114年高雄市立高中教甄聯招-數學詳解

 高雄市 114 學年度市立高級中等學校聯合教師甄選

一、計算證明題(1 至 6 題每題 5 分， 7 至 16 題每題 7 分，共 100 分)
請寫下完整計算過程，否則不予計分

解答:$$\log_9 m=\log_{12} n=\log_{16}(m+n) \Rightarrow {\log m\over \log 9} ={\log n\over \log 12}={\log (m+n)\over \log 16} \\ \Rightarrow \cases{\log 12\log m=\log 9\log n \cdots(1)\\  \log 12 \log(m+n) =\log 16\log n\cdots(2) }  \\\Rightarrow (1)+(2)=(\log 9+\log 16)\log n=\log 12(\log m+\log (m+n))\\ \Rightarrow 2\log 12\log n=\log 12 \log (m(m+n)) \Rightarrow 2\log n= \log(m(m+n)) \Rightarrow n^2=m(m+n) \\ \Rightarrow {n\over m}={m+n\over n} =1+{m\over n} \Rightarrow x=1+{1\over x} , 其中x={n\over m} \Rightarrow x^2-x-1=0 \\ \Rightarrow x=\bbox[red, 2pt]{1+ \sqrt{5} \over 2}\quad (m,n\gt 0 \Rightarrow n/m\gt 0 \Rightarrow {1-\sqrt 5\over 2}不合)$$

解答:$${3\over 5} = {\sum(x_i+\bar x)(y_i+\bar y) \over \sum (y_i+\bar y)^2} ={\sum x_iy_i+{3\over n}(\sum x_i)(\sum y_i) \over \sum y_i^2+{3\over n}(\sum y_i)^2} = {\sum x_iy_i+{3\over 5}\cdot 15 \cdot 10 \over 40+{3\over 5}\cdot 10^2} ={\sum x_iy_i+ 90 \over 100} \\ \Rightarrow \sum x_iy_i=-30 \Rightarrow m={\sum x_iy_i-{1\over n}(\sum x_i)(\sum y_1) \over \sum x_i^2-{1\over n}(\sum x_i)^2} ={-30-{1\over 5}\cdot 15\cdot 10\over 102-{1\over 5} \cdot 15^2} ={-60\over 57} =\bbox[red, 2pt]{-20\over 19}$$

解答:$$假設\cases{\angle ACB=\theta\\ \overline{BC}=x} \Rightarrow {9\over \sin \theta} ={15\over \sin 3\theta} ={15\over -4\sin^3 \theta+3\sin \theta} \Rightarrow \sin^2 \theta={1\over 3} \Rightarrow \sin \theta={\sqrt 3\over 3} \\ \Rightarrow \cos \theta={\sqrt 6\over 3} ={x^2+15^2-9^2 \over 2\cdot x\cdot 15} \Rightarrow x^2-10\sqrt 6 x+144=0 \Rightarrow x= \bbox[red, 2pt]{6\sqrt 6}$$

解答:$$\textbf{Case I }\max(A)=1 \Rightarrow A=\{1\} \Rightarrow n(A)=1\Rightarrow B \subseteq \{2,3,4,5\} \Rightarrow n(B)=2^4-1=15\\ \qquad \Rightarrow A, B組合數=1\times 15=15\\ \textbf{Case II }\max{(A)}=2 \Rightarrow A=\{2\}, \{1,2\} \Rightarrow n(A)=2 \Rightarrow B \subseteq \{3,4,5\} \Rightarrow n(B)=2^3-1=7 \\ \qquad \Rightarrow A, B組合數=2\times 7=14 \\ \textbf{Case III }\max(A)=3 \Rightarrow n(A)=2^2= 4 \Rightarrow B \subseteq \{4,5\} \Rightarrow n(B) =2^2-1=3 \\ \qquad \Rightarrow A, B組合數= 4\times 3=12 \\\textbf{Case IV }\max(A)=4 \Rightarrow n(A)=2^3=8 \Rightarrow B=\{5\} \Rightarrow n(B)=1 \\ \qquad \Rightarrow A, B組合數= 8\times 1=8 \\ \Rightarrow 所有情形的數量總和= 15+14+12+8=\bbox[red, 2pt]{49}$$

