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2021年5月31日 星期一

110年新竹中學教甄-數學詳解

國立新竹高級中學110學年度第1學期第 1 次教師甄選

一、填答題:(50%)

解答:$$六組成對:只有1種情形\\五組成對、1組不成對:不可能,0種情形\\四組成對、2組不成對:C^6_4 \times 2=30種情形\\ 三組成對、3組不成對: C^6_3\times 8 =160種情形\\ 二組成對、4組不成對:C^6_2\times 60=900種情形\\ 共有1+0+30+ 160+ 900 = \bbox[red, 2pt]{1091}種情形\\ 依題意:Aa,aA都算成對;2組不成對的例子,只有2種情形:(Ab,Ba),(AB,ab)$$
解答:$$假設西瓜尺寸分別為1,2,3,4,5,瑪莉由左往右採西瓜,能夠採到最大5的情形如下:\\\begin{array}{}1 & 2 & 3 & 4 & 5 & num \\\hline \bigcirc & \bigcirc & 5 & \bigcirc & \bigcirc & 4!=24\\ \hdashline \bigcirc & 4 &  \bigcirc &5  & \bigcirc & 3!=6 \\\ 4& \bigcirc &  \bigcirc &5  & \bigcirc & 3!=6 \\ \bigcirc & 3 & \bigcirc & 5 & 4 & 2!=2\\ 3 &\bigcirc &   \bigcirc & 5 & 4 & 2!=2\\\hdashline \bigcirc & 4 & \bigcirc & \bigcirc &5 & 3!=6 \\ 4 & \bigcirc   & \bigcirc & \bigcirc &5 & 3!=6 \\ \hline \end{array}\\ 共有24+6\times 4+2\times 2=52種情形,最大值5會被選到;機率為{52\over 5!} ={52\over 120} =\bbox[red, 2pt]{13\over 30}$$
解答:$$S_n=2a_n-1\Rightarrow a_1= S_1=2a_1-1 \Rightarrow a_1=1\\ 又a_n= S_n-S_{n-1}=(2a_n-1) -(2a_{n-1}-1)=2a_n-2a_{n-1} \Rightarrow a_n= 2a_{n-1} \Rightarrow a_n  =2^{n-1}a_1=2^{n-1};\\ 因此b_n=a_{n-1} +b_{n-1} =2^{n-2} +(a_{n-2} +b_{n-2}) =2^{n-2} + 2^{n-3}+ (a_{n-3}+b_{n-3}) = \cdots \\ =2^{n-2} + 2^{n-3}+\cdots +2+(a_1+b_1) =(2^{n-2} + 2^{n-3}+\cdots +2+1) +3 = 2^{n-1}+2\\ \Rightarrow \sum_{k=1}^n b_k =\sum_{k=1}^n (2^{k-1}+2) =2n + \sum_{k=1}^n2^{k-1}= \bbox[red, 2pt]{2n+2^n-1}$$
解答:$$\sin 20^\circ = \sqrt 3 \cos 40^\circ +\sin x^\circ \\\Rightarrow \sin x^\circ =\sin 20^\circ - \sqrt 3 \cos 40^\circ   =\cos 70^\circ - \sqrt 3 \cos 40^\circ =\cos (30^\circ +40^\circ) - \sqrt 3 \cos 40^\circ  \\=\cos 30^\circ \cos 40^\circ -\sin 30^\circ \sin 40^\circ -\sqrt 3\cos 40^\circ ={\sqrt 3\over 2}\cos 40^\circ -{1\over 2}\sin 40^\circ -\sqrt 3\cos 40^\circ \\ =-{\sqrt 3\over 2}\cos 40^\circ -{1\over 2}\sin 40^\circ = \sin 240^\circ\cos 40^\circ + \cos 240^\circ \sin 40^\circ =\sin 280^\circ =\sin 260^\circ\\ \Rightarrow x=\bbox[red, 2pt]{280^\circ 或260^\circ}$$
解答:$$\cases{\vec a=\overrightarrow{OA} =(3,3,1)\\ \vec b= \overrightarrow{OB}=(4,2,0)\\ \vec c=\overrightarrow{OC}=(3,-6,-9) \\ \vec h=\overrightarrow{OH}= (m,n,p)} \Rightarrow \cases{{\vec a\cdot \vec h\times {\vec h\over |\vec h|^2}}=\vec h \\{\vec b\cdot \vec h\times {\vec h\over |\vec h|^2}}=2\vec h \\{\vec c\cdot \vec h\times {\vec h\over |\vec h|^2}}=3\vec h} \Rightarrow \cases{\vec a\cdot \vec h= |\vec h|^2 \\ \vec b\cdot \vec h= 2|\vec h|^2 \\ \vec c\cdot \vec h= 3|\vec h|^2 } \\ \Rightarrow \cases{3m+3n+p=k\\ 4m+2n=2k\\ 3m-6m-9p=3k},\text{where }k=|\vec h|^2 \Rightarrow \cases{m=3k/4\\ n=-k/2\\ p=k/4} \Rightarrow {9\over 16}k^2 +{1\over 4}k^2+{1\over 16}k^2 =k \\ \Rightarrow k={8\over 7} \Rightarrow |\vec h| =\sqrt k=\sqrt{8\over 7} =\bbox[red, 2pt]{4\over \sqrt{14}}$$
解答

