114年台南二中教師甄選-數學詳解

 國立臺南第二高級中學114學年度第一次教師甄選

第一部分： 填充題 (每題 5 分，共 70 分)

解答:$$假設評審分數依序為\;a_1\le a_2\le a_3\le a_4\le a_5 \\ \cases{算術平均數為7 \Rightarrow \sum a_i=35\\ 中位數為8 \Rightarrow a_3=8\\ 眾數為9 \Rightarrow a_4=a_5=9} \Rightarrow a_1+a_2=35-8-9-9=9 \\ \Rightarrow a_1=1,a_2=8 \Rightarrow 8亦為眾數，不合 \Rightarrow (a_1,a_2)=(2,7),(3,6),(4,5) \\ \Rightarrow 可能的分數: \bbox[red, 2pt]{\{2,7,8,9,9\} ,\{3,6,8,9,9\} ,\{4,5,8,9,9\} }$$

解答:$$\lim_{n\to \infty} {n\over (n+4)^2} +{n\over (n+8)^2} +\cdots +{n\over (n+4n)^2} =\lim_{n\to \infty} \sum_{k=1}^n {n\over (n+4k)^2}  =\lim_{n\to \infty} \sum_{k=1}^n {1\over n(1+4(k/n))^2} \\=\int_0^1{1\over (1+4x)^2} \,dx =\int_1^5 {1\over 4u^2}\,du \;(u=1+4x) =\left. \left[ -{1\over 4u} \right] \right|_1^5 = \bbox[red, 2pt]{1\over 5}$$


解答:$$假設\cases{A(2,-2,-1) \\B(4,-4,-3)} \Rightarrow A,B,Q皆在E上 \Rightarrow\cases{\overrightarrow{QA} =(1,-6,-1) \\ \overrightarrow{QB} =(3,-8,-3)} \Rightarrow E的法向量\vec n \parallel  (\overrightarrow{QA} \times \overrightarrow{QB}) \\ \Rightarrow \vec n \parallel (10,0,10)  \Rightarrow \overrightarrow{QP} =(a-1,b-4,c) \parallel \vec n \Rightarrow \cases{a-1=c\\ b=4} \\ \Rightarrow P(a,4,a-1) \Rightarrow \overline{PQ}= \sqrt{(a-1)^2+0+(a-1)^2} =|a-1|\sqrt 2=6\sqrt 2 \Rightarrow \cases{a =7 \Rightarrow c=6 \\a=-5 \Rightarrow c=-6} \\ \Rightarrow P= \bbox[red, 2pt]{(7,4,6),(-5,4,-6)}$$

解答:$$A\cap B=\{a_1,a_4\} \Rightarrow a_1,a_4皆為完全平方數, 又a_1+a_4=10 \Rightarrow \cases{a_1=1\\ a_4=9 =3^2} \\ \textbf{Case I } A=\{1,2,3,9,a_5 \} \Rightarrow B=\{1,4,9,81,a_5^2\} \Rightarrow A\cup B元素和=a_5^2+a_5+100 =224 \Rightarrow a_5 \not \in \mathbb N\\ \textbf{Case II }A=\{1, 3,a_3, 9,a_5 \} \Rightarrow B=\{1,9,a_3^2,81,a_5^2\} \Rightarrow A\cup B元素和=94+a_3+a_5+a_3^2+a_5^2=224 \\\qquad \Rightarrow \cases{a_3=4\\ a_5=10} \Rightarrow A= \bbox[red, 2pt]{\{1,3,4,9,10\}}$$

解答:$${72\over C^9_2} (1\cdot C^4_2+ 2\cdot C^3_2+ 3\cdot C^2_2) =\bbox[red, 2pt]{30}$$

解答:

$$x^3y^2-x^2y^2-xy^4+xy^3 =xy^2(x^2-x-y^2+y)  =xy^2(x^2-y^2-(x-y)) \\=xy^2(x+y-1)(x-y) \ge 0 \Rightarrow x(x+y-1) (x-y)\ge 0 \Rightarrow x(x+y-1)\le 0(\because x\le y) \\ \Rightarrow  \cases{x\ge 0\\ x+y-1\le 0} \Rightarrow 所圍區域為一三角形，頂點為\cases{A(1/2,1/2) \\B(0,1) \\O(0,0)} \Rightarrow 面積=\bbox[red, 2pt]{1\over 4}$$

解答:

$$假設\angle BAD=\theta \Rightarrow \angle CAD=2\theta \Rightarrow \angle B=2\theta \Rightarrow \angle ADC=3\theta \Rightarrow \triangle ABC \sim \triangle DAC (AAA) \\ \Rightarrow a:b:c=b:e:d \Rightarrow \cases{b^2=ae\\ ad=bc} \Rightarrow (d,e) = \bbox[red, 2pt]{({bc\over a},{b^2\over a})}$$