解答:$$\cases{3個蘋果分給3個人有H^3_3= 10種分法\\ 4個梨子分給3個人有H^3_4= 15種分法\\ 5個李子分給3個人有H^3_5=21種分法} \Rightarrow 共有10\times 15\times 21=3150種分法\\ \cases{3個蘋果分給2個人有H^2_3= 4種分法\\ 4個梨子分給2個人有H^2_4= 5種分法\\ 5個李子分給2個人有H^2_5= 6種分法} \Rightarrow 共有C^3_1\times 4\times 5\times 6 =360種分法\\  \cases{3個蘋果分給1個人有1種分法\\ 4個梨子分給1個人有1種分法\\ 5個李子分給1個人有1種分法} \Rightarrow 共有C^3_2\times1\times 1\times 1 =3 種分法 \\ \Rightarrow 每人至少得一個有3150-360+1= \bbox[red, 2pt]{2793} 種分法$$

解答:$$\cases{\vec u=t\vec a+(1-t) \vec b\\ \vec v=(t-{1\over 5}) \vec a+(1-t)\vec b} \Rightarrow \cases{|\vec u|^2 =(t\vec a+(1-t) \vec b) \cdot (t\vec a+(1-t) \vec b)=125t^2-50t+25\\ |\vec v|^2 =((t-{1\over 5}) \vec a+(1-t)\vec b) \cdot ((t-{1\over 5}) \vec a+(1-t)\vec b) =125t^2-90t+29 \\ \vec u\cdot \vec v= 125t^2-70t+25} \\ \Rightarrow \cos \theta ={\vec u\cdot \vec v\over |\vec u||\vec v} ={125t^2-70t+25\over \sqrt{125t^2-50t+25} \cdot \sqrt{125t^2-90t+29}}$$

解答:

$$假設\cases{O=(0,0) \\A(z_1) \\B(z_2) \\C(z_1+z_2) \\ D(z_1-z_2) \\z_1= 3e^{i\theta} } , 由於|z_1|=|z_1+z_3| =3\Rightarrow \cases{z_2= 3e^{i(\theta+2\pi/3) }\\ \overline{OC} \bot \overline{OD}}\\ z_1\cdot \bar z_2 =(3\cos \theta+ 3i\sin \theta ) \cdot (3\cos(\theta+{2\pi\over 3})-3i\sin(\theta+{2\pi\over 3})) =9(\cos{2\pi\over 3}+i\sin{-2\pi\over 3})\\ \Rightarrow (z_1\cdot \bar z_2)^{100}=9^{100} (\cos 200\pi/3+i\sin(-200\pi/3))=9^{100}(-{1\over 2}-{\sqrt 3\over 2}i)\\ \bar z_1\cdot z_2= (3\cos \theta- 3i\sin \theta ) \cdot (3\cos(\theta+{2\pi\over 3})+3i\sin(\theta+{2\pi\over 3})) =9(\cos{2\pi\over 3}+ i\sin{2\pi\over 3}) \\ \Rightarrow (\bar z_1\cdot z_2)^{100}=9^{100}(\cos 200\pi/3 +i\sin 200\pi/3)=9^{100}(-{1\over 2}+{\sqrt 3\over 2}i) \\ \Rightarrow \log_3 |(z_1\cdot \bar z_2)^{100}+(\bar z_1\cdot z_2)^{100}|= \log_3 9^{100} =\bbox[red, 2pt]{200}$$

解答:$$L: {x-3\over 2}={y\over -1} ={z+5\over -2} = \{(3+2t,-t,-5-2t) \mid t\in \mathbb R\} \\ \Rightarrow \overline{PA}+ \overline{PB} =\sqrt{(-7-2t)^2+ (2+t)^2+(10+2t)^2} + \sqrt{(2-2t)^2+ (5+t)^2+ (4+2t)^2} \\= 3 \left( \sqrt{(t+4)^2+1}+ \sqrt{(t+1)^2+4} \right) =3 \left(  \overline{QC}+ \overline{QD} \right), 其中\cases{Q(t,0) \\ C(-4,1) \\D(-1,2)} \\ \Rightarrow D對稱於y=0的對稱點D'(-1,-2) \Rightarrow \overline{QC}+ \overline{QD}的最小值=\overline{CD'} =3\sqrt 2 \\ \Rightarrow \overline{PA}+ \overline{PB} 的最小值=3\cdot 3\sqrt 2= \bbox[red, 2pt]{9\sqrt 2}$$