$${x^2 \over 36 }+{y^2\over 32}=1 \Rightarrow \cases{a=6\\ b=4\sqrt 2} \Rightarrow \cases{c=2\\ 焦點F_1=(-2,0)\\ 焦點F_2(2,0)}\\ \overleftrightarrow{BF_2}交橢圓於P點,即為所求,則\overline{PA}+ \overline{PB} = \overline{PF_1}+ \overline{PF_2}- \overline{BF_2}=2a-\overline{BF_2} =12-5=\bbox[red, 2pt]{7}$$
解答:$$令\cases{\vec b=\overrightarrow{AB} \\\vec c=\overrightarrow{AC} \\\vec a=\overrightarrow{AO}  };由於O是\triangle ABC 外心\Rightarrow \cases{\vec a \cdot \vec b ={1\over 2}|\vec b|^2 \\ \vec a \cdot \vec c ={1\over 2}|\vec c|^2} \Rightarrow \cases{(\vec b +2\vec c) \cdot \vec b ={1\over 2}|\vec b|^2 \\ (\vec b +2\vec c) \cdot \vec c ={1\over 2}|\vec c|^2} \\ \Rightarrow \cases{\vec b\cdot \vec c=-{1\over 4}|\vec b|^2 \\ \vec b\cdot \vec c=-{3\over 2}|\vec c|^2} \Rightarrow {|\vec b|\over |\vec c|}=\sqrt 6\\餘弦定理: \vec b\cdot \vec c=|\vec b||\vec c|\cos \theta \Rightarrow -{1\over 4}|\vec b|^2 =|\vec b||\vec c|\cos \theta \Rightarrow \cos \theta =-{1\over 4}{|\vec b|\over |\vec c|} =-{\sqrt 6\over 4} \\ \Rightarrow \sin \theta = \bbox[red, 2pt]{\sqrt {10}\over 4}$$
解答:$$f(x)=x^3 +ax^2 +bx+c=0,由於 f(x)為三次式,因此f(x)=0至少有一實根;\\又各根的絕對值皆為1,實根為\pm 1;若為三實根,三實根之和可能為\pm 3或\pm 1,不可能是-2;\\因此f(x)=0 的根為一實根,兩虛根;\\若實根為1,虛根為a\pm bi,a,b\in \mathbb{R},且|a\pm bi|=1 \Rightarrow a^2+b^2=1;\\\qquad 三根之和為1+2a=-2 \Rightarrow a=-3/2 \Rightarrow b^2=1-{9\over 4} \lt 0 \Rightarrow b\not \in \mathbb{R},不合;\\若實根為-1,  三根之和為-1+2a=-2 \Rightarrow a=-1/2 \Rightarrow b^2=1-{1\over 4}  \Rightarrow b=\pm{\sqrt 3\over 2}\\ 因此f(x)=0之三根為-1,-{1\over 2}\pm {\sqrt 3\over 2}i \Rightarrow f(x)=(x+1)(x^2+x+1) = x^3+2 x^2 +2x+1\\ \Rightarrow (a,b,c)=\bbox[red, 2pt]{(2,2,1)}$$
解答:$$依題意\cases{Arg({z+k\over z})={\pi\over 6}\\ Arg({z+2k\over z+k})={\pi\over 4}} \Rightarrow \cases{{z+k\over z} =ae^{\pi i/6} \\ {z+2k \over z+k}= 2-{z\over z+k}=be^{\pi i/4}} \Rightarrow 2-(ae^{\pi i/6})^{-1} = be^{\pi i/4} \\ \Rightarrow 2-{1\over a}e^{-\pi i/6} =2-{1\over a}(\cos {\pi \over 6}-i\sin{\pi\over 6}) =(2-{\sqrt 3\over 2a})+ {1\over 2a}i = be^{\pi i/4} \\ \Rightarrow 2-{\sqrt 3\over 2a}={1\over 2a} \Rightarrow a={\sqrt 3+1 \over 4} \Rightarrow {z+k \over z}=1+{k\over z} ={\sqrt 3+1 \over 4}e^{\pi i/6} ={\sqrt 3+1 \over 4}({\sqrt 3\over 2}+{1\over 2}i) \\={3+\sqrt 3\over 8}+ {\sqrt 3+1 \over 8}i  \Rightarrow {k\over z} ={3+\sqrt 3\over 8}+ {\sqrt 3+1 \over 8}i-1 = \bbox[red, 2pt]{{\sqrt 3-5\over 8}+ {\sqrt 3+1 \over 8}i}$$
解答:$$a_{n+2} =\sqrt{a_{n+1}a_n} \Rightarrow \log a_{n+2} ={1\over 2}(\log a_{n+1}+\log a_n)\\ 令b_n=\log_2 a_n,則\cases{b_1=\log_2 1= 0\\ b_2=\log_2 {1\over 2}=-1 \\b_n={1\over 2}(b_{n-1}+ b_{n-2})} \Rightarrow 特徵方程式 x^2-{1\over 2}x-{1\over 2}=0 \Rightarrow 2x^2-x-1=0 \\ \Rightarrow (2x+1)(x-1)=0 \Rightarrow x=1,-{1\over 2} \Rightarrow b_n = C_1\cdot 1^n+C_2(-{1\over 2})^n\\ 將初始值代入,\cases{b_1=0\\ b_2=-1} \Rightarrow \cases{b_1=C_1-{1\over 2}C_2=0 \\ b_2=C_1+{1\over 4}C_2=-1} \Rightarrow \cases{C_1=-2/3\\ C_2=-4/3} \Rightarrow b_n=-{2\over 3}-{4\over 3}(-{1\over 2})^n\\ \Rightarrow \lim_{n\to \infty} a_n= 2^{-2/3} ={1\over \sqrt[3]{4}} = \bbox[red, 2pt]{\sqrt[3] 2\over 2}$$

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  1. https://www.physics.harvard.edu/news/prof-mundy-receives-2021-early-career-research-program-award-doe

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  2. 請問第一題:3組不成對為什麼*8,4組不成對*60?謝謝

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