115年屏東女中教甄-數學詳解

國立屏東女子高級中學115學年度第1次教師甄選

一、填充題： 每題 5 分，共 90 分。 分數均需化簡至最簡分數；比例需以最簡整數比呈現。 每題全對才給分。

解答:$$f(-3+t)=f(1-t) \Rightarrow y=f(x) 圖形對稱於 x={(-3+t)+(1-t) \over 2} =-1 \Rightarrow f(x)=a(x+1)^2+b\\ 若圖形為凹向上(a\gt 0) \Rightarrow f(5)\gt f(-3)與(III)矛盾,因此圖形為凹向下,即a\lt 0\\ 因此在區間[-3,5]中,\cases{最大值24發生在x=-1 \Rightarrow b=24\\ 最小值-84發生在x=5 \Rightarrow f(5)=36a+24=-84 \Rightarrow a=-3} \\ \Rightarrow f(x)=-3(x+1)^2+24 \Rightarrow f(6)=-3\cdot 7^2+24= \bbox[red, 2pt]{-123}$$

解答:$$假設f(x)=ax^3+bx^2+cx+d \Rightarrow \cases{a+c=b+d\\ f(0)=d=12\\ f(1)=a+b+c+d=12} \Rightarrow \cases{b=-6\\ a+c=6} \\ \Rightarrow f(x)=ax^3-6x^2+cx+12 \Rightarrow f'(x)=3ax^2-12x+c \Rightarrow f''(x)=6ax-12=0 \Rightarrow x={12\over 6a}=2 \\ \Rightarrow a=1 \Rightarrow c=5 \Rightarrow f(x)=x^3-6x^2+5x+12 \Rightarrow f(2)=8-24+10+12= \bbox[red, 2pt]6$$

解答:$$平面E:2x-2y+z=0的法向量為\vec n=(2,-2,1), 又\cases{A(0,0,0) \\ B(1,2,3)} \Rightarrow 正立方體稜長s=\overline{AB}=3\\ \cases{\overrightarrow{AD} \bot \overrightarrow{AB} \\\overrightarrow{AD} \bot \vec n} \Rightarrow \overrightarrow{AD} \parallel (\overrightarrow{AB} \times \vec n) \Rightarrow \overrightarrow{AD} \parallel (-6,-3,6) \Rightarrow \overrightarrow{AD} = \pm {3\over ||(-6,-3,6)||} (-6,-3,6) \\ = \pm(-2,-1,2)\Rightarrow D= A+ \overrightarrow{AD}=(-2,-1,2) \Rightarrow C=B+\overrightarrow{AD}=(-1,1,4) \\ \overrightarrow{CG} \bot E \Rightarrow \overrightarrow{CG}=(2,-2,1) \Rightarrow G=C+ \overrightarrow{CG}= \bbox[red, 2pt]{(1,-1,5)}$$

解答:$$假設\cases{a=\overline{BC} \\b=\overline{AC} \\c= \overline{AB}} \Rightarrow {a\over \sin A}={b\over \sin B}={c\over \sin C} \Rightarrow a:b:c= \sin A:\sin B: \sin C=5:7:8 \Rightarrow \cases{a=5k\\ b=7k\\ c=8k} \\ \Rightarrow \cases{\cos A=(b^2+c^2-a^2)/2bc=11/14\\ \cos B=(a^2+c^2-b^2)/2ac=1/2\\ \cos C=(a^2+b^2-c^2)/2ab=1/7} \Rightarrow \cos A:\cos B:\cos C={11\over 14}: {1\over 2}:{1\over 7}= \bbox[red, 2pt]{11:7:2}$$

