臺北市立復興高級中學 114 學年度第一次專任教師甄選
第一部分：計算、證明題 (第一題6分；第2、3 題各7分；第4-13 題各8分)
解答:
$$\cases{A(2,4)\\ B(0,0) \\C(6,0)} \Rightarrow \cases{\triangle ABC= 6\cdot 4/2=12\\ L=\overleftrightarrow{AC}: x+y=6} \Rightarrow Q(t,6-t)\in L \Rightarrow \triangle PQC=12/2=6 \\ \Rightarrow {1\over 2} \begin{Vmatrix}1&0&1\\ t&6-t& 1\\ 6&0& 1 \end{Vmatrix} =6 \Rightarrow |5t-30|=12 \Rightarrow t={18\over 5} \;(t={42\over 5} \Rightarrow Q \not \in \overline{AC}) \\ \Rightarrow Q({18\over 5},{12\over 5}) \Rightarrow \overleftrightarrow{PQ}: y={12\over 13}(x-1) \Rightarrow \bbox[red, 2pt]{12x-13y=12}$$
解答:$$\cases{n=1: 若樓梯只有1階,只有1種方法\\ n=2:若樓梯有2階,有2(2或1+1)種方法 \\n=3:若樓梯有3階,有4(1+1+1,1+2, 2+1,3)種方法} \\ \Rightarrow f(n)=f(n-1)+f(n-2)+ f(n-3),n\ge 4 \Rightarrow f(4)=7 \Rightarrow f(5)=13 \Rightarrow f(6)=24 \\ \Rightarrow f(7)=44 \Rightarrow f(8)=81 \Rightarrow f(9)=149 \Rightarrow f(10)= \bbox[red, 2pt]{274}$$
解答:$$f(x)=x^{40} =(x^2+1)(x+1)^2 p(x)+ ax^3+bx^2+cx+d \\ \Rightarrow f'(x)=40x^{39} =(2x(x+1)^2+2(x^2+1)(x+1))p(x) +(x^2+1)(x+1)^2p'(x)+3ax^2+2bx+c\\ \Rightarrow \cases{f(-1)=1=-a+b-c+d\\ f'(-1)=-40=3a-2b+c \\f(i)=1=-ai-b+ci+d\\ f(-i)=1=ai-b-ci+d} \Rightarrow \cases{a=b=c\\ d=b+1} \Rightarrow \cases{a=b=c=-20\\ d=-19} \\ \Rightarrow 餘式:\bbox[red, 2pt]{-20x^3-20x^2-20x-19}$$
解答:$$f(x)={x-x^3\over 1+2x^2+x^4} ={x-x^3\over (x^2+1)^2} \Rightarrow f'(x)={1-3x^2 \over (x^2+1)^2}-{4x(x-x^3) \over (x^2+1)^3} =0 \\ \Rightarrow x^4-6x^2+1=0 \Rightarrow x^2=3+2\sqrt 2 \Rightarrow x=\pm(\sqrt 2+1) \\ \Rightarrow f(-(\sqrt 2+1)) ={(\sqrt 2+1)(2+2\sqrt 2) \over (4+2\sqrt 2)^2} ={6+4\sqrt 2\over 24+16\sqrt 2} = \bbox[red, 2pt]{1\over 4}$$
解答:$$假設等差數列首項a, 公差d \Rightarrow \cases{S_n=(2a+(n-1)d)n/2 =m \\ S_m=(2a+(m-1)d)m/2=n} \\ \Rightarrow \cases{2a+(n-1)d=2m/n\\ 2a+(m-1)d =2n/m} \Rightarrow \cases{ a=(m^2+(n-1)(m+n))/(mn)\\ d=-2(m+n)/mn} \\ \Rightarrow S_{m+n} ={2a+(m+n-1)d \over 2}\cdot (m+n) =(m+n) \left( {m^2+(n-1)(m+n) \over mn}-{(m+n-1)(m+n) \over mn} \right) \\= (m+n) \cdot {-mn\over mn} = \bbox[red, 2pt]{-(m+n)}$$
解答:$$\log x -1\lt[\log x] \le \log x \Rightarrow \log x -1\lt (\log x)^2 -3 \le \log x \\ \Rightarrow \cases{(\log x)^3-\log x-3 \le 0 \Rightarrow (\log x-{1+\sqrt{13}\over 2}) (\log x-{1-\sqrt{13}\over 2}) \le 0 \\ (\log x)^2-\log x-2\gt 0 \Rightarrow (\log x-2)(\log x+1)\gt 0 } \\ \Rightarrow \cases{(1-\sqrt{13})/2 \approx 2.3\le \log x\le (1+\sqrt{13})/2 \approx 2.3\\ \log x\gt 2 或\log x\lt -1} \Rightarrow \cases{ (1-\sqrt{13})/2 \le \log x\lt -1 \\ 2\lt \log x\le (1+\sqrt{13})/2} \\ (\log x)^2=[\log x]+3 \in \mathbb Z \Rightarrow \log x=\sqrt 5 \approx 2.24 \Rightarrow x=10^{\sqrt 5} \Rightarrow m=\bbox[red, 2pt]{\sqrt 5}$$
解答:
$$C \xrightarrow{順時針旋轉30度}A \Rightarrow 半碗水剩下著色面積的水 \Rightarrow 體積=\int_{r/2}^r (r^2-y^2)\pi\, dy= \bbox[red, 2pt]{{5\over 24}\pi r^3}$$
解答:
$$f(x,y)=5x^2-6xy+5y^2-4x-4y-4=0 \Rightarrow 36-4\cdot 5\cdot 5 \lt 0 \Rightarrow f(x,y)=0為一橢圓\Gamma \\ 又f(x,y)=f(y,x) \Rightarrow 橢圓長軸為直線L:y=x \Rightarrow f(x,x)=0 \Rightarrow x^2-2x-1=0 \\ \Rightarrow x=1\pm \sqrt 2 \Rightarrow 橢圓頂點\cases{A(1+\sqrt 2, 1+\sqrt 2) \\B(1-\sqrt 2,1-\sqrt 2)} \Rightarrow 中心點O={1\over 2}(A+B) =(1,1) \\ \Rightarrow (x-1)^2+(y-1)^2= \overline{PO}^2, P\in \Gamma \Rightarrow P=A 或P=B,則\overline{PO}^2有最大值(\sqrt 2)^2+(\sqrt 2)^2= \bbox[red, 2pt]4$$
解答:$$x \lt y \Rightarrow n\ge m 且m^2+n^2=10 \Rightarrow \cases{n=3 \\m=1} \Rightarrow \cases{\log x=1+\alpha\\ \log y=3+\beta} \Rightarrow \cases{\log x+\log y=4+\alpha+ \beta \\ 10\lt x\lt 100\\ 1000\lt y\lt 10000}\\\Rightarrow \log xy=5 \Rightarrow xy=10^5 \Rightarrow \cases{x=16 \Rightarrow y=6250\\x=20 \Rightarrow y=5000\\ x=25 \Rightarrow y=4000\\ x=32 \Rightarrow y=3125\\ x=40 \Rightarrow y=2500\\ x= 50 \Rightarrow y=2000 \\ x=80 \Rightarrow y=1250} \\ \Rightarrow (x,y)= \bbox[red, 2pt]{(16,6250), (20,5000), (25,4000), (32, 3125), (40,2500),(50,2000), (80,1250)}$$
解答:
$$假設\cases{C(0,0) \\A(a,0) \\B(0,b)} \Rightarrow \overline{AB}^2 =a^2+b^2=15^2 =225\\ 又\cases{G到\overline{BC}=G到y軸 \\ G到\overline{AC}= G到x軸}\Rightarrow G(t,6-t) ={1\over 3}(A+ B+C) =(a/3,b/3) \Rightarrow {a+ b\over 3}=6 \\ \Rightarrow a+b=18 \Rightarrow (a+b)^2= a^2+b^2+ 2ab=18^2 \Rightarrow 225+2ab=324 \Rightarrow 2ab=99 \\ \Rightarrow \triangle ABC ={1\over 2}ab = \bbox[red, 2pt]{99\over 4}$$
解答:$$正弦定理: {a\over \sin A} ={b\over \sin B} ={c\over \sin C} =k \Rightarrow b^2-c^2=ac \Rightarrow k^2\sin^2B -k^2\sin^2 C=k^2\sin A\sin C \\ \Rightarrow \sin^2B-\sin^2C = (\sin B+\sin C)(\sin B+ \sin(-C)) =\sin A\sin C \\ \Rightarrow (\sin B+\sin C)(\sin B+ \sin(-C))=2\sin{B+C\over 2} \cos {B-C\over 2} \cdot 2\sin{B-C\over 2} \cos{B+C\over 2} \\=\sin(B+C) \sin(B-C) = \sin A\sin(B-C)= \sin A\sin C \Rightarrow \sin(B-C)= \sin C \\ \Rightarrow B-C=C \Rightarrow B=2C \Rightarrow 42^\circ+2C+C=180^\circ \Rightarrow C={180^\circ-42^\circ\over 3}= \bbox[red, 2pt]{46^\circ}$$
解答:$$\mu_n={1\over n} \sum_{k=1}^n \sqrt{2k-1} ={1\over n} \sum_{k=1}^n \sqrt{ 2n \left( {2k-1\over 2n} \right) } =\sqrt{2n}\cdot {1\over n} \sum_{k=1}^n \sqrt{ \left( {k-1/2\over n} \right) } \\ \Rightarrow \lim_{n\to \infty} \mu_n \approx \sqrt{2n} \int_0^1 \sqrt x\,dx ={2\over 3}\sqrt{2n} ={2\sqrt 2\over 3} \sqrt n \\ E(X^2) ={1\over n} \sum_{k=1}^n (2k-1) =n \Rightarrow \sigma_n^2 =E(X^2)-\mu_n^2 =n -\mu_n^2 \\ \Rightarrow \lim_{n\to \infty}{\sigma_n\over \mu_n} = \lim_{n\to \infty} \sqrt{n -\mu_n^2 \over \mu_n^2} = \lim_{n\to \infty} \sqrt{{n \over \mu_n^2}-1} = \lim_{n\to \infty} \sqrt{{n \over 8n/9}-1} = \sqrt{1\over 8} = \bbox[red, 2pt]{\sqrt 2\over 4}$$
解答:$$a_1=1 \Rightarrow a_2={-3-\sqrt 5\over 2a_1+2} = {-3-\sqrt 5\over 4} \Rightarrow a_3=4+2\sqrt 5 \Rightarrow a_4={-5+\sqrt 5\over 10} \Rightarrow a_5={-5-\sqrt 5\over 2} \\ \Rightarrow a_6=a_1=1 \Rightarrow 循環數為5 \Rightarrow 2025=5\times 405 \Rightarrow a_{2025}=a_5= \bbox[red, 2pt]{-5-\sqrt 5\over 2}$$
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