110年竹科實中教甄-數學詳解

國立新竹科學園區實驗高級中等學校110學年度第1次教師甄選

一、填充題

解答： $$正\triangle ABC三頂點及P分別為\cases{A(z_1) \\ B(z_2)\\ C(z_3) \\ P(z_0)}，其中 \cases{z_1=\cos 0^\circ+i\sin 0^\circ \\z_2 =\cos 120^\circ +i\sin 120^\circ \\ z_3= \cos 240^\circ +i\sin 240^\circ \\ z_0 =\cos \theta +i\sin \theta} \\ \Rightarrow z_1,z_2,z_3 為x^3-1=0的三根 \Rightarrow f(x)= x^3-1 =(x-z_1)(x-z_2)(x-z_3) \\ \Rightarrow f(z_0) =(z_0-z_1) (z_0-z_2) (z_0-z_3) =z_0^3-1 =\cos 3\theta+i\sin 3\theta -1 \\ \Rightarrow |(x-z_1)(x-z_2)(x-z_3)| =|\cos 3\theta+i\sin 3\theta -1| \\\Rightarrow |(x-z_1)||(x-z_2)||(x-z_3)| = \sqrt{(\cos 3\theta-1)^2 +\sin^2 3\theta } \\ \Rightarrow \overline{PA} \times \overline{PB} \times \overline{PC} = \sqrt{2-2\cos 3\theta} \Rightarrow 當\cos 3\theta =-1時，有最大值 \sqrt{2-(-2)} =\bbox[red, 2pt]{2}$$

解答： $$S=\{1,2,\dots,2021\}有2021個元素，因此S的子集合個數有2^{2021}，\\其中一半的子集合元素和為奇數，另一半為偶數；\\因此元素和為奇數的子集合個數為2^{2021}\div 2 =\bbox[red, 2pt]{2^{2020}}$$

解答： $$\langle a_n\rangle 為等比，假設  \cases{a_1:首項\\ r:公比}；依題意{1\over a_1}- {1\over a_2} ={2\over a_3} \Rightarrow  {1\over a_1}- {1\over a_1r} ={2\over a_1r^2} \Rightarrow r^2-r-2=0 \\ \Rightarrow (r-2)(r+1)=0 \Rightarrow r=2 (r=-1 \Rightarrow S_6= a_1-a_1+a_1-a_1+a_1-a_1=0\ne 63)  \\\Rightarrow S_6={a_1(1-r^6)\over 1-r}=63  \Rightarrow  a_1=1 \Rightarrow \langle a_n\rangle =\langle 2^{n-1}\rangle；\\ b_n = (\log_2 a_n +\log_2 a_{n+1})\div 2 = (n-1+n)\div 2 =n-{1\over 2} \\ \langle (-1)^nb_n^2 \rangle 前2n項的和= \left( b_2^2+b_4^2 + \cdots +b_{2n}^2\right)-  \left( b_1^2+b_3^2 +\cdots +b_{2n-1}^2\right) \\ =\sum_{k=1}^n b_{2k}^2 -\sum_{k=1}^n b_{2k-1}^2 =\sum_{k=1}^n \left(b_{2k}^2 -  b_{2k-1}^2  \right)= {1\over 4}\sum_{k=1}^n \left((4k-1)^2 -  (4k-3)^2  \right) \\= {1\over 4}\sum_{k=1}^n \left( (8k-4)\cdot 2  \right) =  \sum_{k=1}^n (4k-2) =2n(n+1)-2n = \bbox[red, 2pt]{2n^2}$$

解答： $$g(x)=\int_x^{x+1} f(t)dt = x^3 \Rightarrow g'(x) = f(x+1)-f(x)= 3x^2 \cdots(1)\\ 將(1)代入f(x)=ax^3+bx^2 +cx +d \Rightarrow 3ax^2+(3a+2b)x +(a+b+c) = 3x^2\\ \Rightarrow \cases{a=1\\ b=-3/2\\ c=1/2}；又g(0) =\int_0^1 f(t)\;dt=0 \Rightarrow \left. \left[ {1\over 4}x^4-{1\over 2}x^3 +{1\over 4}x^2 +dx \right] \right|_0^1=0 \Rightarrow d-0=0 \\ \Rightarrow d=0 \Rightarrow f(x)= \bbox[red, 2pt]{x^3-{3\over 2}x^2 +{1\over 2}x}$$

解答： $$\cases{2022!! =2022\cdot 2020\cdot 2018\cdots 2 \\n!! 整除 2022!!} \Rightarrow  n=2022,2020,\dots, 2，共1011個\\ 又1011!!= 1011\cdot 1009 \cdot 1007 \cdots 1 \Rightarrow 2\cdot 1011!! = 2022 \cdot 2018 \cdot 2014\cdots 2\\ \Rightarrow 1011!!整除2022!! \Rightarrow n=1011,1009,\dots,1共506個，也符合要求\\ 因此共有1011+506= \bbox[red, 2pt]{1517}個正整數n符合要求$$

解答： $$小立方體有3個面在大立方體的外面，也就是在裡面的機率為{1\over 2}；\\ 總共有4個小立方體的一個面是染色的，因此所有染色面在大立方體內的機率為({1\over 2})^4= \bbox[red, 2pt]{1\over 16}$$

解答： $$\left(S_n\right)^2 =\sum_{k=1}^n\left(a_k\right)^3 \Rightarrow (a_1+a_2 +\cdots +a_n)^2 = a_1^3 + a_2^3 + \cdots + a_n^3\\有一個公式: 1^3+2^3 +\cdots +n^3 = \left({n(n+1)\over 2}\right)^2 =(1+2+\cdots +n)^2\\因此，  a_n=n \Rightarrow S_{109} = 109\times 110 \div 2 = \bbox[red, 2pt]{5995}$$

解答：

$$令\overline{EF}中點為G、\overline{PH} \bot \overline{GD}，見上圖；因此 \overline{EG}= {1\over 2}\overline{EF}= {1\over 2}\overline{PD}= {1\over 2}\overline{BC} = {1\over 2}\sqrt{5}；\\ 直角\triangle PGE \Rightarrow \overline{PG}= \sqrt{5^2-(\sqrt 5/2)^2} ={\sqrt{95} \over 2} ；同理\overline{DG}={\sqrt {95}\over 2}；\\ \triangle DPB\Rightarrow \cos \angle PDG ={\overline{PD}^2 + \overline{DG}^2 -\overline{PG}^2 \over 2\times \overline{DP}\times \overline{DG}} ={5\over  \sqrt 5\times \sqrt{95}} ={1\over \sqrt{19}} \\ \Rightarrow \sin \angle PDG = {3\sqrt 2\over \sqrt{19}}\Rightarrow h= \overline{PH} =\overline{PD}\sin \angle PDG =\sqrt 5\times {3\sqrt 2\over \sqrt{19}} ={3\sqrt{10}\over \sqrt{19}} \\ \Rightarrow 四面體P-DEF體積= {1\over 3}\times \triangle DEF \times h ={1\over 3}\times \left( {1\over 2}\times \overline{EF}\times \overline{DG}\right) \times {3\sqrt{10}\over \sqrt{19}} \\ = {1\over 3}\times \left( {1\over 2}\times \sqrt{5} \times {\sqrt {95}\over 2}\right) \times {3\sqrt{10}\over \sqrt{19}} =\bbox[red, 2pt]{5 \sqrt{10}\over 4}$$

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