110年新竹市香山高中教甄-數學詳解

新竹市立香山高級中學110學年度教師甄選

一、單選題

解答： $$f(x)={1\over 1-x} = 1+x+x^2+x^3+\cdots \Rightarrow f'(x)={1\over (1-x)^2} =1+2x +3x^2+4x^3+\cdots\\ \Rightarrow f'({1\over 2021})={1\over (1-1/2021)^2} =1+2\cdot {1\over 2021}+ 3\cdot ({1\over 2021})^2+ 4\cdot ({1\over 2021})^3+\cdots \\ \Rightarrow 1+2\cdot {1\over 2021}+ 3\cdot ({1\over 2021})^2+ 4\cdot ({1\over 2021})^3+\cdots= ({2021\over 2020})^2，故選\bbox[red,2pt]{(D)}$$

解答： $$\log_9 a= \log_{12} b=\log_{16}(a+b) \Rightarrow {\log a\over \log 9} ={\log b\over \log 12} ={\log (a+b)\over \log 16}  \Rightarrow \cases{\log 12\log a=\log 9\log b\\ \log 16\log b=\log 12\log(a+b)}\\，上二式交叉相加\Rightarrow (\log 12)(\log a+\log(a+b))=(\log 9+\log 16)\log b\\ \Rightarrow (\log 3+\log 4)(\log a+\log(a+b))=2(\log 3+\log 4)\log b \Rightarrow \log a+\log(a+b) =2\log b\\ \Rightarrow a(a+b)= b^2 \Rightarrow {a+b\over b} ={b\over a} \Rightarrow 1+{a\over b}={b\over a} \Rightarrow 1+{1\over x}=x,\text{where }x={b\over a} \\ \Rightarrow x^2-x-1=0 \Rightarrow x={1+\sqrt 5\over 2}，故選\bbox[red,2pt]{(E)}$$

解答： $$f(x)=\int_0^{3x} {1\over \sqrt{1+t^2}}\;dt \Rightarrow f'(x)={1\over \sqrt{1+(3x)^2}} \cdot (3x)'= {3\over \sqrt{1+9x^2}} \Rightarrow f'(1)={3\over \sqrt{10}}\\，故選\bbox[red,2pt]{(E)}$$

解答： $$\lim_{n\to \infty} \left( {\sin{\pi\over n}\over n} + {\sin{2\pi\over n}\over n} + {\sin{3\pi\over n}\over n} +\cdots + {\sin{n\pi\over n}\over n} \right) =\lim_{n\to \infty}\sum_{k=1}^n{1\over n}\sin ({k\over n}\pi) =\int_0^1\sin x\pi\;dx \\ =\left. \left[-{1\over \pi} \cos x\pi \right]\right|_0^1 ={1\over \pi} +{1\over \pi}={2\over \pi}，故選\bbox[red,2pt]{(E)}$$

解答： $$f(x)= (1+x)^n = 1+C^n_1x +C^n_2x^2 +\cdots +C^n_nx^n \\\Rightarrow {f(x)\over x}= {(1+x)^n\over x}={1\over x} +C^n_1 +C^n_2x +C^n_3x^2 +\cdots +C^n_n x^{n-1} \\ \Rightarrow {f(3)\over 3}= {4^n\over 3} ={1\over 3}+C^n_1 +3C^n_2 + 3^2C^n_3+ \cdots +3^{n-1}C^n_n \\ \Rightarrow C^n_1 +3C^n_2 + 3^2C^n_3+ \cdots +3^{n-1}C^n_n ={4^n\over 3} -{1\over 3}={4^n-1\over 3}，故選\bbox[red,2pt]{(A)}$$

解答： $$令S(n)=f(1)+f(2)+\cdots +f(n) \Rightarrow f(n)=S(n)-S(n-1)= n^2f(n)-(n-1)^2f(n-1)\\ \Rightarrow f(n)={(n-1)^2\over n^2-1}f(n-1) ={n-1\over n+1}f(n-1)  = {n-1\over n+1}\cdot {n-2\over n}f(n-2) = {n-1\over n+1}\cdot {n-2\over n} \cdot {n-3\over n-1}f(n-3)\\ =\cdots ={n-1\over n+1}\cdot {n-2\over n} \cdot {n-3\over n-1}\cdots{1\over 3}f(1) ={(n-1)! \over (n+1)!/2}f(1) ={2\over (n+1)n}\cdot 2021 \\ \Rightarrow f(2021) ={2\over 2022\cdot 2021}\cdot 2021={1\over 1011}，故選\bbox[red,2pt]{(B)}$$