解答:$$\cases{\cos B=3/4\\ \cos C=9/16} \Rightarrow \cases{ \sin B= \sqrt 7/4\\ \sin C=5\sqrt 7/16  } \Rightarrow \cos A=-\cos(B+C) =-\cos B\cos C+\sin B\sin C={1\over 8}\\ 又\cases{\overline{AE} =\overline{AB} \cos A\\ \overline{AF} =\overline{AC} \cos A} \Rightarrow {\triangle AEF\over \triangle ABC} ={{1\over 2}\overline{AE}\cdot \overline{AF} \sin A\over {1\over 2} \overline{AB} \cdot \overline{AC} \sin A} =\cos^2 A =\bbox[red, 2pt]{1\over 64}$$

解答:$$假設Z_3=x+yi, x,y\in \mathbb R \Rightarrow (x+yi)(-1+\sqrt 3 ai) = {\sqrt 3(a-1)\over 3a+1}+i \\ \Rightarrow \cases{-x-\sqrt 3ay= {\sqrt 3(a-1)\over 3a+1} \cdots(1) \\\sqrt 3ax-y=1 \cdots(2)}, 由(2)可得y=\sqrt 3ax-1 代入(1) \Rightarrow x={\sqrt 3\over 3a+1}\\ \Rightarrow y=-{1\over 3a+1} \Rightarrow Z_3={\sqrt 3\over 3a+1}-{i\over 3a+1} ={2\over 3a+1} \left({\sqrt 3\over 2}-{1\over 2}i \right) \\={2\over 3a+1} \left(\cos{11\pi\over 6}+ \sin{11\pi\over 6}i \right) \Rightarrow \bbox[red, 2pt]{Arg(Z_3) = \begin{cases} 11\pi/6, & a\gt -1/3\\ 5\pi/6, &a \lt -1/3\end{cases}}$$

解答:$$假設\cases{\angle A=30^\circ \\ \angle B=45^\circ} \Rightarrow \angle C=105^\circ \Rightarrow {\overline{BC} \over \sin A} ={\overline{AC} \over \sin B} =2R \Rightarrow {\overline{BC} \over 1/2} ={\overline{AC} \over \sqrt 2/2} =8 \\ \Rightarrow \cases{\overline{BC}=4\\ \overline{AC}=4\sqrt 2} \Rightarrow \triangle ABC面積={1\over 2}\cdot 4\cdot 4\sqrt 2\cdot \sin(105^\circ)=8\sqrt 2\cdot {\sqrt 6+\sqrt 2\over 4} = \bbox[red, 2pt]{4+4\sqrt 3}$$

解答:$$骰子出現5或6的機率p={1\over 3} \\ \textbf{Case I }喬治擲出0個(5或6)的機率:(1-p)^3={8\over 27}\\\qquad  \Rightarrow 瑪莎擲剩下3個骰子，恰出現1個(5或6)的機率:C^3_1p(1-p)^2 ={12\over 27} \\\textbf{Case II }喬治擲出1個(5或6)的機率:C^3_1(1-p)^2p={12\over 27}\\\qquad  \Rightarrow 瑪莎擲剩下2個骰子，恰出現1個(5或6)的機率:C^2_1p(1-p) ={4\over 9} \\\textbf{Case III }喬治擲出2個(5或6)的機率:C^3_2(1-p) p^2={6 \over 27} \\\qquad  \Rightarrow 瑪莎擲剩下1個骰子，恰出現1個(5或6)的機率:p ={1\over 3} \\\textbf{Case IV }喬治擲出3個(5或6)  \Rightarrow 瑪莎擲剩下0個骰子，恰出現1個(5或6)的機率:0 \\ 因此機率和={8\over 27}\cdot {12\over 27}+{12\over 27}\cdot {4\over 9}+ {6\over 27}\cdot {1\over 3} ={32+48+18\over 243} = \bbox[red, 2pt]{98\over 243}$$

解答:

$$\overline{PS}為角平分線 \Rightarrow \angle RPS=\angle SPQ=\theta, 又\overline{ST} \parallel \overline{PR} \Rightarrow \angle PST=\angle RPS = \theta \Rightarrow  \overline{TS} =\overline{PT}=7 \\ \Rightarrow \cos \angle Q =-{1\over 2} ={5^2+ \overline{SQ}^2-7^2\over 2\cdot 5\cdot \overline{SQ}} \Rightarrow \overline{SQ}^2+5\overline{SQ}-24 =0 \Rightarrow \overline{SQ}= \bbox[red, 2pt]3$$

解答:$$公差d=\log_3{x\over 3}-\log_3 x=-1 \Rightarrow S_6=6\log_3 x-15 \\ 公比r={2\log_3 x\over 3\log_3 x} ={2\over 3} \Rightarrow 無窮等比數列之和= 3\log_3 x\cdot {1\over 1-2/3} =9\log_3 x  \\ \Rightarrow 6\log_3 x-15={1\over 3} \cdot 9\log_3 x = 3\log_3 x \Rightarrow \log_3 x=5 \Rightarrow x=3^5= \bbox[red, 2pt]{243}$$