解答:

$$四面體OABC體積V={1\over 6}\cdot 1\cdot 2\cdot 3=1, 截平面E:{x\over 3}+{ y\over 2}+{z\over k}=1顯然通過\cases{B(0,2,0) \\ D(0,0,k)} \\又\overleftrightarrow{AC}:\cases{{x-1\over -1}={z\over 3} \\ y=0} \Rightarrow E= \overleftrightarrow{AC} \cap E = \left( {3k-9\over k-9} ,0,-{6k\over k-9}\right) \\ \Rightarrow \triangle CDE面積={1\over 2} (3-k) \cdot {3k-9\over k-9} ={3(3-k)^2\over 2(9-k)} \Rightarrow 四面體BCDE體積={V\over 2} \\ \Rightarrow {1\over 3}\cdot {3(3-k)^2\over 2(9-k)}\cdot 2={1\over 2} \Rightarrow 2(3-k)^2=9-k \Rightarrow 2k^2-11k+9=0\Rightarrow (2k-9)(k-1)=0 \\ \Rightarrow k=\bbox[red, 2pt]1 \quad  \left( k={9\over 2}\gt 3\Rightarrow D\not \in \overline{OC}, 不合 \right)$$

解答:$$g(x,y) =(x+y)^5 = \sum_{k=0}^5 C^5_k x^ky^{5-k} \Rightarrow g(\cos 9^\circ, i\sin 9^\circ) =\cos 45^\circ+i \sin 45^\circ ={\sqrt 2\over 2}+{\sqrt 2\over 2}i \\ \Rightarrow Re\{g(\cos 9^\circ, i\sin 9^\circ)\} ={\sqrt 2\over 2} =C^5_1xy^4+C^5_3x^3y^2+ C^5_5 x^5 =5xy^4-10x^3y^2+x^5 \\=5x(1-x^2)^2-10x^3(1-x^2)+x^5 \Rightarrow (32x^5-40x^3+10x)^2=(\sqrt 2)^2 \\ \Rightarrow 1024x^{10}-2560x^8+2240x^6-800x^4+100x^2=2 \\\Rightarrow f(x)= \bbox[red, 2pt]{512x^{10}-1280x^8 +1120x^6 -400x^4+50x^2-1}$$

解答:$$取{x\over x-1}={1\over n} \Rightarrow x ={-1\over n-1} \Rightarrow f({x\over x-1})=xf(x) \Rightarrow f({1\over n})={-1\over n-1}f({-1\over n-1}) ={1\over n-1}f({1\over n-1})\\ \Rightarrow \cases{n=1 \Rightarrow 已知f(1)=1\\ n=2 \Rightarrow f(1/2) =f(1)=1 \\ n=3 \Rightarrow f(1/3)=(1/2)f(1/2) =1/2!\\ n=4 \Rightarrow f(1/4)=(1/3)f(1/3)=1/3!} \Rightarrow f({1\over n}) ={1\over (n-1)!} \\ 欲求  f(1) f({1\over 100})+ f({1\over 2}) f({1\over 99}) + f({1\over 3}) f({1\over 98}) + .... + f({1\over 50}) f({1\over 51}) = \sum_{k=1}^{50} f({1\over k})f({1\over 101-k}) \\= \sum_{k=1}^{50}{1\over (k-1)! (100-k)!} = \sum_{n=0}^{49}{1\over n! (99-n)!} ={1\over 99!} \sum_{n=0}^{49} {99\choose n} ={1\over 2\cdot 99!} \sum_{n=0}^{99} {99\choose n} \\={1\over 2\cdot 99!}\cdot 2^{99} = \bbox[red, 2pt]{2^{98} \over 99!}$$

解答:$$f(x)={\sin(\pi x)-\cos (\pi x)+2 \over \sqrt x} = {\sqrt 2\sin(\pi x-{\pi\over 4})+2 \over \sqrt x}, {1\over 4}\le x\le {5\over 4} \\ 分子g(x)= \sqrt 2\sin(\pi x-{\pi\over 4})+2 \begin{cases} 遞增& 1/4\le x\le 3/4\\ 遞減& 3/4\le x\le 5/4\end{cases} \\ \Rightarrow f(x)的最小值發生兩端點 \Rightarrow \cases{f(1/4)= 4\\ f(5/4)= 4\sqrt 5/5} \Rightarrow f(1/4)\gt f(5/4) \Rightarrow 最小值\bbox[red, 2pt]{4\sqrt 5\over 5}$$

