+c^2-b^2 \over 2ac}=10 \\\overrightarrow{CA} \cdot \overrightarrow{CB} =ab\cos C =ab\cdot \displaystyle {a^2+b^2-c^2\over 2ab} =90} \\ \Rightarrow \cases{b^2+c^2-a^2=108\\ a^2+c^2-b^2=20\\ a^2+b^2-c^2=180} \Rightarrow a^2+b^2+c^2=308 \Rightarrow \cases{308-2a^2=108\\ 308-2b^2=20\\ 308-2c^2=180} \Rightarrow \cases{a=10\\ b=12\\ c=8} \\ \cos A={54\over bc} ={9\over 16} \Rightarrow \sin A={5\sqrt 7\over 16} \Rightarrow \triangle ABC={1\over 2}bc\sin A={1\over 2}\cdot 96\cdot {5\sqrt 7\over 16} = \bbox[red, 2pt]{15\sqrt 7}$$
解答:$$(x^2+x+1)+(2x^2+x+1) +2 \sqrt{(x^2+x+1)(2x^2+x+1)} =x^2-3x+13 \\ \Rightarrow 2\sqrt{(x^2+x+1)(2x^2+x+1)}=-2x^2-5x+ 7\\ \Rightarrow 4(x^2+x+1)(2x^2+x+1) =(-2x^2-5x+ 7)^2 \\ \Rightarrow 4x^4-8x^3+35x^2+94x-29=0 \Rightarrow (2x^2-7x+29)(2x^2+3x-1)=0 \\ 2x^2-7x+29=0無實數解\Rightarrow 2x^2+3x-1=0 \Rightarrow x= \bbox[red, 2pt]{-3+\sqrt{17} \over 4} \; ({-3-\sqrt{17}\over 4}\lt 0不合)$$
解答:$$假設P=所有相異兩數乘積的總和 \Rightarrow (1+2+\cdots +n)^2=(1^2+2^2+ \cdots+ n^2)+2P \\ \Rightarrow \left( {n(n+1)\over 2} \right)^2 = {n(n+1)(2n+1) \over 6}+2P \Rightarrow P= {n(n-1)(n+1)(3n+2) \over 24} \\ 已知{P\over {n(n-1)\over 2}} =55 \Rightarrow P={55\over 2}n(n-1) = {n(n-1)(n+1)(3n+2) \over 24} \\ \Rightarrow {(n+1)(3n+2) \over 12}=55 \Rightarrow 3n^2+5n-658=0 \Rightarrow (n-14)(3n+47)=0 \\ \Rightarrow n= \bbox[red, 2pt]{14}$$
解答:$$假設\cases{h_n= \overline{P_{2n-1}P_{2n}} :第n次向左的水平長度\\ v_n= \overline{P_{2n}P_{2n+1}}:第n次向上的垂直長度} \Rightarrow 已知h_1= \overline{P_1P_2}=60 \\ 假設P_{2n-1}=(x,y) \in L_1 \Rightarrow x+3y=k_1 \Rightarrow 向左走到P_{2n}(x-h_n,y) \in L_2 \Rightarrow 2(x-h_n)+y=k_2 \\ \Rightarrow P_{2n}(x-h_n,y)向上走到P_{2n+1}(x-h_n, y+v_n) \in L_1 \Rightarrow (x-h_n)+3(y+ v_n)=k_1 \\ \Rightarrow (x+3y)-h_n+3v_n=k_1 =x+3y \Rightarrow v_n={1\over 3}h_n \\P_{2n+1}(x-h_n, y+v_n)向左走h_{n+1}到P_{2n+2} (x-h_n-h_{n+1}, y+v_n) \in L_2 \\\Rightarrow 2(x-h_n-h_{n+1})+( y+v_n)=k_2 \Rightarrow 2(x-h_n)+y-2h_{n+1}+v_n=k_2 =2(x-h_n)+y \\ \Rightarrow h_{n+1}={1\over 2}v_n ={1\over 2} \left( {1\over 3}h_n \right) = {1\over 6}h_n; 同理可得 v_{n+1}={1\over 6}v_n\\ \langle h_n\rangle, \langle v_n\rangle 皆為公比{1\over 6}的等比級數, 其中\cases{h_1=60\\ v_1={1\over 3}h_1=20} \Rightarrow \cases{ \sum_{n=1}^\infty h_n ={60\over 1-1/6}=72\\ \sum_{n=1}^\infty v_n={20\over 1-1/6}=24} \\ \Rightarrow \sum_{k=1}^\infty \overline{P_{k}P_{k+1}} =72+24= \bbox[red, 2pt]{96}$$
解答:$$四個頂點共有C^4_2=6條連線,每條連線有2種情況:連線/不連線 \Rightarrow 共有2^6=64種情況\\每台電腦都能夠從其他所有電腦收到到訊息，稱為\textbf{連通}\\ \textbf{連線數為6條(連通)}:只有C^6_6=1種情形\\ \textbf{連線數為5條(連通)}:有C^6_5=6種情形\\ \textbf{連線數為4條(連通)}:有C^6_4=15種情形\\\textbf{連線數為3條(不一定)}:若三條線構成一個三角形，則有一頂點不連通，共C^4_3=4種三角形\\ \qquad 連通情形共有C^6_3-4=16種\\\textbf{連線數2條以下(皆不連通)}:0\\連通總數=1+6+15+16=38 \Rightarrow 機率={38 \over 64} = \bbox[red, 2pt]{19\over 32}$$
解答:$$\cases{O(0,0,0) \\A(1,0,0) \\B(1,1,0) \\C(0,1,0) \\ D(0,0,1)\\E(1,0,1) \\F(1,1,1)\\ G(0,1,1)} \Rightarrow P={1\over 3}(2C+G) =(0,1,1/3) \Rightarrow \cases{E_1= 平面ADF: x-y+z=1\\ E_2=平面OABC:z=0} \\ \Rightarrow H=({5\over 9},{4\over 9},{8\over 9}) \Rightarrow d(H,E_2)= \bbox[red, 2pt]{8\over 9}$$
