111年台中一中教甄-數學詳解

臺中第一高級中等學校111學年度第1次教師甄選 

一、填充題(甲)(每題 4 分，共 24 分。 )

解答： $$當[x]=1或2或3時，[x]\cdot \{x\} \not \ge 3；\\ 考慮[x]=4時，[x]\cdot \{x\}=4 \cdot \{x\} =3 \Rightarrow \{x\}={3\over 4}符合條件；\\因此x=4{3\over 4} =\bbox[red, 2pt]{19\over 4}為符合條件之最小正數$$

解答：

$$作\cases{\overline{MN} \parallel \overline{AB} \\\overline{ST} \parallel \overline{BC} \\ \overline{UV} \parallel \overline{AC} }，見上圖；則\triangle PUM、\triangle PSV、 及\triangle PTN皆為正\triangle，且1,2,3分別為正\triangle 的高；\\ 因此\cases{\overline{PU} =\overline{PM} =\overline{UM}=2/\sqrt 3\\ \overline{PV}=\overline{PS} =\overline{VS} =4/\sqrt 3\\ \overline{PN} = \overline{PT} =\overline{NT} = 6/\sqrt 3} \Rightarrow \cases{ \overline{PA}^2 =2^2 + (6/\sqrt 3+2/\sqrt 3)^2 = 76/3\\ \overline{PB}^2 = 1^2 +(4/\sqrt 3+1/\sqrt 3)^2 =28/3\\ \overline{PC}^2 = 3^2 +(2/\sqrt 3+3/\sqrt 3)^2 =52/3} \\ \Rightarrow \overline{PA}^2 :\overline{PB}^2 :\overline{PC}^2 = \bbox[red, 2pt]{19:7:13}$$

解答： $$a_n+ 3S_nS_{n-1} =S_n-S_{n-1}+ 3S_nS_{n-1} =0 \Rightarrow {1\over S_{n-1}}-{1\over S_n}+3=0 \Rightarrow b_n=b_{n-1}+3, b_n= {1\over S_n} \\ \Rightarrow b_{2022} = b_{2021}+3 = b_{2020} +2\cdot 3 = \cdots = b_1+2021\cdot 3= 4+ 2021\cdot 3 = \bbox[red, 2pt]{6067}$$

解答： $$依題意\cases{A(0,16)\\ B(5,0)\\ C(0,0)\\ D(x,y)}，並取E(0,1)，則滿足{1\over 4}\overline{AD} =\overline{DE}，即{1\over 16}(x^2+(y-16)^2) = x^2+(y-1)^2\\ \Rightarrow x^2+y^2 =16，也就是D的軌跡剛好就是圓\Gamma；因此{1\over 4}\overline{AD} +\overline{BD} =\overline{DE} +\overline{BD} \ge \overline{BE} =\sqrt{26} \\ \Rightarrow 最小值為\bbox[red, 2pt]{\sqrt{26}}$$

解答： $$\cases{f(x+3)\le f(x)+3 \\ f(x)+2 \le f(x+2)} \Rightarrow f(x+3)-3 \le f(x)\le f(x+2)-2 \Rightarrow f(x+3)\le f(x+2)+1\\ \Rightarrow  f(2)\le f(1)+1 =4 \Rightarrow f(3)\le f(2)+1= 5  \cdots(1) \\ 又 f(x)\le f(x+2)-2 \Rightarrow f(1)\le f(3)-2 \Rightarrow 5\le f(3) \cdots(2)\\ 由(1)及(2)可得f(3)=5；同理 \cases{f(2)\le f(1)+1 \\ f(3)\le f(2)+1} \Rightarrow 4\le f(2)\le 4 \Rightarrow f(2)=4\\ 因此f(n)=n+2 \Rightarrow f(2022)= \bbox[red, 2pt]{2024}$$

解答： $$令\cases{u=\ln x\\ dv = x^2\,dx} \Rightarrow \cases{du ={1\over x}\,dx\\[1ex] v={1\over 3}x^3} \Rightarrow \int x^2\ln x\,dx = {1\over 3}x^3\ln x-{1\over 3}\int x^2\, dx \\= \bbox[red, 2pt]{{1\over 3}x^3\ln x-{1\over 9}x^3+C}，其中C 為常數$$

二、填充題(乙)(每題 6 分，共 54 分。 )

解答： $$在複數平面上，A_i 為z^{111}=1的根，i=1-111 \Rightarrow z^{111}-1= (z-A_1)(z-A_2)\cdots (z-A_{111})\\ \Rightarrow |z^{111}-1|= |\cos 111\theta +i\sin 111\theta-1|= \overline{PA_1} \times \overline{PA_2}\times \cdots \times \overline{PA_{111}} \\ 而|\cos 111\theta +i\sin 111\theta-1| \le 2，因此最大值為\bbox[red, 2pt]{2}，此時\theta = {\pi \over 111}$$

