112年嘉義高中教甄-數學詳解

國立嘉義高級中學 112 學年度第 1 學期第 1 次教師甄選

解答: $$從m開始,連續n個整數相加=m+(m+1)+\cdots +(m+n-1)=nm+{n(n-1)\over 2} \\ ={2nm+n(n-1)\over 2}=2023 \Rightarrow n(2m+n-1)=4046\\ 由於4046=2\times 7\times 17^2 \Rightarrow \cases{4046 =1\times 4046 \Rightarrow \cases{n=1\\ m=2023},不符n\ge 2\\ 4046= 2\times 2023 \Rightarrow \cases{n=2\\ m=1011}\\ 4046=7\times 578,14\times 289,17\times 238,34\times 119} \\ \Rightarrow 共有\bbox[red, 2pt]5組解$$

解答:

$$|{1\over z}-1|\ge 1 \Rightarrow |z||{1\over z}-1|\ge |z| \Rightarrow |z-1|\ge |z| \Rightarrow (x-1)^2+y^2 \ge x^2+ y^2 \\ \Rightarrow -2x+1 \ge 0 \Rightarrow x\le {1\over 2} \Rightarrow \Omega_1=\{x\le {1\over 2}\} \\ 又|z-1|\le 1 \Rightarrow \Omega_2=\{ (x-1)^2 +y^2 \le 1\}; 因此\Omega_1\cap \Omega_2為一弓形,如上圖 \\ x={1\over 2}與圓(x-1)^2+y^2=1交於\cases{A(1/2,\sqrt 3/2) \\ B(1/2,-\sqrt 3/2)} \Rightarrow \overline{AB}=\sqrt 3 \\ \Rightarrow \cos \angle AOB= {1+1-3\over 2}=-{1\over 2} \Rightarrow \angle AOB=120^\circ \Rightarrow \cases{\triangle OAB= \sqrt 3/4\\ 扇形OAB=\pi/3} \\ \Rightarrow \Omega_1 \cap \Omega_2面積={\pi\over 3}-{\sqrt 3\over 4} =\bbox[red, 2pt]{4\pi -3\sqrt 3\over 12}$$

解答: $$\cases{A(2,4,5)\\ C(4,4,7)\\ G(3,5,6)\\ F(a,b,c)} \Rightarrow \cases{\overrightarrow{AC}=(2,0,2)\\ \overrightarrow{AG} =(1,1,1)\\   H=\overline{AC}中點＝ (3,4,6)} \Rightarrow \vec n=\overrightarrow{AC} \times \overrightarrow{AG} =(-2,0,2) \\ \Rightarrow 通過A法向量為\vec n的 平面E'＝-x+z=3 又\overline{AC}=2\sqrt{2} \Rightarrow 正八面體稜長=2\\ 又\cases{\overrightarrow{AC}\bot \overrightarrow{HF} \\ \overrightarrow{AG}\bot  \overrightarrow{HF}} \Rightarrow \cases{(2,0,2)\cdot (a-3,b-4,c-6)=0 \\(1,1,1)\cdot (a-3,b-4,c-6)=0} \Rightarrow \cases{a+c=9\\ b=4}\\ \overline{HF} =d(H,E')=\sqrt 2 \Rightarrow {|-a+c-3|\over \sqrt 2}=\sqrt 2 \Rightarrow -a+c=5或1\\ \Rightarrow \cases{a=2,c=7\\ a=4,c=5} \Rightarrow \cases{F=(2,4,7)\\F=(4,4,5)} \Rightarrow \cases{\overrightarrow {FH}=(1,0,-1)=-2\vec n\\ \overrightarrow{FH}=(-1,0,1)=2\vec n} \\ \Rightarrow F=\bbox[red, 2pt]{(2,4,7)}$$

解答: $$n^5=aC^n_5 +bC^n_4 +cC^n_3+ dC^n_2 +eC^n_1\\ n=1 \Rightarrow 1=0+0+0+0+e \Rightarrow e=1\\ n=2 \Rightarrow 32=d+2e \Rightarrow d=30\\ n=3 \Rightarrow 243=c+3d+3e \Rightarrow c=150\\ n=4 \Rightarrow 1024= b+4c+6d+4e \Rightarrow b=240 \\ \Rightarrow n=5 \Rightarrow 3125=a+5b+10c+10d+ 5e \Rightarrow a= 120\\ \Rightarrow a+b+c+d+ e=120+240+150+30+1=\bbox[red, 2pt]{541}$$

