110年鳳山高中教甄-數學詳解

110 年國立鳳山高級中學教師甄選初試筆試試題

一、填充題:(每題 5 分，共 70 分)

解答：

$$過P(1,2)直線L_1:y=m(x-1)+2 \Rightarrow \cases{A({m-2\over m},0)\\ B(0,2-m)} \Rightarrow \overrightarrow{AB}=({2-m\over m},2-m)\\ L_2:3x+y+3=0的方向量\vec u=(-1,3) \Rightarrow \overline{CD}=\overrightarrow{AB} 在\vec u的投影長= {\overrightarrow{AB} \cdot \vec u \over |\vec u|} = {7-(3m+2/m)\over \sqrt{10}} \\ \Rightarrow 3m={2\over m}(m\lt 0)有極值\Rightarrow m=-\sqrt{2\over 3} \Rightarrow \overline{CD}={7+4\sqrt{3/2}\over \sqrt{10}} = \bbox[red,2pt]{7\sqrt{10}+4\sqrt{15}\over 10}$$

解答：$$\cases{f(0)=d\\ f(1)=a+b+c+d} \Rightarrow \cases{f(0-2)=4-d\\ f(-2-1)=4-a-b-c-d} \\ \Rightarrow \cases{-8a+4b-2c+d=4-d\\ -27a+9b-3c+d=4-a-b-c-d} \Rightarrow \cases{4a-2b+c-d=-2\\ 13a-5b+c-d=-2} \Rightarrow 9a=3b \\ \Rightarrow \cases{a=3,b=9 \Rightarrow d=1(\because ad=3) \Rightarrow c=5\\ a=-3,b=-9 \Rightarrow d=-1 \Rightarrow c=-9} \Rightarrow (a,b,c)=(3,9,5,1) 或(-3,-9,-9,-1)\\ y=f(x)與水平線有三交點 \Rightarrow f'(x)= 3ax^2 +2bx+c=0有二相異實根 \Rightarrow 4b^2-12ac \gt 0\\ \Rightarrow b^3\gt 3ac \Rightarrow 共有(3,9,5,1)符合此要求 \Rightarrow \bbox[red,2pt]{f(x)=3x^3+9x^2+5x+1}$$

解答：

$$令\triangle OAB為邊長5的正\triangle \Rightarrow \cases{O(0,0)\\ B(5,0)\\ A({5\over 2},{5\over 2}\sqrt 3)}，又\cases{\overrightarrow{OM}={1\over 2}\overrightarrow{MA} \\\overrightarrow{ON}={3\over 2}\overrightarrow{NB}} \Rightarrow \cases{M=A/3=({5\over 6},{5\over 6}\sqrt 3) \\ N=3B/5=(3,0)} \\\Rightarrow \cases{L_1=\overleftrightarrow{AN}:y=-5\sqrt 3x+15\sqrt 3\\ L_2=\overleftrightarrow{BM}: y=-{\sqrt 3\over 5}x+\sqrt 3} \Rightarrow P({35\over 12},{5\sqrt 3\over 12}) \Rightarrow \overline{OP} ={5\sqrt{13}\over 6} \\ 因此\overrightarrow{OA}\cdot \overrightarrow{OP} = |\overrightarrow{OA}| |\overrightarrow{OP}| \cos \angle AOP \Rightarrow {175\over 24}+{75\over 24} =5\cdot {5\sqrt{13}\over 6}\cos \angle AOP \\ \Rightarrow \cos \angle AOP ={250\over 24}\times {6\over 25\sqrt{13}} ={5\over 2\sqrt{13}} \Rightarrow \sin \angle AOP = \bbox[red,2pt]{3\sqrt{39}\over 26}$$

解答：

$$20個點任取3點，有C^{20}_3=1140種情形；其中需扣除在一直線上的三點，才能形成一個\triangle；\\ 不能形成三角形的情況如下表:\\\begin{array}{l|ccc}類型 & 長度 & 數量 & 情形\\\hline 水平線& 4 & 5 & 5C^4_3=20\\\hdashline 垂直線& 5 & 4 & 4C^5_3=40\\ \hdashline 斜率=1& 3 & 2 & 2C^3_3=2\\ & 4 & 2 & 2C^4_3=8\\\hdashline 斜率=-1 & 3 & 2 & 2C^3_3=2\\ & 4 & 2 & 2C^4_3=8\\\hdashline 斜率=2& 3 & 2 & 2C^3_3=2\\\hdashline 斜率=-2& 3 & 2 & 2C^3_3=2\end{array}\\ \Rightarrow 共有20+40 +2+8+2+8+2+2 = 84種情形不能形成三角形\\ \Rightarrow 構成三角形的機率=1-{84 \over 1140} ={1056\over 1140} =\bbox[red,2pt]{88\over 95}$$