解答:$$a_n+a_{n-1}+2 a_{n-2}=0 \Rightarrow \lambda^2+\lambda+2=0 \Rightarrow \lambda_1 ={-1+ \sqrt 7i\over 2} , \lambda_2= {-1-\sqrt 7 i\over 2}\\ \Rightarrow a_n= c_1\lambda_1^n+ c_2\lambda_2^n \Rightarrow \cases{a_2+a_1+2a_0=0 \Rightarrow a_0=0 =c_1+c_2 \\a_1= c_1\lambda_1+ c_2\lambda_2=1  } \Rightarrow \cases{c_1 =1/\sqrt 7 i \\ c_2=-1/\sqrt 7 i} \\ \Rightarrow a_n={1\over \sqrt 7 i}(\lambda_1^n -\lambda_2^n) \Rightarrow 2^{n+2}-7a_n^2 =2^{n+2}+ (\lambda_1^n-\lambda_2^n)^2 =2^{n+2}+ \lambda_1^{2n}-2\cdot 2^n+ \lambda_2^{2n}  \\=2^{n+1}+ \lambda_1^{2n} + \lambda_2^{2n} =(\lambda_1^n+ \lambda_2^n)^2\\ 現在要證明b_n=\lambda_1^n +\lambda_2^n為整數,由於b_n也是\lambda^2+\lambda+2=0的解的線性組合,\\因此b_n+b_{n-1}+2b_{n-2}=0 \Rightarrow b_n=-b_{n-1}-2b_{n-2} \\ 由數學歸納法: n=0 \Rightarrow b_0=\lambda_1^0+\lambda_2^0=2為整數, b_1=\lambda_1+\lambda_2=-1也是整數\\ 假設n=k時, b_k=\lambda_1^k+\lambda_2^k為整數\\ n=k+1時, b_{k+1} =\lambda_1^{k+ 1}+\lambda_2^{k +1} =(\lambda_1^k+\lambda_2^k)(\lambda_1+ \lambda_2)-\lambda_1 \lambda_2 (\lambda_1^{k-1}+ \lambda_2^{k-1}) \\ \qquad =b_k\cdot b_1-2b_{k-1} 也是整數, 因此b_n=\lambda_1^n +\lambda_2^n為整數\\ \Rightarrow 2^{n+2}-7a_n^2 =(\lambda_1^n+ \lambda_2^n)^2 為一完全平方數\quad\bbox[red, 2pt] {故得證}$$

解答:

$$\triangle A'BC:\cases{\overline{A'D} =\overline{A'B}=1 \\ \overline{BD}= \sqrt 2} \Rightarrow \angle DA'B=90^\circ \Rightarrow \cases{A'(0,0,0) \\ B(0,1,0) \\ D(1,0,0)}, 又\angle A'DC=90^\circ \Rightarrow C(1,0,1) \\ \Rightarrow \cases{\overrightarrow{A'C} =(1,0,1) \\ \overrightarrow {BD}=(1,-1,0)} \Rightarrow \cos \alpha ={\overrightarrow{A'C} \cdot \overrightarrow{BD} \over |\overrightarrow{A'C} ||\overrightarrow{BD} |} ={1+0+0 \over \sqrt 2\cdot \sqrt 2} = \bbox[red, 2pt]{1\over 2}$$

解答:

$$\cases{\triangle APB: \cos \alpha= \displaystyle {9+a^2-32\over 6a} =  {a^2-23\over 6a} \\ \triangle APD:\cos (90^\circ-\alpha)= \displaystyle {9+a^2-50\over 6a} \Rightarrow \sin \alpha={a^2-41\over 6a}} \\ \Rightarrow \cos^2\alpha+\sin^2\alpha=\left( {a^2-23\over 6a} \right)^2 + \left( {a^2-41\over 6a}\right)^2 =1 \Rightarrow a^4-82a^2+1105=0 \\ \Rightarrow (a^2-65)(a^2-17) \Rightarrow 面積=a^2=\bbox[red, 2pt]{65} \;(a^2=17   \Rightarrow \sin \alpha\lt 0 , 不合)$$

解答:$$假設\cases{x: 骰子點數\\ a:要移動的距離} \Rightarrow \cases{x=1 \Rightarrow a=0\\ x=2 \Rightarrow a=1\\ x=3 \Rightarrow a=-2\\ x=4 \Rightarrow a=2 \\ x=5\Rightarrow a=-4\\ x=6\Rightarrow a=3} \\ \Rightarrow 投擲骰子三次在原點的情形:(1,1,1), (1,3,4), (2,2,3), (2,5,6), (4,4,5)及其排列 \\ \Rightarrow 排列數總和=1+6+3+6+3=19 \Rightarrow 機率為{19\over 6^3} =\bbox[red, 2pt]{19\over 216}$$

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