解答: $$f(n)=4f(n-1)+4^{n-2} =4(4f(n-2)+4^{n-3})+4^{n-2} =4^2f(n-2)+2\cdot 4^{n-2} \\ =\cdots = 4^{n-2}f(2)+(n-2)4^{n-2} =4^{n-2}(n-1)(\because f(2)=1) \\ \Rightarrow f(9)=4^7\cdot 8=2^{17}有17＋1＝\bbox[red, 2pt]{18}個正因數$$

解答: $$\begin{array}{} n^2 & a\cdot b\cdot c\cdot d & 排列數\\\hline 1& 1,1,1,1 & 1\\\hdashline 2^2 & 4,1,1,1 & 4\\ & 2,2,1,1& 6\\\hdashline 3^2 & 3,3,1,1& 6\\\hdashline 4^2 & 4,4,1,1& 6\\& 4,2,2,1& 12\\ & 2,2,2,2& 1\\ \hdashline 5^2& 5,5,1,1& 6\\\hdashline 6^2& 6,6,1,1 & 6\\ & 6,3,2,1& 24\\ & 3,3,4,1 & 12\\ & 3,3,2,2& 6\\\hdashline 8^2& 4,4,2,2 & 6\\ & 4,4,4,1& 4\\\hdashline  9^2& 3,3,3,3 & 1\\\hdashline 10^2 &5,5,2,2& 6\end{array}\qquad \begin{array}{} & 5,5,4,1& 12 \\  \hdashline 12^2 & 6,6,4,1& 12\\& 6,6,2,2& 6\\ & 6,4,3,2& 24\\&4,4,3,3& 6\\\hdashline 15^2 & 5,5,3,3 & 6\\ 16^2 & 4,4,4,4& 1\\ 18^2 & 6,6,3,3 & 6\\ 20^2 & 5,5,4,4& 6\\ 24^2& 6,6,4,4& 6\\ 25^2 & 5,5,5,5& 1\\ 30^2& 6,6,5,5 & 6\\ 36^2 & 6,6,6,6& 1\end{array}\\ 合計200 \Rightarrow 機率p={200\over 6^4}={25\over 162} \Rightarrow 幾何分配的期望值＝{1\over p}= \bbox[red, 2pt]{162\over 25}$$

解答:

$$\cases{3\le x\le 5 \Rightarrow f(x)=\int_3^x 5-t\,dt=-{1\over 2}x^2+5x {21\over 2} \\ x\ge 5 \Rightarrow g(x)=\int_3^5 5-t\,dt+\int_5^x t-5\,dt= {1\over 2}x^2-5x+{29\over 2}} \\ 令直線L：y=2x-{2023\over k} ,則L介於兩圖形\cases{y=f(x)\\ y=g(x)}切點之間有三相異交點\\ 因此\cases{2x-{2023\over k}=f(x),判別式＝0 \Rightarrow k=4046/12 \approx 337.2\\ 2x-{2023\over k}=g(x),判別式=0 \Rightarrow k=4046/20=202.3} \\ \Rightarrow k=203-337,共有337-203+1= \bbox[red,2pt]{135}個$$

解答:

$$取\cases{A(1,0,1)\\B(0,1,1) \\C(1,1,0)\\ O(0,0,0)}，則OABC為正四面體，且稜邊長=\sqrt 2\\ 再取\cases{D=\overline{AO}中點=(1/2,0,1/2)\\ E=(2A+B)/3=(2/3,1/3,1)\\ F=(2B+A)/3 =(1/3,2/3,1)\\ G=\overline{BC}中點=(1/2,1,1/2)}，則\triangle CDE \parallel \triangle OFG,也就是\cases{E_2=\triangle CDE\\ E_3=\triangle OFG}\\ 因此\cases{\overrightarrow{CD}=(-1/2,-1,1/2) \\ \overrightarrow{CE} =(-1/3,-2/3,1)} \Rightarrow \vec n=\overrightarrow{CD}\times \overrightarrow{CE} =(-2/3,1/3,0)\\ \Rightarrow \triangle CDE面積＝{1\over 2}|\vec n| ={\sqrt 5\over 6}\\ 又\overrightarrow{AB} =(-1,1,0) \Rightarrow \cos \theta ={\overrightarrow{AB} \cdot \vec n\over |\overrightarrow{AB}||\vec n|} ={3\over \sqrt{10}}\\ 而d(E_1,E_4)= {3\over 5\sqrt 2} \Rightarrow 原四面體稜長={3\over 5\sqrt 2} / {3\over \sqrt{10}} ={1\over \sqrt 5}\\ \Rightarrow 截面積＝\triangle CDE\cdot {1\over 2}({1\over \sqrt 5})^2 = \bbox[red,2pt]{\sqrt 5\over 60}$$