解答:$$1-\sqrt 3i為一根\Rightarrow 1+\sqrt 3 i為另一根 \Rightarrow \cases{兩根之和=2\\ 兩根之積= 4} \Rightarrow 1\pm \sqrt 3i為x^2-2x+4=0的兩根\\ 假設x^3+ ax^2+bx+c=0的另一實根為r \Rightarrow x^3+ ax^2+bx+c=(x^2-2x+4)(x-r) \\=x^3-(r+2)x^2+(2r+4)x-4r \Rightarrow \cases{a=-(r+2) \\b=2r+4\\ c=-4r}\\ x=r也是x^2+ax+2=0的一根 \Rightarrow r^2+ar+2=0 \Rightarrow r^2-(r+2)r+2=0 \Rightarrow r=1\\ \Rightarrow (a,b,c) = \bbox[red, 2pt]{(-3,6,-4)}$$

解答:$$A= \begin{bmatrix}a& b\\c& d \end{bmatrix} \Rightarrow A^2= \begin{bmatrix}a^2+bc& ab+bd\\ ac+cd& bc+d^2 \end{bmatrix}\\ \Rightarrow A^2+3A+2I_2= \begin{bmatrix}a^2+bc+3a+2& b(a+d+3) \\ c(a+d+3) & bc+d^2+3d+2 \end{bmatrix} = \begin{bmatrix} 6& 0 \\14 &20\end{bmatrix} \Rightarrow \cases{a^2 +bc+ 3a+2=6\\b(a+d+3)=0\\c(a+d+3)=14 \\ bc+d^2+3d+2=20}\\ \Rightarrow b=0 \Rightarrow \cases{a^2+3a-4=0\\ d^2+3d-18=0} \Rightarrow \cases{(a+4)(a-1)=0\\ (d+6)(d-3)=0} \Rightarrow \cases{a=1,-4\\ d=3,-6} \\ \textbf{Case I }\cases{a=1\\ d=3} \Rightarrow c={14\over a+d+3}=2 \Rightarrow A= \begin{bmatrix}1&0\\2& 3 \end{bmatrix} \\ \textbf{Case II }\cases{a=1\\ d=-6} \Rightarrow c=-7 \Rightarrow A= \begin{bmatrix}1&0\\-7& -6 \end{bmatrix} \\ \textbf{Case III }\cases{a=-4\\ d=3} \Rightarrow c=7 \Rightarrow A= \begin{bmatrix}-4&0\\7&3 \end{bmatrix} \\ \textbf{ Case IV }\cases{a=-4\\ d=-6} \Rightarrow c=2 \Rightarrow A= \begin{bmatrix} -4&0\\-2& -6 \end{bmatrix} \\ 因此A有四種可能: \bbox[red, 2pt]{\begin{bmatrix} 1&0\\2& 3 \end{bmatrix},\begin{bmatrix}1&0\\-7& -6 \end{bmatrix}, \begin{bmatrix} -4&0\\7&3 \end{bmatrix}, \begin{bmatrix}-4&0\\-2& -6 \end{bmatrix}}$$

解答:$$\cases{A(-2,7,15)\\ B(1,16,3) \\C(10,7,3)} \Rightarrow \cases{ \overrightarrow{AB}=(3,9,-12) \\ \overrightarrow{AC} =(12,0,-12)} \Rightarrow \cases{ |\overrightarrow{AB}|^2= 234 \\ |\overrightarrow{AC}|^2 = 288 \\ \overrightarrow{AB}\cdot \overrightarrow{AC} = 180} \Rightarrow \cases{ \overrightarrow{AO} \cdot \overrightarrow{AB}={1\over 2}|\overrightarrow{AB}|^2= 117 \\ \overrightarrow{AO} \cdot \overrightarrow{AC}={1\over 2} |\overrightarrow{AC} |^2 = 144 } \\ 假設\overrightarrow{AO} = x\overrightarrow{AB} +y\overrightarrow{AC} \Rightarrow \cases{( x\overrightarrow{AB}+ y\overrightarrow{AC}) \cdot \overrightarrow{AB}=117\\ ( x\overrightarrow{AB}+ y\overrightarrow{AC}) \cdot \overrightarrow{AC}=144} \Rightarrow \cases{234x+ 180y=117 \\ 180x+ 288y = 144} \Rightarrow \cases{x=2/9\\ y=13/36}\\ \Rightarrow \overrightarrow{AO} ={2\over 9}(3,9,-12)+ {13\over 36}(12,0,-12)=(5,2,-7) \Rightarrow O=A+\overrightarrow{AO} = \bbox[red, 2pt]{(3,9,8)}$$

