111年新北樟樹高中教甄-數學詳解

新北市立樟樹國際實創高級中等學校 111 學年度教師甄選筆試

【數學科專業能力】試題

解答：$$若我們將三角陣列左右各加一個1，則第n列的數字就是(x+y)^{n+1}的係數，\\因此第2022列的數字就是(x+y)^{2023}的係數，而原三角陣列第2022列總和=2^{2023}-2;\\ 而2^{20}=576 \mod 1000、2^{40} = 776 \mod 1000、2^{80}=176 \mod 1000、\\2^{160} =976 \mod 1000、2^{320}= 576\mod 1000 \Rightarrow 循環數=4\\ 因此2^{2023}= 2^{1280}\cdot 2^{640}\cdot 2^{80}\cdot 2^{20}\cdot 2^3 = 176\cdot 776\cdot 176  \cdot 576\cdot 8 \mod 1000 =608\mod 1000\\ \Rightarrow 2^{2023}-2 =\bbox[red, 2pt]{606} \mod 1000$$

解答：$$a_n= \sum_{k=1}^n\cfrac{3k}{\sqrt{n^4+4k^2n^2}} = \sum_{k=1}^n\cfrac{1}{ n}\cdot \cfrac{3k }{\sqrt{n^2+4k^2 }} = \sum_{k=1}^n\cfrac{1}{ n}\cdot \cfrac{3k/n }{\sqrt{1+4(k/n)^2 }} \\ \Rightarrow \lim_{n\to \infty} a_n = \int_0^1 \cfrac{3x}{\sqrt{1+4x^2}}\,dx =\left.\left[ {3\over 4} \sqrt{1+4x^2}\right] \right|_0^1 = \bbox[red, 2pt]{{3\over 4}(\sqrt 5-1)}$$

解答：$$x=2022: 剩下1-2021作取2個，有C^{2021}_2取法;\\ x=2021: 剩下1-2020作取2個，有C^{2020}_2取法;\\ \cdots\\x=3: 剩下1-2作取2個，有C^{2}_2取法;\\因此期望值=\sum_{k=2}^{2021} (k+1)C^k_2/C^{2022}_3 =\cfrac{1}{C^{2022}_3}\sum_{k=2}^{2021} (k+1)\cfrac{k(k-1)}{2} =\cfrac{1}{2C^{2022}_3}\sum_{k=2}^{2021}(k^3-k)\\= \cfrac{3}{ 2022\cdot 2021\cdot 2020}\left(\cfrac{2021^2\cdot 2022^2}{4} -\cfrac{2021\cdot 2022}{2}\right) \\=\cfrac{3\cdot 2021\cdot 2022}{4\cdot 2020} -\cfrac{3}{2\cdot 2020} =\cfrac{3\cdot 2021\cdot 1011-3}{2\cdot 2020} =\bbox[red,2pt]{6069\over 4}$$

解答：

$$假設 圓心O ，則\angle AOB: \angle BOC: \angle COD: \angle DOE: \angle EOF:\angle FOA= 5:5: 7:7:5:7  \\ \Rightarrow \cases{ \angle AOB=\angle BOC = \angle EOF=50^\circ \\ \angle COD=\angle DOE =\angle FOA=70^\circ} \Rightarrow \angle BAF=  120^\circ \\\Rightarrow \cos \angle BAF= \cos 120^\circ = \cfrac{5^2+7^2- \overline{BF}^2}{2\cdot 5\cdot 7} =-{1\over 2} \Rightarrow \overline{BF}^2=109  \\ \Rightarrow 六角形ABCDEF面積= 正\triangle BDF+ 3\times \triangle ABF ={\sqrt 3\over 4}\cdot \overline{BF}^2 +{3\over 2} \cdot 5\cdot 7\cdot\sin 120^\circ \\ ={109\over 4}\sqrt 3 +{105\over 4}\sqrt 3 =\bbox[red,2pt]{{107\over 2}\sqrt 3}$$

解答：$$A=\begin{bmatrix} -\sqrt 3& 1\\ -1 & -\sqrt 3\end{bmatrix} =-2\begin{bmatrix} \sqrt 3/2 & -1/2 \\ 1/2 & \sqrt 3/2 \end{bmatrix} =-2\begin{bmatrix} \cos 30^\circ & -\sin 30^\circ \\ \sin 30^\circ & \cos 30^\circ \end{bmatrix} \\ \Rightarrow A^{99} = (-2)^{99}\begin{bmatrix} \cos 2970^\circ & -\sin 2970^\circ \\ \sin 2970^\circ & \cos 2970^\circ \end{bmatrix} = (-2)^{99}\begin{bmatrix} \cos 90^\circ & -\sin 90^\circ \\ \sin 90^\circ & \cos 90^\circ \end{bmatrix}\\ = (-2)^{99}\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2^{99} \\ -2^{99} & 0 \end{bmatrix} =\begin{bmatrix} a & b \\ c & d \end{bmatrix} \Rightarrow \cases{a=d=0\\ b=2^{99}\\ c=-2^{99}} \\\Rightarrow \log_4{ad-bc\over a+b-c-d} = \log_4{2^{198} \over 2^{100}} =\log_4 2^{98} =\log_4 4^{49} = \bbox[red,2pt]{49}$$

