113年彰化高中教甄-數學詳解

 國立彰化高級中學113學年度第 1 次教師甄選

填充題

解答:$$\cases{a_1=d\\ b_1=d^2} \Rightarrow {a_1^2+a_2^2 +a_3^2 \over b_1+b_2+ b_3} ={d^2+ 4d^2+9d^2\over d^2+d^2r+d^2r^2} ={14\over 1+r+r^2} =n \in \mathbb N  \\ 0\lt r\lt 1 \Rightarrow 1\lt 1+r+r^2\lt 3 \Rightarrow 5\le n\le 13\\ \Rightarrow r^2+r+1-{14\over n}=0 \Rightarrow 判別式D=1-4(1-{14\over n})={56-3n \over n} 必須是完全平方數\\\Rightarrow n=8 \Rightarrow D=4 \Rightarrow r= \bbox[red, 2pt]{1\over 2}$$

解答:$${2y\over x}= {{2\over 3}\cdot (x+3y)\over x}-{2\over 3} \\ \Rightarrow {6x\over x+3y} +{2y\over x} ={6x\over x+3y} + {{2\over 3}\cdot (x+3y)\over x}-{2\over 3} \ge 2\sqrt{{2\over 3}\cdot 6} -{2\over 3} =4-{2\over 3}= \bbox[red, 2pt]{10\over 3}$$

解答:$$a_{n-1}=a_n+a_{n-2} \Rightarrow \begin{array}{} &a_2&=&a_1+a_3\\ & a_3&= & a_2+a_4\\ & \cdots\\+ & a_{n-1} & = &a_{n-2}+a_n\\\hline  & s_{n-1}-a_1 & = &S_{n-2}+ S_n-a_1-a_2\end{array} \\ \Rightarrow S_{n-1}= S_{n-2}+a_{n-1}-a_{n-1}+S_n-a_2 \Rightarrow 0=-a_{n-1}+S_n-a_2 \Rightarrow S_n=a_{n-1}+a_2 \cdots(1)\\ 又\cases{a_{n-1}=a_{n-2}+ a_n\\ a_n= a_{n-1}+ a_{n+1}} \Rightarrow a_{n-1}-a_{n-2}=a_n=a_{n-1}+a_{n+1} \Rightarrow a_{n+1}=-a_{n-2} \\\Rightarrow a_n=-a_{n-3} \Rightarrow a_n =a_{n-6} \cdots(2)\\ S_{2024}=S_{2023}+a_{2024} \Rightarrow 2023= 2024+a_{2024} \Rightarrow a_{2024}=-1又2024= 2 \text{ mod }6\\ 因此a_{2024}=a_2=-1 \Rightarrow  S_{2025}=a_{2024}+a_2 =a_2+a_2= -1-1=\bbox[red, 2pt]{-2}$$

解答:$$\cases{\cos(ax+b)週期大於2\pi \Rightarrow 0\lt a\lt 1 \\|b| \lt \pi  \Rightarrow -\pi \lt b \lt \pi} \\ f\left({5\pi\over 8} \right)=0 \Rightarrow 3\cos\left({5a\pi\over 8} +b\right)= 0 \Rightarrow {5a\pi\over 8} +b =-{\pi\over 2},{\pi\over 2},{3\pi\over 2} \\f\left({11\pi\over 8} \right)=3 \Rightarrow 3\cos\left({11a\pi\over 8} +b\right)= 3 \Rightarrow {11a\pi\over 8} +b =0,\pi,2\pi\\ 有3\times 3=9種配對,只有:\cases{{5a\pi\over 8} +b =-{\pi\over 2}\\ {11a\pi\over 8} +b =0} \Rightarrow \cases{a=2/3\\ b=\bbox[red, 2pt]{-11\pi/ 12}}符合a與b的要求$$

解答:$$12|z|^2 =2|z+2|^2+|z^2+1|^2+31  \Rightarrow 12z\bar z=2(z+2)(\bar z+2)+ (z^2+1)(\bar z^2+1)+31 \\ \Rightarrow 12z\bar z= (2z\bar z+4z+4\bar z+8)+ (z^2\bar z^2+z^2+ \bar z^2+1)+31 \\ \Rightarrow (z^2\bar z^2-12z\bar z+36)+(z^2+\bar z^2+2^2+ 2(z\bar z+2z+ 2\bar z))=0\\ \Rightarrow (z\bar z-6)^2+(z+\bar z+2)^2=0 \Rightarrow \cases{z\bar z=6 \Rightarrow \bar z=6/z\\ z+\bar z=-2} \\ \Rightarrow z+{6\over z}=z+\bar z=\bbox[red, 2pt]{-2}$$

