朱式幸福: 113年彰化女中第2次教甄-數學詳解

113 學年度國立彰化女中第二次教師甄選初試

解答:

$假設\angle ACD=\theta \Rightarrow \cases{\angle ADC=120^\circ+\theta\\ \angle B=60^\circ+\theta}\\ \triangle ADC: {1\over \sin \theta} ={3\over \sin(120^\circ+\theta)} \Rightarrow \tan \theta={\sqrt 3\over 7} \Rightarrow \cases{\sin \theta=\sqrt 3/2\sqrt{13} \\ \cos \theta =7/2\sqrt{13}}\\ \triangle BCD: {\overline{BC} \over \sin(60^\circ-\theta)} ={2\over \sin 60^\circ} ={4\over \sqrt 3} \Rightarrow \overline{BC} ={4\over \sqrt 3} \sin(60^\circ-\theta) = {6\over \sqrt{13}} \\ 因此\cos A={9+9-{36\over 13} \over 2\cdot 9\cdot 9} = \bbox[red, 2pt]{11\over 13}$

解答:

$|z_1-(3+3i)|=2 \Rightarrow P(z_1) 在圓C_1:(x-3)^2+ (y-3)^2=2^2上\\ z_2=a+bi \Rightarrow iz_2-1=(-1-b)+ai \Rightarrow |iz_2-1|=1 \Rightarrow Q(z_2)在圓C_2:x^2+(y+1)^2=1上 \\ \Rightarrow |z_1-z_2|= \overline{PQ} 的最小值=兩圓心距離-兩圓半徑和=\sqrt{3^2+4^2}-(2+1)= \bbox[red, 2pt]2$

解答:

$令\cases{a=\log_x 3\\ b=\log_y 2} \Rightarrow \cases{2^{\log_x 81} =2^{4\log_x 3} =2^{4a}=k \\ 3^{\log_y 16} =3^{4\log_y 2} =3^{4b} =k} \Rightarrow \cases{4a=\log_2 k =\log k/\log 2\\ 4b=\log_3 k =\log k/\log 3} \Rightarrow \cases{a=\log k/4\log 2\\ b=\log k/4\log 3} \\ \Rightarrow \log_3 x+\log_2 y={1\over a} +{1\over b} ={4\log 2\over \log k}+{4\log 3\over \log k} ={\log (2^4\cdot 3^4)\over \log k} =1 \Rightarrow k=2^4\cdot 3^4= \bbox[red, 2pt]{1296}$

解答:

$\lim_{n\to \infty} \left({1\over \sqrt{n^2+2n}} +{1\over \sqrt{n^2+4n}} + \cdots +{1\over \sqrt{n^2+2n^2}} \right) = \lim_{n\to \infty} \sum_{k=1}^n {1\over \sqrt{n^2+2kn}} \\ = \lim_{n\to \infty} \sum_{k=1}^n {1\over n\sqrt{1+2{k\over n}}} =\int_0^1 {1\over \sqrt{1+2x}} \,dx =\left. \left[ \sqrt{1+2x}\right] \right|_0^1 =\bbox[red, 2pt]{\sqrt 3-1}$

解答:

$a^2-a=a(a-1) =奇數\times 偶數=1000k= (125\times 8)k \Rightarrow a=125p=8q\pm 1 \\ \Rightarrow \cases{p=1,2,不合\\ p=3 \Rightarrow a=125\times 3=375 \Rightarrow a+1=8\times 47=376 \Rightarrow 取a=376 \Rightarrow 376\times 375=1000\times 141 \\ p=4,不合\\ p=5 \Rightarrow a=125\times 5=625 \Rightarrow a-1=624=8\times 78 \Rightarrow 取a=625 \Rightarrow 625\times 624=1000\times 390 \\ p=6,7,8,不合} \\ \Rightarrow a=\bbox[red, 2pt]{376,625}$

解答:

$f(t)=\sqrt{(t-1)^2+ (2t-1)^2+(2t-3)^2} +\sqrt{(t-2)^2+ (2t)^2 +(2t+1)^2} \\\qquad =\sqrt{9t^2-18t+11} +\sqrt{9t^2+5} \Rightarrow f'(t)=0 \Rightarrow {9t-9\over \sqrt{9t^2-18t+11}} +{9t\over \sqrt{9t^2+5}} =0 \\ \Rightarrow 3t^2-10t+5=0 \Rightarrow t={5\pm \sqrt{10} \over 3} \Rightarrow f({5- \sqrt{10} \over 3}) \lt f({5+ \sqrt{10} \over 3}) \Rightarrow t= \bbox[red, 2pt]{5- \sqrt{10} \over 3} 有最小值$

解答:

$x^2+y^2-2x-6y+8=0 \Rightarrow (x-1)^2+ (y-3)^2=2 \Rightarrow y=3-\sqrt{2-(x-1)^2} \\ 兩圖形交於(0,0)及(2,2),因此旋轉體積=\pi\int_0^2 \left( 3-\sqrt{2-(x-1)^2}\right)^2\,dx -\pi\int_0^2 x^2\,dx \\=\pi \left({46 \over 3}-3\pi \right) -{8\over 3}\pi =\bbox[red, 2pt]{ {38\over 3}\pi-3\pi^2}$

解答:

$假設\cases{小圓(月亮)圓心O_1,半徑r=1\\ 大圓(太陽)圓心O_2, 半徑R\\ C= \overline{O_1Q}與\overline{AB}的交點} \Rightarrow \overline{BC}={1\over 2}\overline{AB}= {1\over 2}\sqrt 3 \Rightarrow \cases{\angle CO_1B= 60^\circ\\ \angle O_1BC =30^\circ} \\ 又\overline{CQ} =\overline{O_1P}-\overline{PQ}-\overline{O_1C} =\sqrt 3-{3\over 2} \Rightarrow \tan \angle QBC={\overline{CQ} \over \overline{BC}} =2-\sqrt 3 \Rightarrow \cases{\angle QBC= 15^\circ \\ \angle CQB=75^\circ} \\ 由於\overline{O_2B}=\overline{O_2Q} =R \Rightarrow \angle BO_2C =180^\circ-75^\circ\times 2=30^\circ \Rightarrow R=\overline{BC}\times 2= \sqrt 3 \\ \Rightarrow \cases{扇形AQBO_2面積={1\over 6} (\sqrt 3)^2\pi={\pi\over 2} \\ \triangle ABO_2面積={1\over 2} \sqrt 3\cdot \sqrt 3\sin 60^\circ={3\sqrt 3\over 4}} \Rightarrow 弓形AQB面積={ \pi\over 2} -{3\sqrt 3\over 4} \\ 同理,\cases{扇形APBO_1面積={1\over 3}\pi\\ \triangle ABO_1面積 ={1\over 2}\sin 120^\circ=\sqrt 3/4} \Rightarrow 弓形APB面積={\pi\over 3}-{\sqrt 3\over 4} \\ \Rightarrow 月偏食亮面面積=\left( {\pi\over 3}-{\sqrt 3\over 4}\right)-\left({\pi\over 2} -{3\sqrt 3\over 4}\right)=-{\pi\over 6} +{\sqrt 3\over 2} \\ \Rightarrow {月偏食亮面面積 \over 滿月圓面積} = \left(-{\pi\over 6}\pi +{\sqrt 3\over 2} \right)/ \pi= \bbox[red, 2pt] {{\sqrt 3\over 2\pi}-{1\over 6}}$

解答:

$\lim_{n\to \infty} \left({1\over \sqrt{3n^2}} +{1\over \sqrt{3n^2}} +\cdots+{1\over \sqrt{3n^2}} \right) \lt 原式 \lt \lim_{n\to \infty} \left({1\over \sqrt{3n^2+2n}} +{1\over \sqrt{3n^2 +2n}} +\cdots+{1\over \sqrt{3n^2+2n}} \right) \\ \Rightarrow \lim_{n\to \infty} \left({2n\over \sqrt{3n^2}} \right) \lt 原式 \lt \lim_{n\to \infty} \left({2n\over \sqrt{3n^2+2n}} \right)\Rightarrow {2\over \sqrt 3} \lt 原式 \lt {2\over \sqrt 3} \\ \Rightarrow 原式= \bbox[red, 2pt]{2\sqrt 3\over 3}$

解答:

$依題意取\cases{C(0,0)\\ A(-1,-1)\\ B(1,-1)\\ P({1\over 2}\cos \theta, {1\over 2}\sin \theta)} \Rightarrow \cases{\overrightarrow{AB} =(2,0) \\ \overrightarrow{BC}=(-1,1) \\ \overrightarrow{AP} =({1\over 2}\cos \theta+1, {1\over 2}\sin \theta+1)} \\ \overrightarrow{AP}=\alpha \overrightarrow{AB} +\beta \overrightarrow{BC} \Rightarrow ({1\over 2}\cos \theta+1, {1\over 2}\sin \theta+1)=(2\alpha-\beta,\beta) \Rightarrow \cases{\alpha=1+{1 \over 4}(\sin \theta+\cos \theta) \\ \beta={1\over 2}\sin \theta+1} \\ \Rightarrow 4\alpha-\beta =3+{1\over 2}\sin \theta+\cos \theta =3+{\sqrt 5\over 2} \sin(\theta+\delta) 最大值= \bbox[red, 2pt]{ 3+{\sqrt 5\over 2}}$

解答:

$假設正\triangle的邊長為a，其面積為{\sqrt 3\over 4}a^2 \Rightarrow 體積=2\int_0^2 {\sqrt 3\over 4}a^2\,da = \bbox[red, 2pt]{4\sqrt 3\over 3}$

解答:

$假設\cases{A(0,8) \in L_1 \\B(9,2) \in L_4\\ C(6,0) \in L_2\\ D(-5,4) \in L_3} ,其中\cases{L_1 \parallel L_2\\ L_3 \parallel L_4\\ L_1\bot L_3 } \Rightarrow \cases{L_1: y=mx +a_1 \Rightarrow a_1=8 \Rightarrow y=mx+8\\ L_2: y=mx+ a_2 \Rightarrow 0=6m+a_2 \Rightarrow y=mx-6m\\ L_3: x+my=b_1 \Rightarrow -5+4m=b_1 \Rightarrow x+my=4m-5\\ L_4: x+my=b_2 \Rightarrow 9+2m=b_2 \Rightarrow x+my=2m+9} \\ 又d(L_1,L_2) =d(L3,L_4) \Rightarrow {|6m+8| \over \sqrt{m^2+1}} ={|2m-14| \over \sqrt{m^2+1}} \Rightarrow (6m+8)^2= (2m-14)^2 \\ \Rightarrow \cases{m=3/4 \Rightarrow d(L_1,L_2)= 10\\ m=-11/2 \Rightarrow d(L_1,L_2) =2\sqrt 5} \Rightarrow 最大面積=10^2= \bbox[red, 2pt]{100}$

解答:

$由於\cases{(x-1)^2 \ge 0\\ x^2+2x+3=(x+1)^2+2 \gt 0},因此僅需考慮 (x-k)(x-6)\le 0\\ 若\cases{k\lt 6 \Rightarrow k\lt x \lt 6 有10個整數解 \Rightarrow x=5,4,\dots,-4 \Rightarrow -5\le k\lt -4 \\ k\gt 6 \Rightarrow 6\lt x\lt k有9個整數解(還有x=1,共10個) \Rightarrow x=7,8,\dots,15 \Rightarrow 15\lt k\le 16} \\ \Rightarrow \cases{a=15\\ b=16\\ c=-5\\ d=-4} \Rightarrow a+b+c+d =\bbox[red, 2pt]{22}$

解答:

$\cases{O(-1,2) \\ I(2,2)\\ A(2,8)} \Rightarrow \cases{\triangle ABC外接圓\Gamma_1: (x+1)^2+ (y-2)^2= 45 \\ L=\overleftrightarrow{IA}: x=2} \Rightarrow D= \Gamma \cap L=(2,-4) \\ 以D為圓心,\overline{DI}=6為半徑的圓\Gamma_2:(x-2)^2+ (y+4)^2=36\\ \Gamma_1\cap \Gamma_2 \Rightarrow (x+1)^2+ (y-2)^2- 45=(x-2)^2+ (y+4)^2-36 \\ \Rightarrow \bbox[red, 2pt]{x-2y=4}\\ 依雞爪定理:\Gamma_1\cap \Gamma_2 的兩點即為B,C\; \href{http://w.mathsgreat.com/SC/geom_th/geom_th_123.pdf}{參考資料}$

解答:

$y=x^2+bx+c= (x-\alpha)^2-\beta^2\; (\beta\gt 0)\\ \Rightarrow \cases{A(\alpha-\beta,0)\\ B(\alpha+\beta,0)\\ C(0,\alpha^2-\beta^2\lt 0) \Rightarrow \alpha\lt \beta\\ M(\alpha,-\beta^2)} \Rightarrow S_{\triangle ABM} ={1\over 2}\cdot 2\beta\cdot \beta^2 =\beta^3 \\ 由阿基米德某性質(\href{https://www.cnblogs.com/james-wangx/p/16111454.html}{參考資料})可知:S_{\triangle ACM}={1\over 8} S_{\triangle ABM}\\ 現在 S_{\triangle ABM}+S_{\triangle ACM} =9 \Rightarrow S_{\triangle ABM}= \beta^3=8 \Rightarrow \beta=2 \\又通過(-2,5) \Rightarrow 5=(-2-\alpha)^2-4 \Rightarrow \alpha=1 \Rightarrow y=(x-1)^2-4\\ \Rightarrow y=x^2-2x-3 =x^2+bx+c \Rightarrow (b,c)=\bbox[red, 2pt]{(-2,-3)}$

解答:

$令\cases{P(x,y)\\ Q(0,a)},且\overline{PQ}={2\over 3}\overline{PB} \Rightarrow x^2+(y-a)^2={4\over 9}\left( x^2+ (y-6)^2\right) \\ \Rightarrow 5x^2+5y^2-18ay+48y+9a^2-144=0 \Rightarrow 5\cdot 16-6y(3a-8)+9a^2-144=0 \\ \Rightarrow 9a^2-64-6y(3a-8)=0 \Rightarrow (3a-8)(3a+8-6y)=0 \Rightarrow a={8\over 3} \Rightarrow Q(0,{8\over 3}) \\ 因此3\overline{PA}+ 2\overline{PB}=3\overline{PA}+ 2\cdot {3\over 2}\overline{PQ}= 3(\overline{PA}+ \overline{PQ})最小值出現在Q,P,A在一直線\\ \Rightarrow 最小值 =3\overline{AQ} =3\cdot \sqrt{8^2+(8/3)^2} = 3\cdot \sqrt {640\over 9} = \bbox[red, 2pt]{8\sqrt{10}}$

解答:

$將\cases{k=6\\ n=7} 代入公式 {(k-1)\left[ (k-1)^{n-1}+(-1)^n\right] \over k} ={5(5^6-1)\over 6} = \bbox[red, 2pt]{13020}$

解答:

$令\cases{\angle EAF=\theta \\ \overline{AB}=\overline{AE}=a}\Rightarrow \cos \theta ={a^2-3\over 2a} \Rightarrow \sin \theta={\sqrt{-a^4+10a^2-9} \over 2a} \\ \Rightarrow S_{\triangle ABC} ={1\over 2}\overline{AB} \cdot \overline{AC}\sin \angle BAC ={1\over 2}a\sin (240^\circ-\theta) =-{1\over 2}a\sin(60^\circ-\theta) =-{1\over 4}(\sqrt 3\cos \theta-\sin \theta) \\=f(a)=-{1\over 8}(\sqrt 3a^2-3\sqrt 3-\sqrt{-a^4+10a^2-9}) \Rightarrow f'(a)=0 \Rightarrow 2\sqrt 3a ={-2a^3+10a \over \sqrt{-a^4+10a^2-9}} \\ \Rightarrow 16a^6-160a^4+208a^2=0 \Rightarrow 16a^2(a^4-10a^2+10) =0 \Rightarrow a^2=5-2\sqrt 3 \\ \Rightarrow f(a^2=5-2\sqrt 3) =-{1\over 8}(\sqrt 3(5-2\sqrt 3)-3\sqrt 3-\sqrt{-(5-2\sqrt 3)^2+10(5-2\sqrt 3)-9} \\= -{1\over 8}(2\sqrt 3-8) = \bbox[red, 2pt]{1-{\sqrt 3\over 4}}$

解答:

$\cases{\cos^2 A+\cos^2 B+ 2\sin A\sin B \cos C={15\over 8} \\ \cos^2 B+ \cos^2 C+ 2\sin B\sin C \cos A={14\over 9}}\\兩式相加 \Rightarrow \cos^2 A+ 2\cos^2B +\cos^2 C+2\sin B(\sin A\cos C+ \sin C \cos A) ={15\over 8} +{14\over 9} \\ \Rightarrow \cos^2 A+ 2\cos^2B +\cos^2 C+2\sin B\sin(A+C) =\cos^2 A+ 2\cos^2B +\cos^2 C+2\sin^2 B ={247\over 72}\\\Rightarrow \cos^2A+ \cos^2C ={247\over 72}-2 \Rightarrow \cos^2A+ \cos^2C={103\over 72} \cdots(1)\\ 兩式相減\Rightarrow \cos^2 A-\cos^2C +2\sin B(\sin A\cos C-\sin C\cos A) ={15\over 8}-{14\over 9} \\ \Rightarrow \cos^2A-\cos^2 C+2 \sin (A+C)\sin (A-C) =\cos^2A-\cos^2 C+ \cos(2C)-\cos (2A) ={23\over 72} \\ \Rightarrow \cos^2 A-\cos^2C+ (2\cos^2 C-1)-(2\cos^2A -1) ={23\over 72} \Rightarrow -\cos^2 A+ \cos^2 C={23\over 72} \cdots(2) \\ 由(1)及(2)可得\cases{\cos^2 A=5/9\\ \cos^2 C=7/8} \\因此\cos^2C+ \cos^2 A+2\sin C\sin A\cos B={103\over 72}+ 2\sin C\sin A(-\cos(A+C)) \\={103\over 72}+ 2\sin C\sin A(\sin A\sin C- \cos A\cos C) \\={103\over 72}+2\sin^2 A\sin ^2 C-2\sin C\cos C\sin A\cos A \\={103\over 72}+2\cdot {4\over 9}\cdot {1\over 8}\pm 2\cdot {1\over \sqrt 8} \cdot {\sqrt 7\over \sqrt 8}\cdot {2\over 3}\cdot {\sqrt 5\over 3} = \bbox[red, 2pt]{{111\over 72}\pm {4\sqrt{35}\over 72}}$