解答：$$令\cases{L_1:x-y=0\\ L_2: x-y= 8\\ L_3:3x+y= 0 \\L_3:3x+y = k} \Rightarrow d(L_1,L_2) =d(L_3,L_4) \Rightarrow {8\over \sqrt 2} ={|k|\over \sqrt {10}} =|k|=8\sqrt 5\\ \Rightarrow k=-8\sqrt 5(菱形包含原點)\\ 又\cases{A=L_1\cap L3 =(0,0)\\ B=L_2 \cap L_3 = (2,-6)} \Rightarrow 邊長a=\overline{AB}= 2\sqrt{10} \Rightarrow {k\over a}={-8\sqrt 5 \over 2\sqrt{10}} =\bbox[red, 2pt]{-2\sqrt 2}$$

解答：

(1)$$\cases{A(1,2,0)\\ B(4,2,-3)\\ C(1,8,3)\\ H(a,b,c)} \Rightarrow \cases{\overrightarrow{AB}= (3,0,-3)\\ \overrightarrow{AC} =(0,6,3) \\ \overrightarrow{AH}=(a-1,b-2,c)} \Rightarrow \vec n= \overrightarrow{AB} \times \overrightarrow{AC} = (18,-9,18)\\ \overrightarrow{AH}= {2\over 3}\overrightarrow{AB}-{1\over 3}\overrightarrow{AC}+ 2 (\overrightarrow{AB} \times \overrightarrow{AC}) \Rightarrow (a-1,b-2,c)= (2,0,-2)-(0,2,1)+ (36,-18,36)\\ =(38,-20,33) \Rightarrow 四面體體積= {1\over 6}\begin{Vmatrix} 3 & 0 & -3\\ 0 & 6 & 3\\ 38 & -20 & 33\end{Vmatrix} ={1458\over 6} =\bbox[red,2pt]{243}$$(2)$$\overrightarrow{AH}=(a-1,b-2,c)= (38,-20,33) \Rightarrow \cases{a=39\\ b=-18 \\ c=33} \Rightarrow H(39,-18,33) \\E:18(x-1)-9(y-2)+18z) =0 \Rightarrow 2x-y+2z=0 \\ 令H'(d,e,f)，則\cases{\overrightarrow{HH'}= (39-d,-18-e,33-f) \parallel \vec n\\ H''=\overline{HH'}中點= ((39+d)/2,(-18+e)/2, (33+f)/2)) \in E} \\ \Rightarrow \cases{39-d:-18-e:33-f = 2:-1:2 \cdots(1)\\ 39+d-(-18+e)/2+33+f=0 \cdots(2)} ，由(1)可得\cases{d=39-2t\\ e=t-18\\ f=33-2t} 代入(2)\Rightarrow t=36\\ \Rightarrow \cases{d=-33\\ e=18\\ f=-39} \Rightarrow H'=\bbox[red,2pt]{(-33,18,-39)}$$(3)$$投影點H''= (H+H')/2 =(3,0,-3) \Rightarrow \overrightarrow{AH''}= (2,-2,-3) =\alpha \overrightarrow{AB}+ \beta\overrightarrow{AC} =(3\alpha,6\beta,-3\alpha+ 3\beta)\\ \Rightarrow \cases{\alpha=2/3\\ \beta=-1/3}; \bbox[red, 2pt]{由於-1/3 \lt 0，因此投影點不在\triangle 內部；}$$

解答：

$$(1) D(x)= S(x) \Rightarrow -{x^2\over 10}+270 = {x^2\over 20}+3x+45 \Rightarrow {3\over 20}x^2+3x-225=0\\ \Rightarrow x^2+20x -1500=0 \Rightarrow (x-30)(x+50) =0 \Rightarrow x=30 \Rightarrow D(30)=S(30) =180 \\ \Rightarrow 均衡點= \bbox[red, 2pt]{(30,180)}\\(2) 上圖著色區域面積=\int_0^{30} 180-S(x)\,dx =5400-3150= \bbox[red,2pt]{2250}$$ 

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