解答:$$假設\cases{O為外心\\G為重心 \\H為垂心}\Rightarrow \overrightarrow{GH} = 2\overrightarrow{OG} \cdots(1)\\現在\left| \overrightarrow{OA} +\overrightarrow{OB} +\overrightarrow{OC} \right|=\sqrt 3 \Rightarrow \left| \overrightarrow{OG} +\overrightarrow{GA} + \overrightarrow{OG}+ \overrightarrow{GB} +\overrightarrow{OG} +\overrightarrow{GC} \right| =\left| 3\overrightarrow{OG} +\vec 0\right| =\sqrt 3 \\ \Rightarrow \left| \overrightarrow{OG} \right|= {\sqrt 3\over 3}\\ 同理 \left| \overrightarrow{HA} +\overrightarrow{HB} +\overrightarrow{HC} \right| = \left| 3\overrightarrow{HG} +\vec 0\right| =3\left| \overrightarrow{GH} \right|=3\cdot 2\left| \overrightarrow{OG} \right| \text{ by (1)} \\=3\cdot 2\cdot {\sqrt 3\over 3}=\bbox[red, 2pt]{ 2\sqrt 3}$$

解答:$$題目有\bbox[cyan,2pt]{疑義}$$

解答:$$z^7=1 \Rightarrow (1-z)(1+z+z^2+\dots+z^6)=0 \Rightarrow 1+z+\cdots+z^6=0 \Rightarrow z+z^2+\cdots +z^6=-1\\假設\cases{\alpha=z+z^2+z^4\\ \beta=z^3+z^5+z^6} \Rightarrow \alpha+\beta=z+ z^2+ \cdots+ z^6=-1 \\ 又\alpha \beta =(z+z^2+z^4) (z^3+z^5+z^6) = z^4(1+z+z^2+ \cdots+z^6+2z^3) =z^4(0+2z^3)=2z^7=2\\ 因此我們有\cases{\alpha+\beta=-1\\ \alpha\beta=2} \Rightarrow \alpha,\beta 是x^2+x +2=0的兩根 \Rightarrow \alpha=z+z^2+z^4 =\bbox[red, 2pt]{-1\pm \sqrt 7i\over 2}$$

解答:$$(a+b+c)\left({b+c-a\over bc} +{c+a-b\over ca}+{a+b-c\over ab} \right)= {1\over 4}\cdot 32=8 \\ \Rightarrow {(b+c)^2-a^a\over bc} +{(c+a)^2-b^2\over ca}+{(a+b)^2-c^2\over ab} =8 \\ \Rightarrow {(b+c)^2-a^a\over bc} -4+{(c+a)^2-b^2\over ca}-4+{(a+b)^2-c^2\over ab} =0 \\ \Rightarrow {(b-c)^2-a^a\over bc}  +{(c-a)^2-b^2\over ca}+{(a+b)^2-c^2\over ab} =0 \\ \Rightarrow {(b-c+a)(b-c-a)\over bc}  +{(c-a-b)(c-a+b)\over ca}+{(a+b+c)(a+b-c)\over ab} =0 \\ \Rightarrow (a+b-c)\left( {b-c-a\over bc} +{a-b-c \over ca} +{a+b+c \over ab}\right)=0 \\ \Rightarrow {a+b-c\over abc} (ab-ac-a^2+ab-b^2-bc+ac+bc+c^2)=0 \\ \Rightarrow {a+b-c\over abc}\cdot {}(c^2-(a-b)^2) ={(a+b-c)(c+a-b)(c-a+b)\over abc}=0 \\ \Rightarrow \cases{a+b=c\\ b+c=a\\ c+a=b} \Rightarrow \cases{(\sqrt a)^2+(\sqrt b)^2=(\sqrt c)^2 \\ (\sqrt b)^2+(\sqrt c)^2=(\sqrt a)^2 \\ (\sqrt c)^2+(\sqrt a)^2=(\sqrt b)^2 }$$