解答:$$\sqrt{\sum_{k=1}^n (2k-1)} -\sqrt{\sum_{k=1}^n (2k)} =\sqrt{n(n+1)-n} -\sqrt{n(n+1)} = \bbox[red, 2pt] {n-\sqrt{n(n+1)}}$$

解答:$$假設箱子 A、B、C 中的球數分別為 x_A、x_B、x_C \Rightarrow x_A+x_B+ x_C=12,\\ 其中\cases{x_A\in \{1,3,5,7,9,11\} \\ x_B\in \{0,2,4,6,8,10,12\}} \\ \Rightarrow \cases{x_A=1 \Rightarrow x_B+x_C=11 \Rightarrow x_B=0,2,4,6,8,10 \Rightarrow 6種可能\\ x_A=3 \Rightarrow x_B+x_C= 9 \Rightarrow x_B= 0,2,4,6,8 \Rightarrow 5種可能\\ x_A=5 \Rightarrow x_B+x_C= 7 \Rightarrow x_B=0,2,4,6  \Rightarrow 4種可能\\ x_A=7 \Rightarrow x_B+x_C= 5 \Rightarrow x_B=0,2,4  \Rightarrow 3種可能\\ x_A=9 \Rightarrow x_B+x_C= 3 \Rightarrow x_B=0,2  \Rightarrow 2種可能\\ x_A=11 \Rightarrow x_B+x_C= 1 \Rightarrow x_B=0   \Rightarrow 1種可能} \\ \Rightarrow 合計: 6+5+4+3+2+1= \bbox[red, 2pt]{21}$$

解答:$$a_n=S_n-S_{n-1} \Rightarrow S_n=2(S_n-S_{n-1})+2026 \Rightarrow S_n= 2S_{n-1}-2026 \\ \Rightarrow S_n-2026=2(S_{n-1}-2026) \Rightarrow \langle S_n-2026\rangle 為公比r=2的等比數列\\ S_1= 2a_1+2026 =2S_1+2026 \Rightarrow S_1=-2026 \Rightarrow  等比首項S_1-2026=-4052 \\ \Rightarrow S_n-2026 =(-4052)2^{n-1} \Rightarrow S_n=2026+(-4052)2^{n-1} =2026(1-2^n) \\ \Rightarrow {S_{20} \over S_{10}} = {1-2^{20} \over 1-2^{10}}= 1+2^{10}= \bbox[red, 2pt]{1025}$$

解答:$$x=1代入原式:{1\over 3}+\int_1^x f(t)\,dt={1\over 3}xf(x) \Rightarrow {1\over 3}+0={1\over 3}f(1) \Rightarrow f(1)=1 \\ 又{d\over dx} \left( {1\over 3}+\int_1^x f(t)\,dt \right) ={d\over dx} \left(  {1\over 3}xf(x)\right) \Rightarrow f(x)={1\over 3}f(x)+{1\over 3}xf'(x) \Rightarrow 2f(x)=xf'(x) \\ 可分離一階微分方程 \Rightarrow {dy\over y}={2\over x}\,dx \Rightarrow \ln y=2\ln x+c_1=\ln (c_2x^2) \Rightarrow y=c_2x^2\\ 將f(1)=1(即y(1)=1)代入\Rightarrow 1=c_2 \Rightarrow y=f(x) =x^2 \Rightarrow \int_0^3 f(x)\,dx = \int_0^3 x^2\,dx= \bbox[red, 2pt]9$$