解答:

解答: $$利用公式:{2\over 3}R^3\tan \theta ={2\over 3}\cdot 6^3\tan 30^\circ =\bbox[red,2pt]{48\sqrt 3}\\ \href{https://chu246.blogspot.com/2020/12/blog-post.html}{公式來源}$$

解答: $$欲求之原式=\sum_{k=1}^{16} k(k+2)C^{16}_k x^ky^{16-k},其中x={3\over 4},y={1\over 4}\\ 令f(x,y)=(x+y)^{16} =\sum_{k=0}^{16}C^{16}_kx^ky^{16-k} \Rightarrow f_x= \sum_{k=1}^{16}k C^{16}_kx^{k-1} y^{16-k} \\ \Rightarrow x^3f_x =\sum_{k=1}^{16}k C^{16}_kx^{k+2} y^{16-k} \Rightarrow {\partial \over \partial x}\left(x^3f_x \right) =3x^2 f_x+ x^3f_{xx}=\sum_{k=1}^{16}k (k+2)C^{16}_kx^{k+1} y^{16-k} \\ \Rightarrow {1\over x}\left( 3x^2 f_x+ x^3f_{xx}\right) =3xf_x+ x^2f_{xx} =\sum_{k=1}^{16}k (k+2)C^{16}_kx^{k} y^{16-k} \triangleq 原式\\ f=(x+y)^{16} \Rightarrow f_x=16(x+y)^{15} \Rightarrow \cases{f_{xx}= 240(x+y)^{14} \\ 3xf_x= 48x(x+y)^{15}} \\ \Rightarrow g(x,y)=3xf_x+x^2f_{xx}= 48x(x+y)^{15} +240x^2(x+y)^{14} \\\Rightarrow g({3\over 4},{1\over 4}) =36+135= \bbox[red, 2pt]{171}$$

解答: $$\begin{array}{} 前4球&前4球排列&變色數&後4球&後4球排列數\\\hline 1B3W & BWWW & 1 & 4B & 1\\ & WBWW & 2\\ & WWBW& 2\\ & WWWB& 1&&\\\hdashline 2B2W & BBWW & 1& 3B1W& 4\\ & BWBW & 3\\ & BWWB & 2\\ & WBBW & 2\\ & WBWB& 3\\ & WWBB& 1&& \\\hdashline 3B1W & BBBW & 1& 2B2W & 6\\ & BBWB & 2\\ & BWBB& 2\\ & WBBB& 1\\\hline\end{array} \\ 總變色數=1\cdot (1+2+2+1)+ 4\cdot(1+3+2+2+3+1)+ 6\cdot(1+2+2+1) =90 \\ \Rightarrow 期望值={90\over C^8_3}={90\over 56}=\bbox[red, 2pt]{45\over 28}$$

解答: $$P(1,a) \Rightarrow a\sin({\pi\over 3}+\theta) =a \Rightarrow {\pi\over 3}+\theta={\pi \over 2} \Rightarrow \theta={\pi\over 6} \Rightarrow f(x)=a\sin({\pi\over 3}x+ {\pi\over 6}) \\又Q在最低點\Rightarrow {\pi\over 3}x+ {\pi\over 6}={3\over 2}\pi \Rightarrow x=4 \Rightarrow Q(4,-a)\\ 因此我們有\cases{P(1,a)\\ R(1,0)\\ Q(4,-a)} \Rightarrow \cases{\overrightarrow{RP} =(0,a) \\ \overrightarrow{RQ}=(3,-a)} \Rightarrow \cos \angle PRQ ={\overrightarrow{RP} \cdot \overrightarrow{RQ} \over |\overrightarrow{RP}||\overrightarrow{RQ}} \\ \Rightarrow {-a^2\over a\cdot \sqrt{9+a^2}} =-{1\over 2} \Rightarrow a^2=3 \Rightarrow a=\bbox[red, 2pt] {\sqrt 3}$$