解答：

$$不失一般性，假設\cases{D(0,0)\\ B(-5,0)\\ C(5,0) \\ \angle ACB=\theta} \Rightarrow \overline{BN}= {4\over 5}\times \overline{BC} =8 \Rightarrow \overline{CN}=6\Rightarrow \cases{\sin \theta = 4/5\\ \cos \theta = 3/5} \\ \Rightarrow N=(\overline{CD}-\overline{CN}\cos \theta,\overline{CN}\sin \theta) = (7/5,24/5)\\ 又\triangle ABC \sim \triangle BNC (AAA) \Rightarrow \cfrac{\overline{AB}}{ \overline{BC}} =\cfrac{\overline{BN}}{\overline{CN}} \Rightarrow   \overline{AB}= 10\times 8\div 6 =40/3 \Rightarrow A(-5,{40\over 3})\\ \Rightarrow \cases{L_1= \overleftrightarrow{AD}: 8x+3y=0\\ L_2= \overleftrightarrow{BN}: 4y=3x+15} \Rightarrow M= L_1\cap L_2 = (-{45\over 41},{120\over 41}) \Rightarrow \overline{BM} ={200\over 41} \\\Rightarrow \overline{MN} =8-\overline{BM}={128\over 41} \Rightarrow \cfrac{\overline{BM}}{ \overline{MN} } ={200\over 128} =\bbox[red, 2pt]{25\over 16}$$

解答： $$\cfrac{k\cdot 2^k}{(k+1)(k+2)} =\cfrac{a}{k+1}+ \cfrac{b}{k+2} \Rightarrow a(k+2)+b(k+1) = (a+b)k+ 2a+b = k\cdot 2^k \\ \Rightarrow \cases{a+b= 2^k \\ 2a+b=0} \Rightarrow \cases{a=-2^k\\ b=2\cdot 2^k = 2^{k+1}} \Rightarrow \cfrac{k\cdot 2^k}{(k+1)(k+2)} = \cfrac{2^{k+1}}{k+2} -\cfrac{2^k}{k+1} \\ \Rightarrow A_n = \sum_{k=1}^n \left(\cfrac{2^{k+1}}{k+2} -\cfrac{2^k}{k+1} \right)= \left(\cfrac{2^{2}}{3} -1 \right)+  \left(\cfrac{2^{3}}{4} -\cfrac{2^{2}}{3} \right)+ \cdots +\left(\cfrac{2^{n+1}}{n+2} -\cfrac{2^{n}}{n+1} \right)\\ =\cfrac{2^{n+1}}{n+2}-1；又B_n= \sum_{k=1}^n 2^k =2^{n+1}-2，因此|(n+2)A_n -B_n| = |2^{n+1}-(n+2)-2^{n+1}+2| \\ =|-n| \gt 2022 \Rightarrow n=\bbox[red, 2pt]{2023}$$

解答： $$(a_{n+1})^2 +(a_{n})^2+1 = 2(a_{n+1}\cdot a_n+ a_{n+1}+ a_n)\\ \Rightarrow (a_{n+1})^2 +(a_{n})^2+1-2a_{n+1}\cdot a_n +2a_n -2a_{n+1} =4a_n \Rightarrow (a_{n+1}-a_n-1)^2 =4a_n\\ \Rightarrow a_{n+1}-a_n-1=2\sqrt{a_n} \Rightarrow a_{n+1} =a_n+2\sqrt{a_n}+1 =(\sqrt a_n+1)^2 \Rightarrow \sqrt {a_{n+1}} =\sqrt a_n+1 \\ \Rightarrow \sqrt a_n = \sqrt{a_{n-1}}+1 =\sqrt{a_{n-2}}+2 =\sqrt a_1 +(n-1) =n \Rightarrow a_n = n^2\\ \Rightarrow S_n = \sum_{k=1}^n a_k = \sum_{k=1}^n k^2 ={1\over 6}n(n+1)(2n+1) \Rightarrow \lim_{n\to \infty }\cfrac{S_n}{na_n} =\lim_{n\to \infty }\cfrac{n(n+1)(2n+1)}{6n^3} \\ =\bbox[red, 2pt]{1\over 3}$$