解答:$$P\in L_1:{x-3\over 2} ={y+1\over 2} =z-4 \Rightarrow P(2t+3,2t-1,t+4) 代入L_2 \Rightarrow {2t-1+2\over 3} ={t+4-7\over -2} \\ \Rightarrow {2t+1\over 3}={t-3\over -2} \Rightarrow t=1 \Rightarrow P(5,1,5) 代入L_2 \Rightarrow {5-a\over 6}={1+2\over 3} \Rightarrow a=-1 \\ \cases{L_1的方向向量\vec v_1=(2,2,1) 且|\vec v_1|=3\\ L_2的方向向量\vec v_2=(6,3,-2)且|\vec v_2| =7} \Rightarrow 取\cases{\vec u_1=7\vec v_1=(14,14,7) \\ \vec u_2=3\vec v_2=(18,9,-6)}\\ \Rightarrow 角平分線\cases{\vec w_1=\vec u_1+\vec u_2 =(32,23,1)\\ \vec w_2=\vec u_1-\vec u_2 =(-4,5, 13)} \Rightarrow 角平分線 \bbox[red, 2pt]{\cases{{x-5\over 32}={y-1\over 23} ={z-5\over 1} \\ {x-5\over -4}={y-1\over 5} ={z-5\over 13}}}$$

解答:$$取u=z-w \Rightarrow (z-w)^3=u^3=8 \Rightarrow \cases{u_1=2\\ u_2= 2(\cos 120^\circ+i\sin 120^\circ) =-1+\sqrt 3i\\ u_3=2(\cos 240^\circ+i \sin 240^\circ)=-1-\sqrt 3i} \\ \Rightarrow z=u+w \Rightarrow \cases{z_1= 2+(1+\sqrt 3i)=3+\sqrt 3i\\ z_2=(-1+\sqrt 3i)+(1+\sqrt3 i)=2\sqrt 3i \\ z_3=(-1-\sqrt 3i)+(1+\sqrt 3i)=0} \Rightarrow \cases{|z_1|=2\sqrt 3\\ |z_2|=2\sqrt 3\\ |z_3|=0} \Rightarrow k\ge 2\sqrt 3\\ \Rightarrow k的最小值為\bbox[red, 2pt]{2\sqrt 3}$$

解答:$$\Gamma'順時針旋轉45^\circ 變成\Gamma \Rightarrow \begin{bmatrix}1/\sqrt 2& 1/\sqrt 2\\ -1/\sqrt 2& 1/\sqrt 2 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} = \begin{bmatrix}{1\over \sqrt 2}(x+y) \\ {1\over \sqrt 2} (-x+y)\end{bmatrix} \\ \Rightarrow  \left( {1\over \sqrt 2}(x+y) \right)^2/2+ \left( {1\over \sqrt 2} (-x+y) \right)^2/3=1 \Rightarrow 3(x+y)^2+2(-x+y)^2=12\\ \Rightarrow \bbox[red, 2pt]{5x^2+2xy+5y^2=12}$$

解答:$$假設\cases{球心(0,0,0) \\ 半徑R=2\sqrt 2} \Rightarrow 水面與球心距離=R\sin 45^\circ=2 \\ \Rightarrow 流出的水體積=\int_{-2}^0 \pi(R^2-z^2)\,dz=  \int_{-2}^0 \pi(8-z^2)\,dz= \pi \left. \left[ 8z-{1\over 3}z^3 \right] \right|_{-2}^0 = \bbox[red, 2pt]{{40\over 3}\pi}$$

解答:$$x^2+y^2+axy-1= (x+by+c)(x+dy+e) =x^2+(b+d)xy+ bdy^2 +(c+e)x +(be+cd)y+ ce\\ \Rightarrow \cases{bd=1  \cdots(1)\\ ce=-1  \cdots(2)\\ c+e=0 \cdots(3)\\ be+cd=0  \cdots(4)\\ b+d=a \cdots(5)}, 由(2)及(3)可得c=\pm 1\Rightarrow e=\mp1; 又(1) \Rightarrow d={1\over b}代入(4) \\ \Rightarrow -bc+{c\over b}=0\Rightarrow c \left( {1\over b}-b\right) =0 \Rightarrow  b=\pm 1 \\ \Rightarrow \cases{b=1 \Rightarrow a=2 \Rightarrow a+b+c=\cases{2+1+1=4\\ 2+1-1=2} \\ b=-1 \Rightarrow a=-2 \Rightarrow a+b+c=\cases{-2-1+1=-2\\-2-1-1=-4}} \Rightarrow a+b+c= \bbox[red, 2pt]{\pm 2,\pm 4}$$

