113年大直高中教甄-數學詳解

 臺北市立大直高級中學 113 學年度第一次專任教師甄選

一、填充題（48 分）

解答: $$2310=2\times 3\times 5\times 7\times 11, 在5個因數中,\\ \text{Case I }挑3個連乘積當作a,剩下2個作為b及c,有C^5_3=10種可能 \\\text{Case II }挑2個連乘積當作a,剩下3個再挑2個乘積作為b,有C^5_2C^3_2=30種可能 \\共有10+30=\bbox[red, 2pt]{40}種可能$$

解答: $$a_n = S_n-S_{n-1} =(\sqrt{S_n} +\sqrt{S_{n-1}})(\sqrt{S_n} -\sqrt{S_{n-1}}) =\sqrt{S_n} +\sqrt{S_{n-1}} \\ \Rightarrow (\sqrt{S_n} +\sqrt{S_{n-1}})(\sqrt{S_n} -\sqrt{S_{n-1}}-1)=0 \Rightarrow  \sqrt{S_n} =\sqrt{S_{n-1}} +1\\ 取b_n= \sqrt{S_n}, 則b_n=b_{n-1}+1 = b_{n-2}+2 =\cdots = b_1+(n-1) =n \Rightarrow S_n=n^2 \\ \Rightarrow S_{20}-S_{19}+S_{18} =20^2-19^2+18^2 =39+18^2= \bbox[red, 2pt]{363}$$

解答: $$a_n= \begin{vmatrix} n& n+1& 0\\ n+2 & n+1& n+2 \\n+2 & 0 & n+2\end{vmatrix} = \begin{vmatrix} n& n+1& 0\\ 2 & 0& n+2 \\n+2 & 0 & n+2\end{vmatrix} =(n+1)(n+2)^2-2(n+1)(n+2) \\ =n(n+1)(n+1) \Rightarrow {1\over a_k} ={1\over k(k+1)(k+2)} ={1\over 2} \left({1\over k(k+1)}-{1\over (k+1)(k+2)} \right) \\ \Rightarrow \lim_{n\to \infty} \sum_{k=1}^n{1\over a_k} =\lim_{n\to \infty} {1\over 2} \sum_{k=1}^n \left({1\over k(k+1)}-{1\over (k+1)(k+2)} \right) \\=\lim_{n\to \infty} {1\over 2} \left({1\over 2} -{1\over 6}+ {1\over 6}-{1\over 12}+ \cdots-{1\over (n+1)(n+2)} \right)=\lim_{n\to \infty} {1\over 2} \left( {1\over 2}-{1\over (n+1)(n+2)}\right) \\= \bbox[red, 2pt]{1\over 4}$$

解答: $$取\cases{O(0,0)\\ A(0,4) \\B(4,0)} \Rightarrow C(0,2) \Rightarrow D= (4C+3B)/7=({12\over 7},{8\over 7}) \Rightarrow \cases{L_1= \overleftrightarrow{AD}: y=-{5\over 3}x+4 \\ L_2=  \overleftrightarrow{AB}: x+y=4 \\ L_3= \overleftrightarrow{OD}: y={2\over 3}x} \\ \Rightarrow \cases{E=L_1\cap x軸= ({12\over 5},0) \\ F= L_2\cap L_3=({12\over 5}, {24\over 15})} \Rightarrow {S_{\triangle CEF} \over S_{\triangle OAB}} = {\begin{Vmatrix} 0& 2& 1\\ 12/5 & 0 & 1\\ 12/5& 24/15 & 1\end{Vmatrix} \over \begin{Vmatrix} 0& 4& 1\\ 0 & 0 & 1\\ 4& 0 & 1\end{Vmatrix}}={{96\over 25}\over 16} = \bbox[red, 2pt]{6\over 25}$$

解答:

解答:

$$取\cases{A(0,20) \\B(0,0)\\ C(5,0) \\P(5,a)} \Rightarrow L_1=\overleftrightarrow{AP}: y={a-20\over 5}x+20 \Rightarrow Q({100\over 20-a},0) \\ \Rightarrow S_{\triangle ADP} +S_{\triangle PQC} ={1\over 2} \left(5(20-a)+ ({100\over 20-a}-5)a \right) \\ 取f(a)= 5(20-a)+ ({100\over 20-a}-5)a \Rightarrow f'(a)=0 \Rightarrow -10+{100\over 20-a} +{100a\over (20-a)^2} =0 \\ \Rightarrow a^2-40a+200=0 \Rightarrow a= \bbox[red, 2pt]{20-10\sqrt 2}$$

解答: $${x^2\over 9}+{y^2\over 25}=1 \Rightarrow \cases{a=5\\ b=3} \Rightarrow c=4 \\ \Rightarrow 繞長軸(y軸) 旋轉體積= 2\pi \int_0^4 \left( \sqrt{9(1-{y^2\over 25})}\right)^2 \,dy = 2\pi \left. \left[ 9y-{3y^4\over 25} \right] \right|_0^4 = \bbox[red, 2pt]{{1416\over 25} \pi}$$

解答: $$\cases{A(-1,3,2) \\B(3,3,4)} \Rightarrow \cases{\overrightarrow{AB} =(4,0,2) \\ P=\overline{AB}中點=(1,3,3)} \Rightarrow 過P且法向量為\overrightarrow{AB}的平面E_p:2x+z=5 \\ \Rightarrow 球心O在直線L=E_p \cap E上 \Rightarrow O(t, 10-{5\over 2}t,5-2t), t\in \mathbb R \\\Rightarrow 球半徑R= \overline{OA} =\sqrt{ (t+1)^2+(7-{5\over 2}t)^2+ (3-2t)^2}\\ 令f(t)=(t+1)^2+(7-{5\over 2}t)^2+ (3-2t)^2 \Rightarrow f'(t)=0 \Rightarrow t=2 \Rightarrow \cases{R=\sqrt{14} \\O(2,5,1)} \\ \Rightarrow 球方程式: \bbox[red, 2pt]{(x-2)^2+ (y-5)^2+ (z-1)^2=14}$$

二、非選題（52 分）

解答: $$柯西不等式:xy= ((x-z)+z) (z+(y-z)) \ge \left(\sqrt{z(x-z)} +\sqrt{z(y-z)} \right)^2 \\ \Rightarrow \sqrt{xy}\ge \sqrt{z(x-z)} +\sqrt{z(y-z)}, \bbox[red, 2pt]{故得證}$$

解答:

$$直線L:\cases{\sqrt 3x-z-6=0\\ y=0} \Rightarrow \cases{A=L\cap xy平面=(2\sqrt 3, 0,0) \\ B=L\cap z軸= (0,0,-6)}\\ 假設\cases{球心P(0,0,-r), r為球半徑\\ Q為球與L的切點} \Rightarrow \triangle ABO \sim \triangle PBQ (AAA) \Rightarrow {\overline{OA} \over \overline{AB}}  ={\overline{PQ} \over \overline{ PB}} \\ \Rightarrow {2\sqrt 3\over 4\sqrt 3} ={r\over 6-r} \Rightarrow r=2 \Rightarrow P\bbox[red, 2pt]{(0,0,-2)}$$

解答: $$\bbox[red, 2pt]{n=7}, 參考資料:\href{https://www.sec.ntnu.edu.tw/uploads/asset/data/625640e3381784d09345bce2/09-97050.pdf}{中學生通訊解題第五十九期}$$

解答: $$略$$

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解題僅供參考，教甄歷年試題及詳解