解答:$$\overline{AC}=a \Rightarrow \overline{AB} =2a \Rightarrow S={1\over 2}(a+2a+6)=3+{3\over 2}a \Rightarrow \triangle ABC面積= \sqrt{s(s-a)(s-2a)(s-6)}\\ \Rightarrow f(s)=s(s-a)(s-2a)(s-6) =-{9\over 16}a^4+{90\over 4}a^2-81 \Rightarrow f'(s)=0 \Rightarrow 180a-9a^3=0 \\ \Rightarrow 9a(20-a^2)=0 \Rightarrow a=2\sqrt 5 \Rightarrow \cos A={a^2+4a^2-36\over 4a^2} ={64\over 80} =\bbox[red, 2pt]{4\over 5}$$
解答:$$\cases{\theta=45^\circ\\ R=3} \Rightarrow 代入公式 V={2\over 3}R^3 \tan \theta={2\over 3}\cdot 3^3 \tan 45^\circ= \bbox[red, 2pt]{18} \\ \href{https://chu246.blogspot.com/2020/12/blog-post.html}{公式證明}$$
第貳部分、 計算題(共 4 題， 占 40 分)
解答:$$\cases{f(a,b,c)=(a+{1\over a})^2+(b+{1\over b})^2 +(c+{1\over c})^2 \\g(a,b,c)=a+b+c-4} \Rightarrow \cases{f_a=\lambda g_a \\f_b=\lambda g_b \\f_c=\lambda g_c \\g=0} \Rightarrow \cases{2(a-1/a^3) =\lambda\\ 2(b-1/b^3) =\lambda\\ 2(c-1/c^3) =\lambda\\ a+b+c=4} \\ \Rightarrow a=b=c={4\over 3} \Rightarrow f(4/3,4/3,4/3) =3 \left( {4\over 3}+{3\over 4} \right)^2= \bbox[red, 2pt]{625\over 48} \\ 這題根本不用計算, a,b,c完全對稱, 極值發生在a=b=c$$
解答:$$假設\cases{狀態1(S_1):當天有車\\ 狀態2(S_2):當天無車,但前一天有車\\ 狀態3(S_3):連續兩天無車} \Rightarrow 轉移矩陣P= \begin{bmatrix}3/4&1/4 & 0\\ 1/2& 0& 1/2\\ 0& 0& 1 \end{bmatrix} = \begin{bmatrix}Q&R\\0& I \end{bmatrix} \\ \Rightarrow Q= \begin{bmatrix}3/4& 1/4\\ 1/2&0 \end{bmatrix} \Rightarrow N=(I-Q)^{-1}= \begin{bmatrix}1/3& -1/4\\-1/2& 1 \end{bmatrix}^{-1} = \begin{bmatrix}8& 2\\4& 2 \end{bmatrix} \\ \Rightarrow \begin{bmatrix}8& 2\\4& 2 \end{bmatrix} \begin{bmatrix}1\\ 1 \end{bmatrix} = \begin{bmatrix}10\\ 6 \end{bmatrix} \Rightarrow \cases{從S_1(有車)開始達到連續兩天無車的期望步數是10步\\ 從S_2(無車)開始達到連續兩天無車的期望步數是6步} \\ \Rightarrow 依題意欲求之X期望值為\bbox[red, 2pt]{10}$$
解答:$$\textbf{(1) }\cases{\overrightarrow{AB} \cdot \overrightarrow{AC} =6\times 4\times \cos 60^\circ =12\\ \overrightarrow{AB} \cdot \overrightarrow{AP} =6\times 2\times \cos 30^\circ =6\sqrt 3\\ \overrightarrow{AC} \cdot \overrightarrow{AP} =4\times 2\times \cos 30^\circ = 4\sqrt 3\\ |\overrightarrow{AP}|^2 =2^2=4} \Rightarrow \cases{\overrightarrow{PB} =\overrightarrow{AB}-\overrightarrow{AP} \\\overrightarrow{PC} =\overrightarrow{AC}-\overrightarrow{AP} } \\ \Rightarrow \overrightarrow{PB} \cdot \overrightarrow{PC} = (\overrightarrow{AB}-\overrightarrow{AP}) \cdot (\overrightarrow{AC} -\overrightarrow{AP}) =\overrightarrow{AB} \cdot \overrightarrow{AC}- \overrightarrow{AB} \cdot \overrightarrow{AP}- \overrightarrow{AP} \cdot \overrightarrow{AC}+ |\overrightarrow{AP}|^2 \\=12-6\sqrt 3-4\sqrt 3+4= \bbox[red, 2pt]{16-10\sqrt 3} \\\textbf{(2) }由於x,y,z \ge 0且x+y+z=2 \Rightarrow 動點K所成的集合的面積=\triangle BPC \\ \triangle BPC= \triangle ABC-\triangle ABP-\triangle ACP ={1\over 2}\cdot 6\cdot 4\sin 60^\circ-{1\over 2}\cdot 6\cdot 2\sin 30^\circ-{1\over 2}\cdot4\cdot 2\cdot \sin 30^\circ \\=6\sqrt 3-3-2= \bbox[red, 2pt]{6\sqrt 3-5}$$
解題僅供參考，其他教甄試題及詳解
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