解答:$$取t= \sin x+\cos x =\sqrt 2\sin(x+45^\circ) \Rightarrow -\sqrt 2\le t\le \sqrt 2\\ 又t^2=1+2\sin x\cos x \Rightarrow \sin x\cos x={t^2-1\over 2} \Rightarrow y=f(t)={(t^2-1)/2\over t+2} ={t^2-1\over 2(t+2)} \\ \Rightarrow f'(t)={t^2+4t+1\over 2(t+2)^2}=0 \Rightarrow t^2+4t+1=0 \Rightarrow t=-2+ \sqrt 3 (-2-\sqrt 3\lt -\sqrt 2,不合) \\ \Rightarrow f(-2+\sqrt 3)= {(-2+\sqrt 3)^2-1\over 2(-2+\sqrt 3+2)}= \bbox[red, 2pt]{\sqrt 3-2}$$

解答:$$1+z^k+ z^{2k} =(z^k-\omega) (z^k-\omega^2), 其中\omega^2+\omega+1=0 \\ \Rightarrow {1\over 1+z^k+z^{2k}} ={1\over (z^k-\omega) (z^k-\omega^2)} ={1\over \omega-\omega^2} \left( {1\over z^k-\omega}-{1\over z^k-\omega^2} \right)\\ 取g(x)={x^{23}-1\over x-1}=1+x+\cdots+x^{22} \Rightarrow z,z^2,\dots,z^{22}為g(x)=0的根 \\ \Rightarrow g(x)=(x-z)(x-z^2) \cdots(x-z^{22}) \Rightarrow {g'(x)\over g(x)} ={1\over x-z}+{1\over x-z^2}+\cdots+{1\over x-z^{22}} \\ \Rightarrow \sum_{k=1}^{22} {1\over 1+z^k+ z^{2k}}  ={1\over \omega-\omega^2} \sum_{k=1}^{22} \left( {1\over z^k-\omega}-{1\over z^k-\omega^2} \right)   ={1\over \omega-\omega^2}  \left( -{g'(\omega)\over g(\omega)} +{g'(\omega^2) \over g(\omega^2)}\right)\\= {1\over \omega-\omega^2} \left( (15\omega+7)-(-15\omega-8) \right) ={30\omega+15\over \omega-\omega^2} = {30\omega+15\over \omega-(-\omega-1)} =15 \\ \Rightarrow \sum_{k=0}^{22} {1\over 1+z^k+ z^{2k}}={1\over 3}+15= \bbox[red, 2pt]{46\over 3}\\ 註: g(x)={x^{23}-1\over x-1} \Rightarrow g'(x)={23x^{22}(x-1)-(x^{23}-1) \over (x-1)^2} \Rightarrow \cases{g(\omega)=-\omega^2\\ g'(\omega)=(22\omega-1)/(\omega-1)}$$

 二、 證明題：每題 10 分，共 10 分。

解答:$$X\sim B(n,p) \Rightarrow P(X=k)= {n\choose k}p^kq^{n-k}, k=0,1,2,\dots,n \Rightarrow E(X)= \sum_{k=0}^n k{n\choose k}p^kq^{n-k} \\ 由於(p+q)^n = \sum_{k=0}^n{n\choose k}p^kq^{n-k} \Rightarrow {d\over dp} \left( (p+q)^n \right) =n(p+q)^{n-1} =\sum_{k=0}^n k{n\choose k}p^{k-1}q^{n-k} \\ \Rightarrow np(p+q)^{n-1} =np =\sum_{k=0}^n k{n\choose k}p^{k}q^{n-k} =E(X) \Rightarrow E(X)=np, \bbox[red, 2pt]{QED.}$$

====================== END ==========================
解題僅供參考，其他教甄試題及詳解