解答：$$F(x)=\int_0^x (x-t)\sin^2 t\;dt =\int_0^x x\sin^2 t\;dt-\int_0^x t\sin^2 t\;dt \\ \Rightarrow F'(x)={\partial\over \partial x}\int_0^x x\sin^2 t\;dt-x\sin^2x={\partial\over \partial x}\left[{x^2\over 2}-{x\sin(2x)\over 4}\right]-x\sin^2x \\ =x-{\sin(2x)+2x\cos(2x)\over 4}-x\sin^2 x \Rightarrow F'({\pi\over 3}) ={\pi\over 3}-({\sqrt 3\over 2}-{\pi\over 3})\cdot {1\over 4}-{\pi\over 4} =\bbox[red,2pt]{{\pi\over 6}-{\sqrt 3\over 8}}$$

解答：

$$f(x)= 2\sin x+3\cos x=\sqrt{13}\left({2\over \sqrt{13}} \sin x +{3\over \sqrt{13}}\cos x \right) =\sqrt{13}\sin(x+\theta),其中\cases{\sin \theta=3/\sqrt{13}\\ \cos \theta=2/\sqrt{13}} \\ \Rightarrow \cases{f(0)=\sqrt{13}\sin \theta= 3\\ f(\pi/2)=\sqrt{13}\sin(\pi/2+\theta)=\sqrt{13}\cos \theta=2 \\ f(x)最大值=\sqrt{13},x+\theta=\pi/2,此時0\lt x\lt \pi/2}\\ 因此兩圖形\cases{y=f(x)\\ y=k} 有兩交點 \Rightarrow \bbox[red,2pt]{3\le k\lt \sqrt{13}}$$

解答：$$\sum_{k=1}^\infty  {1\over k(k+2)(k+3)} =\sum_{k=1}^\infty  \left({1\over k(k+2)}-{1\over k(k+3)} \right)= \sum_{k=1}^\infty  \left({1\over 2}({1\over k}-{1\over k+2})-{1\over 3}({1\over k}-{1\over k+3}) \right) \\ ={1\over 2}\sum_{k=1}^\infty \left({1\over k}-{1\over k+2}\right)- {1\over 3}\sum_{k=1}^\infty \left({1\over k}-{1\over k+3} \right)= {1\over 2}\left( 1+{1\over 2}\right) -{1\over 3}\left( 1+{1\over 2}+{1\over 3}\right) \\ ={1\over 2}\cdot {3\over 2} -{1\over 3}\cdot {11\over 6} =\bbox[red,2pt]{5\over 36}$$

解答：$$C^{20}_4+ C^{20}_5=C^{21}_5=C^{21}_{16}= {21!\over 5!16!} =20349=C^{20349}_1= C^{20349}_{20348} \\ \Rightarrow (n,k)=\bbox[red, 2pt]{(21,5),(21,16),(20349,1),(20349,20348)}$$

解答：

$$用手算(如上圖)可求得\cases{\sqrt{433}\approx 20.808\\ \sqrt{434}\approx 20.832}\\只考慮小數部分，即0.80\lt s\le 0.83 ，而  {2\over 3}(0.67),{3\over 4}(0.75),{4\over 5}(0.8), {5\over 6}(0.833),{5\over 7}(0.71),{6\over 7}(0.857),\\{7\over 8}(0.875), {7\over 9}(0.777),{8\over 9}(0.888)不是太大就是太小；\\而{9\over 11}=0.818符合要求 \Rightarrow 20{9\over 11}=\bbox[red, 2pt]{229\over 11}符合要求$$

解答：

$$格線紙有5\times 6=30方格，假設被剪去後的左半部方格數量，由上往下分別是2、x、y、z、w；\\依題意需滿足條件:\cases{2+x+y+z+w=15\\1\le x,y,z,w\le 5 \\x,y,z,w\in \mathbb{N}} \Rightarrow x'+y'+z'+w'=9，其中\cases{x=x'+1\\ y=y'+1\\ z=z'+1\\ w=w'+1} \\ 因此共有H^4_9組解，但需扣除x,y,z,w=6,7,8,9，C^4_1H^4_4 \Rightarrow H^4_9-C^4_1H^4_4=\bbox[red,2pt]{80}$$