解答： $$顯然p=q=2 \Rightarrow \cases{2+2=2^2\\ 2+7\times 2=4^2} \Rightarrow p^2+q^2=8，故選\bbox[red,2pt]{(A)}$$

解答： $$\cases{\sin x+\sin y={\sqrt 2/2}  \\ \cos x+\cos y=\sqrt 6/2  }  \Rightarrow \cases{2\sin {x+y\over 2} \cos {x-y\over 2} =\sqrt 2/2\\ 2\cos{x+y\over 2}\cos {x-y\over 2}=\sqrt 6/2}\Rightarrow \cases{\sin {x+y\over 2} \cos {x-y\over 2} =\sqrt 2 /4\\ \cos{x+y\over 2}\cos {x-y\over 2}=\sqrt 6/4}\\ 兩式相除\Rightarrow \tan {x+y\over 2} =\sqrt{2\over 6} ={\sqrt 3\over 3} \Rightarrow \tan (x+y)={2\sqrt 3/3\over 1-1/3} =\sqrt 3 \Rightarrow \sin(x+y)={\sqrt 3\over 2}\\，故選\bbox[red,2pt]{(E)}$$

解答： $$S=z+2z^2+ 3z^3+\cdots +36z^{36} \Rightarrow zS=z^2+2z^3+ 3z^4+\cdots +36z^{37}\\ \Rightarrow S-zS=z+z^2+\cdots +z^{36}-36z^{37} \Rightarrow (1-z)S= z(1+z+\cdots +z^{35})-36z^{37}\\ =z\cdot {1-z^{36}\over 1-z}-36z^{37} \Rightarrow S={z(1-z^{36})\over (1-z)^2}-36{z^{37}\over 1-z} \cdots(1)\\ z=\cos 10^\circ +i\sin 10^\circ \Rightarrow z^{36}=1 代入(1) \Rightarrow S=-36{z\over 1-z} \Rightarrow |S|^{-1}={1\over 36}\cdot {|1-z|\over |z|} \\ ={1\over 36}|1-z| ={1\over 36}|1-\cos 10^\circ+i\sin 10^\circ|={1\over 36}\sqrt{(1+\cos 10^\circ)^2+\sin^2 10^\circ} ={1\over 18}\sin 5^\circ\\，故選\bbox[red,2pt]{(C)}$$

解答： $$令\cases{\alpha =a-b\\ \beta=b-c \\ \gamma =c-a} \Rightarrow \cases{\alpha+\beta +\gamma=0 \\ \sqrt 3\alpha +3\beta +\gamma=0} \Rightarrow \cases{\gamma =-(\alpha+\beta)\\ \alpha+\beta=\sqrt 3\alpha+3\beta\Rightarrow \beta ={1-\sqrt 3\over 2}\alpha} \\ 因此{(a-b)(a-c)\over (b-c)^2} ={-\alpha \gamma \over \beta^2} ={\alpha(\alpha+\beta) \over \left( 1-\sqrt 3\over 2 \right)^2 \alpha^2} ={{3-\sqrt 3\over 2}\alpha^2 \over {4-2\sqrt 3\over 4}\alpha^2} ={3-\sqrt 3\over 2-\sqrt 3} ={(3-\sqrt 3)(2+\sqrt 3) \over (2-\sqrt 3)(2+\sqrt 3)}\\ =3+\sqrt 3，故選\bbox[red,2pt]{(D)}$$