解答: $$假設\overline{AC}=a \Rightarrow \overline{AB}=10-a \Rightarrow 三角柱體積=f(a)={1\over 2}a(10-a)\sin 60^\circ\cdot 2a\\ \Rightarrow f(a)={\sqrt 3\over 2}(10a^2-a^3) \Rightarrow f'(a)={\sqrt 3\over 2}(20a-3a^2) \Rightarrow f''(a)={\sqrt 3\over 2}(20-6a)\\ f'(a)=0 \Rightarrow \cases{a=0\\ a=20/3} \Rightarrow \cases{f''(0)\gt 0\\ f''(20/3)\lt 0} \Rightarrow f({20\over 3})= {\sqrt 3\over 2}\cdot {400\over 9}\cdot(10-{20\over 3}) =\bbox[red,2pt]{2000\sqrt 3\over 27}$$

解答: $$80(1+3.53\%)^n \ge 100(1+1.5\%)^n \Rightarrow \left({ 1.0353\over 1.015}\right)^n \ge {100\over 80} \Rightarrow \left({102\over 100}\right) ^n \ge {10\over 8}\\ \Rightarrow n\log 1.02 \ge 1-3\log 2=1-3\cdot 0.301=0.097 \Rightarrow n\ge {0.097\over 0.0086} \approx 11.3\\ \Rightarrow 最少\bbox[red,2pt]{12}年後$$

解答:

$$兩圖形皆對稱y軸,取\cases{A(a,-a^2+4)\\ B(-a,-a^2+4)\\ C(-a,2a^2-8)\\ D(a,2a^2-8)} \Rightarrow \cases{\overline{AB}=2a\\ \overline{AD}= -a^2+4+8-2a^2=-3a^2+12} \\ \Rightarrow 矩形面積= f(a)=2a(-3a^2+12) =-6a^3+24a \Rightarrow f'(x)=-18a^2+24 =0 \\ \Rightarrow a={2\over \sqrt 3} \Rightarrow f({2\over \sqrt 3})=-{16\over \sqrt 3}+{48\over \sqrt 3}=\bbox[red, 2pt]{32\sqrt 3\over 3}$$

解答: $$I為內心,X為任一點 \Rightarrow \overrightarrow{XI}={a\over a+b+c}\overrightarrow{XA} +{b\over a+b+c}\overrightarrow{XB} +{c\over a+b+c}\overrightarrow{XC} \href{https://web.math.sinica.edu.tw/math_media/d341/34103.pdf}{公式來源} \\ 因此\overrightarrow{AI} = {3\over 2+3+4}\overrightarrow{AB} +{4\over 2+3+4}\overrightarrow{AC} ={1\over 3}\overrightarrow{AB} +{4\over 9} \overrightarrow{AC} \\ \Rightarrow \overrightarrow{AO} \cdot \overrightarrow{AI} ={1\over 3}\overrightarrow{AO} \cdot \overrightarrow{AB} +{4\over 9}\overrightarrow{AO}\cdot \overrightarrow{AC}  ={1\over 3} \cdot {1\over 2}\overline{AB}^2 +{4\over 9} \cdot {1\over 2}  \overline{AC}^2 \\={1\over 6}\cdot 16+{2\over 9}\cdot 9 =\bbox[red, 2pt]{14\over 3}$$

解答: $$\cases{E_1:2x+y+2z+3=0\\ E_2:3x+4y-5=0} \Rightarrow \cases{E_1法向量\vec u=(2,1,2)\\ E_2法向量\vec v=(3,4,0)} \Rightarrow \cases{\vec n_1=\vec u/|\vec u| =(2/3,1/3,2/3)\\ \vec n_2=\vec v/|\vec v|=(3/5,4/5,0)} \\ \Rightarrow \cos \theta ={\vec n_1\cdot \vec n_2\over |\vec n_1||\vec n_2|} \gt 0 \Rightarrow \vec n_1與\vec n_2的夾角為銳角\\ 因此取 \vec s= -\vec n_1+\vec n_2 =(-1/15,7/15,-10/15)為角平分面之法向量，\\並任找一點P(-17/5,19/5,0)\in E_1\cap E_2 \\\Rightarrow 角平分面方程式：-{1\over 15}(x+{17\over 5})+{7\over 15}(y-{19\over 5})-{10\over 15}z=0 \Rightarrow \bbox[red, 2pt]{x-7y+10z+30=0}$$

