114年湖口高中教甄-數學詳解

新竹縣立湖口高中 114 年 第 1 次教師甄選

一、填充題：每格 7 分，共 84 分（答案請以最簡形式表示）

解答:$$\cases{(1,2,4-20):17個, \\\cases{ (2,3,5-20):16個\\ (3,4,6-20),(1,3,4):16個\\ (4,5,7-20), (1-2,4,5):16個\\ ....\\ (1-16, 18,19):16個} \Rightarrow 共17\times 16=272個\\(1-17,19,20): 17個} \Rightarrow 共17+272+17=306 \\ \Rightarrow 機率={306\over {20\choose 3}} ={306\over 1140} =\bbox[red, 2pt]{51 \over 190}$$

解答:$$\log \left( a^{1/x}b^{1/(x+3)} c^{1/(x+6)} \right) =\log 10 \Rightarrow {\log a\over x}+{\log b\over x+3}+ {\log c\over x+6} =1 \\ \Rightarrow (x+3)(x+6)\log a +x(x+6)\log b+ x(x+3)\log c=x(x+3)(x+6) \\ \Rightarrow x^3 +(9-(\log a+\log b+\log c))x^2+ (18-9\log a-6\log b-3\log c)x-18\log a=0 \\ \Rightarrow 三根之和=\log abc-9=11+21+31=63 \Rightarrow \log abc=72 \Rightarrow abc= \bbox[red, 2pt] {10^{72}}$$

解答:$$\cases{\displaystyle \lim_{x\to 2^+} f(x)=2a+2b \\\displaystyle \lim_{x\to 2^-} f(x)=2^2=4} \Rightarrow 2a+2b=4\Rightarrow a+b=2 \\ \cases{\lim_{h\to 0^+} {f(2+h)-f(2) \over h} = \lim_{h\to 0^+} {a(2+h)+b[2+h]-4 \over h} = \lim_{h\to 0^+}{2a+ah+2b-4\over h} =\lim_{h\to 0^+}{ah\over h}=a\\ \lim_{h\to 0^-} {f(2+h)-f(2) \over h} =\lim_{h\to 0^-}{(2+h)^2-4 \over h}  =\lim_{h\to 0^-}{h^2+4h\over h} =4} \Rightarrow a=4 \\ \Rightarrow b=2-4=-2 \Rightarrow (a,b) = \bbox[red, 2pt]{(4,-2)}$$

解答:$$\triangle ABC=2\sqrt 3 \Rightarrow {1\over 2}\cdot \overline{AB}\cdot \overline{AC} \sin {\pi\over 3} = 2\sqrt 3 \Rightarrow \overline{AB}\cdot \overline{AC}=8 \\ \cases{M是\overline{BC}中點\\ N是\overline{BM}中點} \Rightarrow \cases{\overline{BM}:\overline{MC} =1:1\\ \overline{BN}:\overline{NC} =1:3} \Rightarrow \cases{\overrightarrow{AM} ={1\over 2} \overrightarrow{AB} +{1\over 2} \overrightarrow{AC} \\ \overrightarrow{AN} ={3\over 4} \overrightarrow{AB} +{1\over 4} \overrightarrow{AC}} \\ \Rightarrow \overrightarrow{AM} \cdot \overrightarrow{AN} = \left( {1\over 2} \overrightarrow{AB} +{1\over 2} \overrightarrow{AC} \right) \cdot \left( {3\over 4} \overrightarrow{AB} +{1\over 4} \overrightarrow{AC} \right) ={3\over 8}\overline{AB}^2+ {1\over 2} \overrightarrow{AB} \cdot \overrightarrow {AC}+ {1\over 8} \overline{AC}^2 \\ ={3\over 8}\overline{AB}^2+ {1\over 2} \overline{AB} \cdot \overline {AC} \cos \angle A+{1\over 8} \overline{AC}^2= {3\over 8}\overline{AB}^2+ 2  +{1\over 8} \overline{AC}^2 \\={1\over 8}\left( 3\overline{AB}^2+ \overline{AC}^2\right) +2  \ge {1\over 4} \sqrt{3\overline{AB}^2 \cdot \overline{AC}^2} +2= \bbox[red, 2pt]{2\sqrt 3+2}$$