解答：$$\left|\overrightarrow{CP}-\alpha\overrightarrow{AP}-\beta \overrightarrow{BP} \right| =\left| (4\alpha-\beta-6, -5\alpha+2\beta-5, 2\alpha-2\beta+4)\right| \\令f(\alpha,\beta)=(4\alpha-\beta-6)^2+ (-5\alpha+2\beta-5)^2+( 2\alpha-2\beta+4)^2\\ 則 \cases{f_\alpha=0 \\ f_\beta=0} \Rightarrow \cases{5\alpha-2\beta+1=0 \\ 6\alpha-3\beta+4=0} \Rightarrow \cases{\alpha=5/4\\ \beta=14/3} \Rightarrow \alpha-\beta= \bbox[red,2pt]{-3}$$

解答：

$$由長除法(見上圖)可得\cases{a+e=d+1\\ f+b=d+1\\ g+c=d+1}；由於a-g\in \{0-9\}，且互異\\ \Rightarrow \begin{array}{} (a,e),(b,f),(c,g) & d+1(d) & 數量\\\hline  (8,2),(7,3),(6,4) & 10(9) & 2^3\times 3!=48 \\ (9,0),(7,2),(6,3),(5,4) & 9(8) & C^4_3\times 2^3\times 3!=192 \\ (8,0),(6,2),(5,3) & 8(7) & 2^3\times 3!=48\\ (7,0),(5,2),(4,3) & 7(6) & 2^3\times 3!= 48\\\hline \end{array}\\ 因此共有48+192+48+48 = \bbox[red,2pt]{336}個多項式$$

解答：$$ { \sin(3x)\sin^3 x +\cos (3x)\cos^3x\over 1+\cos (4x)} ={ \sin^2 x(\sin(3x)\sin x) +\cos^2 x(\cos (3x)\cos x)\over 1+(2\cos^2 (2x)-1)}  \\={{1\over 2}\sin^2 x(\cos(2x)-\cos(4x))+{1\over 2}\cos^2 x(\cos (2x)+\cos(4x))\over 2\cos^2(2x)} \\={ \cos(2x)(\sin^2 x+\cos^2 x)+\cos(4x)(\cos^2 x-\sin^2x) \over 4\cos^2(2x)}= { \cos(2x) +\cos(4x)\cos(2x) \over 4\cos^2(2x)}\\ = {1+\cos(4x)\over 4\cos (2x)}\\ \Rightarrow 原式= {1+\cos(4x)\over 4\cos (2x)}+{1\over 6\cos(2x)} ={ 5+ 3\cos (4x)\over 12\cos(2x)} = { 5+ 3(2\cos^2 (2x)-1)\over 12\cos(2x)} ={2+6\cos^2(2x)\over 12\cos(2x)} \\={1\over 6\cos(2x)} +{\cos(2x)\over 2} \ge 2\sqrt{{1\over 6\cos(2x)}\cdot {\cos(2x)\over 2}} =2\sqrt{1\over 12}={1\over \sqrt 3}\\ \Rightarrow 最大值為\bbox[red, 2pt]{\sqrt 3\over 3}$$

解答：$$令\cases{\overline{AB}=c\\ \overline{BC}= a\\ \overline{CA}=b } \Rightarrow \cases{6\overrightarrow{HA}\cdot \overrightarrow{AB} =-6\overrightarrow{AH}\cdot \overrightarrow{AB} =-6(b^2+c^2-a^2)/2\\ 3\overrightarrow{HB}\cdot \overrightarrow{BC} =-3\overrightarrow{BH}\cdot \overrightarrow{BC} =-3(a^2+c^2-b^2)/2\\ 4\overrightarrow{HC}\cdot \overrightarrow{CA} =-4\overrightarrow{CH}\cdot \overrightarrow{CA} =-4(a^2+b^2-c^2)/2 }\\ 依題意:6\overrightarrow{HA}\cdot \overrightarrow{AB} =3\overrightarrow{HB}\cdot \overrightarrow{BC} =4\overrightarrow{HC}\cdot \overrightarrow{CA}\\ \Rightarrow 3(b^2+c^2-a^2)={3\over 2}(a^2+c^2-b^2)=2(a^2+b^2-c^2) \Rightarrow \cases{3a^2-3b^2-c^2=0 \\ 5a^2-b^2-5c^2=0} \\ \Rightarrow \cases{a^2=b^2+c^2/3\\ 12b^2=10c^2} \Rightarrow \cos  A= {b^2+c^2-a^2 \over 2bc} ={2c^2/3 \over 2bc} ={c\over 3b} ={1\over 3}\cdot \sqrt{6\over 5} ={\sqrt{30}\over 15} \\ \Rightarrow \sin A=\sqrt{1-{30\over 225}} = \bbox[red,2pt]{\sqrt{195}\over 15}$$