解答:$$\cases{\overline{AB}=\sqrt 3\\ \overline{AC}=\sqrt 7\\ \overline{BC}= \sqrt{10}} \Rightarrow \overline{BC}^2= \overline{AB}^2+ \overline{AC}^2 \Rightarrow \angle BAC=90^\circ \Rightarrow \cases{A(0,0,0)\\ B(\sqrt 3,0,0)\\ C(0,\sqrt 7,0) \\ D(a,b,c)} \\ \Rightarrow \cases{\overline{AD}=\sqrt{10} \\ \overline{BD} =\sqrt 7 \\ \overline{CD}=\sqrt{7}} \Rightarrow \cases{a^2+b^2+c^2=10\\ (a-\sqrt 3)^2+b^2+c^2=7\\ a^2+(b-\sqrt 7)^2+c^2=7} \Rightarrow \cases{a=3/\sqrt 3\\ b=5/\sqrt 7} \Rightarrow c= \sqrt{24\over 7} \\ \Rightarrow D(\sqrt 3,{5\sqrt 7\over 7},{2\sqrt{42} \over 7}) \Rightarrow 體積={1\over3}\left({1\over 2} \sqrt 3\cdot \sqrt 7 \right)\cdot {2\sqrt{42}\over 7} =\bbox[red, 2pt]{\sqrt 2}$$

解答:$$E=2\times{1\over 4^2} +(1+E)(1-{1\over 4})+ (2+E)({1\over 4}\cdot (1-{1\over 4})) \Rightarrow {5\over 4}+{15\over 16}E=E\\ \Rightarrow E=\bbox[red, 2pt]{20}$$

解答:$$\left(\left({\sin^{1012} \theta\over \cos^{1011} \phi} \right)^2+ \left({\cos^{1012} \theta\over \sin^{1011} \phi} \right)^2\right) \left( \left({\cos^{1011} \phi\over \sin^{1010} \theta} \right)^2+ \left({\sin^{1011} \phi\over \cos^{1010} \theta} \right)^2 \right) \ge \left( \sin^2 \theta+ \cos^2 \theta\right)^2\\ \Rightarrow 1\cdot \left( \left({\cos^{1011} \phi\over \sin^{1010} \theta} \right)^2+ \left({\sin^{1011} \phi\over \cos^{1010} \theta} \right)^2 \right) \ge 1 \Rightarrow  \left({\cos^{1011} \phi\over \sin^{1010} \theta} \right)^2+ \left({\sin^{1011} \phi\over \cos^{1010} \theta} \right)^2=1\\ 此時 {\sin ^{2022}\theta \over \cos^{2022}\phi} ={\cos^{2022} \theta \over \sin^{2022} \phi} \Rightarrow \sin\theta \sin \phi=\cos \theta\cos \phi \Rightarrow \cos(\theta+\phi)=0 \Rightarrow \theta+\phi ={\pi\over 2} \\ \Rightarrow \sin \theta= \sin({\pi\over 2}-\phi)= \cos \phi \Rightarrow \sin^{2023} \theta-\cos^{2023} \phi =\bbox[red, 2pt]0$$

解答:$$\cases{a+b+c=5 \Rightarrow a+b=5-c\\ ab+bc+ca=7 \Rightarrow ab=7-c(a+b) } \Rightarrow ab=7-c(5-c)=c^2-5c+7 \\\Rightarrow abc= c^3-5c^2+7c\\ 欲求f(c)=c^3-5c^2+7c之極值, f'(c)=3c^2-10c+7=0 \Rightarrow (3c-7)(c-1)=0 \\ \Rightarrow \cases{c=7/3 \Rightarrow a+b=5-c=8/3 \Rightarrow ab=7-c(a+b)= 7-{7\over 3}\cdot {8\over 3}={7\over 9}\\ c=1 \Rightarrow a+b=5-c=4 \Rightarrow ab=7-c(a+b)=7-1\cdot 4=3} \\ \Rightarrow \cases{abc={7\over 9}\cdot {7\over 3}={49\over 27} \\abc=3\cdot 1=3} \Rightarrow \cases{M=3\\ m=49/27} \Rightarrow M+m=\bbox[red, 2pt]{130\over 27}$$

解答:$$a,b,c,d相異且介於1與17之間 \Rightarrow (1+2)\le a+c\le (16+17) \Rightarrow 3\le a+c\le 33\\ 又17\mid (a-b+c-d) \Rightarrow a+c-(b+d)=0,\pm 17\\ 沒有好方法,只能列出計算\Rightarrow \cases{(a+c)-(b+d)=0 有308個\\ (a+c)-(b+d)=\pm 17有168個},合計\bbox[red, 2pt]{476}$$