解答： $$x^2-(y-1)^2=1 \Rightarrow 2x-2(y-1)y'=0 \Rightarrow y'={x\over y-1}\\過原點的切線L:y=mx \Rightarrow 切點P(t,mt) \Rightarrow \cases{P在雙曲線上\\ m=y'(P)} \Rightarrow \cases{t^2-(mt-1)^2=1 \cdots(1)\\ m=t/(mt-1) \cdots(2)}\\ 式(2) \Rightarrow t={m\over m^2-1} 代入(1)\Rightarrow {1\over m^2-1} =1 \Rightarrow m=\sqrt 2 (m\gt 0) \Rightarrow t=\sqrt 2 \\ \Rightarrow P(\sqrt 2,2)=(a,b) \Rightarrow \sin^{-1}{a\over b} =\sin^{-1}{1\over \sqrt 2} ={\pi\over 4}，故選\bbox[red,2pt]{(B)}$$

解答：

$$假設\cases{C(0,0) \\\overline{BC}=6} \Rightarrow \cases{P(2,0)\\ L_1=\overleftrightarrow{AC} :y=x\\ L_2= \overleftrightarrow{AB}: y=-{\sqrt 6+\sqrt 2\over \sqrt 6-\sqrt 2}(x-6)} \Rightarrow A=L_1\cap L_2=(3+\sqrt 3,3+\sqrt 3)\\ 作\overline{AQ} \bot \overline{BC}且Q在x軸上 \Rightarrow Q(3+\sqrt 3,0) \Rightarrow \cases{\overline{AQ}=3+\sqrt 3\\ \overline{PQ}=1+\sqrt 3} \Rightarrow {\overline{AQ}\over \overline{PQ}} ={3+\sqrt 3\over 1+\sqrt 3} =\sqrt 3\\ \Rightarrow \angle APB=60^\circ，故選\bbox[red,2pt]{(B)}$$

解答： $$若n\gt 1 \Rightarrow n\cdot 2^{n-1}+1為奇數 \Rightarrow n\cdot 2^{n-1}+1=(2k-1)^2,k\in \mathbb{N}\\ \Rightarrow n\cdot 2^{n-1}=(2k-1)^2-1 =(2k-2)2k = 2^2k(k-1) \Rightarrow n\cdot 2^{n-3}= k(k-1)\\而\;n\cdot 2^{n-3}=一奇一偶相乘；而2^{n-3}的唯一質因數是2，所以n是奇數；\\\\先討論這兩個數n,2^{n-3}的大小，顯然當n\ge 6時，2^{n-3}\gt n \Rightarrow \cases{k=2^{n-3} \\k-1=n }\Rightarrow k\gt k-1矛盾\\ 因此只要考慮1\le n\le 5，只有n=5時，n\cdot 2^{n-1}+1=81=9^2符合要求，故選\bbox[red,2pt]{(A)}$$

解答： $$f(2x)=3f(x) \Rightarrow f(x)={1\over 3}f(2x) \Rightarrow \int_0^1 f(x)\;dx= \int_0^1{1\over 3}f(2x)\;dx =1\cdots(1)\\ 令u=2x,則du=2dx\;代入(1) \Rightarrow \int_0^2 {1\over 6}f(u)\;du =1 \Rightarrow \int_0^2 f(u)\;du=6 \equiv \int_0^2 f(x)\;dx=6\\ \Rightarrow \int_0^1 f(x)\;dx +\int_1^2 f(x)\;dx =6 \Rightarrow 1+\int_1^2 f(x)\;dx =6 \Rightarrow \int_1^2 f(x)\;dx =5，故選\bbox[red,2pt]{(C)}$$

解答： $$令\cases{\overline{AD}=h \\ \angle BAD=\theta_1\\ \angle DAC=\theta_2} \Rightarrow \cases{\tan \theta_1=3/h\\ \tan \theta_2=17/h} \Rightarrow \tan(\theta_1+\theta_2) =\tan \angle BAC={\tan \theta_1+\tan \theta_2\over 1-\tan \theta_1\tan \theta_2} \\ \Rightarrow {22\over 7}={20/h\over 1-{51/h^2}} ={20h\over h^2-51} \Rightarrow 11h^2-70h-561=0 \Rightarrow (11h+51)(h-11)=0 \Rightarrow h=11 \\ \triangle ABC={1\over 2}\overline{BC}\cdot h={1\over 2}\cdot 20\cdot 11=110，故選\bbox[red,2pt]{(A)}$$