解答： $$由\cases{x\gt 0\\ x,y,z\in \mathbb{R}}可假設\cases{x=\tan A\\ y=\tan B \\ z=\tan C}，其中\cases{ 0\lt A\lt \pi/2 \\ -\pi/2 \lt B,C\lt \pi/2} \\ 因此\cases{5\left(x+{1\over x}\right) = 12\left(y+{1\over y}\right) =13\left(z+{1\over z}\right) \\ xy +yz +zx=1} \Rightarrow \cases{5\cdot {2\over \sin 2A} =12\cdot {2\over \sin 2B} =13\cdot {2\over \sin 2C} \\ \tan A\tan B+ \tan B\tan C+ \tan C\tan A=1} \\ \Rightarrow \cases{{5\over \sin 2A} ={12\over \sin 2B} ={13\over \sin 2C} \\ A+B+C = 90^\circ(或\pm 180^\circ)} \Rightarrow \cases{\tan 2A= 5/12\\ \tan 2B=12/5\\ \tan 2C=\infty} \Rightarrow \cases{\tan A=1/5 \\ \tan B= 2/3\\\tan C=1} \\ \Rightarrow (x,y,z)= \bbox[red, 2pt]{\left({1\over 5},{2\over 3},1\right)}$$

解答： $$\left((4\sqrt 6)^2-(2\sqrt 6)^2\right)\cdot {\sqrt 3\over 4}\cdot 4= \bbox[red, 2pt]{72\sqrt 3}\\ \href{https://m.tgjsw.com/k12/exam/detail/2747654.html}{這裡有圖及詳解}$$

解答： $$令\cases{a+ 2b+c = x\\ a+b+2c = y\\ a+b+3c= z} \Rightarrow \cases{a= -x+5y-3z\\ b=x-2y+z \\ c=-y+z} \Rightarrow \cases{a+3c=-x+2y \\ 4b= 4x-8y+4z\\ 8c = -8y+8z} \\ \Rightarrow {a+3c \over a+2b+c} +{4b\over a+b+2c}-{8c\over a+b+3c}+17 ={-x+2y\over x} +{4x-8y+4z\over y}-{-8y+8z\over z}+17 \\=-1+{2y\over x} -8+{4(x+z)\over y}-8+{8y\over z}+17 = \left( {2y\over x}+{4x\over y} \right)+ \left({4z\over y}+{8y\over z} \right) \\\ge 2\sqrt{{2y\over x}\cdot {4x\over y}} +2\sqrt{{4z\over y} \cdot {8y\over z}}  =4\sqrt 2+8\sqrt 2=12\sqrt 2 \Rightarrow 最小值為\bbox[red, 2pt]{12\sqrt 2}$$

解答：

$$\cases{直角\triangle OAB =\sqrt 3/2\\ 扇形OBC = \pi/6} \Rightarrow 4\times{\sqrt 3\over 2}+2 \times {\pi \over 6} = \bbox[red, 2pt]{2\sqrt 3+{\pi \over 3}}$$

解答： $$看英文比較容易懂，\href{https://math.stackexchange.com/questions/2052209/least-positive-integer-n-such-that-any-set-of-n-pairwise-relatively-prime}{說明在此}，該題不含1，所以n=15，本題含1，故n=\bbox[red,2pt]{16}\\第i個質數a_i及其平方數:\\ \begin{array}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c}i&1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15\\ \hline a_i &2 & 3 & 5 & 7 & 11 & 13 & 17 & 19 & 23 & 29 & 31 & 37 & 41 & 43 & 47\\ \hline a_i^2 &4 & 9 & 25 & 49 & 121 & 169 & 289 & 361 & 529 & 841 & 961 & 1369 & 1681 & 1849 & 2209\end{array} \\ 從上表中，我們可以找到15個數:1,a_1^2, a_2^2,\dots, a_{14}^2，滿足兩兩互質且小於等於2022；\\若再加上a_{15}^2，就有16個數滿足兩兩互質，但a_{15}^2 \gt 2022，因此符合題意的n值可能是16；\\現在要來證明n=16，也就是任取16個兩兩互質的數，其中至少有一個質數。\\若此16個數依序為1,x_1,x_2,\dots , x_{15}\le 2022，滿足兩兩互質且都不是質數，\\因為x_1\sim x_{15}都不是質數，因此每一個x_i = p_{i} \times k_i，其中p_{i } 是質數,k_i\ge 2;\\將15個不同的質數依序排列p_1\lt p_2\lt \cdots \lt p_{15}，則p_{15}\ge 47，最大的x_{15}\ge 47^2 \gt 2022，矛盾$$

三、 計算與證明題 (請先標示題號，再詳列計算/證明過程，未有過程僅有答案者不給分。 共 22 分。 )