解答:$$取\cases{B為原點\\ \overleftrightarrow{BC}為x軸} \Rightarrow \cases{A(0,0,6) \\B(0,0,0)\\ C(12\sqrt 2, 0,0) \\D(6\sqrt 2, 6\sqrt 6,0)} \Rightarrow \cases{E=(C+D)/2=(9\sqrt 2,3\sqrt 6,0) \\F=(B+D)/2= (3\sqrt 2, 3\sqrt 6,0)} \\ \Rightarrow \cases{\overrightarrow{AF}= (3\sqrt 2, 3\sqrt 6, -6) \\ \overrightarrow{BE} =(9\sqrt 2,3\sqrt 6,0) \\ \overrightarrow{AB} =(0,0,-6)} \Rightarrow \vec n= \overrightarrow{AF} \times \overrightarrow{BE} =(18\sqrt 6, -54\sqrt 2, -18\sqrt {12}) \\ \Rightarrow d(\overleftrightarrow{AF}, \overleftrightarrow{BE}) ={\overrightarrow{AB} \cdot \vec n \over |\vec n|}={108\sqrt {12}\over 108} =\bbox[red, 2pt]{2\sqrt 3}$$

解答:$$\log_9 \alpha=\log_{12} \beta= \log_{16}(\alpha+\beta) =k \Rightarrow \cases{\alpha=9^k\\ \beta=12^k\\ \alpha+ \beta=16^k} \Rightarrow 9^k+12^k=16^k \Rightarrow \left( {3\over 4} \right)^{2k} +\left( {3\over 4} \right)^k =1 \\ \Rightarrow \left( {3\over 4} \right)^k ={-1+ \sqrt 5\over 2}   \Rightarrow {\alpha^2+\beta^2\over \alpha\beta} ={\alpha\over \beta}+{\beta\over \alpha} = \left( {3\over 4} \right)^k +\left( {4\over 3} \right)^k={-1+ \sqrt 5\over 2}+{2\over -1+\sqrt 5} =\sqrt 5 \\ \Rightarrow \log_{25}{\alpha^2+\beta^2\over \alpha\beta} =\log_{25} \sqrt 5={\log_ 5\sqrt 5\over \log_5 25} ={1/2\over 2} = \bbox[red, 2pt]{1\over 4}$$

解答:$$\cases{\alpha^3+6\alpha^2+14\alpha+11=(\alpha+2)^3+2\alpha+3=0 \\ \beta^3+6\beta^2+14\beta+13=(\beta+2)^3+ 2\beta+5=0},取\cases{u=\alpha+2\\ v=\beta+2} \Rightarrow \cases{ u^3+2u=1\\v^3+2v=-1} \\ 令f(t)=t^3+2t \Rightarrow \cases{f(-t)=-f(t) \Rightarrow f為奇函數 \\f'(t)=3t^2+2 \gt 0 \Rightarrow f為嚴格遞增函數}\\現在\cases{f(u)=1\\ f(v)=-1 \Rightarrow f(-v)=1} \Rightarrow f(u)=f(-v) \Rightarrow u=-v \Rightarrow u+v=0\\ \Rightarrow \alpha+2+\beta+2=0 \Rightarrow \alpha+\beta= \bbox[red, 2pt]{-4}$$

解答:$$\lim_{n\to \infty} {\left( \sum_{k=1}^n  k^2\right) \left( \sum_{k=1}^n k^5\right)\over \left( \sum_{k=1}^n k^3\right)\left( \sum_{k=1}^n k^4\right)}= \lim_{n\to \infty} {\left( \sum_{k=1}^n {1\over n}({k\over n})^2\right) \left( \sum_{k=1}^n {1\over n}({k\over n})^5\right)\over \left( \sum_{k=1}^n {1\over n}({k\over n})^3\right) \left( \sum_{k=1}^n {1\over n}({k\over n})^4\right)} \\={ \left( \int_0^1 x^2\,dx \right) \left( \int_0^1 x^5\,dx \right) \over \left( \int_0^1 x^3\,dx \right) \left( \int_0^1 x^4\,dx \right)} ={{1\over 3} \cdot {1\over 6} \over {1\over 4} \cdot {1\over 5}} ={20\over 18} =\bbox[red, 2pt]{10\over 9}$$

解答:$$y^2=8x \Rightarrow F(2,0), 又\cases{\overline{FA}=4\\ \overline{FB}=10} \Rightarrow \cases{A(2,4)\\ B(8,-8)} 或\cases{A(2,-4) \\B(8,8)} \\ 取 \cases{A(2,4)\\ B(8,-8) } \Rightarrow P(t^2/8, -t) \Rightarrow \cases{\overrightarrow{AP} =(t^2/8-2,-t-4) \\\overrightarrow{AB} =(6,-12)} \\\Rightarrow \triangle APB面積={1\over 2}|(t^2/8-2,-t-4, 1) \times (6,-12,1)|={1\over 2}|(0,0,-{3\over 2}t^2+6t+48)|\\={1\over 2} |-{3\over 2}t^2+6t+48| \\ 令f(t)=-{3\over 2}t^2+6t+48 \Rightarrow f'(t)=-3t+6=0 \Rightarrow t=2 \Rightarrow f(t)=54 \\\Rightarrow \triangle APB面積={1\over 2}\cdot 54= \bbox[red, 2pt]{27}$$