二、多選題

解答： $$(A)\bigcirc: 2,2^2,2^3,2^4,2^5,\dots的個位數字為2,4,8,6,2,\dots \Rightarrow 循環數為4，而2021=4\times 505+1\\ \qquad \Rightarrow 2^{2021}的個位數字為2\Rightarrow 2^{2021}-1的個位數字為1\\(B)\times: 同理，17^n的個位數字為7,9,3,1,7,\dots \Rightarrow 循環數為4，而110=4\times 27+2 \\ \qquad \Rightarrow 17^{110}的個位數字是9 \Rightarrow 17^{110}+3的個位數字是2\ne 6\\(C)\times: \log_{2021}{1\over 2}\cdot \log_{2020}{1\over 3}\cdot \log_{2019}{1\over 4}\cdot \cdots\log_{2}{1 \over 2021} ={-\log 2\over \log 2021} \cdot {-\log 3\over \log 2020} \cdot \cdots {-\log 2021\over \log 2 }\\ \qquad =(-1)^{2020}\cdot {\log 2021\cdot \log 2020\cdot \cdots \log 2\over \log 2021\cdot \log 2020\cdot \cdots \log 2}=1 \ne -1\\ (D) \bigcirc: 7=1+6= 6+1 =2+5= 5+2=3+4=4+3，共有6種情形，機率為6/36=1/6\\(E)\times: \cases{\sin x為奇函數\\ 1+x^2為偶函數} \Rightarrow {\sin x\over 1+x^2}為奇函數\Rightarrow \int_{-\pi}^\pi {\sin x\over 1+x^2}\;dx=0\\，故選\bbox[red,2pt]{(AD)}$$

解答： $$(A)\times: \cases{\langle a_n \rangle =1,-1,0,0,\dots\\ \langle b_n \rangle =1, 1,1,1,\dots} \Rightarrow \cases{\sum a_n=0 收斂\\ \sum b_n=\infty發散}\\(B)\times: \cases{\sum 1/n 發散\\ \sum 1/n^2=\pi^2/6 收斂} \\(D)\times: {1\over n\sqrt{n+1} +(n+1)\sqrt n} ={(n+1)\sqrt n-n\sqrt{n+1}\over (n\sqrt{n+1} +(n+1)\sqrt n)((n+1)\sqrt n-n\sqrt{n+1})} \\\qquad ={(n+1)\sqrt n-n\sqrt{n+1}\over n(n+1)}  ={\sqrt n\over n} -{\sqrt{n+1}\over n+1} ={1\over \sqrt n}-{1\over \sqrt{n+1}} \\\qquad \Rightarrow \sum_{n=1}^\infty\left( {1\over \sqrt n}-{1\over \sqrt{n+1}} \right)=1 \ne 2\\，故選\bbox[red,2pt]{(CE)}$$

解答： $$(B)\times:A=I \Rightarrow AB=BA，B即為任意矩陣，不一定能對角化\\(E)\times: (A+B)(A-B) =A^2-AB+BA-B^2 \ne A^2-B^2(除非AB=BA)\\，故選\bbox[red,2pt]{(ACD)}$$

解答： $${a\over \sin A}=2R={2a \sqrt{bc}\over b+c} \Rightarrow \sin A={b+c\over 2\sqrt{bc}} \Rightarrow \cos^2 A=1-\sin^2 A=1-{(b+c)^2\over 4bc} =-{(b-c)^2\over 4bc} \ge 0 \\ \Rightarrow b=c \Rightarrow \sin A={b+b\over 2\sqrt {b^2}} =1 \Rightarrow A=90^\circ \Rightarrow {a\over \sin A}=a=2R，故選\bbox[red,2pt]{(BCE)}$$

解答： $$(A)\times: 顯然錯誤\\(B)\bigcirc: 均值定理\\(C)\times: \int_a^b |f(x)|\;dx \gt 0 \\(D)\bigcirc: \left|\int_a^b f(x)\;dx\right|= |0|=0 \\(E) 反例:\cases{f(x)=x\\ a=-2\\ b=2\\ c=-1} \Rightarrow \cases{\int_a^b f(x)\;dx =0\\ \int_a^c f(x)\;dx \lt 0}\\，故選\bbox[red,2pt]{(BD)}$$

============= END ================

解題僅供參考，其他教甄試題及詳解