解答: $$橢圓{x^2\over m}+y^2=1 \Rightarrow \cases{a=\sqrt m\\ b=1} \Rightarrow c=\sqrt{m-1} \Rightarrow 焦點\cases{F_1(-\sqrt{m-1},0) \\ F_1(\sqrt{m-1},0)} \\ 雙曲線{x^2\over n}-{y^2\over 3}=1 \Rightarrow \cases{a=\sqrt n\\ b=\sqrt 3} \Rightarrow c=\sqrt{n+3} \Rightarrow 焦點\cases{F_1(-\sqrt{n+3},0) \\ F_2(\sqrt{n+3},0)}\\ 兩焦點相同\Rightarrow m-1=n+3 \Rightarrow m-n=4 \Rightarrow 取\cases{n=1\\ m=5} \Rightarrow \cases{F_1(-2,0) \\ F_2(2,0)}\\ 兩圖形交點：{x^2\over 5}+y^2=x^2-{y^2\over 3}\Rightarrow x^2={5\over 3}y^2 \Rightarrow y^2={3\over 4} \Rightarrow 取y={\sqrt 3\over 2} \\ \Rightarrow x={\sqrt 5\over 2} \Rightarrow P(\sqrt 5/2,\sqrt 3/2) \Rightarrow \cases{\overline{PF_1}=\sqrt 5+1 \\ \overline{PF_2}=\sqrt 5-1} \Rightarrow \cos \angle F_1PF_2={(\sqrt 5+1)^2+ (\sqrt 5-1)^2-4^2 \over 2(\sqrt 5+1)(\sqrt 5-1)}\\ =-{1\over 2} \Rightarrow \angle F_1PF_2=120^\circ \Rightarrow \tan \angle F_1PF_2=\bbox[red, 2pt]{-\sqrt 3}$$

解答:

$$假設A(R\cos \alpha,R\sin \alpha)對稱軸為直線L:y=\tan \theta x,則對稱點A'(R\cos(2\theta-\alpha), R\sin (2\theta-\alpha))\\ 也就是\cases{x=R\cos \alpha\\ y=R\sin \alpha} 對應到\cases{x'=R\cos(2\theta-\alpha)= R\cos 2\theta \cos \alpha +R\sin 2\theta \sin \alpha=x\cos 2\theta+y\sin 2\theta\\ y'=R\sin(2\theta-\alpha) =R(\sin 2\theta \cos \alpha-\sin\alpha \cos 2\theta)= x\sin 2\theta -y\cos 2\theta} \\ 即\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ \sin 2\theta &-\cos 2\theta\end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix} =\begin{bmatrix} x'\\ y'\end{bmatrix} \Rightarrow 對直線L的鏡射矩陣為\bbox[red,2pt]{\begin{bmatrix} \cos 2\theta & \sin 2\theta\\ \sin 2\theta &-\cos 2\theta\end{bmatrix}}$$

解答: $$假設兩鏡射矩陣分別為\cases{A=\begin{bmatrix}\cos 2\alpha & \sin 2\alpha\\ \sin 2\alpha & -\cos 2\alpha \end{bmatrix} \\[1ex] B=\begin{bmatrix}\cos 2\beta & \sin 2\beta\\ \sin 2\beta & -\cos 2\beta \end{bmatrix}} \\ \Rightarrow AB=\begin{bmatrix}\cos 2\alpha \cos 2\beta+ \sin 2\alpha \sin 2\beta& \cos 2\alpha \sin 2\beta-\sin 2\alpha \cos 2\beta\\ \sin 2\alpha \cos 2\beta-\cos 2\beta \sin 2\alpha & \sin 2\alpha \sin 2\beta+ \cos 2\alpha \cos 2\beta \end{bmatrix} \\ =\begin{bmatrix} \cos(2\alpha-2\beta) & -\sin(2\alpha-2\beta) \\ \sin (2\alpha-2\beta) & \cos (2\alpha-2\beta)\end{bmatrix}為一\bbox[red,2pt]{旋轉}矩陣，旋轉角度為兩對稱軸角度差的2倍$$
 =============== END =============

解題僅供參考，其他教甄試題及詳解