解答:$$取g(x)=f(x)-x^2 \Rightarrow g(1)=g(2)=g(3)=0 \Rightarrow g(x)=(x-1)(x-2)(x-3)(x-a) \\ \Rightarrow f(x)= (x-1)(x-2)(x-3)(x-a)+x^2 \Rightarrow \cases{f(-7) =720(a+7)+49 =720a+ 5089 \\f(11)= 720(11-a) +121=8041-720a} \\ \Rightarrow f(-7)+ f(11)= 5089+ 8041= \bbox[red, 2pt]{13130}$$

解答:$$z=1+\cos20^\circ +i\sin 20^\circ=2 \cos^2 10^\circ +i 2\sin10^\circ \cos10^\circ =2\cos 10^\circ (\cos 10^\circ+i\sin 10^\circ) \\ \Rightarrow z^n =(2\cos 10^\circ)^n (\cos 10n^\circ+i\sin 10n^\circ) \in \mathbb R \Rightarrow \sin 10n^\circ=0\\ \Rightarrow 10n=180k, k\in \mathbb N \Rightarrow n=18k \Rightarrow 30\lt 18k\lt 300 \Rightarrow k=2,3,\dots,16 \\ \Rightarrow \sum_{k=2}^{16}18k =18\cdot {18\cdot 15\over 2} =\bbox[red, 2pt]{2430}$$

解答:$$(\sqrt 2+\sqrt 3)^{2004} =(5+2\sqrt 6)^{1002} \Rightarrow 假設\cases{\alpha=5+2\sqrt 6\\ \beta=5-2\sqrt 6 \lt 1} \Rightarrow \cases{\alpha+ \beta=10\\ \alpha \beta=1} \\ \Rightarrow \alpha, \beta 是t^2-10t+1=0的兩根 \Rightarrow a_n= \alpha^n+\beta^n 滿足a_{n+2} =10a_{n+1}-a_n \\ \Rightarrow a_{n+2} \equiv -a_n{\mod 10} \Rightarrow \cases{a_0= \alpha^0+\beta^0 =2 \equiv 2 {\mod 10} \\ a_1= \alpha +\beta=10 \equiv 0 {\mod 10} \\a_2= 10a_1-a_0=-2 \equiv 8 {\mod 10} \\ a_3=10a_2-a_1 \equiv -a_1 \equiv 0 {\mod 10} \\ a_4 \equiv -a_2 \equiv -8 \equiv 2 {\mod 10}} \Rightarrow 循環數為4\\ n=1002 \equiv 2 {\mod 4} \Rightarrow a_{1002}的個位數與a_2相同, 即8\\ \Rightarrow \alpha^{1002}=a_{1002}-\beta^{1002} \Rightarrow (\sqrt 2+\sqrt 3)^{2004} 的個位數=8-1= \bbox[red, 2pt]7 (\because 0\lt \beta^{1002}\lt 1)$$

二、計算證明題：每題 8 分，共 16 分（請標明題號）

解答:$$X\sim B(n,p) \Rightarrow P(X=k) ={n\choose k }p^kq^{n-k}, p+q=1\\取f(p)=(p+q)^n= \sum_{k=0}^n {n\choose k}p^k q^{n-k} \Rightarrow f'(p)=n(p+q)^{n-1} =\sum_{k=0}^n k{n\choose k}p^{k-1} q^{n-k} \\ \Rightarrow pf'(p)=np(p+q)^{n-1}= np= \sum_{k=0}^n k{n\choose k}p^{k} q^{n-k} =E(X) \Rightarrow E(X)=np \quad \bbox[red, 2pt]{QED.}$$

 

解答:$$\cases{\alpha_n+\beta_n=-{1\over 2}n-1\\ \alpha_n \beta_n=n^2-2} \Rightarrow {1\over (\alpha_n+2)(\beta_n+2)} ={1\over \alpha_n \beta_n+2(\alpha_n+ \beta_n)+4} ={1\over n^2-2-n-2+4} \\={1\over n^2-n} ={1\over n-1}-{1\over n} \Rightarrow \sum_{n=3}^{2011}{1\over (\alpha_n+2)(\beta_n+2)} =\sum_{n=3}^{2011} \left( {1\over n-1}-{1\over n} \right) \\= \left( {1\over 2}-{1\over 3} \right) + \left({1\over 3}-{1\over 4}  \right) + \cdots + \left( {1\over 2010}-{1\over 2011} \right) ={1\over 2}-{1\over 2011} = \bbox[red, 2pt]{2009\over 4022}$$

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