二、計算證明題:(每題 10 分，共 30 分)

解答：$$\cases{a_1= a_2=1\\a_3= 2\\ a_n={1\over a_{n-3}}(a_{n-2}\cdot a_{n-1}+3)} \Rightarrow a_4={1\over a_1}(a_2\cdot a_3+3)  =(2+3)/1 =5\\ 又 a_na_{n-3} =a_{n-2}a_{n-1}+3 \cdots(1) \Rightarrow  a_{n-1}a_{n-4}= a_{n-3}a_{n-2}+3 \cdots(2)\\ 式(1)-式(2) \Rightarrow a_na_{n-3}-a_{n-1}a_{n-4} =a_{n-2}a_{n-1}- a_{n-3}a_{n-2} \\\Rightarrow a_na_{n-3}+ a_{n-3}a_{n-2}=a_{n-2}a_{n-1}+a_{n-1}a_{n-4} \Rightarrow a_{n-3}(a_n+a_{n-2}) =a_{n-1}(a_{n-2}+ a_{n-4}) \\ \Rightarrow {a_n+a_{n-2}\over a_{n-1}} ={a_{n-2} +a_{n-4}\over a_{n-3}} ={a_{n-4}+a_{n-6}\over a_{n-5}}= \cdots =\begin{cases} {a_4-a_2 \over a_3}={5-1\over 2}=2 & n是偶數\\ {a_3-a_1\over a_2}={2-1\over 1}=1& n是奇數\end{cases} \\ \Rightarrow a_n 為整數,n\ge 1，\bbox[red,2pt]{故得證}$$

解答：$$\Gamma: {x^2\over 25}+ {y^2\over 16}=1 \Rightarrow {2x\over 25}+{yy'\over 8}=0 \Rightarrow y'=-{16\over 25}\cdot {x\over y}\\ 因此切點A\in \Gamma \Rightarrow A(5\cos \theta,4\sin \theta) \Rightarrow 切線L:y=-{16\over 25}\cdot {5\cos \theta \over 4\sin \theta}(x-5\cos\theta)+4\sin\theta\\ \Rightarrow L:4\cos \theta x+5\sin\theta y=20 \Rightarrow R=d(O,L) ={20 \over \sqrt{16\cos^2\theta +25\sin^2 \theta}} \\ \Rightarrow {400\over R^2} =16\cos^2\theta +25\sin^2 \theta \cdots(1);又 \overline{OA}^2 = 25\cos^2\theta +16\sin^2\theta \cdots(2)\\ 式(1)+式(2) \Rightarrow {400\over R^2} +\overline{OA}^2 =41 \Rightarrow \overline{AB}^2=\overline{OA}^2-R^2=(41-{400\over  R^2})-R^2 =41-(R^2+{400\over R^2})\\算機不等式: R^2+{400\over R^2} \ge 2\sqrt{R^2\cdot {400\over R^2}} =40 \Rightarrow \overline{AB}^2 \le 41-40=1\\ \Rightarrow \overline{AB}最大值為1，此時R^2=20，即R=\sqrt{20}=2\sqrt 5；也就是\bbox[red,2pt]{R=2\sqrt 5時，\overline{AB}有最大值1}$$

解答：

$$\left[ 3a(f(x))^2+ 2bf(x)+c\right](f(x)-x_2)=0 \Rightarrow \cases{f(x)-x_2=0 \cdots(1)\\  3a(f(x))^2+ 2bf(x)+c=0\cdots(2) }\\ 式(1) \Rightarrow f(x)=x_2，由於x_1 \lt y=f(x)=x_2\lt x_3 \Rightarrow 兩圖形\cases{y=x_2\\ y=f(x)}有三相異交點\\ 式(2) \Rightarrow 3a(f(x))^2+ 2bf(x)+c=0 \Rightarrow 6af(x)f'(x)+2bf'(x)=0 \Rightarrow f'(x)(3af(x)+b)=0\\ \Rightarrow \cases{f'(x)=0 \Rightarrow x_1,x_2為其根\\ f(x)=-b/3a \Rightarrow 反曲點x_0為其根，因為f''(x)=6ax+2b且\cases{f''(x_1)\lt 0\\ f''(x_3)\gt 0\\ f''(x_0)=0}} \\ \Rightarrow 式(2)有三個相異極值 \Rightarrow 式(2)有四個相異根\Rightarrow 式(1)+式(2)共有3+4=\bbox[red, 2pt]{7}個根$$

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