解答:$$\left({7\over 5},-{1\over 5} \right) \xrightarrow{\text{rotate}} \left({7\over 5}\cos \theta+{1\over 5} \sin \theta,-{1\over 5}\cos \theta+{7\over 5}\sin \theta \right) \\\xrightarrow{\text{enlarge}} \left({14\over 5}\cos \theta+{2\over 5} \sin \theta,-{2\over 5}\cos \theta+{14\over 5}\sin \theta \right) \\\xrightarrow{\text{translate}}\left({14\over 5}\cos \theta+{2\over 5} \sin \theta+1,-{2\over 5}\cos \theta+{14\over 5}\sin \theta -3\right) =\left({7\over 5},-{1\over 5} \right)  \\ \Rightarrow \cases{\displaystyle {14\over 5}\cos \theta+{2\over 5} \sin \theta={2\over 5} \\-\displaystyle {2\over 5}\cos \theta+{14\over 5}\sin \theta ={14\over 5}} \Rightarrow \cases{\sin \theta=1 \\ \cos \theta=0} \Rightarrow \theta = \bbox[red, 2pt]{\pi \over 2}$$

第二部分： 計算證明題 (每題 10 分，共 30 分)

解答:$${S_n\over S_n+n^2+n}= k為一定值 \Rightarrow S_n= {k\over 1-k}(n^2+n) \Rightarrow a_n=S_n-S_{n-1} ={k\over 1-k}\cdot 2n \\ \Rightarrow {a_{n} \over a_{n-1}} ={2n\over 2(n-1)} ={n\over n-1} \text{ is dependent on }n \Rightarrow \langle a_n \rangle 不是等比數列 \;\bbox[red, 2pt]{故得證}$$

解答:

$$假設\cases{\triangle ABC在z=0平面上\\ C=C'\\ A'在z=b平面上\\ B'在z=b+c平面上} \Rightarrow \cases{c^2=16-a^2 \\ b^2=9-a^2\\ (b+c)^2=24-a^2} \Rightarrow \sqrt{16-a^2}+\sqrt{9-a^2} =\sqrt{24-a^2} \\ \Rightarrow 3a^4-98a^2+575=0 \Rightarrow (a^2-25)(3a^2-23) =0 \Rightarrow a=\sqrt{23\over 3} (a\lt 斜邊4 \Rightarrow a=5不合) \\ \Rightarrow \cases{b=2/\sqrt 3\\ c=5/\sqrt 3}, 假設\cases{A(0,0,0) \\ B(a, 0,0)} \Rightarrow \cases{C(a/2, \sqrt 3a/2,0)=C'\\ A'(0,0,b) \\B'(a,0,b+c)} \Rightarrow \cases{\overrightarrow{C'B'} =(a/2,-\sqrt 3a/2, b+c) \\ \overrightarrow{C'A'} =(-a/2,-\sqrt 3a/2,b)} \\ \Rightarrow \vec n= \overrightarrow{C'B'} \times \overrightarrow{C'A'} =({\sqrt 3\over 2}ac, -ab-{1\over 2}ac,-{\sqrt 3\over 2}a^2) \parallel ({\sqrt 3\over 2}c, -b-{1\over 2}c,-{\sqrt 3\over 2}a) \\=({5\over 2},-{3\sqrt 3\over 2},-{\sqrt{23}\over 2}) ,又 \triangle ABC的法向量為\vec u(0,0,1) \Rightarrow \cos \theta={\vec n\cdot \vec u\over |\vec n||\vec u|} =-{\sqrt{69} \over 15} \\ \Rightarrow 兩平面之\cos \theta=\bbox[red, 2pt]{\pm {\sqrt{69} \over 15}}$$

解答:

$$取y={x\over x+1} \Rightarrow{ \color{blue}{xy+y-x=0}} \Rightarrow (x+1)(y-1)=-1 \\ 欲求兩圖形\cases{\Gamma_1: (x+1)(y-1)=1\\ \Gamma_2: x^2+y^2=1} 之交點, 由於兩交點對稱於直線y=-x\\ 假設其中一交點為(x,{x\over x+1}) \Rightarrow 另一交點為(-{x\over x+1},-x) \Rightarrow \cases{a=x\\ b=-{x\over x+1}}\\ \Rightarrow ab=-{x^2\over x+1} ={\color{blue}{-xy=-(x-y)}}\Rightarrow (xy)^2=(x-y)^2=x^2+y^2-2xy =1-2xy \\ \Rightarrow (xy)^2+2xy-1=0 \Rightarrow xy=-1+ \sqrt 2 (交點在一、三象限 \Rightarrow xy\gt 0) \\ \Rightarrow ab=-xy= \bbox[red, 2pt]{1-\sqrt 2}$$

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