解答：

$$(1)\cases{A(0,0,0)\\ B(2,0,0)\\ P(0,0,2)\\ C(a,b,0)} \Rightarrow \cases{F=(P+B)\div 2=(1,0,1) \\ \overrightarrow{CA}=(-a,-b,0)\\ \overrightarrow{CB} =(2-a,-b,0) \\ \overleftrightarrow{PC}:{x\over a} ={y\over b}= {z-2\over -2}}\Rightarrow \cases{  \overrightarrow{CA} \cdot \overrightarrow{CB} =0 \Rightarrow a^2+b^2 =2a\\ E=(at,bt,-2t+2,t\in \mathbb{R}}\\ \Rightarrow \overrightarrow{EA} =(-at,-bt,2t-2)  \Rightarrow \overrightarrow{EA} \cdot \overrightarrow{PC} = -t(a^2+b^2)-4t+4 =0 \Rightarrow -t(2a)=4t-4\\ \Rightarrow a=-2+{2\over t} =-2+2m(取m=1/t)\Rightarrow b^2=2a-a^2 = -4(m^2-3m+2)\\ \Rightarrow b= 2\sqrt{-m^2+3m-2} (1\lt m\lt 2) \Rightarrow E=(-2t+2,2\sqrt{-1+3t-2t^2},-2t+2)\\ \Rightarrow \cases{\overrightarrow{AE}= (-2t+2,2\sqrt{-1+3t-2t^2},-2t+2)\\ \overrightarrow{AF}=(1,0,1)} \Rightarrow f(t)= |\overrightarrow{AE}|^2|\overrightarrow{AF}|^2-(\overrightarrow{AE}\cdot \overrightarrow{AF})^2 \\ =-16t^2+24t-8 \Rightarrow t={24\over 32}={3\over 4} 有極大值f(3/4)=1 \Rightarrow \triangle AEF面積={1\over 2}f(2/4)= \bbox[red,2pt]{1\over 2}\\ (2)t=3/4 \Rightarrow E=({1\over 2}, {1\over \sqrt 2},{1\over 2}) \Rightarrow \cases{\overrightarrow{FP} =(-1,0,1)\\ \overrightarrow{FE}=(-1/2,\sqrt 2/2,-1/2)} \Rightarrow \overrightarrow{FP} \cdot \overrightarrow{FE}=0 \\ \Rightarrow \angle PFE=90^\circ \Rightarrow \tan \theta =\overline{FE}/\overline{FP} ={1\over \sqrt 2} =\bbox[red,2pt]{\sqrt 2\over 2}$$

解答：

(1) $$假設\cases{\Gamma: y^2=6x \\ L=\overleftrightarrow{AB}: y=mx+b}，將L代入\Gamma可得(mx+b)^2=6x \Rightarrow m^2x^2+ (2mb-6)x +b^2=0 \\ \Rightarrow x_1+x_2 = {6-2mb \over m^2} =4 \Rightarrow b= \cfrac{4m^2-6}{-2m}={3\over m}-2m \Rightarrow L: y=mx +{3\over m}-2m \\ \Rightarrow \overline{AB}中點M(2,{3\over m}) \Rightarrow \overline{AB}中垂線L'，其斜率為(-{1\over m})且通過點D \Rightarrow L': y=-{1\over m}(x-2)+{3\over m} \\ \Rightarrow L'與x軸交於\bbox[red, 2pt]{C(5,0)}$$ (2) $$由於\Gamma 對稱x軸且 0\le x_1,x_2，因此取2個端點，當\cases{A=(0,0) \Rightarrow B=(4,\pm 2\sqrt 6) \Rightarrow M=(2,\pm \sqrt 6)\\ A=B=(2,\pm 2\sqrt 3) \Rightarrow M=(2,\pm 2\sqrt 3)} \\ \Rightarrow \bbox[red, 2pt]{-2\sqrt 3\lt y_0\lt 2\sqrt 3}(因為A、B相異，沒有等號)$$ (3)

$$L:y=m x+b ，由(1)知:b={3\over m}-2m\Rightarrow x={y-b\over m} 代回\Gamma \Rightarrow my^2-6y+6b=0 \\ \Rightarrow \cases{兩根之和:y_1+y_2= 6/m\\ 兩根之積:y_1y_2= 6b/m} \Rightarrow (y_1-y_2)^2 =(y_1+y_2)^2 -4y_1y_2 = 36/m^2-24b/m \\ ={48m^2-36\over m^2} \Rightarrow y_1-y_2= \sqrt{48m^2-36\over m^2}\\ 令D=L\cap x軸=(-{b\over m},0) \Rightarrow \overline{CD} =5+{b\over m} =3+{3\over m^2} \\ 因此\triangle ABC = {1\over 2}\overline{CD}\times |y_1-y_1|={1\over 2}(3+{3\over m^2}) \sqrt{ 48m^2-36\over m^2} =3(1+{1\over m^2})\sqrt{12-{9\over m^2}}\\ 令a=\sqrt{12-{9\over m^2}}，則\triangle ABC=g(a)= 7a-{1\over 3}a^3 \Rightarrow g'(a)=0 \Rightarrow a^2=7 \\\Rightarrow g(\sqrt 7)= 7\sqrt 7-{7\over 3}\sqrt 7 = \bbox[red, 2pt]{{14\over 3}\sqrt{7}}為面積最大值$$

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