解答:$$已知條件只能求角度,無法得知三邊長,\bbox[cyan,2pt]{題目有疑義}$$

解答:$$a+b+c=29的正整數解有H^3_{29-3} =C^{28}_{26} =378組解，但需要扣除「非相異」的解\\ 29=27+1+1 = 25+2+ 2=\cdots=1+14+14,有14組非相異解再加其排列，\\共有14\cdot 3=42組解，因此a+b+c=29的相異正整數解有378-42=336組解\\ 再考慮d+e=29 \Rightarrow 29=1+28=2+27 =\cdots =15+14，有14組解，但d,e需與a,b,c相異\\，因此只剩下14-3=11組解,再加上排列共11\times 2=22組(d,e); 總共有336\times 22=\bbox[red, 2pt]{7392}$$

解答:

$$y=x^2+bx+c=  (x-\alpha)^2-\beta^2\; (\beta\gt 0)\\ \Rightarrow \cases{A(\alpha-\beta,0)\\ B(\alpha+\beta,0)\\ C(0,\alpha^2-\beta^2\lt 0) \Rightarrow \alpha\lt \beta\\ M(\alpha,-\beta^2)} \Rightarrow  S_{\triangle ABM} ={1\over 2}\cdot 2\beta\cdot \beta^2 =\beta^3 \\ 由阿基米德某性質(參考資料)可知:S_{\triangle ACM}={1\over 8} S_{\triangle ABM}\\ 現在 S_{\triangle ABM}+S_{\triangle ACM} =9 \Rightarrow S_{\triangle ABM}= \beta^3=8 \Rightarrow \beta=2 \\又 通過(-2,5) \Rightarrow 5=(-2-\alpha)^2-4 \Rightarrow \alpha=1 \Rightarrow y=(x-1)^2-4\\ \Rightarrow  y=x^2-2x-3 =x^2+bx+c  \Rightarrow (b,c)=\bbox[red, 2pt]{(-2,-3)}$$

計算證明題

解答:$$待續$$

解答:$$利用\text{ Lagrange }算子求極值,假設\cases{f(x,y,z)=(x+y)^2+ (y+z)^2+ (z+x)^2 \\ g(x,y,z)=(x+y+z)xyz-9} \\ \Rightarrow \cases{f_x=\lambda g_x \\f_y=\lambda g_y \\f_z=\lambda g_z \\ g=0} \Rightarrow \cases{2(2x+y+z)= \lambda(yz(2x+y+z))  \cdots(1)\\ 2(x+2y+z)= \lambda(xz (x+2y+z))  \cdots(2)\\ 2(x+y+2z) =\lambda(xy(x+y+2z))  \cdots(3)\\ (x+y+z)xyz=9 \cdots(4)} \\ \Rightarrow \cases{{(1)\over (2)} ={2x+y+z\over x+2y+z} ={y(2x+y+z) \over x(x+2y+z)} \\ {(2)\over (3)} ={x+2y+z\over x+y+2z} ={z(x+2y+z)\over y(x+y+2z}} \Rightarrow x=y=z 代入(4) \Rightarrow 3x^4=9 \\ \Rightarrow x=y=z=\sqrt [4]3 \Rightarrow f的極小值=f(\sqrt [4]3,\sqrt [4]3,\sqrt [4]3) = 12\sqrt 3 \ge 18,\bbox[red, 2pt]{故得證}$$

解答:$$假設\cases{p為彰數\\ q為彰數} \Rightarrow \cases{p=\det(A),其中A=\begin{bmatrix}a & c & b \\b & a & c \\c & b & a \\  \end{bmatrix} \\[1ex] q=\det(B), 其中B= \begin{bmatrix}d & f & e \\e & d & f \\f & e & d \end{bmatrix}} \\ \Rightarrow p\times q=\det(A)\det(B) =\det(AB)=\begin{vmatrix}ad+bf+ce & cd+af+be & bd+cf+ae \\bd+cf+ ae & ad+bf+ce & cd+af+be \\cd+af+be & bd+cf+ae & ad+bf+ce \end{vmatrix} \\ =(ad+bf+ce)^3+ (bd+cf+ ae)^3+ (cd+af+be)^3 -3(ad+bf+ce)(bd+cf+ ae)(cd+af+be) \\ \Rightarrow p\times q 亦為彰數, \bbox[red, 2pt]{故得證}$$

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解題僅供參考，其他歷年試題及詳解